信号与系统练习与解答-英文.pdf
信号与系统奥本海姆英文版课后答案chapter10
∞
−n
n =−∞
∑ (1/ 4)
∞
z −n zn
n =−3
∑ (1/ 4)
= ∑ (1/ 4) − n +3 z n −3
n=0
= (1/ 64) z −3 /(1 − 4 z ), z < 1/ 4 = (1/16) z −4 /(1 − (1/ 4) z −1 ), z < 1/ 4
The Fourier transform does not exist because the ROC includes the unit circle. (g) Consider x1 ( z ) = 2n μ[ − n].
X ( z) =
n =−∞
jw
r> 1 . The second summation
2
10.2
∑ ( 5 ) u[n − 3]z
n
∞
1
z −3 ∞ 1 n − n ]∑ ( ) z =[ 125 n = 0 5
∞ 1 = ∑ ( )n z − n n =3 5 z −3 1 ] =[ 125 1 − 1 z −1 5 −n
X ( Z ) = ∑ x[n]z − n
n =∞ ∞
−5
= ∑ (−1) n z − n
n =0
∞
= 1/(1 + z −1 ), z > 1
The Fourier transform does not exist because the ROC does not include the unit circle (d) For x [n]= ( 1 ) n +1 μ[n + 3],
x( z ) =
英文版《信号与系统》第九章习题解答
2 H (s ) = s (s + 4 )
Re{s} > 0
The system is unstable.
15
Chapter 9 9.33 A causal LTI system H (s ) =
x(t ) = e
t
Problem Solution
s +1 s 2 + 2s + 2
, ∞ < t < +∞
Determine and sketch the response y (t )
2 X (s ) = 2 s 1
1 < Re{s} < 1 Re{s} > 1
2(s + 1) (s + 1)(s 1) s 2 + 2s + 2
s +1 H (s ) = 2 s + 2s + 2 Y (s ) = X (s )H (s ) =
2 2 4 y (t ) = + e t u ( t ) + e t (cos t )u (t ) + e t (sin t )u (t ) 5 5 5
17
Chapter 9
Problem Solution
9.35 Consider a causal LTI system with the input x(t ) and output y (t ) . (b) Is this system stable?
2
②
1
③
1
④
2
σ
① Re{s} < 2 anticausal , unstable ② -2 < Re{s} < 1 noncausal , unstable
英文版《信号与系统》第三章习题解答
ck
1 e jk 2
sin
k / 2 k
e jk / 2
bk
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jk
2 1
dx1t
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ck
1
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4t
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Chapter 3
Problem Solution
0
ak
1
jk
2
jk
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k is even k is odd
(c) x t
ake jk0t
ak 0 , k is odd
for each of the following inputs :
(a) xt t n n
(b) xt 1n t n n
(c) xt is the periodic wave depicted in Figure P3.34
1/ 2 1 xt
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1
2
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Chapter 3
(c) ak 0 , k is odd xt is even harmonic
Could T be the fundamental period for such a signal?
(d) If one of two things happens,T is fundamental period.
xt xt T / 2
Chapter 3
Problem Solution
(b) Solution 1 xt x1t x1t 1
xt
x1 t FSbk
1
2 1 0 1 2 3
ak bk 1 e jk0t0 0 t0 1
1
x1 t
1
ak bk 1 e jk
英文版《信号与系统》第二章习题解答
Problems Solution
dx(t ) ∵ → 3 y (t ) + e 2t u (t ) dt
2e 3 h(t 1) = e 2t u (t )
the impulse 1 3 2(t +1) u (t + 1) response h(t ) = 2 e e
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Chapter 2
∫
t
∞
h (τ )d τ =
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t<0,t>2
sin (πτ )d τ =
1
∫
t
1
0
π
cos πτ | 0t
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=
π
(1 cos π t )[u (t ) u (t 2 )]
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Chapter 2
y (t ) = 2
Problems Solution
π
2
(1 cos π (t 1 ))[u (t 1 ) u (t 3 )] (1 cos π (t 3 ))[u (t 3 ) u (t 5 )]
∞
+∞
2
t
y (t ) = ∫ e (t τ 2 )u (t τ 2 )x(τ )dτ
∞
+∞
h(t ) = e
(t 2 )
u (t 2)
h(t τ )
(b ) y(t ) = x(t ) h(t ) = x′(t ) h (1) (t )
= h (1) (t + 1) h (1) (t 2)
0
e 3τ dτ
1 e 3(t 3 ) = 3 e 3(t 5 ) e 3(t 3 ) = 3
x(t τ )
t 3
信号与系统(习题课)
∴ y(t) = e-3t + t e-3t = (1+ t) e-3t
by wky
习题 3-6 (1)
已知系统的微分方程为 y’’(t) +5 y’(t) + 4 y(t) =2 f ’(t) + 5f(t), t >0; 初始状态y(0-) =1，y’(0-) =5， 求系统的零输入响应yx(t)。 解：系统特征方程为 s2+5s+4=0 , 解得特征根 s1=-1, s2=-4
特解 （强迫响应）
比较：完全响应=零输入响应 + 零状态响应 = e－t + (1 - 1/2e－t -1/2e－3t)
by wky
习题 3-4
已知微分方程为 y’(t) + 3 y (t) = f(t)，t >0; y(0) =1，
求系统的固有响应(齐次解) yh(t)、强迫响应 (特解) yp(t)和完全响应(全解) y(t) 解：系统特征方程为 s+3=0,
f(t)
f(-t)
2
2
1
1
-3 -2 -1 0 1 2 3 t -3 -2 -1 0 1 2 3 t
2 f(t+2)
f(-3t)
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-3 -2 -1 0 1 2 3 t -3 -2 -1 0 1 2 3 t by wky
2-10 已知信号波形, 绘出下列信号波形
f(t)
f(-t)
2
2
1
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-3 -2 -1 0 1 2 3 t -3 -2 -1 0 1 2 3 t
信号与系统奥本海姆英文版课后答案chapter
Chapter 1 Answers1.1 Converting from polar to Cartesian coordinates: 1.2 converting from Cartesian to polar coordinates:55j=, 22j e π-=,233jj eπ--=212je π--=, 41j j π+=, ()2221jj eπ-=-4(1)j j eπ-=, 411j jeπ+=- 12e π-= 1.3. (a) E ∞=4014tdt e∞-=⎰, P ∞=0, because E ∞<∞ (b) (2)42()j t t x eπ+=, 2()1t x =.Therefore, E ∞=22()dt t x +∞-∞⎰=dt +∞-∞⎰=∞,P ∞=211limlim222()TTTTT T dt dt TTt x --→∞→∞==⎰⎰lim11T →∞=(c) 2()t x =cos(t). Therefore, E∞=23()dt t x +∞-∞⎰=2cos()dt t +∞-∞⎰=∞,P ∞=2111(2)1lim lim 2222cos()TTTT T T COS t dt dt T Tt --→∞→∞+==⎰⎰(d)1[][]12nn u n x =⎛⎫⎪⎝⎭,2[]11[]4nu n n x =⎛⎫ ⎪⎝⎭. Therefore, E ∞=24131[]4nn n x +∞∞-∞===⎛⎫∑∑⎪⎝⎭P ∞=0，because E ∞<∞. (e) 2[]n x =()28n j e ππ-+, 22[]n x =1. therefore, E ∞=22[]n x +∞-∞∑=∞,P ∞=211limlim1122121[]NNN N n Nn NN N n x →∞→∞=-=-==++∑∑.(f) 3[]n x =cos 4nπ⎛⎫ ⎪⎝⎭. Therefore, E ∞=23[]n x +∞-∞∑=2cos()4n π+∞-∞∑=2cos()4n π+∞-∞∑,P ∞=1limcos 214nNN n NN π→∞=-=+⎛⎫∑ ⎪⎝⎭1cos()112lim ()2122NN n Nn N π→∞=-+=+∑ 1.4. (a) The signal x[n] is shifted by 3 to the right. The shifted signal will be zero for n<1, And n>7. (b) The signal x[n] is shifted by 4 to the left. The shifted signal will be zero for n<-6. And n>0. (c) The signal x[n] is flipped signal will be zero for n<-1 and n>2.(d) The signal x[n] is flipped and the flipped signal is shifted by 2 to the right. The new Signal will be zero for n<-2 and n>4.(e) The signal x[n] is flipped and the flipped and the flipped signal is shifted by 2 to the left. This new signal will be zero for n<-6 and n>0.1.5. (a) x(1-t) is obtained by flipping x(t) and shifting the flipped signal by 1 to the right. Therefore, x (1-t) will be zero for t>-2. (b) From (a), we know that x(1-t) is zero for t>-2. Similarly, x(2-t) is zero for t>-1, Therefore, x (1-t) +x(2-t) will be zero for t>-2. (c) x(3t) is obtained by linearly compression x(t) by a factor of3. Therefore, x(3t) will be zero for t<1. (d) x(t/3) is obtained by linearly compression x(t) by a factor of 3. Therefore, x(3t) will be zero for t<9.1.6 (a) x 1(t ) is not periodic because it is zero for t<0. (b) x 2[n ]=1 for all n. Therefore, it is periodic with a fundamental period of 1. (c) x 3[n1.7. (a)v ε[4])n --Therefore,()1[]vn xεis zero for 1[]n x >3. (b) Since x 1(t ) is an odd signal, ()2[]vn x εis zero for all values of t.(c) (){}11311[][][][3][3]221122v nnn n n u n u n x x x ε-⎡⎤⎢⎥=+-=----⎢⎥⎢⎥⎣⎦⎛⎫⎛⎫ ⎪ ⎪⎝⎭⎝⎭Therefore, ()3[]vn x εis zero whenn <3 and when n →∞.(d)()1554411()(()())(2)(2)22vttt t t u t u t x x x ee ε-⎡⎤=+-=---+⎣⎦Therefore,()4()vt x εis zero only whent →∞.1.8. (a) ()01{()}22cos(0)tt t x eπℜ=-=+l (b) ()02{()}cos()cos(32)cos(3)cos(30)4t t t t t x e ππℜ+==+l(c) ()3{()}sin(3)sin(32t t t t t x e e ππ--ℜ=+=+l(d) ()224{()}sin(100)sin(100)cos(1002t t t t t t t x e e e ππ---ℜ=-=+=+l1.9. (a) 1()t xis a periodic complex exponential.(b) 2()t x is a complex exponential multiplied by a decaying exponential. Therefore,2()t x is not periodic.（c ）3[]n x is a periodic signal. 3[]n x =7j n e π=j n e π.3[]n x is a complex exponential with a fundamental period of22ππ=. (d) 4[]n x is a periodic signal. The fundamental period is given by N=m(23/5ππ)=10().3m By choosing m=3. We obtain the fundamental period to be 10.(e) 5[]n x is not periodic. 5[]n x is a complex exponential with 0w =3/5. We cannot find any integer msuch that m(02wπ ) is also an integer. Therefore, 5[]n xis not periodic.1.10. x (t )=2cos(10t ＋1)-sin(4t-1)Period of first term in the RHS =2105ππ=.Period of first term in the RHS =242ππ= .Therefore, the overall signal is periodic with a period which the least commonmultiple of the periods of the first and second terms. This is equal to π . 1.11.x[n] = 1+74j n e π?25j n e πPeriod of first term in the RHS =1. Period of second term in the RHS =⎪⎭⎫ ⎝⎛7/42π=7 （when m=2）Period of second term in the RHS =⎪⎭⎫ ⎝⎛5/22ππ=5 (when m=1)Therefore, the overall signal x[n] is periodic with a period which is the least common Multiple of the periods of the three terms inn x[n].This is equal to 35.1.12. The signal x[n] is as shown in figure S1.12. x[n] can be obtained by flipping u[n] and thenShifting the flipped signal by 3 to the right. Therefore, x[n]=u[-n+3]. This implies that 1.13y (t)= ⎰∞-tdt x )(τ =dt t))2()2((--+⎰∞-τδτδ=⎪⎩⎪⎨⎧>≤≤--<2,022,12,0,t t tTherefore ⎰-==∞224dt E=2x 1[n-2]+ 5x 1[n-3] + 2x 1[n-4] The input-output relationship for S isy[n]=2x[n-2]+ 5x [n-3] + 2x [n-4](b) The input-output relationship does not change if the order in which S 1and S 2 are connected series reversed. . We can easily prove this assuming that S 1 follows S 2. In this case , the signal x 1[n], which is the input to S 1 is the same as y 2[n].Therefore y 1[n] =2x 1[n]+ 4x 1[n-1]= 2y 2[n]+4 y 2[n-1]=2( x 2[n-2]+21 x 2[n-3] )+4(x 2[n-3]+21x 2[n-4]) =2 x 2[n-2]+5x 2[n-3]+ 2 x 2[n-4]The input-output relationship for S is once againy[n]=2x[n-2]+ 5x [n-3] + 2x [n-4]1.16 (a)The system is not memory less because y[n] depends on past values of x[n].(b)The output of the system will be y[n]= ]2[][-n n δδ=0(c)From the result of part (b), we may conclude that the system output is always zero for inputs of the form ][k n -δ, k ∈ ?. Therefore , the system is not invertible .1.17 (a) The system is not causal because the output y(t) at some time may depend on future values of x(t). For instance , y(-π)=x(0).(b) Consider two arbitrary inputs x 1(t)and x 2(t).x 1(t) →y 1(t)= x 1(sin(t)) x 2(t) → y 2(t)= x 2(sin(t))Let x 3(t) be a linear combination of x 1(t) and x 2(t).That is , x 3(t)=a x 1(t)+b x 2(t)Where a and b are arbitrary scalars .If x 3(t) is the input to the given system ,then the corresponding output y 3(t) is y 3(t)= x 3( sin(t))=a x 1(sin(t))+ x 2(sin(t))=a y 1(t)+ by 2(t)Therefore , the system is linear.1.18.(a) Consider two arbitrary inputs x 1[n]and x 2[n].x 1[n] → y 1[n] =][001k x n n n n k ∑+-=x 2[n ] → y 2[n] =][02k x n n n n k ∑+-=Let x 3[n] be a linear combination of x 1[n] and x 2[n]. That is :x 3[n]= ax 1[n]+b x 2[n]where a and b are arbitrary scalars. If x 3[n] is the input to the given system, then the corresponding outputy 3[n] is y 3[n]=][03k x n n n n k ∑+-==])[][(2100k bx k ax n n n n k +∑+-==a ][001k x n n n n k ∑+-=+b ][02k x n n n n k ∑+-== ay 1[n]+b y 2[n]Therefore the system is linear.(b) Consider an arbitrary input x 1[n].Lety 1[n] =][01k x n n n n k ∑+-=be the corresponding output .Consider a second input x 2[n] obtained by shifting x 1[n] in time:x 2[n]= x 1[n-n 1]The output corresponding to this input isy 2[n]=][02k x n n n n k ∑+-== ]n [1100-∑+-=k x n n n n k = ][01011k x n n n n n n k ∑+---=Also note that y 1[n- n 1]=][01011k x n n n n n n k ∑+---=.Therefore , y 2[n]= y 1[n- n 1] This implies that the system is time-invariant.(c) If ][n x <B, then y[n]≤(2 n 0+1)B. Therefore ,C ≤(2 n 0+1)B.1.19 (a) (i) Consider two arbitrary inputs x 1(t) and x 2(t). x 1(t) → y 1(t)= t 2x 1(t-1)x 2(t) → y 2(t)= t 2x 2(t-1)Let x 3(t) be a linear combination of x 1(t) and x 2(t).That is x 3(t)=a x 1(t)+b x 2(t)where a and b are arbitrary scalars. If x 3(t) is the input to the given system, then the corresponding output y 3(t) is y 3(t)= t 2x 3 (t-1)= t 2(ax 1(t-1)+b x 2(t-1))= ay 1(t)+b y 2(t)Therefore , the system is linear.(ii) Consider an arbitrary inputs x 1(t).Let y 1(t)= t 2x 1(t-1)be the corresponding output .Consider a second input x 2(t) obtained by shifting x 1(t) in time:x 2(t)= x 1(t-t 0)The output corresponding to this input is y2(t)= t2x2(t-1)= t2x1(t- 1- t)Also note that y1(t-t)= (t-t)2x1(t- 1- t)≠y2(t)Therefore the system is not time-invariant.(b) (i) Consider two arbitrary inputs x1[n]and x2[n]. x1[n] →y1[n] = x12[n-2]x2[n ] →y2[n] = x22[n-2].Let x3(t) be a linear combination of x1[n]and x2[n].That is x3[n]= ax1[n]+b x2[n]where a and b are arbitrary scalars. If x3[n] is the input to the given system, then the corresponding outputy3[n] is y3[n] = x32[n-2]=(a x1[n-2] +b x2[n-2])2=a2x12[n-2]+b2x22[n-2]+2ab x1[n-2] x2[n-2] ≠ay1[n]+b y2[n]Therefore the system is not linear.(ii) Consider an arbitrary input x1[n]. Let y1[n] = x12[n-2]be the corresponding output .Consider a second input x2[n] obtained by shifting x1[n] in time:x 2[n]= x1[n- n]The output corresponding to this input isy 2[n] = x22[n-2].= x12[n-2- n]Also note that y1[n- n]= x12[n-2- n]Therefore , y2[n]= y1[n- n]This implies that the system is time-invariant.(c) (i) Consider two arbitrary inputs x1[n]and x2[n].x 1[n] →y1[n] = x1[n+1]- x1[n-1]x2[n ]→y2[n] = x2[n+1 ]- x2[n -1]Let x3[n] be a linear combination of x1[n] and x2[n]. That is :x3[n]= ax1[n]+b x2[n]where a and b are arbitrary scalars. If x3[n] is the input to the given system, then thecorresponding output y3[n] is y3[n]= x3[n+1]- x3[n-1]=a x1[n+1]+b x2[n +1]-a x1[n-1]-b x2[n -1]=a(x1[n+1]- x1[n-1])+b(x2[n +1]- x2[n -1])= ay1[n]+b y2[n]Therefore the system is linear.(ii) Consider an arbitrary input x1[n].Let y1[n]= x1[n+1]- x1[n-1]be the corresponding output .Consider a second input x2[n] obtained by shifting x1[n] in time: x2[n]=x 1[n-n]The output corresponding to this input isy 2[n]= x2[n +1]- x2[n -1]= x1[n+1- n]- x1[n-1- n]Also note that y1[n-n]= x1[n+1- n]- x1[n-1- n]Therefore , y2[n]= y1[n-n]This implies that the system is time-invariant.(d) (i) Consider two arbitrary inputs x1(t) and x2(t).x 1(t) →y1(t)= dO{}(t)x1x 2(t) →y2(t)= {}(t)x2dOLet x 3(t) be a linear combination of x 1(t) and x 2(t).That is x 3(t)=a x 1(t)+b x 2(t)where a and b are arbitrary scalars. If x 3(t) is the input to the given system, then the corresponding output y 3(t) is y 3(t)= d O {}(t) x 3={}(t) x b +(t) ax 21d O=a d O {}(t) x 1+b {}(t) x 2d O = ay 1(t)+b y 2(t)Therefore the system is linear.(ii) Consider an arbitrary inputs x 1(t).Lety 1(t)= d O {}(t) x 1=2)(x -(t) x 11t -be the corresponding output .Consider a second input x 2(t) obtained by shifting x 1(t) in time:x 2(t)= x 1(t-t 0)The output corresponding to this input isy 2(t)= {}(t) x 2d O =2)(x -(t) x 22t -=2)(x -)t -(t x 0101t t --Also note that y 1(t-t 0)= 2)(x -)t -(t x 0101t t --≠ y 2(t)Therefore the system is not time-invariant. 1.20 (a) Givenx )(t =jt e 2 y(t)=tj e 3x )(t =jte2- y(t)=tj e3-Since the system liner+=tj et x 21(2/1)(jt e 2-) )(1t y =1/2(tj e3+tj e3-)Thereforex 1(t)=cos(2t))(1t y =cos(3t)(b) we know thatx 2(t)=cos(2(t-1/2))= (j e -jte 2+jejt e 2-)/2Using the linearity property, we may once again write x 1(t)=21( j e -jt e 2+j e jt e 2-))(1t y =(je-jt e 3+je jt e 3-)= cos(3t-1)Therefore,x 1(t)=cos(2(t-1/2)))(1t y =cos(3t-1)1.21.The signals are sketched in figure S1.21.(f) Not period.1.26 (a) periodic, period=7.(b) Not period.(c) periodic, period=8.(d) x[n]=(1/2)[cos(3πn/4+cos(πn/4)). periodic, period=8. (e) periodic, period=16. 1.27 (a) Linear, stable(b) Not period. (c) Linear(d) Linear, causal, stable(e) Time invariant, linear, causal, stable (f) Linear, stable(g) Time invariant, linear, causal 1.28 (a) Linear, stable(b) Time invariant, linear, causal, stable (c)Memoryless, linear, causal (d) Linear, stable (e) Linear, stable(f) Memoryless, linear, causal, stable (g) Linear, stable1.29 (a) Consider two inputs to the system such that[][][]{}111.S e x n y n x n −−→=ℜand [][][]{}221.Se x n y n x n −−→=ℜNow consider a third inputx 3[n]= x2[n]+x 1[n]. The corresponding system outputWill be [][]{}[][]{}[]{}[]{}[][]33121212e e e e y n x n x n x n x n x n y n y n ==+=+=+ℜℜℜℜtherefore, we may conclude that the system is additive Let us now assume that inputs to the system such that andNow consider a third input x 3 [n]= x 2 [n]+ x 1 [n]. The corresponding system output Will betherefore, we may conclude that the system is additive (b) (i) Consider two inputs to the system such that()()()()211111Sdx t x t y t x t dt ⎡⎤−−→=⎢⎥⎣⎦and ()()()()222211S dx t x t y t x t dt ⎡⎤−−→=⎢⎥⎣⎦Now consider a third inputx 3[t]= x2[t]+x 1[t]. The corresponding system outputWill betherefore, we may conclude that the system is not additiveNow consider a third input x 4 [t]= a x 1 [t]. The corresponding system output Will beTherefore, the system is homogeneous.(ii) This system is not additive. Consider the fowling example .Let δ[n]=2δ[n+2]+ 2δ[n+1]+2δ[n] andx2[n]=δ[n+1]+ 2δ[n+1]+ 3δ[n]. The corresponding outputs evaluated at n=0 areNow consider a third input x 3 [n]= x 2 [n]+ x 1 [n].= 3δ[n+2]+4δ[n+1]+5δ[n]The corresponding outputs evaluated at n=0 is y 3[0]=15/4. Gnarly, y 3[0]≠]0[][21y y n +.ThisTherefore, the system is homogenous.1.30 (a) Invertible. Inverse system y(t)=x(t+4)(b)Non invertible. The signals x(t) and x 1(t)=x(t)+2πgive the same output (c) δ[n] and 2δ[n] give the same output d) Invertible. Inverse system; y(t)=dx(t)/dt(e) Invertible. Inverse system y(n)=x(n+1) for n ≥0 and y[n]=x[n] for n<0 (f) Non invertible. x (n) and –x(n) give the same result (g)Invertible. Inverse system y(n)=x(1-n) (h) Invertible. Inverse system y(t)=dx(t)/dt(i) Invertible. Inverse system y(n) = x(n)-(1/2)x[n-1] (j) Non invertible. If x(t) is any constant, then y(t)=0 (k) δ[n] and 2δ[n] result in y[n]=0 (l) Invertible. Inverse system: y(t)=x(t/2)(m) Non invertible x 1 [n]= δ[n]+ δ[n-1]and x 2 [n]= δ[n] give y[n]= δ[n] (n) Invertible. Inverse system: y[n]=x[2n]1.31 (a) Note that x 2[t]= x 1 [t]- x 1 [t-2]. Therefore, using linearity we get y 2 (t)= y 1 (t)- y 1 (t-2).this is shown in Figure S1.31(b)Note that x3 (t)= x1 [t]+ x1 [t+1]. .Therefore, using linearity we get Y3 (t)= y1 (t)+ y1 (t+2). this is2220201.34. (a) ConsiderIf x[n] is odd, x[n] +x [-n] =0. Therefore, the given summation evaluates to zero. (b) Let y[n] =x 1[n]x 2[n] .Theny [-n] =x 1[-n] x 2[-n] =-x 1[n]x 2[n] =-y[n]. This implies that y[n] is odd.(c)ConsiderUsing the result of part (b), we know that x e [n]x o [n] is an odd signal .Therefore, using {}1[][0][][]n n x n x x n x n ∞∞=-∞==++-∑∑22[][]e o n n n n x x ∞∞=-∞=-∞=+∑∑222[][][]e on n n n n n x x x ∞∞∞=-∞=-∞=-∞==+∑∑∑the result of part (a) we may conclude thatTherefore,(d)ConsiderAgain, since x e (t) x o (t) is odd,Therefore,1.35. We want to find the smallest N 0 such that m(2π /N) N 0 =2πk or N 0 =kN/m,where k is an integer, then N must be a multiple of m/k and m/k must be an integer .this implies that m/k is a divisor of both m and N .Also, if we want the smallest possible N 0, then m/k should be the GCD of m and N. Therefore, N 0=N/gcd(m,N).1.36．(a)If x[n] is periodic 0(),0..2/j n N T o ewhere T ωωπ+= This implies that 022o T kNT k T T Nππ=⇒==a rational number . (b)T/T 0 =p/q then x[n] =2(/)j n p q e π,The fundamental period is q/gcd(p,q) and the fundmental frequencyis(c) p/gcd(p,q) periods of x(t) are needed .1.37.(a) From the definition of ().xy t φWe have (b) Note from part(a) that()().xx xx t t φφ=-This implies that ()xy t φis even .Therefore,the odd part of ().xx t φis zero.(c) Here, ()().xy xx t t T φφ=-and ()().yy xx t t φφ= 1.38.(a) We know that /22(2)().t t δδ=V V Therefore This implies that(b)The plot are as shown in Figure s3.18.1.39 We have Also,2[][]0eon n n x x ∞=-∞=∑222[][][].e on n n n n n xx x ∞∞∞=-∞=-∞=-∞==+∑∑∑2220()()()2()().eoet dt t dt t dt t t dt x x x x x ∞∞∞∞-∞-∞-∞-∞=++⎰⎰⎰⎰0()()0.et t dt x x ∞-∞=⎰222()()().e ot dt t dt t dt xx x ∞∞∞-∞-∞-∞=+⎰⎰⎰0022gcd(,)gcd(,)gcd(,)gcd(,).T pp q p q p q p q q p q p pωωππ===/21lim (2)lim ().2t t δδ→∞→∞=V V V V(b) y(t)=x 2(t) is such a systerm . (c) No.For example,consider y(t) ()()ty t x d ττ-∞=⎰with ()()(1).x t u t u t =--Then x(t)=0for t>1,but y(t)=1 for t>1.1.41. (a) y[n]=2x[n].Therefore, the system is time invariant.(b) y[n]=(2n-1)x[n].This is not time-invariant because y[n- N 0]≠(2n-1)2x [n- N 0]. (c) y[n]=x[n]{1+(-1)n +1+(-1)n-1}=2x[n].Therefore, the system is time invariant .1.42.(a) Consider two system S 1 and S 2 connected in series .Assume that if x 1(t) and x 2(t) arethe inputs to S 1..then y 1(t) and y 2(t) are the outputs.respectively .Also,assume thatif y 1(t) and y 2(t) are the input to S 2 ,then z 1(t) and z 2(t) are the outputs, respectively . Since S 1 is linear ,we may writewhere a and b are constants. Since S 2 is also linear ,we may write We may therefore conclude thatTherefore ,the series combination of S 1 and S 2 is linear. Since S 1 is time invariant, we may write andTherefore,Therefore, the series combination of S 1 and S 2 is time invariant.(b) False, Let y(t)=x(t)+1 and z(t)=y(t)-1.These corresponds to two nonlinear systems. If these systems are connected in series ,then z(t)=x(t) which is a linear system.(c) Let us name the output of system 1 as w[n] and the output of system 2 as z[n] .Then The overall system is linear and time-invariant. 1.43. (a) We haveSince S is time-invariant.Now if x (t) is periodic with period T. x{t}=x(t-T). Therefore, we may conclude that y(t)=y(t-T).This implies that y(t) is also periodic with T .A similar argument may be made in discrete time . (b)1.44 (a) Assumption : If x(t)=0 for t<t 0 ,then y(t)=0 for t< t 0.To prove That : The system is causal.Let us consider an arbitrary signal x 1(t) .Let us consider another signal x 2(t) which is the same as x 1(t)fort< t 0. But for t> t 0 , x 2(t) ≠x 1(t),Since the system is linear, Since ()()120x t x t -=for t< t 0 ,by our assumption =()()120y t y t -=for t< t 0 .This implies that()()12y t y t =for t< t 0 . In other words, t he output is not affected by input values for 0t t ≥. Therefore, thesystem is causal .Assumption: the system is causal . To prove that :If x(t)=0 for t< t 0 .then y(t)=0 for t< t 0 .Let us assume that the signal x(t)=0 for t< t 0 .Then we may express x(t) as ()()12()x t x t x t =-, Where()()12x t x t = for t< t 0 . the system is linear .the output to x(t) will be()()12()y t y t y t =-.Now ,since the system is causal . ()()12y t y t = for t< t 0 .implies that ()()12y t y t = for t< t 0 .Therefore y(t)=0 for t< t 0 .(b) Consider y(t)=x(t)x(t+1) .Now , x(t)=0 for t< t 0 implies that y(t)=0 for t< t 0 .Note that the system is nonlinear and non-causal .(c) Consider y(t)=x(t)+1. the system is nonlinear and causal .This does not satisfy the condition of part(a). (d) Assumption: the system is invertible. To prove that :y[n]=0 for all n only if x[n]=0 for all n . Consider[]0[]x n y n =→. Since the system is linear :2[]02[]x n y n =→.Since the input has not changed in the two above equations ,we require that y[n]= 2y[n].This implies that y[n]=0. Since we have assumed that the system is invertible , only one input could have led to this particular output .That input must be x[n]=0 .Assumption: y[n]=0 for all n if x[n]=0 for all n . To prove that : The system is invertible . Suppose that andSince the system is linear ,By the original assumption ,we must conclude that 12[][]x n x n =.That is ,any particular y 1[n] can be produced that by only one distinct input x 1[n] .Therefore , the system isinvertible.(e) y[n]=x 2[n].1.45. (a) Consider ,and()222()()s hx x t y t t φ→=.Now, consider ()()()312x t ax t bx t =+. The corresponding system output will be Therefore, S is linear .Now ,consider x 4(t)=x 1(t-T).The corresponding system output will beClearly, y 4(t)≠ y 1(t-T).Therefore ,the system is not time-invariant.The system is definitely not causal because the output at any time depends on future values of the input signal x(t).(b) The system will then be linear ,time invariant and non-causal.1.46. The plots are in Figure S1.46.1.47.(a) The overall response of the system of Figure P1.47.(a)=(the response of the system to x[n]+x 1[n])-the response of the system to x 1[n]=(Response of a linear system L to x[n]+x 1[n]+ zero input response of S)- (Response of a linear system L to x 1[n]+zero input response of S)=( (Response of a linear system L to x[n]).。
信号与系统习题集英文版
信号与系统习题集第一部分：时域分析 一、填空题 1. =---)3()()2(t t u et δ()。
2. The unit step response )(t g is the zero-state response when the input signal is ( ).3. Given two continuous – time signals x(t) and h(t), if their convolution is denoted by y(t), then the convolution of )1(-t x and )1(+t h is ( ).4. The convolution =+-)(*)(21t t t t x δ( ).5. The unit impulse response )(t h is the zero-state response when the input signal is ( ).6. A continuous – time LTI system is stable if its unit impulse response satisfies the condition: ( ).7. A continuous – time LTI system can be completely determined by its ( ).8. =⎰∞∞-(t)dt 2sin 2 δt t ( ).9. Given two sequences }1,2,2,1{][=n x and }5,6,3{][=n h , their convolution=][*][n h n x ( ).10. Given three LTI systems S1, S2 and S3, their unit impulse responses are )(1t h ,)(2t h and )(3t h respectively. Now, construct an LTI system S using these threesystems: S1 parallelinterconnected by S2, then series interconnected by S3. the unit impulse response of the system S is ( ).11. It is known that the zero-stat response of a system to the input signal x(t) is⎰∞-=td x t y ττ)()(, then the unit impulse response h(t) is ( ).12. The complete response of an LTI system can be expressed as a sum of its zero-state response and its ( ) response.13. It is known that the unit step response of an LTI system is )(2t u et-, then the unit impulseresponse h(t) is ( ).14. =++-=⎰∞dt t t t t x ))1()1((2sin )(0δδπ( ).15. We can build a continuous-time LTI system using the following three basic operations: ( ) , ( ), and ( ). 16. The zero-state response of an LTI system to the input signal )1()()(--=t u t u t xis )1()(--t s t s , where s(t) is the unit step response of the system, then the unit impulse response h(t) is ( ).17. The block diagram of a continuous-time LTI system is illustrated in the following figure. The differential equation describing the input-output relationship of the system is ( )。
[信号与系统][第二版][奥本海姆][课后答案][英文][文字版]
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Signals & Systems
(Second Edition)
—Learning Instructions
(Exercises Answers)
Department of Computer Engineering 2005.12
Contents
Chapter 1 ······················································· 2 Chapter 2 ······················································· 17 Chapter 3 ······················································· 35 Chapter 4 ······················································· 62 Chapter 5 ······················································· 83 Chapter 6 ······················································· 109 Chapter 7 ······················································· 119 Chapter 8 ······················································· 132 Chapter 9 ······················································· 140 Chapter 10 ·······················································160
英文版《信号与系统》第六七八章习题解答
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cos 5Wt
cos 3Wt
Figure (a)
W j 1 2 X 1 j X 2 j
W j 0 , 1 2
Nyquist rate
maximum sampling interval
s 2 1 2
Tmax 2
s
1 2
Chapter 7 7.9 Consider the signal
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2
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Chapter 6
Problem Solution
2 y t x t
6.27 The output y t of a causal LTI system is related to the input x t by the differential equation dy t
Chapter 6 (c)
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e jπ / 2 0 H j e -j π / 2 0
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Problem Solution
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信号与系统奥本海姆英文版课后答案chapter6
6.12. Using the Bode magnitude plot, specified in Figure P6.12(a). we may obtain an expression For H1 (j ω ). The figure shows that H1 (j ω ) has the break frequencies ω1 =1, ω2 =8,And
G ( jω ) = FT {2h(t ) cos(4000π t )} = H ( j (ω − 4000π )) + H ( j (ω + 4000π ))
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-4000
-2000
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2000
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k =0 N k k =1
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信号与系统奥本海姆英文版课后答案chapter9
Chapter 9 Answers（a ）The given integral may be written as(5)0t j t e e dt σω∞-+⎰If σ<-5 ,then the function (5)te σ-+ grows towards ∞ with increasing t and the given integral does not converge .but if >-5,then the integral does converge (b) The given integral may be written as0(5)t j tee d σω-+-∞⎰t If σ>-5 ,then the function (5)te σ-+ grows towards ∞as t decreases towards -∞and the given integral doesnot converge .but if σ<-5,then the integral does converge (c) The given integral may be written as5(5)5t j t e e d σω-+-⎰t Clearly this integral has a finite value for all finite values of σ. (d) The given integral may be written as(5)t j t e e d σω∞-+-∞⎰tIfσ>-5 ,then the function (5)t e σ-+ grows towards ∞as t decreases towards -∞and the given integraldoes not converge If σ<-5, ,then function (5)te σ-+ grows towards ∞ with increasing t and the given integral does not converge If σ=5, then the integral stilldoes not have a finite value. therefore, the integral does not converge for any value of σ. (e) The given integral may be written as0(5)t j tee d σω-+-∞⎰t+ (5)0t j t e e d σω∞-+⎰t The first integral converges for σ<-5, the second internal converges if σ>-5,therefore, the given internal converges whenσ<5.(f) The given integral may be written as0(5)t j t e e d σω-+-∞⎰tIf σ>5 ,then the function (5)te σ--+grows towards ∞ as t decrease towards -∞ and the given integral does not converge .but if σ<5,then the integral does converge. (a)X(s)= 5(1)t dt eu t e dt ∞---∞⎰- =(5)0s tedt ∞-+⎰ =(5)5s es -++As shown in Example the ROC will be {}Re s >-5. (b) By using eg., we can easily show that g(t)=A 5te-u(-t-0t ) has the Laplace transformG(s)= 0(5)5s t Ae s ++The ROC is specified as {}Re s <-5 . Therefore ,A=1 and 0t =-1Using an analysis similar to that used in Example we known that given signal has a Laplace transform of the formX(s)115s s β+++The corresponding ROC is {}Re s >max(-5,Re{β}). Since we are given that the ROC isRe{s}>-3, we know that Re{β}=3 . there are no constraints on the imaginary part of β. We know form Table that111()sin(2)()()()Lt x t e t u t X s X s -=-←−→=-, Re{s}>-1 We also know form Table thatx(t)= 1()Lx t -←−→X(s)= 1()X s -The ROC of X(s) is such that if 0s was in the ROC of 1()X s , then -0s will be in the ROC of X(s). Putting the two above equations together ,we havex(t)= 1x (-t) =sin(2)()t e t u t --L ←−→X(s)= 1()X s -=-222(1)2s -+, {}Re s <1the denominator of the form 2s -2s+5. Therefore, the poles of X(s) are 1+2j and 1-2j.(a) the given Laplace transform may be written as ()X s =24(1)(3)s s s +++.Clearly ,X(s) has a zero at s=-2 .since in X(s) the order of the denominator polynomial exceeds the order of the numerator polynomial by 1 ,X(s) has a zero at ∞. Therefore ,X(s) has one zero in finite s-plane and one zero at infinity.(b) The given Laplance transform may be written asX(s)=1(1)(1)s s s +-+= 11s -Clearly ,X(s) has no zero in the finite s-plane .Since in X(s) the of the denominator polynomial exceedsthe order the numerator polynomial by 1,X(s) has a zero at ∞.therefore X(s) has no zero in the finite s-plane and one zero at infinity.(c) The given Laplace transform may be written as22(1)(1)()1(1)s s s X s s s s -++==-++Clearly ，X （s ）has a zero at s= in X(s) the order of the numerator polynomial exceeds the order of the denominator polynomial by 1,X(s) has zeros at ∞ .therefore , X(s) has one zero in the s-plane and no zero at infinity .(a) No. From property 3 in Section we know that for a finite-length signal .the ROC is the entire s-plane .therefore .there can be no poles in the finite s-plane for a finite length signal . Clearly in this problem this not the case.(b) Yes. Since the signal is absolutely integrable, The ROC must include, the j ω-axis . Furthermore ,X(s) has a pole at s=2 .therefore, one valid ROC for the signal would be Re{s}<2. From property 5 in section we know that this would correspond to a left-sided signal(C) No . Since the signal is absolutely integrable, The ROC must include , the j ω-axis . Furthermore ,X(s) has a pole at s=2. therefore ,we can never have an ROC of the form Re{s}> α. From property 5 in section we knew that x(t) can not be a right-side signal(d) Yes . Since the signal is absolutely integrable, The ROC must include , the j ω-axis . Furthermore ,X(s) has a pole at s=2 .therefore, one valid ROC for the signal could be α<Re{s}<2 such that α<0 .From property 6 in section ,we know that this would correspond to a two side signalWe may find different signal with the given Laplace transform by choosing different regions of 02s =- 13s =- 2132s j =- 3132s j =-Based on the locations of the locations of these poles , we my choose form the following regions of convergence: (i) Re{s}>- 12(ii)-2< Re{s}<- 12(iii)-3<Re{s}<-2 (iv)Re{s}<-3Therefore ,we may find four different signals the given Laplace transform. From Table ,we know thatG(t)= 2()()(2)L te x t G s X s ←−→=-. The ROC of G(s) is the ROC of X(s) shifted to the right by 2We are also given that X(s) has exactly 2 poles at s=-1 and s=-3. since G(s)=X(s-2), G(s)also has exactly two poles ,located at s=-1+2=1 and s=-3+2=-1 since we are given G(j ω) exists , we may infer that j ω-axis lies in the ROC of G(s). Given this fact and the locations of the poles ,we may conclude that g(t) is a two sidesequence .Obviously x(t)= 2te g(t) will also be two sidedUsing partial fraction expansion X(s)= 4243s s -++Taking the inverse Laplace transform, X(t)=443()2()t t e u t e u t ---The pole-zero plots for each of the three Laplace transforms is as shown in Figure(a) form Section we knew that the magnitude of the Fourier transform may be expressed aswe se that the right-hand side of the above expression is maximum for ω=0 and decreases as ω becomesincreasing more positive or more negative . Therefore 1()H j ω is approximately lowpass (b) From Section we know that the magnitude of the Fourier transform may be express as 1313(length of vector from to -+j )(Length of vector from to --j )ωωwe see that the right-hand side of the above expression is zero for ω= then increams withincreasing |ω| until |ω| reach 1/2. Then it starts decreasing as |ω| increase even further. Therefore | 2H (j )ω| is approximately bandpass.(c) From Section we know that the magnitude of the Fourier transform may be express as21313(length of vector from to -+j )(Length of vector from to --j ) ωωWe see that the right-hand side of the above expression is zero for ω=0. It then increases withincreasing |ω| until |ω| reaches 21. Then |ω| increases,| 3()H j ω| decreases towards a value of1(because all the vector lengths became almost identical and the ratio become 1) .Therefore |3()H j ω| is approximately highpass.X(s) has poles at s=13-213-2(s) has zeros at s=132and 132.From Section we know that |X(j ω)| is1313(Length of vector from toto 22221313(length of vector from to -+j )(Length of vector from to --j )ωωωωThe terms in the numerator and denominator of the right-band side of above expression cancel ourgiving us |X(j ω)|=1.(a) If X(s) has only one pole, then x(t) would be of the form ate - such a signal violates condition 2. Therefore , this statement is inconsistent with the given information.(b) If X(s) has only two poles, then x(t) would be of the form A 0sin()ate t ω- .Clearly such a signal could be made to satisfy all three conditions(Example:0ω=80π,α=19200). Therefore, this statement is consistent with the given information. (c) If X(s) has more than two poles (say 4 poles), then x(t) could be assumed to be of theform 00sin()sin()at btAe t Be t ωω--+. Clearly such a signal could still be made to satisfy all three conditions. Therefore, this statement is consistent with the given information. We have1}Re{,1)(->+=s s s X β.Also,(Length of vector form ω to -1)(Length of vector form ω to 1 I m-1 -3 Re -4 Re 1 Im Im Re1}Re{1),()()(<<--+=s s X s X s G αTherefore, ].11[)(2s s s s G -++-=ααβComparing with the given equation for G(s), ,1-=α .21=β. Since X(s) has 4 poles and no zero in the finite s-plane, we many assume that X(s) is of the form .))()()(()(d s c s b s a s As X ----=Since x(t) is real ,the poles of X(s) must occur in conjugate reciprocal pairs. Therefore, we mayassume that b=*a and d=*c . This result in .))()()(()(**c s c s a s a s As X ----=Since the signal x (t) is also even , the Laplace transform X(s) must also be even . This implies thatthe poles have to be symmetric about the j ω-axis. Therefore, we may assume that c=*a -. This results in .))()()(()(**a S a s a s a s As X ++--=We are given that the location of one of the poles is (1/2)4πj e . If we assume that this pole is a, we have 4444AX(s)=.1111(s-)(s-)(s+)(s+)2222j j j j e e e e ππππ-- This gives us22().11()()4422AX s s s s s =Also ,we are give that()(0)4x t dt X ∞-∞==⎰Substituting in the above expression for X(s), we have A=1/4. Therefore, 221/4().11()()4422X s s s s s =. Taking the Laplace transform of both sides of the two differential equations, we haves X(s)=1)(2+-s Y and s Y(s)=2X(s) . Solving for X(s) and Y(s), we obtain4)(2+=s s s X and Y(s)= 22s 4+.The region of convergence for both X(s) and Y(s) is Re{s}>0 because both are right-hand signals. . Taking the Laplace transform of both sides of the given differential equations ,we obtain ).(])1()1()[(223s X s s s s Y =+++++αααα therefore,.)1()1(1)()()(223αααα+++++==s s s s X s Y s H(a) Taking the Laplace transform of both sides of the given equation, we haveG(s) = s H(s)+ H(s). Substituting for H(s) from above,.1)1()1()1()(22223αααααα++=++++++=s s s s s s s GTherefore, G(s) has 2 poles.(b) we know that H(s) =.))(1(122αα+++s s s Therefore, H(s) has poles at and j ),2321(,1+--α ).2321(j --α If the system has to be stable,then the real part of the poles has to be less than zero. For this to be true, we require that ,02/<-α .,0>α.The overall system show in Figure may be treated as two feedback system of the form shown in figure connected in parallel. By carrying out an analysis similar to that described in Section we find the system function of the upper feedback system to be.82)/2(41/2)(1+=+=s s s s HSimilarly, the system function of the lower feedback system is .21)2/1(21/1)(2+=+=s s s HThe system function of the overall system is now.1610123)()()(221+++=+=s s s s H s H s HSince H(s)=Y(s)/X(s), we may write]123)[(]1610)[(2+=++s s X s s s Y . Taking the inverse Laplace transform, we obtaindtt dx t x t y dt t dy dt t y d )(3)(12)(16)(10)(2+=++. ( a) From problem , we know that differential equation relating the input and output of the RLC circuit is2()()()().d y t dy t y t x t dtdt++=Taking the Laplace transform of this (while nothing that the system is causal and stable), we obtain 2()[1]().Y s s s X s ++= Therefore ,2()1(),()1Y s H s X s s s ==++ 1{}.2e s ℜ>-(b) We note that H(s) has two poles at132s =--132s =-+From Section we know that the magnitude of the Fourier transform may be expressed as1313(Length of vector from to -+j )(Length of vector from to --j )ωω We see that the right hand side of the above expression Increases with increasing |ω| until |ω| reaches 12. Then it starts decreasing as |ω| increasing even further. It finally reaches 0 for |ω|=∞.Therefore 2|()|H j ω is approximately lowpass.(c) By repeating the analysis carried out in Problem and part (a) of this problem with R =310-Ω, wecan show that2()1(),()1Y s H s X s s s ==++ {}0.0005.e s ℜ>-(d) We have33(Vect.Len.from to -0.0005+j )(Vect.Len.from to -0.0005-j )ωωWe see that when |ω| is in he vicinity , the right-hand side of the above equation takes onextremely large value. On either side of this value of |ω| the value of |H (j ω)| rolls off rapidly. Therefore, H(s) may be considered to be approximately bandpass. . (a) The unilateral Laplace transform isX(s) = 20(1)t st e u t e dt -∞--+⎰= 20t st e e dt -∞--⎰=21+s {} 2.e s ℜ>-(b) The unilateral Laplace transform is2(3)0()[(1)()(1)]t st X s t t e u t e dt δδ-∞-+-=++++⎰2(3)0[()]t st t e e dt δ-∞-+-=+⎰612e s -=++ {} 2.e s ℜ>- (c) The unilateral Laplace transform is240()[()()]t t st X s e u t e u t e dt -∞---=⎰240[]t t st e e e dt -∞---=+⎰1124s s =+++ {} 2.e s ℜ>-. In Problem , we know that the input of the RL circuit are related by ).()()(t x t y dtt dy =+Applying the unilateral Laplace transform to this equation, we have ).()()0()(s x s y y s sy =+--(a) For the zero-state response, set (0)0y -=.Also we have u s x =)(L{)(2t u et-}=21+s .Therefore,y(s)(s+1)=.21+sComputing the partial fraction expansion of the right-hand side of the above equation and then taking its inverse unilateral Laplace transform, we have ).()()(2t u e t u e t y t t ---=(b) For the zero-state response, assume that x(t) = we are given that (0)1y -=,.11)(0)(1)(+=⇒=+-s s y s y s sy Taking the inverse unilateral Laplace transform, we have ()().t y t e u t -=Figure2()2()().t t y t e u t e u t --=- . The pole zero plots for all the subparts are shown in figure . (a) The Laplace transform of x(t) isX(s)= 230()t t st e e e dt ∞---+⎰= (2)(3)00[/(2)]|[/(3)]|s t s te s e s -+∞-+∞-++-+ =211252356s s s s s ++=++++(b) Using an approach similar to that show in part (a), we have41(),4L t e u t s -←−→+ {} 4.e s ℜ>-Also,551(),55L t j t e e u t s j -←−→+-and(){}551,555LT t j t e e u t e s s j --←−→ℜ>-++.From this we obtain()()()()55555215sin 52525LTt t j t t j t e t u t e e e e u t js ----⎡⎤=-←−→⎣⎦++ ,where {}5e s ℜ>- .Therefore,()()(){}245321570sin 5,51490100LTt t s s e u t e t u t e s s s s --+++←−→ℜ>-+++. -2 -3Im R a -2 R Im e Im R f Im Rg Im R h-2 2 4 -4 ImRd R b Im c R Im(c)The Laplace transform of ()x t is ()()023t t st X s e e e dt --∞=+⎰()()()()2300/2|/3|s t s t e s e s ----∞-∞⎡⎤⎡⎤=--+--⎣⎦⎣⎦ 211252356s s s s s -=+=---+.The region of convergence (ROC) is {}2e s ℜ<.(d)Using an approach along the lines of part (a),we obtain(){}21,22LT t e u t e s s -←−→ℜ>-+. Using an approach along the lines of part (c) ,we obtain(){}21,22LT t e u t e s s -←−→ℜ<-.From these we obtain()()222224t LT t t s e e u t e u t s --=+-←−→-, {}22e s -<ℜ<. Using the differentiation in the s-domain property , we obtain(){}22222228,2244t LT d s s te e s ds s s -+⎡⎤←−→-=--<ℜ<⎢⎥-⎣⎦-. (e)Using the differentiation in the s-domain property on eq.,we get()(){}2211,222LT t d te u t e s ds s s -⎡⎤←−→-=ℜ>-⎢⎥+⎣⎦+.Using the differentiation in the s-domain property on eq ,we get ()(){}2211,222LT t d te u t e s ds s s ⎡⎤--←−→=-ℜ<⎢⎥-⎣⎦-.Therefore,()()()(){}222224,2222t LT t t st e te u t te u t e s s s ---=--←−→-<ℜ<+-.(f)From the previous part ,we have ()()(){}2221,22LT t t t e u t te u t e s s -=--←−→-ℜ<-.(g)Note that the given signal may be written as ()()()1x t u t u t =-- .Note that (){}1,0LTu t e s s←−→ℜ>.Using the time shifting property ,we get(){}1,0s LT e u t e s s--←−→ℜ>.Therefore ,()1x t()()11,sLT e u t u t s----←−→ All s . Note that in this case ,since the signal is finite duration ,the ROC is the entire s-plane.(h)Consider the ()()()11x t t u t u t =--⎡⎤⎣⎦that the signal ()x t may beexpressed as ()()()112x t x t x t =+-+ . We have from the previous part()()11sLT e u t u t s----←−→, All s . Using the differentiation in s-domain property ,we have()()()12111s s s LT d e se e x t t u t u t ds ss ---⎡⎤--+=--←−→=⎡⎤⎢⎥⎣⎦⎣⎦, All s . Using the time-scaling property ,we obtain()121s s LT se e x t s --+-←−→, All s .Then ,using the shift property ,we have()21212s sLT s se e x t es ---+-+←−→ ,All s . Therefore ,()()()21122112s s s sLT s se e se e x t x t x t e s s----+--+=+-+←−→+, All s. (i) The Laplace transform of ()()()x t t u t δ=+ is (){}11/,0X s s e s =+ℜ>.(j) Note that ()()()()33t u t t u t δδ+=+.Therefore ,the Laplace transform is the same as the result of the previous part. (a)From Table ,we have()()()1sin 33x t t u t =.(b)From Table we know that()(){}2cos 3,09LT st u t e s s ←−→ℜ>+. Using the time scaling property ,we obtain()(){}2cos 3,09LT s t u t e s s -←−→-ℜ<+Therefore ,the inverse Laplace transform of ()X s is()()()cos 3x t t u t =--.(c)From Table we know that ()()(){}21cos 3,119LTt s e t u t e s s -←−→ℜ>-+.Using the time scaling property ,we obtain ()()(){}21cos 3,119LTt s e t u t e s s -+-←−→-ℜ<-++. Therefore ,the inverse Laplace transform of ()X s is()()()cos 3t x t e t u t -=--.(d)Using partial fraction expansion on ()X s ,we obtain ()2143X s s s =-++ .From the given ROC ,we know that ()x t must be a two-sided signal .Therefore ()()()432t t x t e u t e u t --=+-.(e)Using partial fraction expansion on ()X s ,we obtain()2132X s s s =-++. From the given ROC ,we know that ()x t must be a two-sided signal ,Therefore,()()()332ttx t e u t e u t --=+-.(f)We may rewrite ()X s as ()2311s X s s s =+-+()()2211/23/2s =-+()()()()222211/23/21/23/2s s =+-+-+Using Table ,we obtain()())())()/2/23cos 3/23sin3/2t t x t t e t u t e t u t δ--=+.(g)We may rewrite ()X s as ()()2311s X s s =-+.From Table ,we know that(){}21,0LT tu t e s s ←−→ℜ>.Using the shifting property ,we obtain()(){}21,11LT t e tu t e s s -←−→ℜ>-+.Using the differentiation property ,()()()(){}2,11LT t t t d s e tu t e u t te u t e s dt s ---⎡⎤=-←−→ℜ>-⎣⎦+. Therefore,()()()()33t t x t t e u t te u t δ--=--.four pole-zero plots shown may have the following possible ROCs:·Plot (a): {}2e s ℜ<- or {}22e s -<ℜ< or {}2e s ℜ>.·Plot (b): {}2e s ℜ<- or {}2e s ℜ>-. ·Plot (c): {}2e s ℜ< or {}2e s ℜ>. ·Plot (d): Entire s-plane.Also, suppose that the signal ()x t has a Laplace transform ()X s with ROC R . (1).We know from Table that()()33LT te x t X s -←−→+.The ROC 1R of this new Laplace transform is R shifted by 3 to the left .If ()3t x t e - is absolutely integrable, then 1R must include the jw -axis.·For plot (a), this is possible only if R was {}2e s ℜ> . ·For plot (b), this is possible only if R was {}2e s ℜ>-. ·For plot (c), this is possible only if R was {}2e s ℜ> . ·For plot (d),R is the entire s-plane. (2)We know from Table that(){}1,11LT t e u t e s s -←−→ℜ>-+.Also ,from Table we obtain()()(){}2,11LT t X s x t e u t R R e s s -⎡⎤*←−→=ℜ>-⎡⎤⎣⎦⎣⎦+ If ()()te u t x t -*is absolutely integrable, then 2R must include the jw -axis.·For plot (a), this is possible only if R was {}22e s -<ℜ<. ·For plot (b), this is possible only if R was {}2e s ℜ>-. ·For plot (c), this is possible only if R was {}2e s ℜ< . ·For plot (d),R is the entire s-plane.(3)If ()0x t = for 1t > ,then the signal is a left-sided signal or a finite-duration signal . ·For plot (a), this is possible only if R was {}2e s ℜ<-. ·For plot (b), this is possible only if R was {}2e s ℜ<-. ·For plot (c), this is possible only if R was {}2e s ℜ< . ·For plot (d),R is the entire s-plane.(4)If ()0x t =for 1t <-,then the signal is a right-sided signal or a finite-duration signal ·For plot (a), this is possible only if R was {}2e s ℜ>.·For plot (b), this is possible only if R was {}2e s ℜ>- . ·For plot (c), this is possible only if R was {}2e s ℜ>.·For plot (d),R is the entire s-plane..(a)The pole-zero diagram with the appropriate markings is shown Figure .(b)By inspecting the pole-zero diagram of part (a), it is clear that the pole-zero diagram shown in Figure will also result in the same ()X jw .This would correspond to the Laplace transform()112X s s =-, {}12e s ℜ<.(c)≮()X jw π=-≮()1X jw .(d)()2X s with the pole-zero diagram shown below in Figure would have the property that ≮()2X jw =≮()X jw .Here ,()211/2X s s -=-. (e) ()()21/X jw X jw =.(f)From the result of part (b),it is clear that ()1X s may be obtained by reflecting the poles and zeros in the right-half of the s-plane to the left-half of the s-plane .Therefore, ()11/22s X s s +=+.From part (d),it is clear that ()2X s may be obtained by reflecting the poles (zeros) in the right-half of the s-plane to the left-half and simultaneously changing them to zeros (poles).Therefore,()()()()2211/22s X s s s +=++plots are as shown in Figure . Table we have()()(){}2111,22LT t x t e u t X s e s s -=←−→=ℜ>-+and()()(){}3111,33LTt x t e u t X s e s s -=←−→=ℜ>-+.Using the time-shifting time-scaling properties from Table ,we obtain()(){}22112,22s LT s e x t e X s e s s ---←−→=ℜ>-+and()(){}33223,33s LT s e x t e X s e s s---+←−→-=ℜ>--.Therefore, using the convolution property we obtain ()()()()23122323s s LTe e y t x t x t Y s s s --⎡⎤⎡⎤=-*-+←−→=⎢⎥⎢⎥+-⎣⎦⎣⎦. clues 1 and 2,we know that ()X s is of the form()()()AX s s a s b =++. Furthermore , we are given that one of the poles of ()X s is 1j -+.Since ()x t is real, the poles of ()X s must occur in conjugate reciprocal pairs .Therefore, 1a j =-and 1b j =+and ()()()11AH s s j s j =+-++. From clue 5,we know that ()08X =.Therefore, we may deduce that 16A = and ()21622H s s s =++ .Let R denote the ROC of ()X s .From the pole locations we know that there are two possible choices of R .R may either be {}1e s ℜ<-or {}1e s ℜ>-.We will now useclue 4 to pick one .Note that()()()()22LTt y t e x t Y s X s =←−→=-.The ROC of ()Y s is R shifted by 2 to the right .Since it is given that ()y t is not absolutely integrable ,the ROC of ()Y s should not include the jw axis -.This is possible only ofR is {}1e s ℜ>-..(a) The possible ROCs are(i) {}2e s ℜ<-.(ii) {}21e s -<ℜ<-. (iii) {}11e s -<ℜ<.( iv) {}1e s ℜ>.(b)(i)Unstable and anticausal. (ii) Unstable and non causal. （iii ）Stable and non causal. (iv) Unstable and causal. .(a)Using Table ,we obtain(){}1,11X s e s s =ℜ>-+and(){}1, 2.2H s e s s =ℜ>-+(b) Since ()()()y t x t h t =*,we may use the convolution property to obtain()()()()()112Y s X s H s s s ==++.The ROC of ()Y s is {}1e s ℜ>-.(c) Performing partial fraction expansion on ()Y s ,we obtain . ()1112Y s s s =-++.Taking the inverse Laplace transform, we get()()()2t t y t e u t e u t --=-. (d)Explicit convolution of ()x t and ()h t gives us()()()y t h x t d τττ∞-∞=-⎰()()20t e e u t d ττττ∞---=-⎰t t e e d ττ--=⎰ for0t >()2.t t e e u t --⎡⎤=-⎣⎦ the input ()()x t u t =, the Laplace transform is (){}1,0.X s e s s=ℜ>The corresponding output ()()1t t y t e te u t --⎡⎤=--⎣⎦ has the Laplace transform()()(){}221111,0111Y s e s s s s s s =--=ℜ>+++. Therefore,()()()(){}21,0.1Y s H s e s X s s ==ℜ>+ Now ,the output ()()3123t t y t e e u t --⎡⎤=-+⎣⎦has the Laplace transform()()(){}12316,0.1313Y s e s s s s s s s =-+=ℜ>++++ Therefore , the Laplace transform of the corresponding input will be()()()()(){}1161,0.3Y s s X s e s H s s s +==ℜ>+ Taking the inverse Laplace transform of the partial fraction expansion of ()1,X s we obtain ()()()3124.t x t u t e u t -=+.(a).Taking the Laplace transform of both sides of the given differential equation and simplifying, we obtain ()()()212Y s H s X s s s ==--. The pole-zero plot for ()H s is as shown in figure .b).The partial fraction expansion of ()H s is()1/31/321H s s s =--+. (i).If the system is stable ,the ROC for ()H s has to be {}12e s -<ℜ< . Therefore ()()()21133t t h t e u t e u t -=---.(ii).If the system is causal, the ROC for ()H s has to be {}2e s ℜ> .Therefore()()()21133t t h t e u t e u t -=-.(iii)If the system is neither stable nor causal ,the ROC for ()H s has to be {}1e s ℜ<-.Therefore ,()()()21133t t h t e u t e u t -=--+-. If ()2t x t e =produces ()()21/6t y t e =,then ()()21/6H =. Also, by taking the Laplace transform of both sides ofthe given differential equation we get ()()()()442s b s H s s s s ++=++.Since ()21/6H = ,we may deduce that 1b = .Therefore()()()()()222424s H s s s s s s +==+++. ()()()t t t x t e e u t e u t --==+-,()()(){}112,111111X s e s s s s s -=-=-<ℜ<+-+-. We are also given that ()2122s H s s s +=++.Since the poles of ()H s are at 1j -±, and since ()h t is causal ,we may conclude that the ROC of()H s is {}1e s ℜ>-.Now()()()()()22221Y s H s X s s s s -==++-. The ROC of ()Y s will be the intersection of the ROCs of ()X s and ()H s .This is {}11e s -<ℜ<. We may obtain the following partial fraction expansion for ()Y s :()22/52/56/5122s Y s s s s +=-+-++. We may rewrite this as ()()()222/521411551111s Y s s s s ⎡⎤⎡⎤+=-++⎢⎥⎢⎥-++++⎢⎥⎢⎥⎣⎦⎣⎦.0 -1 2 ReImFigureNothing that the ROC of ()Y s is {}11e s -<ℜ<and using ,we obtain ()()()()224cos sin 555t t t y t e u t e tu t e tu t --=-++know that()()(){}111,0LTx t u t X s e s s=←−→=ℜ> Therefore,()1X s has a pole at0s =.Now ,the Laplace transform of the output()1y t of the system with()1x t as the input is()()()11Y s H s X s =Since in clue 2, ()1Y s is given to be absolutely integrable ,()H s must have a zero at 0s =whichcancels out the pole of ()1X s at 0s =.We also know that()()(){}2221,0LT x t tu t X s e s s=←−→=ℜ> Therefore , ()2x s has two poles at 0s =.Now ,the Laplace transform of the output ()2y t of the system with ()2x t as the input is()()()22Y s H s X s =Since in clue 3, ()2Y s is given to be not absolutely integrable ,()H s does not have two zeros at0s =.Therefore ,we conclude that ()H s has exactly one zero at 0s =. From clue 4 we know that the signal ()()()()2222d h t dh t p t h t dt dt=++is finite duration .Taking the Laplace transform of both sides of the above equation ,we get ()()()()222P s s H s sH s H s =++. Therefore,()()222P s H s s s =++.Since ()p t is of finite duration, we know that ()P s will have no poles in the finite s-plane .Therefore, ()H s is of the form()()1222Ni i A s z H s s s =-=++∏,where i z ,1,2,....,i N =represent the zeros of ()P s .Here ,A is some constant.From clue 5 we know that the denominator polynomial of ()H s has to have a degree which is exactly one greater than the degree of the numerator polynomial .Therefore, ()()1222A s s H s s s -=++.Since we already know that ()H s has a zero at 0s = ,we may rewrite this as ()222As H s s s =++ From clue 1 we know that ()1H 0.2 this ,we may easily show that 1A = .Therefore,()222s H s s s =++. Since the poles of ()H s are at 1j -± and since ()h t is causal and stable ,the ROC of ()H s is {}1e s ℜ>-..(a) We may redraw the given block diagram as shown in Figure . From the figure ,it is clear that()()1F s Y s s=. Therefore, ()()1/f t dy t dt =. Similarly, ()()/e t df t dt =.Therefore, ()()221/e t d y t dt =.From the block diagram it is clear that()()()()()()()21111266d y t dy t y t e t f t y t y t dtdt=--=--.。
信号与系统奥本海姆英文版课后答案chapter
Chapter 1 Answers 1.1 1.2 Con vert ing from polar to Cartesia n coord in ates: con vert ing from Cartesia n to polar coord in ates: jo5 =5e, 1 . 3 耳 J e ■2 2 j(1-j)二e 4 , 2, =3e七 =2e七1.3.旳& oe(b) X 2(t )弋心 4)P『砧k(c)dt 1 j^T e - 421-j2 b 2 _2 一1一「3一 eP :.=0, because E:.:：:::,X 2(t )二1 |21 TX 2(t)|dt 斗汁亍..2dt - dt-：： J-oO(d)-/'2TX 2(t)=cos(t). Therefore,P oo =lim — fT存2T L1 "Xl [n]u[n]2P :一=0, becausex 2[n]=e a< ? 8),.Therefore, E - Jx 2(t)| lim1 =1 T _ L :E ：：= jx 3(t)「dt = ：8S (t)2dt -：T 工dt=1 COS(2t11 dt =_2 2.Therefore, ；cos(t 「dt尹丰；'|x1[ n]|l 4丿2=1. therefore,nu[n]jx i [ n]|4 =3X 2[n]_1NP：: = lim - x 2[n]N Y 2N +1 7以2屮」| nE：：'n] 2二：：X 3[n]=cos1 . Therefore, 14丿 21 Nlim -------- '、'N ；：2N E* Sjx 3[n]| 匚cosgn) ^cos(^j- N 1 亠cos( n) 17( 2 )二1 烛 2 2 2n) 匚1 V 伍1 lim --------- \ cos —j2N 4 The signal x[n] is shifted by 3 to the right. The shifted signal lim -— =N：.：2N 1 1.4. (a) will be zero for n <1, And n>7. (b) The signal x[n] is shifted by 4 to the left. The shifted signal will be zero for n<-6. And n>0. (c) The sig nal x[n] is flipped sig nal will be zero for n<-1 and n>2. (d) The signal x[n] is flipped and the flipped signal is shifted by 2 to the right. The new Sig nal will be zero for n v-2 and n>4. (e) The signal x[n] is flipped and the flipped and the flipped signal is shifted by 2 to the left. This new sig nal will be zero for n<-6 and n>0. 1.5 . (a) x(1-t) is obtained by flipping x(t) and shifting the flipped signalby 1 to the right. Therefore, x (1-t) will be zero for t>-2.(b) From (a), we know that x(1-t) is zero for t>-2. Similarly, x(2-t) is zero for t>-1, Therefore, x (1-t) +x(2-t) will be zero for t>-2.(c) x(3t) is obta ined by lin early compressi on x(t) by a factor of 3. Therefore, x(3t) will bezero for t<1.(d) x(t/3) is obta ined by lin early compressi on x(t) by a factor of 3. Therefore, x(3t) will bezero for t<9.1.6 (a) x«t ) is not periodic because it is zero for t<0.(b) X 2[ n ]=1 for all n. Therefore, it is periodic with a fun dame ntal periodof 1. (c) X 3[ n ] is as shown in the Figure S1.6.X3Therefore 1.7 . (a) 邛X」n 】)= /ith a fun -4 1r2Xi [n] X -ame ntal period 11 4=-(u[ n] —u[r 2- 5n—4] +u[—n] —u[ of 4. —n —4]) Therefore, x[n] is zero for (b) Si nee x*t) is an odd sig nal, 昭X 3[n]島如如⑴耳律| X 1【n 】 >3. (c) x [n ] is zero for all values of t. f 1 ) u[n -3] ■ 12丿丄i u[ J • 3]I(d) Therefore, X^n] is zero when n <3 and when 1, 丄，八、1 一 五t . n 「：： 1 , 1 一五t 5t, 1 (xe 尸^(xQ+xZ)*1 e u (t_2)-eu (T+2)]5t 1.8 . (a)(b) (c) ‘ f 2、八4、‘ 八4、‘‘ 2 — - Therefore, ^(t) is zero only when 叹{为⑴}产「2 =2e0t cos(Ot 二) [ •：{X2(t)}二.2 cos 仃)cos(3t 亠2c ) =cos(3t) =e°tcos(3t 亠0) ： T{X3(t)} =e^sin(3 二 t)=e±sin(3t ?)(d)1.9 . (a) (b) Therefore, X 2 • 2t . t -2t 二{X 4(t)} - -e sin(1°0t)=e sin(1°0t 二)=ecos(1°Ct 3) (t) is a periodic complex exp onen tial. (t) is a complex exp onen tial multiplied by a decay ing exp onen tial. X 1(C) X 2(t) is not periodic.X3[n] is a periodic signal.X3[n]=e j7n?= e jn?.xjn] is a complex exp onen tial with a fun dame ntal period (d)加 is a periodic sig nal. The fun dame ntal period isof —.2give n byN=m(2 )3 二 / 5=m(10) By choos ing m=3. Weobta in the fun dame ntal period to be 10. (e) X5【n] is not periodic. cannot find any in teger m such that m(x [n] is a complex exponential with w =3/5. We 2二)is also an integer. W o Therefore, x [n] is not periodic. 1.10 . x (t)=2cos(10t +1)-sin(4t-1) Period of first term in the RHS = Period of first term in the RHS = 2 二 - T =2 Therefore, the overall signal is periodic with a period which the least com mon multiple of the periods of the first and second terms. This is equal to 二. 1.11 . .7 2x[ n] = 1+ e F?e 3Period of first term in the RHS =1. Period of second term in the RHS = =7 (whe n m=2) 2 一 =5 (whe n m=1) Therefore, the overall signal x[n] is periodic with a period which is the least com mon Multiple of the periods of the three terms inn x[n].This is equal to 35. 1.12 . The signal x[n] is as shown in figure S1.12. x[n] can be obtained by flipp ing u[n] and the n Shifting the flipped signal by 3 to the right.This implies that M=-1 and no=-3.Period of second term in the RHS = Therefore, x[n]二 u[-n+3].1.13 y (t)=E ::Figure S 1.12二 42二 dt (2)Therefore x (t) and its derivativeX[n ]g (t) are show n in FigureS1.14.1.14 The signalLet x 3(t) be a lin ear comb in ati on of x 1 (t) x 1 (t)+b x 2 (t)Where a and b are arbitrary scalars .If x system ,the n the corresp onding output y X 3( sin(t))This implies that A 1=3, t 1 =0, A 2 =-3, and t 2=1.1.15 (a) The signal x 2 [n], which is the in put to S 2 , is the same asy 1[ n] .Therefore ,y 2【n]= x 2【n-2]+ 12 x 2〔n-3]=y 11【n-2]+ 1 2y 1【n-3]=2x 1 [n-2] +4x 1 [n-3] + 12( 2x1[n-3]+ 4x1[n-4])=2x 』n-2]+ 5x Jn-3] + 2x Jn-4]The in put-output relatio nship for S isy[n ]=2x [n-2]+ 5x [n-3] + 2x [n-4](b) The in put-output relatio nship does not change if the order in which SpndJn], which is the in put to S 1 isS 2 are conn ected series reversed. . We can easily prove this assu mingthat S 1 follows S 2 • In this case , the sig nal x the same as y 2[n]. Therefore y =2( x =2 x i [n]+ 4x i [n-1] 2【n]+4 y 2【n-1] 1 12 [n-2]+ - x 2 [n-3] )+4(x 2 [n-3]+ 12 2 2 [n-2]+5x 2 [n-3]+ 2 x 2 [n-4] Jn] =2x =2y 2 [n-4]) The in put-output relatio nship for S is once aga in y[n ]=2x[ n-2]+ 5x [n-3] + 2x [n-4] 1.16 (a)The system is not memory less because y[n] depends on past values of x[n]. (b) The output of the system will be y[n]= 、. [ n ]、 [ n — 2(c) F rom the result of part (b), we may con clude that the system output is always zero for in puts of the form [ n - k], k ?. Therefore , the system is not in vertible . 1.17 (a) The system is not causal because the output y(t) at some time may depend on future values of x(t). For instanee , y(- ■: )=x(0). (b) Con sider two arbitrary in puts x 1 (t)a nd x x 1 (t) > y 1 (t)= x 2(t). i (sin (t)) =0 X 2(t) > y 2 (t)= x 2 (sin(t)) and x 2(t).That is , x 3(t)=a3 (t) is the in put to the give n3(t) is y3(t)==a x 1(si n( t))+ x 2(si n( t))=a y 1 (t)+ by 2 (t) Therefore , the system is lin ear. 1.18 .(a) Con sider two arbitrary in puts x 』n]and x 2 [n]. n*io X 』n] > y 1 [n] = '、xdk]k =n _n °n *oX 2【n ] > y 2[n] = 、X 2[k] k ^-no 』n] and x 2【n] is the in put Let x 3 [n] be a lin ear comb in atio n of x X 3[ n]= ax 1[n ]+b x where a and b are arbitrary scalars. If 2 [n]. That is : X 3【n] to the given the n the corresp onding output y 3【n] is 3【n]= system,n n o' X 3【k]k=n 』on n o =ay n n o(ax 1[k ] bx 2[k])=a 二 xjkl +b 二 X 2[k ]k =n -n o 』n]+b y 2【n]心-n o k =n -r i o Therefore the system is lin ear. (b) Con sider an arbitrary in put x y i [n]= Jn ].Let n n o ' x/k] k =n -n obe the corresp onding output .Con sider a sec ond in put x shifti ng x 1[n] in time: 2 [n] obtained by X 2【n]= x 』n-n J The output corresp onding to this in put is n n 0 Also note that y 2【n ］二 n n o ' x 2[k] = ' x 」k - n 1]= k =n .n '0k =n. i [n- n Therefore , This implies that the system is time-i nvaria nt. (c) If x[n] <B, then y[n] -(2 n Therefore ,C 1.19 (a) (i) > y 1 (t)= t 2[n]= y n -n 1 m Z X 1[k ]k -n ■ q -n° 1]= ' X 1[k]. k -n .n 1 -n ° [n- n 1] y[n] -(2 n o +1)B. Con sider two arbitrary in puts 2X 1(t-1) o +1)B. x 1 (t) and x 2(t). i(t)x t 2X 2(t-1) Let x 3(t) be a lin ear comb in atio n of x 1 (t) x 1 (t)+b x 2 (t)2(t)、y 2 (t)=and x 2(t).That is3(t)=a=a 2x 12[n-2]+b 2x 22[n-2]+2ab x 』n-2] x 2[n-2] -ay 』n]+b y 2 [n]Therefore the system is not linear. (ii) Con sider an arbitrary in put x 』n]. Let ybe the corresp onding output .Con sider a sec ond in put x shifti ng x 1[n] in time:xjn]= x 』n-n dThe output corresp onding to this in put is y 2 [n] = xThis implies that the system is time-i nvaria nt. (c) (i) Con sider two arbitrary in puts x』n]and x 2 [n]. x 』n] rjn] = x 』n+1]- x 』n-1] x 2 [n ] - y 2[n] = x 2 [n+1 ]- x 2[n -1]where a and b are arbitrary scalars. If x 3(t) then the corresponding output y 3 (t) is y =t 2(ax 1(t-1)+b x 2(t-1)) =ay 1 (t)+b y 2 (t)Therefore , the system is lin ear. (ii) Con sider an arbitrary in puts x 1 (t).Let is the in put to the give n system,3(t)= t 2X 3 (t-1)be the corresp onding output .Con sider a sec ond in put x shifting x 1 (t) in time: 1(t)= t2X 1(t-1)2 (t) obtained byX 2(t)二 xThe output corresponding to this input t 0) Also note that y 2(t) Therefore the system is not time-i nvaria nt. (b) (i) Con sider two arbitrary in puts x yd n] = x i 2[n-2] x be a lin ear comb in ati on of x 1 1(t-t)is y2(t) = t 2x 2(t-1)= t 2x 1 (t- 1-1(t-to)=(t-t』n]and x 2 [n]. o)2X 1(t-1- t 0)-1[n]—Let X 3(t) X 2【n] where a and b are arbitrary scalars. If the n the corresp onding output y 3[n] is=(a x 』n-2] +b x 2 [n] = x [n]and x 2 [n].That is x2【n ] > y 22 [n-2]. 3【n]二 ax 』n]+b X 3【n] is the in put to the y 3 [n] = x 』n-2])given system, 32[n-2] 1[n] = x 12 [n-2]2[n] obtained by2 [n-2].= x / [n-2- n (J』n- n 0]= x 12[n-2- n 0 2【n]= y 1 [n- n 』Also note thatTherefore ,Let x 3[n] be a lin ear comb in ati on of x Jn] and x 2 [n]. That is :Therefore the system is lin ear.(ii) Con sider an arbitrary in put x be the corresp onding output .Con sider a sec ond in put x shifting x1[n] in time: x 2[n]= x 』n-n 0]The output corresp onding to this in put isy 』n]二 x 2【n +1]- x 2 [n -1]= x 』n+1- n °]- x 』n-1- n °]Also note thaty』n-n 0]= x 』n+1- n 0]- x 』n-1- n 0]Therefore ,y2[n]= y 』n-n 0 ]This implies that the system is time-inva ria nt.(d) (i) Con sider two arbitrary in puts x1 (t) and x 2(t).X 1 (t) t y 1 (t)= od 'x 1 (t) /X 2(t)T y 2(t)=Od * 2 (t) /Let x 3 (t) be a lin ear comb in ati on of x1(t) and x 2(t).That is x3(t) =ax 1 (t)+b x 2 (t)where a and b are arbitrary scalars. If x 3(t) is the in put to the give n system, then thecorresponding output y 3 (t) is y3 (t)= Od 》3 (t) ?二空1 (t) + b X 2 (t)』=a od <x 1 (t) '+bQd \2 (t)上 ay 1 (t)+b y 2(t)Therefore the system is lin ear. (ii) Con sider an arbitrary in puts x1 (t).Lety 1 (t)= Od&1(t)}=x 1⑴-x 1(-t )2be the corresp onding output .Con sider a sec ond in put x 2 (t) obta ined byshifting x1 (t) in time:X 2(t)二 x 1 (t-t 0)The output corresp onding to this in put isy 2 (t)= c )d "x 2 (t) ；= x2(t )- X 2( Y ) 2=X1(t-t o )-X 1(T —t o )2X 3【n]= ax 1[n ]+b x where a and b are arbitrary scalars. the n the corresp onding output y 3 [n] is 2【n] If x 3[n] is the input to the given system, 3【n]= x 3【n+1]- x =a x 1[ n+1]+b X 2【n3【n-1] +1]-a x 』n-1]-b x 2 [n -1] =a(x 』n+1]- x 』n-1])+b(x 2 [n +1]- x 2 [n-1]) =ay i [n]+b y 别]』n].Lety 』n]二 x 』n+1]- x 』n-1]2 [n] obtained byAlso note that y 1 (t-t 0)= X1(t-t o)-x1(T —t。
信号与系统历年习题解答
3t
(a) Determine the system function the ROC of
H (s)
H (s)
, then indicate
(b) Determine the unit impulse response h(t ) of this system causal and stable? (c) For the input x(t ) = e3t ,compute the output y (t ) (d) Determine the differential equation of this system.
j
function
H ( s)
, then
3 -1 2
indicate the ROC of
H (s)
Figure 3
pole-zero pattern
(b) Determine the unit impulse response h(t ) of this system, is this system causal? (c) For the input x(t ) = tet u (t ) ,compute the output y(t ) (d) Draw a block diagram representation of this system.
典型习题解析
1. Consider a LTI system with unit impulse response
dδ (t ) + δ (t ) , the output h(t ) = e u (t ) , if the input x(t ) = dt y (t ) is ( a ).
信号与系统Chapter 10 Answers
Chapter 1010.2. Soluti on:nn zn x Z X -∞-∞=∑=][)(n n n z n u -∞-∞=-=∑]3[)51(nn z )51(13-∞=∑=131315111251511)51(-----⋅=-=zzzzROC:1|51|1<-zi.e. 51>z10.3. Solution:][][)1(][0n n u a n u n x nn--+-=∴11111`1)(00----⋅++=azzazz X n n , ROC: ||||1a z <<and 2||1<<z∴ 2||=a , 0n arbitrary10.6. N, N, Y , Y Solution:Because ][n x can be absolutely summable , the ROC of )(z X contains1||=z .And also )(z X have a pole at 21=z .Then ][n x can be a right-sided signal or a two-sided signal.10.7. 3 Solution:According to the expression of )(z X , we know that )(z X has fourpoles:2j ±,21-,43-, which three of them has the same absolute values. So there arethree different regions of convergence which could correspond to )(z X .10.9. Solution:2||,)21)(1(311)(111>+--=---z zzzz X2||,219719211>++-=--z z z∴ ][)2(97][92)(n u n u n x n-+=10.10 Solution: (a). 92,32,1-(b). 18,6,3-10.11 Solution: 0||,211)21(2111211)21(1211102410241)(110111101110>---=--=--⋅=-------z zzzzzzz z X∴ ]10[)21(][)21(][--=n u n u n x nn= ])10[][()21(--n u n u n= ⎪⎩⎪⎨⎧≤≤others n n,.........090,...)21(10.13 Solution:(a) ]6[][]1[][][--=--=n n n x n x n g δδ∴ 0||,1)(6>-=-z z z G(b) ][][][][][][n u n g k n u k g k g n x k n k *=-==∑∑∞-∞=-∞=∴1||,1111)()(161>--=-⋅=---z zz zz G z X10.16 Solution: (a). No Because the order of the numerator is greater than the order of the denominator. (b). Y es. (c). No.Because the expression has a pole at 34-=z , and the premise tells us thesystem is stable, i.e. the ROC include 1=z . So the Roc can’t be right-sided. 10.17 Solution: (a). Y es. (b). Y es. According to the condition given, it ’s easy to get the conclusions. 10.18 Solution: (a). 21[][1][2][]6[1]8[2]39y n y n y n x n x n x n --+-=--+-(b). Y es. This system is stable 10.23 Solution:(1). 21||,4111)(21>--=--z zz z XMethod 1: )211)(211(14111)(11121------+-=--=zzzzz z X21||,2112/12112/311>--++=--z zz∴][)21(21][)21(23][n u n u n x nn--= Method 2:∴,41]3[,41]2[,1]1[,1]0[-==-==x x x x∴][)21(21][)21(23][n u n u n x nn --=(2).21||,4111)(21<--=--z zz z XMethod 1: )211)(211(14111)(11121------+-=--=zzzzz z X21||,2112/12112/311<--++=--z zz∴]1[)21(21]1[)21(23][--+----=n u n u n x nnMethod 2:∴ ,16]4[,16]3[,4]2[,4]1[,0]0[-=-=--=-=-=x x x x x∴]1[)21(21]1[)21(23][--+----=n u n u n x nn(3). ][)21(23][2][n u n n x n+-=δ (4).]1[)21(23][2][----=n u n n x nδ(5). ]1[)21)(1(23][)21(2][+++-=n u n n u n x nn(6).]2[)21)(1(23]1[)21(2][--+--=n u n n u n x nn10.24 Solution: (a). )211(1)211)(21(2125121)(1111211--------=---=+--=zzzz zzz z X21||>z (because][n x isROC:absolutely summable)][)21(][n u n x n=∴(b).∴ ,41]3[,21]2[,1]1[,1]0[-==-==x x x x∴][)21(2][][n u n n x n-+-=δ(c). )411)(211(381411381413)(1112111-------+-=--=--=zzzzzz zz z X 1141142114--+-+-=zzROC:21||>z (because ][n x is absolutely summable)∴ ][)41(4][)21(4][n u n u n x n n --=10.31 Solution:According to the conditions:Suppose )21)(21()(3/3/2ππj j ez ez kzz X ---=Because 38)1(=X , k=2So )21)(21(2)(3/3/2ππj j ez ez zz X ---=10.47 Solution:(1) (2)According to the condition 1:0)2(|)(][2=-⋅=-=nz z H n ySo 0)2(=-HAccording to the condition 2][)21(][n u n x n=↔12111--z][)41(][][n u a n n y n+=δ↔1114114114111)(-----+=-+=zza za z YSo0)811()411)(811()411()211)(411()2()2()2(2111=++++=---+=--=--=---a zzza X Y H zthen 89-=a(2)∞<<∞--=----===n H n x H n y ,41411)211)(41891()1(][)1(][。
通信工程专业英语习题答案
通信工程专业英语习题答案练习参考答案第1单元信号与系统1. 完形填空(1) band-limited, cutoff frequencies, to, between, the information rate is proportional to the bandwidth of the channel, on, to, by, by simply increasing the number of levels, noise will cause the receiver to mistake one level for another, no matter how elaborately the data is coded, between, theoretical maxima.(2) from, a loss of information , the twisted pair, addition of noise to the signal, depends on the signal frequency, the high-frequency components of the signal, over, between, through, of, with, including.(3) for, to, In some cases, over, (microwave relay, coaxial cable, or fiber), ADC, to, thereby achieving a reduction in required channel bandwidth, of, signal recovery, (Techniques such as filtering, auto-correlation and convolution).2. 英汉互译(1) “电信”（telecommunication）一词来源于希腊语tele（含义为“遥远的”）和拉丁语communicatio（涵义为“连接”）。
信号与系统第一章答案
w0 )*m, and m=3. w0 )*m=10
Because
w0 =3 /5, N=(2 /
m/3 ,
it’s not a rational number.
13/37
5 Exercises Answers
1.11 Solution
x[n ] 1 e e
j 4 n 7 j 4 n 7 j 2 n 5
Then,
y[n] 2 x[n 2] 5x[n 3] 2 x[n 4]
16/37
5 Exercises Answers
(b) No. For it’s linearity.
the relationship between
y1 [ n ]
and x 2 [n]
is the same in-out relationship with (a).
2
9/37
5 Exercises Answers
(e) x 2 [n] e
E
j(
) 2n 8 2 j( ) 2n 8
n
e
12
n -
N 1 1 1 P lim E lim 1 lim 2N+1 1 N 2N 1 N 2N 1 N 2N 1 n -N (f) x 2 [n ] cos( 4 n ) n 1 cos 2 E cos2 ( n ) 4 2 n n 1 cos n N 1 1 1 1 2 P lim E lim lim N N 2N 1 N 2N 1 N 2 2N 1 2 n N
信号与系统奥本海姆英文版课后答案chapter9
s1 = −3
1 3 s2 = − + j 2 2
1 3 s3 = − − j 2 2
Based on the locations of the locations of these poles , we my choose form the following regions of convergence: (i) Re{s}>- 1
( s − 1)( s + 1)
s −1
Clearly ,X(s) has no zero in the finite s-plane .Since in X(s) the of the denominator polynomial exceeds the order the numerator polynomial by 1,X(s) has a zero at ∞ .therefore X(s) has no zero in the finite s-plane and one zero at infinity. (c) The given Laplace transform may be written as
s+5 s+β
Ae( s +5)t0 s+5
β.
158
The ROC of X(s) is such that if s0 was in the ROC of X 1 ( s) , then - s0 will be in the ROC of X(s). Putting the two above equations together ,we have L −t → X(s)= X 1 ( − s ) =2 Re {s} <1 x(t)= x1 (-t) = e sin(2t )u ( −t ) ←⎯ ,
信号与系统奥本海姆英文版课后答案
Chapter 1 Answers1.1 Converting from polar to Cartesian coordinates:111cos 222j e ππ==- 111cos()222j eππ-=-=- 2cos()sin()22jj j eπππ=+=2cos()sin()22jj j eπππ-=-=-522j jj eeππ==4)sin())144j j jπππ=+=+9441j jj ππ=-9441j j j ππ--=-41jj π-=-1.2 converting from Cartesian to polar coordinates:55j=, 22j e π-=, 233jj eπ--=212je π--=, 41j j π+=, ()2221jj eπ-=-4(1)j j e π-=, 411j je π+=-12eπ-. (a) E ∞=4014tdt e∞-=⎰, P ∞=0, because E ∞<∞ (b) (2)42()j t t x e π+=, 2()1t x =.Therefore, E ∞=22()dt t x +∞-∞⎰=dt +∞-∞⎰=∞,P ∞=211lim lim222()TTTTT T dt dt T Tt x --→∞→∞==⎰⎰lim11T →∞=(c) 2()t x =cos(t). Therefore, E∞=23()dt t x +∞-∞⎰=2cos()dt t +∞-∞⎰=∞,P ∞=2111(2)1lim lim 2222cos()TTTT T T COS t dt dt T Tt --→∞→∞+==⎰⎰(d)1[][]12nn u n x =⎛⎫⎪⎝⎭,2[]11[]4nu n n x =⎛⎫ ⎪⎝⎭. Therefore, E ∞=24131[]4nn n x +∞∞-∞===⎛⎫∑∑⎪⎝⎭P ∞=0，because E ∞<∞.(e) 2[]n x =()28n j e ππ-+, 22[]n x =1. therefore, E ∞=22[]n x +∞-∞∑=∞,P ∞=211limlim 1122121[]NNN N n Nn NN N n x →∞→∞=-=-==++∑∑.(f) 3[]n x =cos 4nπ⎛⎫ ⎪⎝⎭. Therefore, E ∞=23[]n x +∞-∞∑=2cos()4n π+∞-∞∑=2cos()4n π+∞-∞∑,P ∞=1limcos 214nNN n NN π→∞=-=+⎛⎫∑ ⎪⎝⎭1cos()112lim ()2122NN n Nn N π→∞=-+=+∑ . (a) The signal x[n] is shifted by 3 to the right. The shifted signal will be zero for n<1, And n>7.(b) The signal x[n] is shifted by 4 to the left. The shifted signal will be zero for n<-6. And n>0.(c) The signal x[n] is flipped signal will be zero for n<-1 and n>2.(d) The signal x[n] is flipped and the flipped signal is shifted by 2 to the right. The new Signal will be zero for n<-2 and n>4.(e) The signal x[n] is flipped and the flipped and the flipped signal is shifted by 2 to the left.This new signal will be zero for n<-6 and n>0.. (a) x(1-t) is obtained by flipping x(t) and shifting the flipped signal by 1 to the right. Therefore, x (1-t) will be zero for t>-2.(b) From (a), we know that x(1-t) is zero for t>-2. Similarly, x(2-t) is zero for t>-1,Therefore, x (1-t) +x(2-t) will be zero for t>-2.(c) x(3t) is obtained by linearly compression x(t) by a factor of 3. Therefore, x(3t) will bezero for t<1.(d) x(t/3) is obtained by linearly compression x(t) by a factor of 3. Therefore, x(3t) will bezero for t<9.(a) x1(t) is not periodic because it is zero for t<0.(b) x2[n]=1 for all n. Therefore, it is periodic with a fundamental period of 1.(c) x3[n] is as shown in the Figure .. (a)()1[]vnxε={}1111[][]([][4][][4])22n n u n u n u n u nx x+-=--+----Therefore, ()1[]vnxεis zero for1[]nx>3.(b) Since x1(t) is an odd signal, ()2[]vnxεis zero for all values of t.(c)(){}11311[][][][3][3]221122vn nn n n u n u nx x xε-⎡⎤⎢⎥=+-=----⎢⎥⎢⎥⎣⎦⎛⎫⎛⎫⎪ ⎪⎝⎭⎝⎭Therefore, ()3[]vnxεis zero when n<3 and when n→∞.(d) ()1554411()(()())(2)(2)22vt tt t t u t u tx x x e eε-⎡⎤=+-=---+⎣⎦Therefore, ()4()vtxεis zero only when t→∞.. (a) ()01{()}22cos(0)tt tx eπℜ=-=+l(b) ()02{()}cos()cos(32)cos(3)cos(30)4tt t t tx eππℜ=+==+l(c) ()3{()}sin(3)sin(3)2t tt t tx e eππ--ℜ=+=+l(d) ()224{()}sin(100)sin(100)cos(100)2t t tt t t tx e e eππ---ℜ=-=+=+l. (a)1()tx is a periodic complex exponential.101021()j t j tt jx e eπ⎛⎫+⎪⎝⎭==(b)2()tx is a complex exponential multiplied by a decaying exponential. Therefore,is not periodic.（c）3[]nx is a periodic signal. 3[]n x=7j neπ=j neπ.3[]nx is a complex exponential with a fundamental period of 22ππ=.(d)4[]nx is a periodic signal. The fundamental period is given by N=m(23/5ππ)=10().3mBy choosing m=3. We obtain the fundamental period to be 10.(e)5[]nx is not periodic. 5[]nx is a complex exponential with 0w=3/5. We cannot find anyinteger m such that m(02wπ ) is also an integer. Therefore,5[]n x is not periodic.. x (t )=2cos(10t ＋1)-sin(4t-1)Period of first term in the RHS =2105ππ=.Period of first term in the RHS =242ππ= .Therefore, the overall signal is periodic with a period which the least commonmultiple of the periods of the first and second terms. This is equal to π . .x[n] = 1+74j n e π−25j n ePeriod of first term in the RHS =1.Period of second term in the RHS =⎪⎭⎫ ⎝⎛7/42π=7 （when m=2）Period of second term in the RHS =⎪⎭⎫ ⎝⎛5/22ππ=5 (when m=1)Therefore, the overall signal x[n] is periodic with a period which is the least common Multiple of the periods of the three terms inn x[n].This is equal to 35.. The signal x[n] is as shown in figure . x[n] can be obtained by flipping u[n] and thenShifting the flipped signal by 3 to the right. Therefore, x[n]=u[-n+3]. This implies thatM=-1 and no=-3.y (t)= ⎰∞-tdt x )(τ =dt t))2()2((--+⎰∞-τδτδ=⎪⎩⎪⎨⎧>≤≤--<2,022,12,0,t t tTherefore ⎰-==∞224dt E∑∑∞-∞=∞-∞=----=k k k t k t t g 12(3)2(3)(δδ)This implies that A 1=3, t 1=0, A 2=-3, and t 2=1.(a) The signal x 2[n], which is the input to S 2, is the same as y 1[n].Therefore ,y 2[n]= x 2[n-2]+21x 2[n-3] = y 1[n-2]+ 21y 1[n-3]=2x 1[n-2] +4x 1[n-3] +21( 2x 1[n-3]+ 4x 1[n-4]) =2x 1[n-2]+ 5x 1[n-3] + 2x 1[n-4] The input-output relationship for S isy[n]=2x[n-2]+ 5x [n-3] + 2x [n-4](b) The input-output relationship does not change if the order in which S 1and S 2 are connected series reversed. . We can easily prove this assuming that S 1 follows S 2. In this case , the signal x 1[n], which is the input to S 1 is the same as y 2[n]. Therefore y 1[n] =2x 1[n]+ 4x 1[n-1]= 2y 2[n]+4 y 2[n-1]=2( x 2[n-2]+21 x 2[n-3] )+4(x 2[n-3]+21x 2[n-4]) =2 x 2[n-2]+5x 2[n-3]+ 2 x 2[n-4]The input-output relationship for S is once againy[n]=2x[n-2]+ 5x [n-3] + 2x [n-4](a)The system is not memory less because y[n] depends on past values of x[n].(b)The output of the system will be y[n]= ]2[][-n n δδ=0(c)From the result of part (b), we may conclude that the system output is always zero for inputs of the form ][k n -δ, k ∈ ґ. Therefore , the system is not invertible .(a) The system is not causal because the output y(t) at some time may depend on future values of x(t). For instance , y(-π)=x(0).(b) Consider two arbitrary inputs x 1(t)and x 2(t).x 1(t) →y 1(t)= x 1(sin(t)) x 2(t) → y 2(t)= x 2(sin(t))Let x 3(t) be a linear combination of x 1(t) and x 2(t).That is , x 3(t)=a x 1(t)+b x 2(t) Where a and b are arbitrary scalars .If x 3(t) is the input to the given system ,then the corresponding output y 3(t) is y 3(t)= x 3( sin(t))=a x 1(sin(t))+ x 2(sin(t))=a y 1(t)+ by 2(t) Therefore , the system is linear..(a) Consider two arbitrary inputs x 1[n]and x 2[n].x 1[n] → y 1[n] =][01k x n n n n k ∑+-=x 2[n ] → y 2[n] =][02k x n n n n k ∑+-=Let x 3[n] be a linear combination of x 1[n] and x 2[n]. That is :x 3[n]= ax 1[n]+b x 2[n]corresponding output y 3[n] is y 3[n]=][03k x n n n n k ∑+-==])[][(21k bx k ax n n n n k +∑+-==a ][01k x n n n n k ∑+-=+b ][02k x n n n n k ∑+-== ay 1[n]+b y 2[n]Therefore the system is linear.(b) Consider an arbitrary input x 1[n].Lety 1[n] =][01k x n n n n k ∑+-=be the corresponding output .Consider a second input x 2[n] obtained by shifting x 1[n] in time: x 2[n]= x 1[n-n 1]The output corresponding to this input isy 2[n]=][02k x n n n n k ∑+-== ]n [1100-∑+-=k x n n n n k = ][01011k x n n n n n n k ∑+---=Also note that y 1[n- n 1]=][01011k x n n n n n n k ∑+---=.Therefore , y 2[n]= y 1[n- n 1] This implies that the system is time-invariant. (c) If ][n x <B, then y[n]≤(2 n 0+1)B. Therefore ,C ≤(2 n 0+1)B.(a) (i) Consider two arbitrary inputs x 1(t) and x 2(t). x 1(t) → y 1(t)= t 2x 1(t-1)x 2(t) → y 2(t)= t 2x 2(t-1)Let x 3(t) be a linear combination of x 1(t) and x 2(t).That is x 3(t)=a x 1(t)+b x 2(t) where a and b are arbitrary scalars. If x 3(t) is the input to the given system, then the corresponding output y 3(t) is y 3(t)= t 2x 3 (t-1)= t 2(ax 1(t-1)+b x 2(t-1))= ay 1(t)+b y 2(t)Therefore , the system is linear.(ii) Consider an arbitrary inputs x 1(t).Let y 1(t)= t 2x 1(t-1)be the corresponding output .Consider a second input x 2(t) obtained by shifting x 1(t) in time:x 2(t)= x 1(t-t 0) The output corresponding to this input is y 2(t)= t 2x 2(t-1)= t 2x 1(t- 1- t 0) Also note that y 1(t-t 0)= (t-t 0)2x 1(t- 1- t 0)≠ y 2(t)Therefore the system is not time-invariant.(b) (i) Consider two arbitrary inputs x 1[n]and x 2[n]. x 1[n] → y 1[n] = x 12[n-2]x 2[n ] → y 2[n] = x 22[n-2].Let x 3(t) be a linear combination of x 1[n]and x 2[n].That is x 3[n]= ax 1[n]+b x 2[n]corresponding output y 3[n] is y 3[n] = x 32[n-2]=(a x 1[n-2] +b x 2[n-2])2=a 2x 12[n-2]+b 2x 22[n-2]+2ab x 1[n-2] x 2[n-2]≠ay 1[n]+b y 2[n]Therefore the system is not linear.(ii) Consider an arbitrary input x 1[n]. Let y 1[n] = x 12[n-2]be the corresponding output .Consider a second input x 2[n] obtained by shifting x 1[n] in time:x 2[n]= x 1[n- n 0]The output corresponding to this input isy 2[n] = x 22[n-2].= x 12[n-2- n 0] Also note that y 1[n- n 0]= x 12[n-2- n 0] Therefore , y 2[n]= y 1[n- n 0] This implies that the system is time-invariant.(c) (i) Consider two arbitrary inputs x 1[n]and x 2[n].x 1[n] →y 1[n] = x 1[n+1]- x 1[n-1] x 2[n ]→y 2[n] = x 2[n+1 ]- x 2[n -1]Let x 3[n] be a linear combination of x 1[n] and x 2[n]. That is :x 3[n]= ax 1[n]+b x 2[n]where a and b are arbitrary scalars. If x 3[n] is the input to the given system, then the corresponding output y 3[n] is y 3[n]= x 3[n+1]- x 3[n-1]=a x 1[n+1]+b x 2[n +1]-a x 1[n-1]-b x 2[n -1]=a(x 1[n+1]- x 1[n-1])+b(x 2[n +1]- x 2[n -1])= ay 1[n]+b y 2[n]Therefore the system is linear.(ii) Consider an arbitrary input x 1[n].Let y 1[n]= x 1[n+1]- x 1[n-1]be the corresponding output .Consider a second input x 2[n] obtained by shifting x 1[n] in time: x 2[n]= x 1[n-n 0]The output corresponding to this input isy 2[n]= x 2[n +1]- x 2[n -1]= x 1[n+1- n 0]- x 1[n-1- n 0] Also note that y 1[n-n 0]= x 1[n+1- n 0]- x 1[n-1- n 0] Therefore , y 2[n]= y 1[n-n 0]This implies that the system is time-invariant.(d) (i) Consider two arbitrary inputs x 1(t) and x 2(t).x 1(t) → y 1(t)= d O {}(t) x 1 x 2(t) → y 2(t)= {}(t) x 2d OLet x 3(t) be a linear combination of x 1(t) and x 2(t).That is x 3(t)=a x 1(t)+b x 2(t) where a and b are arbitrary scalars. If x 3(t) is the input to the given system, then the corresponding output y 3(t) is y 3(t)= d O {}(t) x 3={}(t) x b +(t) ax 21d O=a d O {}(t) x 1+b {}(t) x 2d O = ay 1(t)+b y 2(t)Therefore the system is linear.(ii) Consider an arbitrary inputs x 1(t).Lety 1(t)= d O {}(t) x 1=2)(x -(t) x 11t -be the corresponding output .Consider a second input x 2(t) obtained by shifting x 1(t) in time:x 2(t)= x 1(t-t 0) The output corresponding to this input isy 2(t)= {}(t) x 2d O =2)(x -(t) x 22t -=2)(x -)t -(t x 0101t t --Also note that y 1(t-t 0)= 2)(x -)t -(t x 0101t t --≠ y 2(t)Therefore the system is not time-invariant. (a) Givenx )(t =jt e 2 y(t)=tj e 3x )(t =jte2- y(t)=tj e3-Since the system liner +=t j e t x 21(2/1)(jt e 2-) )(1t y =1/2(t j e 3+tj e3-)Thereforex 1(t)=cos(2t))(1t y =cos(3t)(b) we know thatx 2(t)=cos(2(t-1/2))= (j e -jt e 2+je jt e 2-)/2Using the linearity property, we may once again writex 1(t)=21( j e -jt e 2+j e jt e 2-))(1t y=(je-jt e 3+je jt e 3-)= cos(3t-1)Therefore,x 1(t)=cos(2(t-1/2)))(1t y =cos(3t-1)signals are sketched in figure .The even and odd parts are sketched in Figure (a) periodic period=2π/(4)= π/2 (b) periodic period=2π/(4)= 2(c) x(t)=[1+cos(4t-2π/3)]/2. periodic period=2π/(4)= π/2 (d) x(t)=cos(4πt)/2. periodic period=2π/(4)= 1/2 (e) x(t)=[sin(4πt)u(t)-sin(4πt)u(-t)]/2. Not period. (f) Not period.(a) periodic, period=7.(b) Not period.(c) periodic, period=8.(d) x[n]=(1/2)[cos(3πn/4+cos(πn/4)). periodic, period=8. (e) periodic, period=16. (a) Linear, stable(b) Not period. (c) Linear(d) Linear, causal, stable(e) Time invariant, linear, causal, stable (f) Linear, stable(g) Time invariant, linear, causal (a) Linear, stable(b) Time invariant, linear, causal, stable (c)Memoryless, linear, causal (d) Linear, stable (e) Linear, stable(f) Memoryless, linear, causal, stable (g) Linear, stable(a) Consider two inputs to the system such that[][][]{}111.S e x n y n x n −−→=ℜand [][][]{}221.Se x n y n x n −−→=ℜNow consider a third inputx 3[n]= x2[n]+x 1[n]. The corresponding system outputWill be [][]{}[][]{}[]{}[]{}[][]33121212e e e e y n x n x n x n x n x n y n y n ==+=+=+ℜℜℜℜtherefore, we may conclude that the system is additive Let us now assume that inputs to the system such that[][][]{}/4111.Sj e x n y n e x n π−−→=ℜand[][][]{}/4222.Sj e x n y n e x n π−−→=ℜNow consider a third input x 3 [n]= x 2 [n]+ x 1 [n]. The corresponding system outputWill be[][]{}()[]{}()[]{}()[]{}()[]{}()[]{}()[]{}[]{}[]{}[][]/433331122/4/41212cos /4sin /4cos /4sin /4cos /4sin /4j e m e m e m e j j e e y n e x n n x n n x n n x n n x n n x n n x n e x n e x n y n y n πππππππππ==-+-+-=+=+ℜℜI ℜI ℜI ℜℜ therefore, we may conclude that the system is additive (b) (i) Consider two inputs to the system such that()()()()211111Sdx t x t y t x t dt ⎡⎤−−→=⎢⎥⎣⎦and ()()()()222211Sdx t x t y t x t dt ⎡⎤−−→=⎢⎥⎣⎦ Now consider a third input x 3[t]= x2[t]+x 1[t]. The corresponding system outputWill be()()()()()()()()()2333211111211dx t y t x t dt d x t x t x t x t dt y t y t ⎡⎤=⎢⎥⎣⎦⎡⎤+⎡⎤⎣⎦=⎢⎥+⎢⎥⎣⎦≠+ therefore, we may conclude that the system is not additiveNow consider a third input x 4 [t]= a x 1 [t]. The corresponding system output Will be()()()()()()()()2444211211111dx t y t x t dt d ax t ax t dt dx t a x t dt ay t ⎡⎤=⎢⎥⎣⎦⎡⎤⎡⎤⎣⎦=⎢⎥⎢⎥⎣⎦⎡⎤=⎢⎥⎣⎦=Therefore, the system is homogeneous.(ii) This system is not additive. Consider the fowling example .Let δ[n]=2δ[n+2]+ 2δ[n+1]+2δ[n] and x2[n]= δ[n+1]+ 2δ[n+1]+ 3δ[n]. The corresponding outputs evaluatedat n=0 are[][]120203/2y and y ==Now consider a third input x 3 [n]= x 2 [n]+ x 1 [n].= 3δ[n+2]+4δ[n+1]+5δ[n]The corresponding outputs evaluated at n=0 is y 3[0]=15/4. Gnarly, y 3[0]≠ ]0[][21y y n +.This[][][][][]444442,1010,x n x n x n y n x n otherwise ⎧--≠⎪=-⎨⎪⎩ [][][][][][]4445442,1010,x n x n ax n y n ay n x n otherwise ⎧--≠⎪==-⎨⎪⎩Therefore, the system is homogenous.(a) Invertible. Inverse system y(t)=x(t+4)(b)Non invertible. The signals x(t) and x 1(t)=x(t)+2πgive the same output (c) δ[n] and 2δ[n] give the same output d) Invertible. Inverse system; y(t)=dx(t)/dt(e) Invertible. Inverse system y(n)=x(n+1) for n ≥0 and y[n]=x[n] for n<0 (f) Non invertible. x (n) and –x(n) give the same result (g)Invertible. Inverse system y(n)=x(1-n) (h) Invertible. Inverse system y(t)=dx(t)/dt (i) Invertible. Inverse system y(n) = x(n)-(1/2)x[n-1] (j) Non invertible. If x(t) is any constant, then y(t)=0(k)δ[n] and 2δ[n] result in y[n]=0 (l) Invertible. Inverse system: y(t)=x(t/2)(m) Non invertible x 1 [n]= δ[n]+ δ[n-1]and x 2 [n]= δ[n] give y[n]= δ[n] (n) Invertible. Inverse system: y[n]=x[2n](a) Note that x 2[t]= x 1 [t]- x 1 [t-2]. Therefore, using linearity we get y 2 (t)= y 1 (t)- y 1 (t-2).this is shown in Figure(b)Note that x3 (t)= x1 [t]+ x1 [t+1]. .Therefore, using linearity we get Y3 (t)= y1 (t)+(3) x(t) periodic, period T; y 2 (t) periodic, period2T (4) y 2(t) periodic, period T; x(t) periodic, period T/2; (1) True x[n]=x[n+N ]; y 1 (n)= y 1 (n+ N 0). periodic with N 0=n/2if N is even and with period N 0=n if N is odd.(2)False. y 1 [n] periodic does no imply x[n] is periodic . Let x[n] = g[n]+h[n] where0,1,[][]0,(1/2),nn even n even g n and h n n odd n odd⎧⎧==⎨⎨⎩⎩ Then y 1 [n] = x [2n] is periodic but x[n] is clearly not periodic. (3)True. x [n+N] =x[n]; y 2 [n+N 0] =y 2 [n] where N 0=2N(4) True. y 2 [n+N] =y 2 [n]; y 2 [n+N 0 ]=y 2 [n] where N 0=N/2 . (a) Consider()()()()()().xy yx t x t y d y t x d t φττττττφ∞-∞∞-∞=+=-+=-⎰⎰If x[n] is odd, x[n] +x [-n] =0. Therefore, the given summation evaluates to zero. (b) Let y[n] =x 1[n]x 2[n] .Theny [-n] =x 1[-n] x 2[-n] =-x 1[n]x 2[n] =-y[n]. This implies that y[n] is odd.(c)ConsiderUsing the result of part (b), we know that x e [n]x o [n] is an odd signal .Therefore, using the result of part (a) we may conclude thatTherefore,(d)ConsiderAgain, since x e (t) x o (t) is odd,Therefore,. We want to find the smallest N 0 such that m(2π /N) N 0 =2πk or N 0 =kN/m,where k is an integer, then N must be a multiple of m/k and m/k must be an integer .this implies that m/k is a divisor of both m and N .Also, if we want the smallest possible N 0, then m/k should be the GCD of m and N. Therefore, N 0=N/gcd(m,N).．(a)If x[n] is periodic 0(),0..2/j n N T o ewhere T ωωπ+= This implies that 022o T kNT k T T Nππ=⇒==a rational number . (b)T/T 0 =p/q then x[n] = 2(/)j n p q eπ,The fundamental period is q/gcd(p,q) and the fundmentalfrequency is(c) p/gcd(p,q) periods of x(t) are needed ..(a) From the definition of ().xy t φWe have{}1[][0][][]n n x n x x n x n ∞∞=-∞==++-∑∑22[][]eon n n n xx∞∞=-∞=-∞=+∑∑222[][][]e on n n n n n x xx∞∞∞=-∞=-∞=-∞==+∑∑∑2[][]0eon n n x x ∞=-∞=∑222[][][].eon n n n n n x x x ∞∞∞=-∞=-∞=-∞==+∑∑∑222()()()2()().e o et dt t dt t dt t t dt xx x x x ∞∞∞∞-∞-∞-∞-∞=++⎰⎰⎰⎰()()0.et t dt x x ∞-∞=⎰222()()().e ot dt t dt t dt xx x ∞∞∞-∞-∞-∞=+⎰⎰⎰0022gcd(,)gcd(,)gcd(,)gcd(,).T pp q p q p q p q q p q p pωωππ===(b) Note from part(a) that ()().xx xx t t φφ=-This implies that ()xy t φis even .Therefore,the odd part of ().xx t φis zero. (c) Here, ()().xy xx t t T φφ=-and ()().yy xx t t φφ= .(a) We know that /22(2)().t t δδ=V V Therefore This implies that1(2)().2t t δδ=(b)The plot are as shown in Figure . We havelim ()()lim (0)()0.u t t u t δδ→→==V V V VAlso,⎰⎰∞∞∞--=-=0)()()()()(ττδτττδτd t u d t u t gTherefore,/21lim (2)lim ().2t t δδ→∞→∞=V V V V 01lim ()()().2u t t t δδ→=V V V u Δ'（t ） 1 1/2 Δ-Δ/t0 tu Δ'（t ）12Δ t0 tu Δ'（t ） 1 1/2 Δ-Δttu Δ'（t ） 1 1/2Δ-Δt 0t0,0()1,00t g t t undefined for t >⎧⎪=<⎨⎪=⎩()0()()()t u t t δττδτδτ-=-=-Q Q .(a) If a system is additive ,then also, if a system is homogeneous,then(b) y(t)=x 2(t) is such a systerm . (c) example,consider y(t) ()()ty t x d ττ-∞=⎰with ()()(1).x t u t u t =--Then x(t)=0for t>1,but y(t)=1 for t>1.. (a) y[n]=2x[n].Therefore, the system is time invariant.(b) y[n]=(2n-1)x[n].This is not time-invariant because y[n- N 0]≠(2n-1)2x[n- N 0].(c) y[n]=x[n]{1+(-1)n +1+(-1)n-1}=2x[n].Therefore, the system is time invariant ..(a) Consider two system S 1 and S 2 connected in series .Assume that if x 1(t) and x 2(t) arethe inputs to S 1..then y 1(t) and y 2(t) are the .Also,assume that if y 1(t) and y 2(t) are the input to S 2 ,then z 1(t) and z 2(t) are the outputs, respectively . Since S 1 is linear ,we may write()()()()11212,s ax t bx t ay t by t +→+where a and b are constants. Since S 2 is also linear ,we may write()()()()21212,s ay t by t az t bz t +→+We may therefore conclude that)()()()(212121t b t a t b t a z z x x s s +−→−+Therefore ,the series combination of S 1 and S 2 is linear.Since S 1 is time invariant, we may write()()11010s x t T y t T -→-and()()21010s y t T z t T -→-Therefore,()()121010s s x t T z t T -→-Therefore, the series combination of S 1 and S 2 is time invariant. (b) False, Let y(t)=x(t)+1 and z(t)=y(t) corresponds to two nonlinear systems. If these systems are connected in series ,then z(t)=x(t) which is a linear system.(c) Let us name the output of system 1 as w[n] and the output of system 2 as z[n] .Then11[][2][2][21][22]24y n z n w n w n w n ==+-+-[][][]241121-+-+=n x n x n xThe overall system is linear and time-invariant.. (a) We have())(t y t x s−→−Since S is time-invariant.())(T t y T t x s -−→−-Now if x (t) is periodic with period T. x{t}=x(t-T). Therefore, we may conclude that00.()().00x t y t =→=0()()()()0x t x t y t y t =-→-=y(t)=y(t-T).This implies that y(t) is also periodic with T .A similar argument may be made in discrete time . (b)(a) Assumption : If x(t)=0 for t<t 0 ,then y(t)=0 for t< prove That : The system is causal. Let us consider an arbitrary signal x 1(t) .Let us consider another signal x 2(t) which is thesame as x 1(t)for t< t 0. But for t> t 0 , x 2(t) ≠x 1(t),Since the system is linear,()()()()1212,x t x t y t y t -→-Since ()()120x t x t -=for t< t 0 ,by our assumption =()()120y t y t -=for t< t 0 .This implies that ()()12y t y t =for t< t 0 . In other words, t he output is not affected by input values for0t t ≥. Therefore, the system is causal .Assumption: the system is causal . To prove that :If x(t)=0 for t< t 0 .then y(t)=0 for t< t 0 . Let us assume that the signal x(t)=0 for t< t 0 .Then we may express x(t) as ()()12()x t x t x t =-, Where ()()12x t x t = for t< t 0 . the system is linear .the output to x(t) will be()()12()y t y t y t =-.Now ,since the system is causal . ()()12y t y t = for t< t 0 .implies that ()()12y t y t = for t< t 0 .Therefore y(t)=0 for t< t 0 .(b) Consider y(t)=x(t)x(t+1) .Now , x(t)=0 for t< t 0 implies that y(t)=0 for t< t 0 .Note that the system is nonlinear and non-causal .(c) Consider y(t)=x(t)+1. the system is nonlinear and causal .This does not satisfy the condition of part(a).(d) Assumption: the system is invertible. To prove that :y[n]=0 for all n only if x[n]=0 for all n . Consider[]0[]x n y n =→. Since the system is linear :2[]02[]x n y n =→.Since the input has not changed in the two above equations ,we require that y[n]= 2y[n].This implies that y[n]=0. Since we have assumed that the system is invertible , only one input could have led to this particular output .That input must be x[n]=0 .Assumption: y[n]=0 for all n if x[n]=0 for all n . To prove that : The system is invertible . Suppose that11[][]x n y n → and21[][]x n y n →Since the system is linear ,1212[][][][]0x n x n y n y n -=→-=By the original assumption ,we must conclude that 12[][]x n x n =.That is ,any particular y 1[n] can be produced that by only one distinct input x 1[n] .Therefore , the system is invertible.(e) y[n]=x 2[n]. . (a) Consider ,()111()()shx x t y t t φ→= and()222()()s hx x t y t t φ→=.Now, consider ()()()312x t ax t bx t =+. The corresponding system output will be()()12331212()()()()()()()()()hx hx y t x h t d a x h t d b x t h t d a t b t ay t by t ττττττττφφ∞-∞∞∞-∞-∞=+=+++=+=+⎰⎰⎰Therefore, S is linear .Now ,consider x 4(t)=x 1(t-T).The corresponding system output will be()14411()()()()()()()hx y t x h t d x T h t d x h t T d t T τττττττττφ∞-∞∞-∞∞-∞=+=-+=++=+⎰⎰⎰Clearly, y 4(t)≠ y 1(t-T).Therefore ,the system is not time-invariant.The system is definitely not causal because the output at any time depends on future values of the input signal x(t).(b) The system will then be linear ,time invariant and non-causal. . The plots are in Figure ..(a) The overall response of the system of Figure .(a)=(the response of the system to x[n]+x 1[n])-the response of the system to x 1[n]=(Response of a linear system L to x[n]+x 1[n]+ zero input response of S)- (Response of a linear system L to x 1[n]+zero input response of S)=( (Response of a linear system L to x[n]).。