奥本海姆信号与系统习题参考答案电子科技大学
《信号与系统》考研奥本海姆版配套2021考研真题库
《信号与系统》考研奥本海姆版配套2021考研真题库第一部分考研真题精选一、选择题1下列关于冲激函数性质的表达式不正确的是（）。
[西安电子科技大学2012研]A．f（t）δ′（t）＝f（0）δ′（t）B．f（t）δ（t）＝f（0）δ（t）C．D．【答案】A查看答案【解析】A项，正确结果应该为f（t）δ′（t）＝f（0）δ′（t）－f′（0）δ（t）。
2x（t）＝asint－bsin（3t）的周期是（）。
[西南交通大学研]A．π/2B．πC．2πD．∞【答案】C查看答案【解析】因为asint的周期为T1＝2π/1＝2π，bsin（3t）的周期为T2＝2π/3，因为T1/T2＝3/1为有理数，因此x（t）是周期信号，且x（t）＝asint－bsin （3t）的周期是3T2＝T1＝2π。
3序列f（k）＝e j2πk/3＋e j4πk/3是（）。
[西安电子科技大学2012研]A．非周期序列B．周期N＝3C．周期N＝6D．周期N＝24【答案】B查看答案【解析】f1（k）＝e j2πk/3的周期N1＝2π/（2π/3）＝3，f2（k）＝e j4πk/3的周期N2＝2π/（4π/3）＝3/2，由于N1/N2＝2为有理数，因此f（k）为周期序列，周期为2N2＝N1＝3。
4积分[西安电子科技大学2011研]A．2B．1C．0D．4【答案】A查看答案【解析】5序列乘积δ（k＋1）δ（k－1）＝（）。
[西安电子科技大学研]A．0B．δ（k）C．δ（k＋1）D．δ（k－1）【答案】A查看答案【解析】根据f（k）δ（k－k0）＝f（k0）δ（k－k0），因此δ（k＋1）δ（k－1）＝δ（2）δ（k－1）＝0。
6信号f1（t）＝2，f2（t）的波形如图1-1-1所示，设y（t）＝f1（t）*f2（t），则y（11）＝（）。
[西安电子科技大学2011研]图1-1-1A．1B．0C．2D．3【答案】B查看答案【解析】7已知一连续系统在输入f（t）作用下的零状态响应为y（t）＝f（4t），则该系统为（）。
奥本海姆《信号与系统》第2版上册配套题库
奥本海姆《信号与系统》第2版上册配套题库奥本海姆《信号与系统》（第2版）配套题库【考研真题精选＋章节题库】（上册）目录第一部分考研真题精选一、选择题二、填空题三、判断题四、简答题五、画图题六、证明题七、计算题第二部分章节题库第1章绪论第2章线性时不变系统第3章周期信号的傅里叶级数表示第4章连续时间傅里叶变换第5章离散时间傅里叶变换第6章信号与系统的时域和频域特性•试看部分内容考研真题精选一、选择题1下列关于冲激函数性质的表达式不正确的是（）。
[西安电子科技大学2012研]A．f（t）δ′（t）＝f（0）δ′（t）B．f（t）δ（t）＝f（0）δ（t）C．D．【答案】A查看答案【解析】A项，正确结果应该为f（t）δ′（t）＝f（0）δ′（t）－f′（0）δ（t）。
2x（t）＝asi n t－b si n（3t）的周期是（）。
[西南交通大学研]A．π/2B．πC．2πD．∞【答案】C查看答案【解析】因为asin t的周期为T1＝2π/1＝2π，bsin（3t）的周期为T2＝2π/3，因为T1/T2＝3/1为有理数，因此x（t）是周期信号，且x（t）＝asint－b sin（3t）的周期是3T2＝T1＝2π。
3序列f（k）＝e j2πk/3＋e j4πk/3是（）。
[西安电子科技大学2012研]A．非周期序列B．周期N＝3C．周期N＝6D．周期N＝24【答案】B查看答案【解析】f1（k）＝e j2πk/3的周期N1＝2π/（2π/3）＝3，f 2（k）＝e j4πk/3的周期N2＝2π/（4π/3）＝3/2，由于N1/N2＝2为有理数，因此f（k）为周期序列，周期为2N2＝N1＝3。
4积分[西安电子科技大学2011研] A．2B．1C．0D．4【答案】A查看答案【解析】5序列乘积δ（k＋1）δ（k－1）＝（）。
[西安电子科技大学研]A．0B．δ（k）C．δ（k＋1）D．δ（k－1）【答案】A查看答案【解析】根据f（k）δ（k－k0）＝f（k0）δ（k－k0），因此δ（k＋1）δ（k－1）＝δ（2）δ（k－1）＝0。
《信号与系统》考研奥本海姆版2021考研名校考研真题
《信号与系统》考研奥本海姆版2021考研名校考研真题第一部分考研真题精选一、选择题1已知信号f（t）的频带宽度为Δω，则信号y（t）＝f2（t）的不失真采样间隔（奈奎斯特间隔）T等于（）。
[西南交通大学研]A．π/（Δω）B．π/（2Δω）C．2π/（Δω）D．4π/（Δω）【答案】B查看答案【解析】根据卷积定理可知，y（t）＝f2（t）→[1/（2π）]F（jω）*F（j ω）。
若信号f（t）的频带宽度为Δω，则y（t）的频带宽度为2Δω。
则奈奎斯特采样频率为4Δω，所以不失真采样间隔（奈奎斯特间隔）T等于2π/（4Δω）＝π/（2Δω）。
2已知f（t）↔F（jω），f（t）的频带宽度为ωm，则信号y（t）＝f（t/2－7）的奈奎斯特采样间隔等于（）。
[西南交通大学研]A．2π/ωmB．2π/（2ωm－7）C．4π/ωmD．π/ωm【答案】A查看答案【解析】根据时域和频域之间关系，可知若时域扩展，则频域压缩。
所以若f（t）的频带宽度为ωm，则信号y（t）＝f（t/2－7）的频带宽度为ωm/2。
所以，其奈奎斯特采样频率为（ωm/2）×2＝ωm，即奈奎斯特采样间隔等于2π/ωm。
3有限长序列x（n）的长度为4，欲使x（n）与x（n）的圆卷积和线卷积相同，则长度L的最小值为（）。
[中国科学院研究生院2012研]A．5B．6C．7D．8【答案】C查看答案【解析】x（n）的长度为4，则其线卷积的长度为4＋4－1＝7。
当x（n）与x（n）的圆卷积L≥7时，x（n）与x（n）的圆卷积和线卷积相同，可知L的最小值为7。
4下面给出了几个FIR滤波器的单位函数响应。
其中满足线性相位特性的FIR滤波器是（）。
[东南大学研]A．h（n）＝{1，2，3，4，5，6，7，8}B．h（n）＝{1，2，3，4，1，2，3，4}C．h（n）＝{1，2，3，4，4，3，2，1}D．h（n）＝{1，2，3，4，－1，－2，－3，－4}【答案】C查看答案【解析】线性相位FIR滤波器必满足某种对称性，即h（n）＝h（N－1－n）或者h（n）＝－h（N－1－n）。
奥本海姆《信号与系统》配套题库【名校考研真题】(周期信号的傅里叶级数表示)
【答案】4
【解析】因是周期信号，其角频率
2π T
π
，则： a0
2 T
2 f (t)dt 2
0
T
2 dt 2，k 0
0
2
ak T
2 f (t) cos(kt)dt 2
0
T
2
cos(kπt)dt
sin(kπt)
2
0， k
1, 2,
0
kπ 0
所以：
ak 2 4
k
3 ． x t 是 一 连 续 时 间 周 期信 号 ， 其 基 波 频率 为 1 ， 傅 里 叶 系 数 为 ak ， 现 已 知
y(t) x(1 t ) x(t 1，) 问 y(t) 的基本频率 2 与 1 是什么关系？______； y(t) 的傅里叶级数
系数 bk 与 ak 的关系是什么？______。[华南理工大学 2007 研]
t0 T
1
bk T
T
t0 T x 1 t
x
t 1
e jkt dt
1 T
T
x 1t
t0 T
e jktdt 1 T
T x t 1 e jktdt
t0 T
1 T
T x 1 t ejktd 1 t 1
t0 T
T
T
x
t0 T
t 1
e jkt d
t 1
ak ak
2．一连续时间
LTI
系统的频率响应
H ( j)
1, 0,
≥250 ，当输入基波周期 T＝ π ，
其余
7
傅立叶级数系数为 ak 的周期信号 x t 时，发现输出 y(t) x(t) 。ak 需满足什么条件？（ ）
奥本海姆信号与系统(第二版)复习题参考答案
第一章作业解答1.9解：（b ）jt t t j e e e t x --+-==)1(2)(由于)()(2)1()1())(1(2t x e e e T t x T j t j T t j ≠==++-+-++-，故不是周期信号；（或者：由于该函数的包络随t 增长衰减的指数信号，故其不是周期信号；） （c ）n j e n x π73][= 则πω70= 7220=ωπ是有理数，故其周期为N=2； 1.12解：]4[1][1)1(]1[1][43--=--==+---=∑∑∞=∞=n u m n mk k n n x m k δδ-3 –2 –1 0 1 2 3 4 5 6 n1…减去：-3 –2 –1 0 1 2 3 4 5 6 nu[n-4]等于：-3 –2 –1 0 1 23 4 5 6 n…故：]3[+-n u 即：M=-1，n 0=-3。
1.14解：x(t)的一个周期如图(a)所示，x(t)如图(b)所示：而：g(t)如图(c)所示……dtt dx )(如图（d ）所示：……故：)1(3)(3)(--=t g t g dtt dx 则：1t ,0t 3,32121==-==；A A 1.15解：该系统如下图所示： 2[n]（1）]4[2]3[5]2[2]}4[4]3[2{21]}3[4]2[2{]3[21]2[][][1111111222-+-+-=-+-+-+-=-+-==n x n x n x n x n x n x n x n x n x n y n y即：]4[2]3[5]2[2][-+-+-=n x n x n x n y（2）若系统级联顺序改变，该系统不会改变，因为该系统是线性时不变系统。
（也可以通过改变顺序求取输入、输出关系，与前面做对比）。
1.17解：（a ）因果性：)(sin )(t x t y =举一反例：当)0()y(,0int s x t =-=-=ππ则时输出与以后的输入有关，不是因果的；（b ）线性：按照线性的证明过程（这里略），该系统是线性的。
奥本海姆《信号与系统》配套题库【名校考研真题】(连续时间傅里叶变换)
）。[西安电子科技大学 2010
A． f1 t t0 f2 t t0 f t B． f t t f t
C． f t t f t
D． f1 2t f2 2t f 2t
【答案】D
【解析】根据傅里叶变换性质和卷积定理， f1 2t f2 2t 的傅里叶变换为：
1 2
F1
f
(t)
2t
d dt
cos
2t
π 3
t
的傅里叶变换
F j
等于（
）。[西安电子科
技大学 2008 研]
A．1 j
B．1 j
C．-1
D． ej
【答案】C
【解析】由于
f
(t )
2t
d dt
cos
2t
π 3
t
t t ，根据常用傅里叶变换和时域微分
定理，可知 t j 。再根据频域微分性质，可得 t t 1。
求 cos0t 的傅里叶变换：
cos
0t
cos
0
t
0
FT
j
πe 0
0
0
所以：
1 2π
F
F
cos 0t
ej 0
0
2
F
0
ej 0
0
2
F
0
则其频带宽度为 0 W ，因为 0 W ，所以 0 W 0 。
6．设 f t f1 t f2 t ，则下列卷积等式丌成立的是（
tf (t) j dF ；再由时秱性质，可知 (1 t) f (1 t) j dF() ej 。
d
பைடு நூலகம்
d
10．已知信号 f (t) 的频带宽度为 ，则信号 y(t) f 2 (t) 的丌失真采样间隔（奈奎斯
《信号与系统》考研奥本海姆版配套2021考研真题库
《信号与系统》考研奥本海姆版配套2021考研真题库第一部分考研真题精选一、选择题1下列关于冲激函数性质的表达式不正确的是（）。
[西安电子科技大学2012研]A．f（t）δ′（t）＝f（0）δ′（t）B．f（t）δ（t）＝f（0）δ（t）C．D．【答案】A查看答案【解析】A项，正确结果应该为f（t）δ′（t）＝f（0）δ′（t）－f′（0）δ（t）。
2x（t）＝asint－bsin（3t）的周期是（）。
[西南交通大学研]A．π/2B．πC．2πD．∞【答案】C查看答案【解析】因为asint的周期为T1＝2π/1＝2π，bsin（3t）的周期为T2＝2π/3，因为T1/T2＝3/1为有理数，因此x（t）是周期信号，且x（t）＝asint－bsin （3t）的周期是3T2＝T1＝2π。
3序列f（k）＝e j2πk/3＋e j4πk/3是（）。
[西安电子科技大学2012研]A．非周期序列B．周期N＝3C．周期N＝6D．周期N＝24【答案】B查看答案【解析】f1（k）＝e j2πk/3的周期N1＝2π/（2π/3）＝3，f2（k）＝e j4πk/3的周期N2＝2π/（4π/3）＝3/2，由于N1/N2＝2为有理数，因此f（k）为周期序列，周期为2N2＝N1＝3。
4积分[西安电子科技大学2011研]A．2B．1C．0D．4【答案】A查看答案【解析】5序列乘积δ（k＋1）δ（k－1）＝（）。
[西安电子科技大学研]A．0B．δ（k）C．δ（k＋1）D．δ（k－1）【答案】A查看答案【解析】根据f（k）δ（k－k0）＝f（k0）δ（k－k0），因此δ（k＋1）δ（k－1）＝δ（2）δ（k－1）＝0。
6信号f1（t）＝2，f2（t）的波形如图1-1-1所示，设y（t）＝f1（t）*f2（t），则y（11）＝（）。
[西安电子科技大学2011研]图1-1-1A．1B．0C．2D．3【答案】B查看答案【解析】7已知一连续系统在输入f（t）作用下的零状态响应为y（t）＝f（4t），则该系统为（）。
信号与系统 奥本海姆1-4答案.doc
Signals and SystemChap11.6 Determine whether or not each of the following signals is periodic:(a): (/4)1()2()j t x t e u t π+= (b): 2[][][]x n u n u n =+-(c): 3[]{[4][14]}k x n n k n k δδ∞=-∞=----∑Solution:(a).No 【周期信号无始无终，单边肯定不周期】Because 12cos()2sin(),0()440,0t j t t x t t ππ⎧+++>⎪=⎨⎪<⎩ when t<0, )(1t x =0. (b).No 【注意n ＝0】 Because 21,0[]2,01,0n n n n x >⎧⎪==⎨⎪<⎩(c).Y es 【画图、归纳】 Because∑∞-∞=--+--+=+k k m n k m n m n x ]}414[]44[{]4[3δδ∑∞-∞=------=k m k n m k n )]}(41[)](4[{δδ{[4][14]}k n k n k δδ∞=-∞=----∑N=4.1.9 Determine whether or not each of the following signals is periodic, if a signal is periodic, specify its fundamental period:(a): 101()j tx t je =(b): (1)2()j t x t e -+=(c): 73[]j n x n e π=(d): 3(1/2)/54[]3j n x n e π+= (e): 3/5(1/2)5[]3j n x n e += Solution: (a). T=π/5Because 0w =10, T=2π/10=π/5. (b). Aperiodic.Because jt t e e t x --=)(2, while t e -is not periodic, )(2t x is not periodic. (c). N=2Because 0w =7π, N=(2π/0w )*m, and m=7. (d). N=10Because n j j e e n x )5/3(10/343)(ππ=, that is 0w =3π/5,N=(2π/0w )*m, and m=3. (e). Aperiodic.Because 0w =3/5, N=(2π/0w )*m=10πm/3 , it ’s not a rational number.1.14 consider a periodic signal 1,01()2,12t x t t ≤≤⎧=⎨-<<⎩with periodT=2. The derivative of this signal is related to the “impulsetrain ”()(2)k g t t k δ∞=-∞=-∑, with period T=2. It can be shownthat1122()()()dx t A g t t A g t t dt=-+-. Determine the values of1A , 1t , 2A , 2t .Solution:A 1=3, t 1=0, A 2=-3, t 2=1 or -1 Because∑∞-∞=-=k k t t g )2()(δ,)1(3)(3)(--=t g t g dtt dx1.15. Consider a system S with input x[n] and output y[n].This system is obtained through a series interconnection of a system S 1 followed by a system S2. The input-output relationships for S 1 and S 2 areS 1: ],1[4][2][111-+=n x n x n y S 2: ]3[21]2[][222-+-=n x n x n yWhere ][1n x and ][2n x denote input signals.(a) Determine the input-output relationship for system S.(b)Does the input-output relationship of system S change if the order in which S 1 and S 2 are connected in series is reversed(ie., if S2 follows S 1)? Solution: (a)]3[21]2[][222-+-=n x n x n y]3[21]2[11-+-=n y n y]}4[4]3[2{21]}3[4]2[2{1111-+-+-+-=n x n x n x n x]4[2]3[5]2[2111-+-+-=n x n x n xThen, ]4[2]3[5]2[2][-+-+-=n x n x n x n y【可以考虑先求取单位脉冲响应，再做卷积】(b).No. because it ’s linear, S 1 and S 2 do not diverge.1.16. Consider a discrete-time system with input x[n] and output y[n].The input-output relationship for this system is]2[][][-=n x n x n y(a) Is the system memory less?(b) Determine the system output when the input is ][n A δ, where A is any real or complex number . (c) Is the system invertible? Solution: (a). No.For example, when n=0, y[0]=x[0]x[-2]. So the system is memory. (b). y[n]=0.When the input is ][n A δ,]2[][][2-=n n A n y δδ, so y[n]=0.(c). No.For example, when x[n]=0, y[n]=0; when x[n]=][n A δ, y[n]=0. So the system is not invertible.1.17.Consider a continuous-time system with input x(t) and output y(t) related by ))(sin()(t x t y =， (a) Is this system causal? (b) Is this system linear? Solution: (A). No.For example,)0()(x y =-π. So it ’s not causal.【得到什么启示？】 (b). Y es.Because : ))(sin()(11t x t y = , (sin()(22tx t y =)()())(sin())(sin()(21213t by t ay t bx t ax t y +=+=1.21. A continuous-time signal ()x t is shown in Figure P1.21. Sketch and label carefully each of the following signals:(a): (1)x t - (b): (2)x t - (c): (21)x t + (d): (4/2)x t - (e): [()()]()x t x t u t +-(f): ()[(3/2)(3/2)]x t t t δδ+--Solution: (a).(b).(c). (d).1.22. A discrete-time signal ][n x is shown in as the following. Sketch and label carefully each of the following signals: (a): [4]x n - (b): [3]x n - (c): [3]x n(d): [31]x n + (e): [][3]x n u n -(f): [2][2]x n n δ--(g): 11[](1)[]22nx n x n +-(h): 2[(1)]x n -Solution:(a).(b).(e).(f) ]2[-n δ(g)1.25. Determine whether or not each of the following continuous-time signals is periodic. If the signal is periodic, determine its fundamental period.(a): ()3cos(4)3x t t π=+ (b): (1)()j t x t e π-=(c): 2()[cos(2)]3x t t π=-(d): (){cos(4)()}x t t u t ενπ=(e): (){sin(4)()}x t t u t ενπ= (f): (2)()t n n x t e∞--=-∞=∑Solution:(a).Periodic. T=π/2. Solution: T=2π/4=π/2. (b). Periodic. T=2.Solution: T=2π/π=2.(c). Periodic. T=π/2.【括号内周期，平方后仍然周期，或者做三角变换】 (d). Periodic. T=0.5. Solution: )}()4{cos()(t u t E t x v π= )}())(4cos()()4{cos(21t u t t u t --+=ππ )}()(){4cos(21t u t u t -+=π)4cos(21t π=So, T=2π/4π=0.5【值得商榷】 (e)、(f)非周期信号。
《信号与系统》奥本海姆英文版课后答案
2
2 (f) x [n] = . Therefore, E = = = , 3 cos x cos( n) cos( n) 3[n]
4
4
4
2 )1 lim 1 ( 2 2 N 2 N 1 n N 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 of 3. Therefore, x(3t) will be zero for t<1.
奥本海姆信号与系统习题解答Chapter 4
Chapter 44.1 Solution: (a). ⎰∞∞--=dt e t x jw X jwt )()(⎰∞∞-----=dt e t u e jwt t )1()1(2 ⎰∞∞-----=dt e t u e jwt t )1()1(2⎰⎰∞--∞---==12)2(1)1(2dt e e dt e e t jw jwt tjwe jw e e jw e e jw jw t jw +=---=--=---∞--222)2(21)2(24.2 Solution: (a). ⎰∞∞--=dt e t x jw X jwt )()(⎰∞∞++=-j w t -1)]e -(t 1)(t [dt δδ⎰⎰∞∞∞∞-++=-j w t-j w t-1)e -(t 1)e (tdt dt δδw e e jw jw cos 2=+=-+(b). ⎰∞∞--=dt e t x jw X jwt )()(dt e t u t u dtdjwt -∞∞--+--=⎰)}2()2({ dt e t t jwt -∞∞--+---=⎰)}2()2({δδ dt e t dt e t jwt jwt -∞∞--∞∞--+---=⎰⎰)2()2(δδw j e ew j wj 2s i n 222-=+-=-+4.3 Solution: (a). )42sin(ππ+t =je et j t j 2)42()42(ππππ+-+-=je e ee tj jtj j22424ππππ---t j e π2 −→←FT)2(2ππδ-wt j e π2- −→←FT)2(2ππδ+w∴ )42sin(ππ+t −→←FT)}2()2({44πδπδπππ+---w ew e jjj4.4 Solution: (a). dw e jw X t x jwt)(21)(⎰∞∞-=π dw e w w w jwt )}4()4()(2{21ππδππδπδπ++-+=⎰∞∞- })4()4()(2{21dw e w dw e w dw e w jwt jwtjwt ππδππδπδπ++-+=⎰⎰⎰∞∞-∞∞-∞∞-)4cos(1)}4cos(22{21}2{2144t t e e t j t j ππππππππ+=+=++=- 4.5 Solution:dw e e jw X dw e jw X t x jwt jw X j jwt )()(21)(21)(<∞∞-∞∞-⎰⎰==ππdw e ew u w u jwt w j )23()}3()3({221ππ+-∞∞---+=⎰dw ee dw ee wt j j jwtw j )23(33)23(331--+--⎰⎰==ππππ)23(1)23(1)23(3)23(333)23(--⋅-=-⋅-=-----t j e e t j e t j t j wt j ππ)23()23(3sin 2)23()23(3sin 21---=--⋅-=t t j t j t j ππ If0)(=t x ,then ⎪⎪⎩⎪⎪⎨⎧≠-±±==-0)23(,...2,1,0,.........)23(3t k k t πThat is 0., (2)33≠+=k k t π 4.6 Solution:Accorrding to the properties of the Fourier transform, we ’ll get: (a). )(t x −→←FT)(jw X∴)1(t x - −→←FT jwe jw X --)()1(t x -- −→←FT jwe jw X +-)(then)1()1()(1t x t x t x --+-=−→←FTw jw X e jw X e jw X jw X jw jw cos )(2)()()(1-=-+-=--(b). )(t x −→←FT)(jw X)(b at x + −→←FTwa bj e aw j X a )(1∴)63()(2-=t x t x −→←FTw j e wj X jw X 22)3(31)(-= (c).)(t x −→←FT)(jw X )1(-t x −→←FTjw e jw X -)()(22t x dtd −→←FT)()(2jw X jw∴)1()(223-=t x dtd t x −→←FTjw e jw X w jw X --=)()(234.7. (a). neither,neither Solution: )2()()(1--=w u w u jw X∴ )()(*11jw X jw X -≠,that is )()(*t x t x ≠so )(t x is not real.)()(*11jw X jw X -≠-,that is )()(*t x t x -≠so )(t x is not imaginary.And )()(11jw X jw X -≠,that is )()(t x t x -≠so )(t x is not even.)()(11jw X jw X --≠,that is )()(t x t x --≠so )(t x is not odd.4.10. (a). s olution:suppose t t t f πsin )(= −→←FT⎪⎩⎪⎨⎧><=1,.......01,........1)(w w jw Fthen 21)]([)(t f t f =−→←FT)()(21jw F jw F *π=)(1jw F )()(21)(1jw F jw F jw F '*='π)]1()1([)(21--+*=w w jw F δδπ)]}1([)]1([{21--+=w j F w j F π⎪⎪⎪⎩⎪⎪⎪⎨⎧<≤-<≤-=others w w ..........,.........020...,.........2102...,.........21ππ∴)(1jw F ⎪⎪⎪⎩⎪⎪⎪⎨⎧<≤--<≤-+=others w w w w ..........,.........020..., (22)02...,.........22ππ∴ )()(1t tf t x = −→←FT)()(1jw F dw d j jw X =⎪⎪⎪⎩⎪⎪⎪⎨⎧<≤-<≤-=others w j w j..........,.........020...,.........202...,.........2ππ (b). Solution:dw jw X dt t t t A 222)(21sin ⎰⎰∞∞-∞∞-=⎪⎭⎫⎝⎛=ππ32222214421221πππππ===⎰-dw j 4.11 A=1/3, B=3 Solution: )()()(t h t x t y *= −→←FT)()()(jw H jw X jw Y =and )3()3()(t h t x t g *=)3(t x −→←FT)3(31w j X )3(t h −→←FT)3(31w j H then )3(31)3(31)(w j H w j X jw G ⋅=)3(91wj Y =−→←FT)3(31)(t y t g =∴3,31==B A4.13. Solution:(a). According to the property of FT, )(t tx −→←FT)(jw X dwdj te- −→←FT212w+∴tte- −→←FT()()222221412212w wj w w j w dw d j+-=+⋅-=⎪⎭⎫ ⎝⎛+(b). According to the duality of FT, If)(t x −→←FT)(jw X then )(jt X −→←FT)(2w f -πandtte- −→←FT()2214w wj +-tjte- −→←FT()2214w w+∴()2214t t+ −→←FT wwwej ew j ----=-ππ2)(24.14 Solution: From (1), we know )(t x is real and 0)(≥t x ;From (2), we know : )(2t u Ae t- −→←FT)()1(jw X jw +And we also know )(2t u Ae t - −→←FT2+jw ASo )()1(jw X jw +=2+jw AThat is21)2)(1()(+-++=++=jw Ajw A jw jw A jw X−→←FT )()()(2t u Ae t u Ae t x t t ---=From (3), we know: π2)(2=⎰∞∞-dw jw XBut dt e e A dt t x dw jw X t t 22222)(2)(2)(--∞∞∞-∞∞--==⎰⎰⎰ππdt e e e At t t )2(243202---∞+-=⎰π∞----+---=04322)4322(2tt t e e e A π22122)413221(2A A ππ=+-= So 2122A π=π2, that is 122=A, 12±=AWhile 0)(≥t x ,12=AThen)()()(2t u Ae t u Ae t x t t ---=)()(122t u e e t t ---=4.15 Solution:From (3), we know:tet - −→←FT)}(Re{jw X)}(Re{jw X 2)()(jw X jw X *+=−→←FT2)()(t x t x -+*From (1)、(2), we know: )(t x is real and )()()(t u t x t x =∴2)()(t x t x -+*=2)()(t x t x -+=te t -∴ )()]()([t u t x t x -+*=)()]()([t u t x t x -+===)()()(t x t u t x )(2t u e t t-then=)(t x )(2t u te t -4.16. Solution:(a).)(sin )4()4()4sin()(t g t t k t k kt x k ⎪⎭⎫ ⎝⎛=-=∑∞-∞=ππδππ∴ )4()(πδπk t t g k -=∑∞-∞=(b).)4(sin )(sin )(πδπππk t t t t g t t t x k -⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛=∑∞-∞=⎪⎭⎫ ⎝⎛=t t t f πsin )(−→←FT⎩⎨⎧≤=others w jw F ...,.........01....,.........1)()4()(πδπk t t g k -=∑∞-∞=−→←FT)8(8)4/2(4/2)(k w k w jw G k k -=-=∑∑∞-∞=∞-∞=δπππδπππ∴ ))8((4)()(21)(k w j F jw G jw F jw X k -=*=∑∞-∞=π∴ In one period, ⎪⎩⎪⎨⎧≤<≤=41.,.........01.,.........4)(w w jw X 4.21. (g).Solution: ⎰∞∞--=dt e t x jw X jwt )()(⎰⎰⎰------++-=211112dt e dt tedt ejwt jwtjwt=21111112}{1jw ejw e te jwjwe jwt jwt jwt jwt -+-------------222)(jw e e jw e e e e e e jwjw jw w j jw jw w j jw ----++++-=----222)(jw e e jw e e jwjw w j w j ---+=--jw wjww j w w j jw w 2cos 2sin 2sin 22cos 222-=----=4.22 Solution:(a). According to the duality of FT: )(jt X −→←FT)(2w x -π, we will gett Wt πsin −→←FT⎪⎩⎪⎨⎧>≤=Ww W w jw X ,...0,...1)(∴when )2()]2(3sin[2)(ππ--=w w jw X , ⎪⎩⎪⎨⎧>≤=3.......,.........03.,.........)(2t t e t x tj π(b). 22)34cos()(4343)34()34(wj jwj jw j w j e e ee e ew jw X --+-+-=-=+=πππππ∴2)4()4()(33--+=-t e t e t x jjδδππ(d). )]2()2([3)]1()1([2)(πδπδδδ++-++--=w w w w jw X∴ t t j e e e e t x t j t j jt jt πππππππ2cos 3sin 2)(23)(1)(22+=++-=-- 4.24. (a). Solution:for )(t x is real, we have:(1). 0)}(Re{=jw X02)()()}(Re{=+=*jw X jw X jw X−→←FT02)()(2)()()}({=-+=-+=*t x t x t x t x t x Ev ∴To satisfy 0)}(Re{=jw X , )()(t x t x --=, )(t x is real and odd.∴ figure (a), (d) satisfy the condition.(2). 0)}(Im{=jw X02)()()}(Im{=-=*jw X jw X jw X−→←FT02)()(2)()()}({=--=--=*t x t x t x t x t x Ev ∴To satisfy 0)}(Im{=jw X , )()(t x t x -=, )(t x is real and even.∴ figure (e), (f) satisfy the condition.(3). There exists a real αsuch that )(jw X e w j α is real)(jw X e w j α is real∴ 0)}(Im{=jw X e w j α, and )(jw X e w j α −→←FT)(α+t x ∴ To satisfy 0)}(Im{=jw X e w j α, )(α+t x is real and even. ∴ figure(a) and (b) )0(=α, (e) and (f) )0(≠αsatisfy the condition.(4). 0)(=⎰∞∞-dw jw X)0(x =0)(=⎰∞∞-dw jw X∴ If 0)(=t x when 0=t , then it can satisfy 0)(=⎰∞∞-dw jw X∴ figure(a) , (c), (d) and (f) satisfy the condition.Note: (b) is not satisfies the condition.)1(2-t δ −→←FT2 (5). 0)(=⎰∞∞-dw jw wX)(t x dtd −→←FT)(jw jwX∴ To satisfy0)(=⎰∞∞-dw jw wX , it only needs0)(0==t t x dt d)1(2-t δ −→←FT2, 02=⎰∞∞-wdw∴ figure (b), (c), (e), (f) satisfy this condition.(6). )(jw X is periodicIf )(jw X is periodic, )(t x is discrete in time domain.So only (b) satisfies the condition.4.25. Solution:(a). Suppose )(1t x =)1(+t x , then we ’ll find )(1t x is real and even. So )(1jw X is real and even, which means 0)(1=<jw X .)(1jw X =jw e jw X )(∴ )(jw X =jw e jw X -)(1∴ ⎩⎨⎧≤-≥-=<0)(,......0)(,.........)(11jw X w jw X w jw X π (b). dt t x X j X )()0()0(⎰∞∞-== is the integral area of )(t xthen, 718122124)0(=-=**-*=X (c). πππ422)0(2)(=*==⎰∞∞-x dw jw X (e).dt t x dw jw X 22)(2)(⎰⎰∞∞-∞∞-=π⎪⎭⎫ ⎝⎛++-+=⎰⎰⎰⎰-dt dt t dt t dt 2322212102012)2(22ππ376=4.27. Solution: (1).)(t u −→←FTjww 1)(2+πδ∴ )3()2(2)1()(-+---=t u t u t u t x−→←FT)2(1)(2)(32wj w j jw e e e jw w jw X ---+-⎪⎪⎭⎫ ⎝⎛+=πδ)2(132w j w j jwe e e jw---+-=(2).dt e t x T dt et x T a t T jk Tt Tjk T k ππ22)(1)(~1--⎰⎰==)2(1)(12Tjk X T dte t x T t T jk ππ==-∞∞-⎰4.30. Solution: (a).⎩⎨⎧≤=others w jw G ., (02).,.........1)( ∴ tt t t t t g ππcos sin 22sin )(==t t x t g cos )()(=∴ ttt x πsin 2)(=4.32 Solution: (a).)()()(11t h t x t y *=)1())1(4sin()(--=t t t h π −→←FT ⎩⎨⎧≤=-others w e jw H jw .........,.........04....,.........)()26cos()(1π+=t t x −→←FT)}6()6({)(121-++=w w ejw X jδδππ∴ )()()(11jw H jw X jw Y =0=(b).)()()(22t h t x t y *=)1())1(4sin()(--=t t t h π −→←FT⎩⎨⎧≤=-others w e jw H jw .........,.........04....,.........)()3sin()21()(02kt t x k k ∑∞==−→←FT)}3()3({)21()(02k w k w j jw X k k --+=∑∞=δδπ∴ ⎪⎪⎩⎪⎪⎨⎧=--+=-==--others k w w e j k w w e j jw H jw X jw Y jw jw ........,.........01)},...3()3({210....)},.......()({)()()(22δδπδδπ⎪⎪⎩⎪⎪⎨⎧=--+==-o t h e r s k w e j w e j k j j ........................................................,.........01..),........3(21)3(210.......................................................,.........033δπδπ∴ only when 1k =, 2()Y j ω has output impulse .∴ )1(3sin 212214141)()1(3)1(333332-=--=-=-----t j e e e e j e e j t y t j t j t j t j 4.34 Solution: (a).)()()(jw H jw X jw Y =65)(4564)(22+++=+-+=jw jw jw jw w jw jw H∴ }4){(}65)){((2+=++jw jw X jw jw jw Y∴)(4)()(6)(5)(22t x t x dt dt y t y dt d t y dt d +=++ (b).65)(4564)(22+++=+-+=jw jw jw jw w jw jw H3122)3)(2(4+-++=+++=jw jw jw jw jw∴ )()(2)(32t u e t u e t h t t ---=(c). )()()(44t u te t u e t x t t ---=∴ 22)4(3)4(141)(++=+-+=jw jw jw jw jw X)3)(2(4564)(2+++=+-+=jw jw jw jw w jw jw H∴ )3)(2(4)4(3)()()(2+++++==jw jw jw jw jw jw H jw X jw Y)2(2/1)4(2/1)2)(4(1+++-=++=jw jw jw jw∴ ))()((21)(42t u e t u e t y t t---=4.35 Solution: (a).0........,.........)(>+-=a jwa jwa jw Hthe magnitude is: 1)(2222=++=+-=wa w a jwa jwa jw Hthe phase is: ⎪⎭⎫⎝⎛-=⎪⎭⎫ ⎝⎛-⎪⎭⎫ ⎝⎛-=<a w arctg a w arctg a w arctg jw H 2)( the impulse response is:jwa jwjw a a jw a jw a jw H +-+=+-=)()()(2))(()()(t t u ae t u e dtd t u ae t h at atat δ-=-=--- (b). )3cos(cos )3/cos()(t t t t x ++= +-++=)3/1()3/1({)(w w jw X δδπ)}3()3()1()1(-+++-+++w w w w δδδδandarctgw j e jwjwjw H 211)(-=+-=so )()()(jw H jw X jw y =+-++=---)3/1()3/1({)3/1(2)3/1(2w e w e a r c t g j a r c t gj δδπ)1()1()1(2)1(2-+++---w e w e arctg j arctg j δδ)}3()3()3(2)3(2-+++---w e w e arctg j arctg j δδ+-++=-)3/1()3/1({3/3/w e w e j j δδπππ)1()1(2/2/-+++-w e w e j j δδππ)}3()3(3/23/2-+++-w e w e j j δδππThen )3/23cos()2/cos()3/3/cos()(πππ-+-+-=t t t t yWe can find that y(t) is the sum of a phase shift for each x(t)’s component. 4.36 Solution: (a). )(][)(3t u e e t x t t --+=∴ )3)(1()2(23111)(+++=+++=jw jw jw jw jw jw X )(]22[)(4t u e e t y t t ---=∴ )4)(1(64212)(++=+-++=jw jw jw jw jw Y∴ )2)(4()3(3)()()(+++==jw jw jw jw X jw Y jw H(b). According to (a)’s result,22/342/3)2)(4()3(3)(+++=+++=jw jw jw jw jw jw H∴ )()(23)(24t u e e t h t t--+=(c).86)(93)2)(4()3(3)(2+++=+++=jw jw jw jw jw jw jw H∴ )}93){(}86)){((2+=++jw jw X jw jw jw Y∴ )(9)(3)(8)(6)(22t x t x dt dt y t y dt d t y dtd +=++4.37 Solution:(a). Supposing )(t x −→←FT)(jw X ,and from the differentiation property of FT,we canget22()d x t dt −→←FT 2()()j X j ωω22()(1)2()(1)d x t t t t dt δδδ=+-+-∴ 2/2/222()()2()(2sin(/2))j j j j j X j e e ee j ωωωωωωω--=-+=-=Finally, 22sin(/2)()[]X j ωωω=(b).(c). It is obvious that another signal g(t) sketched as following figure can reduce:~()()(4)k x t g t t k δ∞=-∞=*-∑(d).(4)()22k k t k k ππδδω∞∞=-∞=-∞-↔-∑∑From the convolution property of FT,we can get: ~~()()(4)()()()222k k kx t g t t k X j G jk πππδωδω∞∞=-∞=-∞=*-↔=-∑∑~~()()(4)()()()222k k kx t x t t k X j X jk πππδωδω∞∞=-∞=-∞=*-↔=-∑∑ So, we can know that,although ()G j ω is different from ()X j ω,()()22kkG j X jππ= forall integers k .4.43 Solution:We can draw a system as figure (a) ,where ,sin ttπ −→←FT ()H j ω2tshown in figure(c).Extra problems:We have known )(t x −→←FT)(jw X , prooftt x dt d π1)(* −→←FT)(jw X w ⋅ Proof: )(t x −→←FT)(jw X∴)(t x dtd −→←FT)(jw jwX)(t sign −→←FTjw2According to the duality of FT, we will get t π1 −→←FT)()(w jsign w jsign -=- ∴ tt x dt d π1)(* −→←FT)]([)(w jsign jw jwX -⋅)()()(w sign w jw X w wsign ⋅==Then, we havett x dt d π1)(* −→←FT)(jw X w ⋅)(t s )(t y )(t x t)(a )(b。
奥本海姆 信号与系统 习题参考答案
第五、六、七、八章作业解答5.1解：（a ）ωωωωωωj j j j n j n n j n n n e e e ee e n u F -----∞=--∞=--=-===-∑∑211211212)21(2)21(]}1[)21{(1111 其模为：ωωcos 4111|)(|-+=j e X5.4解：根据:∑∞-∞=-=k k F )2(2}1{πωπδ)2()2(}{c o s 000ωπωπδωπωπδω+-+--=∑∞-∞=k k F k故：n 2cos1)(1π+=t x5.23解：（a ） 因为：nj N j en x eX ωω-+∞-∞=∑=][)(则：6)1(121121)1(][)(0=-+++++++-==∑+∞-∞=N j n x e X（b ）设]2[][1+=n x n x则]2[][1-=n x n x ωωω21)()(j j j e e X e X -=x1[n]是实偶对称信号，则⎪⎩⎪⎨⎧<>=∠0)(0)(0)(111ωωωπj j j e X e X e X故：⎪⎩⎪⎨⎧<->-=∠+-=∠0)(20)(2)(2)(111ωωωωωπωωj j j j e X e X e X e X（c ）因为： ⎰=πωωωπ2)(21][d e e X n x n j j则：ππωππω4]0[2)(==⎰-x d e X j(d) 因为：nj N j en x eX ωω-+∞-∞=∑=][)(2)1()1(12)1(112)1(1)1()1()1(][][)(=-+-⨯++-⨯+++-⨯+-⨯-=-==∑∑+∞-∞=+∞-∞=nN nj N j n x en x e X ππ(e) 2][][][)}(Re{n x n x n x eX e Fj -+=−→←π（图略）(f) I 根据帕斯瓦尔关系式：⎰∑+∞-∞==πωωπ222|)(|21|][|d e X n x j n 则：πππωωππ28)11411411(2|][|2|)(|22=+++++++==∑⎰+∞-∞=-n j n x d e XII 根据：ωωd e dX n nx j FT)(][j −→←-则：πππωωωππ316)4925649119(2|][-j |2|)(|22=++++++==−→←∑⎰∞-∞=-n j FTn nx d d e dX6.5解：而：ttt h c πωsin )(1= 的傅里叶变换为：0 ωc ω -ωc则)()()(1t g t h t h = 而：)2()2()(11c c j H j H j H ωωωωω-++= 故：t e e t g c t j tj c c ωωω2cos 2)(22=+=-6.23(a) ttt h c πωsin )(=(b) ⎩⎨⎧<=othere j H c Tj 0||)(ωωωω则：)()(sin )(T t T t t h c ++=πω（c ）⎪⎪⎪⎩⎪⎪⎪⎨⎧<<-<<=-othere e j H c j c j 000)(22ωωωωωππ故：ωπωπωωπωωπd e e d e e t h t j jtj jcc⎰⎰+=--02022121)(c c t j jt j je t e e t e ωωπωωπππ0202|j 2|j 2+=--c c t j jt j je te e t e ωωπωωπππ0202|j 2|j 2+=-- ]1[2]1[222-+-=--t j jt j jc c e t j e e t j e ωπωπππ]}[]{[21122t j t j j j c c e e e e jt ωωπππ---+-=]}[]{[211)2/()2/(22πωπωπππt j t j j j c c e e e e jt -+--+-=]1[cos 1-=t t c ωπ7.3解：(a))(t x 的最高频率为4000π,故奈奎斯特率为8000π； （b ） 同上；（c ） 时域相乘，频域相卷，故)(t x 的最高频率为8000π，则奈奎斯特率为16000π；7.22解： y(t)为：)()()(ωωωj H j X j Y =πωω1000||0)(>=j Y对y(t)进行采样，则要求采样频率fs>2000π，故采样周期因为ms T 120002=<ππ7.27解： （a ）x (t)e ∑∞-∞=-=n nT t t p )()(δx p (t)因为：ω0如下图图所示，为ω1与ω2的中点位置：X(j ω) ωω1 -ω2 -ω11ω2 ω0（a ）)()(01ωωω+=j X j X)()()(12ωωωj H j X j X =21X2(j ω) ω(ω2-ω1)/21-(ω2-ω1)/2（b ）最大的采样周期为：122ωωπ-(c)该系统为：先通过如下图所示的低通滤波器，得到X2(j ω)，然后将其乘以 e j ω0t ,则信号频移ω0，然后通过一个反折系统，其系统框图图下图所示：H (j ω) ω(ω2-ω1)/2T-(ω2-ω1)/2e j ω0t说明：因为)(X ωj a 为实数，则有：)(X )(X *ωωj j a a =而)()()}(Re{2)(x *t x t x t x t aa a +==)()()(*ωωωj X j X j X a a -+=又因为)(X )(X *ωωj j a a =故：)(X )(X *ωωj j aa -=-则：)()()(ωωωj X j X j X a a -+=得到结论8.3解：定性的理解由于载波信号与同步信号的相位相差为90o ，故输出信号为0； （也可以通过调制、解调的时域表达式计算出结果也为0）。
奥本海姆 信号与系统 习题参考答案 电子科技大学
Chapter 22.1 Solution:Because x[n]=(1 2 0 –1)0, h[n]=(2 0 2)1-, then (a).So, ]4[2]2[2]1[2][4]1[2][1---+-+++=n n n n n n y δδδδδ(b). according to the property of convolutioin:]2[][12+=n y n y(c). ]2[][13+=n y n y2.3 Solution:][*][][n h n x n y =][][k n h k x k -=∑∞-∞=∑∞-∞=-+--=k k k n u k u ]2[]2[)21(2][211)21()21(][)21(12)2(0222n u n u n n k k --==+-++=-∑][])21(1[21n u n +-=the figure of the y[n] is:2.5 Solution:We have known: ⎩⎨⎧≤≤=elsewhere n n x ....090....1][，，, ⎩⎨⎧≤≤=elsewhere N n n h ....00....1][，，,(9≤N)Then, ]10[][][--=n u n u n x , ]1[][][---=N n u n u n h∑∞-∞=-==k k n u k h n h n x n y ][][][*][][∑∞-∞=-------=k k n u k n u N k u k u ])10[][])(1[][(So, y[4] ∑∞-∞=-------=k k u k u N k u k u ])6[]4[])(1[][(⎪⎪⎩⎪⎪⎨⎧≥≤=∑∑==4,...14, (14)N N k Nk =5, then 4≥NAnd y[14] ∑∞-∞=------=k k u k u N k u k u ])4[]14[])(1[][(⎪⎪⎩⎪⎪⎨⎧≥≤=∑∑==14,...114, (114)55N N k Nk =0, then 5<N∴ 4=N2.7 Solution:[][][2]k y n x k g n k ∞=-∞=-∑（a ） [][1]x n n δ=-,[][][2][1][2][2]k k y n x k g n k k g n k g n δ∞∞=-∞=-∞=-=--=-∑∑(b) [][2]x n n δ=-,[][][2][2][2][4]k k y n x k g n k k g n k g n δ∞∞=-∞=-∞=-=--=-∑∑(c) S is not LTI system.. (d) [][]x n u n =,0[][][2][][2][2]k k k y n x k g n k u k g n k g n k ∞∞∞=-∞=-∞==-=-=-∑∑∑2.8 Solution:)]1(2)2([*)()(*)()(+++==t t t x t h t x t y δδ)1(2)2(+++=t x t xThen,That is, ⎪⎪⎪⎩⎪⎪⎪⎨⎧≤<-≤<-+-=-<<-+=others t t t t t t t t y ,........010,....2201,.....41..,.........412,.....3)(2.10 Solution:(a). We know: Then, )()()(αδδ--='t t t h)]()([*)()(*)()(αδδ--='='t t t x t h t x t y)()(α--=t x t xthat is,So, ⎪⎪⎩⎪⎪⎨⎧+≤≤-+≤≤≤≤=others t t t t t t y ,.....011,.....11,....0,.....)(ααααα(b). From the figure of )(t y ', only if 1=α, )(t y ' would contain merely therediscontinuities.2.11 Solution:(a). )(*)]5()3([)(*)()(3t u e t u t u t h t x t y t ----==⎰⎰∞∞---∞∞--------=ττττττττd t u e u d t u eu t t )()5()()3()(3)(3⎰⎰-------=tt t t d e t u d et u 5)(33)(3)5()3(ττττ⎪⎪⎪⎪⎩⎪⎪⎪⎪⎨⎧≥+-=-<≤-=<=---------⎰⎰⎰5,.......353,.....313.........,.........0315395)(33)(3393)(3t e e d e d e t e d e t tt t t t t t t t ττττττ(b). )(*)]5()3([)(*)/)(()(3t u e t t t h dt t dx t g t----==δδ)5()3()5(3)3(3---=----t u e t u e t t(c). It ’s obvious that dt t dy t g /)()(=.2.12 Solution∑∑∞-∞=-∞-∞=--=-=k tk tk t t u ek t t u e t y )]3(*)([)3(*)()(δδ∑∞-∞=---=k k t k t u e)3()3(Considering for 30<≤t ,we can obtain33311])3([)(---∞=-∞-∞=--==-=∑∑e e e ek t u e e t y tk k tk kt.(Because k must be negetive , 1)3(=-k t u for 30<≤t ).2.19 Solution:(a). We have known: ][]1[21][n x n w n w +-=(1)][]1[][n w n y n y βα+-=(2)from (1), 21)(1-=E EE Hfrom (2), αβ-=E EE H )(2then, 212212)21(1)21)(()()()(--++-=--==E E E E E E H E H E H ααβαβ∴ ][]2[2]1[)21(][n x n y n y n y βαα=-+-+-but, ][]1[43]2[81][n x n y n y n y +-+--=∴ ⎪⎩⎪⎨⎧=⎪⎭⎫ ⎝⎛=+=143)21(:....812βααor∴⎪⎩⎪⎨⎧==141βα (b). from (a), we know )21)(41()()()(221--==E E E E H E H E H21241-+--=E E E E∴ ][)41()21(2][n u n h n n ⎥⎦⎤⎢⎣⎡-=2.20 (a). 1⎰⎰∞∞-∞∞-===1)0cos()cos()()cos()(0dt t t dt t t u δ(b). 0 dt t t )3()2sin(5+⎰δπ has value only on 3-=t , but ]5,0[3∉-∴dt t t )3()2sin(5+⎰δπ=0(c). 0⎰⎰---=-641551)2cos()()2cos()1(dt t t u d u πτπττ⎰-'-=64)2cos()(dt t t πδ0|)2(s co ='=t t π0|)2sin(20=-==t t ππ2.23 Solution:∑∞-∞=-==k t h kT t t h t x t y )(*)()(*)()(δ∑∞-∞=-=k kT t h )(∴2.27 Solution()y A y t dt ∞-∞=⎰,()xA x t dt ∞-∞=⎰,()hA h t dt ∞-∞=⎰.()()*()()()y t x t h t x x t d τττ∞-∞==-⎰()()()()()()()()()(){()}y x hA y t dt x x t d dtx x t dtd x x t dtd x x d d x d x d A A ττττττττττξξτττξξ∞∞∞-∞-∞-∞∞∞∞∞-∞-∞-∞-∞∞∞∞∞-∞-∞-∞-∞==-=-=-===⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰2.40 Solution(a) ()()(2)tt y t ex d τττ---∞=-⎰,Let ()()x t t δ=,then ()()y t h t =.So , 2()(2)(2)()(2)()(2)tt t t t h t ed e d e u t τξδττδξξ---------∞-∞=-==-⎰⎰(b) (2)()()*()[(1)(2)]*(2)t y t x t h t u t u t eu t --==+---(2)(2)(1)(2)(2)(2)t t u eu t d u e u t d ττττττττ∞∞-------∞-∞=+------⎰⎰22(2)(2)12(1)(4)t t t t u t ed u te d ττττ---------=---⎰⎰(2)2(2)212(1)[]|(4)[]|t t t t u t e e u t ee ττ-------=--- (1)(4)[1](1)[1](4)t t e u t e u t ----=-----2.46 SolutionBecause)]1([2)1(]2[)(33-+-=--t u dtde t u e dt d t x dt d t t)1(2)(3)1(2)(333-+-=-+-=--t e t x t et x tδδ.From LTI property ,we know)1(2)(3)(3-+-→-t h e t y t x dtdwhere )(t h is the impulse response of the system. So ,following equation can be derived.)()1(223t u e t h e t --=-Finally, )1(21)()1(23+=+-t u e e t h t 2.47 SoliutionAccording to the property of the linear time-invariant system: (a). )(2)(*)(2)(*)()(000t y t h t x t h t x t y ===(b). )(*)]2()([)(*)()(00t h t x t x t h t x t y --==)(*)2()(*)(0000t h t x t h t x --=)2()(00--=t y t y(c). )1()1(*)(*)2()1(*)2()(*)()(00000-=+-=+-==t y t t h t x t h t x t h t x t y δ(d). The condition is not enough.(e). )(*)()(*)()(00t h t x t h t x t y --== τττd t h x )()(00+--=⎰∞∞-)()()(000t y dm m t h m x -=--=⎰∞∞-(f). )()]([)](*)([)(*)()(*)()(000000t y t y t h t x t h t x t h t x t y "=''='--'=-'-'==Extra problems:1. Solute h(t), h[n](1). )()(6)(5)(22t x t y t y dt dt y dtd =++ (2). ]1[][2]1[2]2[+=++++n x n y n y n y012y(t)t4Solution:(1). Because 3121)3)(2(1651)(2+-++=++=++=P P P P P P P H so )()()()3121()(32t u e e t P P t h t t ---=+-++=δ (2). Because )1)(1(1)1(22)(22i E i E EE E E E E E H -+++=++=++=iE Ei i E E i -+-+++=1212so []][)1()1(2][1212][n u i i i k i E E i i E E i n h n n +----=⎪⎪⎪⎪⎭⎫⎝⎛-+-+++=δ。
信号与系统奥本海姆第二版课后答案
信号与系统 奥本海姆第二版 习题解答Department of Computer Engineering2005.12ContentsChapter 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)Chapter 1 Answers1.1 Converting from polar to Cartesian coordinates:111cos 222j eππ==- 111c o s ()222j e ππ-=-=- 2cos()sin()22jj j eπππ=+=2c o s ()s i n ()22jjj eπππ-=-=- 522j jj eeππ==4c o s ()s i n ())144jjj πππ+=+9441j jj ππ=-9441j j j ππ--==-41jj π-=-1.2 055j=, 22j e π-=,233jj e π--=212je π--=, 41j j π+=, ()2221jj eπ-=-4(1)j je π-=, 411j je π+=-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()TTTTT 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 bezero for t<9.1.6(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[n1.7. (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→∞.1.8. (a) ()01{()}22cos(0)tt tx eπℜ=-=+(b) ()02{()}cos()cos(32)cos(3)cos(30)4tt t t tx eππℜ=+==+(c) ()3{()}sin(3)sin(3)2t tt t tx e eππ--ℜ=+=+(d) ()224{()}sin(100)sin(100)cos(100)2t t tt t t tx e e eππ---ℜ=-=+=+1.9. (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,2()tx 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 any integer m such that m(2wπ) is also an integer. Therefore,5[]nxis 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 toπ.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 M=-1 and no=-3.1.13y (t)=⎰∞-tdt x )(τ =dt t))2()2((--+⎰∞-τδτδ=⎪⎩⎪⎨⎧>≤≤--<2,022,12,0,t t tTherefore ⎰-==∞224d t 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=1.1.15 (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]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 ∈ ґ. 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] =][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]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 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]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 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.1.20 (a) Givenx )(t =jt e 2 y(t)=t j e 3x )(t =jt e 2- y(t)=t j e 3- Since the system liner+=tj e t x 21(2/1)(jt e 2-))(1t y =1/2(tj e 3+tj e 3-)Thereforex 1(t)=cos(2t))(1t y =cos(3t)(b) we know thatx 2(t)=cos(2(t-1/2))= (j e -jte 2+je jt e 2-)/2Using the linearity property, we may once again writex 1(t)=21( j e -jt e 2+j e jte 2-))(1t y =(j e -jt e 3+je jte 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 S1.21.1.24 The even and odd parts are sketched in Figure S1.24 1.25 (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.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 inputx3[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 ()()()()222211S dx t x t y t x t dt ⎡⎤−−→=⎢⎥⎣⎦ Now consider a third input x3[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] andx2[n]=δ[n+1]+ 2δ[n+1]+ 3δ[n]. The corresponding outputs evaluated at n=0 are [][]120203/2y andy ==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.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 is2(4) y 2(t) periodic, period T; x(t) periodic, period T/2;1.33(1) True x[n]=x[n+N ]; y 1 (n)= y 1 (n+ N 0)i.e. 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 i.e. 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 1.34. (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 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,{}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∞∞∞=-∞=-∞=-∞==+∑∑∑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 ∞∞∞-∞-∞-∞=+⎰⎰⎰()()()()()().xy yx t x t y d y t x d t φττττττφ∞-∞∞-∞=+=-+=-⎰⎰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 periodic0(),0..2/j n N T o e where T ωωπ+= This implies that022o 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 havepart(a) that()().xx xx t t φφ=-This implies that()xy t φis(b) Note from 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=ThereforeThis implies that1(2)().2t t δδ=(b)The plot are as shown in Figure s3.18. 1.39 We havelim ()()lim (0)()0.u t t u t δδ→→==Also,0022gcd(,)gcd(,)gcd(,)gcd(,).T pp q p q p q p q q p q p pωωππ===/21lim (2)lim ().2t t δδ→∞→∞=01lim ()()().2u t t t δδ→=u Δ'（t ） 1 1/2Δ/2-Δ/2t 0tu Δ'（t ）12Δ t 0tu Δ'（t ） 1 1/2Δ-Δttu Δ'（t ）1 1/2Δ-Δt 0t⎰⎰∞∞∞--=-=0)()()()()(ττδτττδτd t u d t u t gTherefore,0,0()1,00t g t t undefined for t >⎧⎪=<⎨⎪=⎩()0()()()t u t t δττδτδτ-=-=-1.40.(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) 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 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)-1.These corresponds to two nonlinear systems. If these systems are connected in series ,then z(t)=x(t) which is a linear system.00.()().00x t y t =→=0()()()()0x t x t y t y t =-→-=(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.1.43. (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 that y(t)=y(t-T).This impliesthat 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,()()()()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 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 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]. 1.45. (a) Consider ,()111()()shx x t y t t φ→= and()222()()shx 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. 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]).Chapter 2 answers2.1 (a) We have know that 1[]*[][][]k y x n h n h k x n k ∞=-∞==-∑1[][1][1][1][1]y n h x n h x n =-++-2[1]2[1]x n x n =++-This gives1[]2[1]4[]2[1]2[2]2[4]y n n n n n n δδδδδ=+++-+--- (b)We know that2[][2]*[][][2]k y n x n h n h k x n k ∞=-∞=+=+-∑Comparing with eq.(S2.1-1),we see that21[][2]y n y n =+(c) We may rewrite eq.(S2.1-1) as1[][]*[][][]k y n x n h n x k h n k ∞=-∞==-∑Similarly, we may write3[][]*[2][][2]k y n x n h n x k h n k ∞=-∞=+=+-∑Comparing this with eq.(S2.1),we see that31[][2]y n y n =+2.2 Using given definition for the signal h[n], we may write{}11[][3][10]2k h k u k u k -⎛⎫=+-- ⎪⎝⎭The signal h[k] is non zero only in the rang 1[][2]h n h n =+. From this we know that the signal h[-k] is non zero only in the rage 93k -≤≤.If we now shift the signal h[-k] by n to the right, then the resultant signal h[n-k] will be zero in the range (9)(3)n k n -≤≤+. Therefore ,9,A n =- 3B n =+ 2.3 Let us define the signals11[][]2nx n u n ⎛⎫= ⎪⎝⎭and1[][]h n u n =. We note that1[][2]x n x n =- and 1[][2]h n h n =+ Now,。
奥本海姆信号与系统第二章部分习题答案
5
+ + 2.22
概率画出结果。
< 有三种解法，建议用图解法
2.22
概率画出结果。
2.28 下面均为离散时间线性时不变系统的单位脉冲响应，试判定每一系统是 否是因果和/或稳定的。陈述理由。 (a)ℎ ������ = (5)������ ������[������] (c) ℎ ������ = (2)������ ������ −������
4 5
y
长度为 M
y
x
而 N ≤n ≤N 长度为 M ，试用M 和 M 来表示 M 。
h x
(c) 考虑一个离散线性时不变系统，它具有这么一个特点，即若对 全部 n ≥10, x[n] = 0 ，则对所有的 n ≥15 都有 y[n] = 0 。系统单 位脉冲响应 h[n] 必须满足什么条件才有此特性？ (a) (c) (b)
(d) 有一个线性时不变系统的单位脉冲响应如图P2.44所示。为了确定 y (0) ，必须要知道在什么一个区间上的 x(t ) ？
2.50 图P2.50所示为两个系统的级联，其中一个系统 A 是线性时不变的，
y (t ) 而第二个系统 B 是系统 A 的逆系统，若 y (t )是系统 A 对 x (t )的响应，
5 ������ −5 1
1 − ������ cos 2������������ ������������
������0 ������ = ������ ������
(a) (b) (c)
u0 (t ) cos(t )dt (t ) cos(t )dt cos(0) 1
第二章
2.4 计算并画出y[n]=x[n]*h[n]，其中
奥本海姆信 与系统 部分习题答案
(c)若y(t)=x(t)*h(t),则y(-t)=x(-t)*h(-t)
√ห้องสมุดไป่ตู้
（b）
（c）
2.16 对下列说法，判断是对还是错 (d)若t>T1时x(t)=1且t>T2时h(t)=0,则t>T1+T2时x(t)*h(t)=0 √
2.20 求下列积分
(a)
∞ −∞
������0
(������)cos
<
+ + + 2.11
< 有三种解法，建议用图解法
2.16 对下列说法，判断是对还是错 (a) 若n<N1时，x[n}=0且n<N2时h[n]=0,那么n<N1+N2时x[n]*h[n]=0 √
（a）
2.16 对下列说法，判断是对还是错
(b)若y[n]=x[n]*h[n],则y[n-1]=x[n-1]*h[n-1] ×
2
3
x
而
N 4
≤n
≤N 长度为 5
M
y
，试用M h
和M x
来表示 M y
。
(c) 考虑一个离散线性时不变系统，它具有这么一个特点，即若对
全部 n ≥10, x[n] = 0 ，则对所有的 n ≥15 都有 y[n] = 0 。系统单
位脉冲响应 h[n] 必须满足什么条件才有此特性？
(a)
(b)
(c)
(c)ℎ ������ = ������−2������������ ������ + 50 非因果、稳定
(e) h ������ = ������−6 ������
非因果、稳定
������−100
+(g) h t = 2������−������ − ������ 100 ������(������) 因果、非稳定
奥本海姆信号与系统第一章部分习题答案
1.26 判定下列离散时间信号的周期性；若是周期的，确定它的基波周期。
(a )
x[n] sin(
N
6
n 1)
7
2
* m 7, m 3
6 / 7
∴ 是周期的，基波周期为 =7
(d )
x[n] cos( n) cos( n)
2
4
1
积化和差：cos cos = 2 [cos + + cos( − )]
关函数及其有关性质进行简单介绍。
设() 和()两个信号，相关函数定义为
较卷积少翻转
函数 xx (t ) 通常称为信号()的自相关函数，而 xy (t ) 称为互相关函数。
（a） xy (t ) 和
之间是什么关系？
（b）求 xx (t ) 的奇部。
（c）假设
（a）
，将 xy (t ) 和
第一章
1.10 求信号 = 2 cos 10 + 1 − sin 4 − 1 的基波周期。
基波周期0 ： 使 = + 成立的最小正值。
2
1 =
=
10 5
2
2 =
=
4
2
基波周期0 = （最小公倍数）
¤ 不是乘积
1.11 求信号x[n]=1+ 4/7 − 2/5 的基波周期。
3 = 1 + 2
2 t = 1 ( − 0 )
1.19判定下列输入-输出关系的系统是否具有线性性质、时不变性质，或两者俱有。
线
性： 3 − 0 )
(c) = + 1 − − 1
∴ 是线性的
∴ 是时不变的