数字信号处理(英文版)复习重点题型和答案

b. (omit)
pp76 Q1.13 Given the discrete time sinusoid x[n] = 2cos(0.2π n + 0.1π ) and the sampling frequency Fs = 2 kHz a. Determine the corresponding continuous time sinusoid in the frequency range 0 < F0 < F / 2 b. Determine two other continuous time sinusoids with the same sample values
⎞ ⎟ X [k ] ⎠
, where X [k ] are computed in the previous question.
b. x[n] = u[n − 2] Ans:
y[n] = h[n] ∗ x[n] = ∑ h[n − k ]x[k ] = ∑ 0.5n−k u[ n − k ]u[ k − 2]
k k
Since u[n] = 0 for n < 0 , u[n − k ] ≠ 0 for k ≤ n , and u[k − 2] ≠ 0 for k ≥ 2 Therefore, when n < 2 , y[n] = 0 . When n ≥ 2
Ans. a. x[n] = 2cos(0.2π n + 0.1π ) Digital frequency: ω0 = 0.2π = 2π F0Ts Therefore, the continuous time frequency:
F0 =
ω0 0.2π 2 = 0.2 kHz = 2π Ts 2π
π
180
=
π
12
π ⎞ π ⎞ ⎛ ⎛ − j ⎜100π t + ⎟ ⎞ 100π t + ⎟ 3 ⎛ j⎜ 12 ⎠ 12 ⎠ ⎝ ⎝ x(t ) = ⎜ e +e ⎟ ⎟ 2⎜ ⎝ ⎠ π π −j ⎞ 3 ⎛ j12 j100π t 12 − j100π t = ⎜e e +e e ⎟ 2⎝ ⎠
a. x = […,1,2,3,…] Ans: y[n] = 0.8 y[n − 1] + 0.2 x[n]
Y ( z ) = 0.8 z −1Y ( z ) + 0.2 X ( z ) Y ( z) 0.2 H ( z) = = X ( z ) 1 − 0.8 z −1 Let z = e jω , we have
xa [n] = 3cos(−1.9π n − 0.2π ) = 3cos(1.9π n + 0.2π )
pp.75 Q1.8 Write the following sinusoids in terms of complex exponentials
a. x(t ) = 3cos(100π t + 15o ) b. x(t ) = 2cos(10π t − 0.1π ) Ans. a. 15o ⇒ 15 ×
ω1 2.2π 2 = 2.2 kHz = 2π Ts 2π ω2 1.8π 2 = 1.8 kHz = 2π Ts 2π
pp.76 Q1.16 In each of the following systems let x(t ) or x[n] be the input and y (t ) or y[n] be the output. Determine whether each system is (1) linear, (2) time invariant, (3) causal, (4) BIBO stable.
Digital Signal Processing Review Questions Apr 15, 2012
pp. 73, Q1.1 Given the sequences shown, write them in terms of unit impulses
x[n]
1
•
1
•
0.5
•
• •
•
n
•
-0.5
+ 3e
− jk
, k = 0,1, 2
k = 0, k = 1, k = 2,
X [0] = 1 + 2 + 3 = 6 X [1] =பைடு நூலகம்1 + 2e
−j 2π 3 4π 3
+ 3e
j
2π 3 −j 8π 3
X [2] = 1 + 2e = 1 + 2e
j
−j
+ 3e
j
2π 3
+ 3e
4π 3
pp. 81 Q1.39 Consider the linear time invariant system with difference equation y[n] = 0.8 y[n − 1] + 0.2 x[n] . Determine the DFS of the output signal for each of the inputs
Pp77 Q1.17 A linear time invariant system has impulse
response h[n] = 0.5n u[n] . Determine the output sequence y[n] for each of the following input signals:
y[n] = ∑ 0.5n−k = ∑ 0.5k =
k =2 k =0
n
n−2
1 − 0.5n−1 , for n ≥ 2 0.5
In summary:
n<2 ⎧ 0, ⎪ y[n] = ⎨1 − 0.5n−1 , n≥2 ⎪ ⎩ 0.5
The output y[n] is the output of the filter h[n] with input x[n] , shown in the following figure
Filter coefficients h[ n] = 0.5n u[n] x[n] = u[n − 2]
z −1 0.5
z −1 0.53
z −1 0.54 +
••• •••
z −1 0.5k
••• 0.5k +1 • • •
1
y[n]
Pp77 Q1.23. Using partial fraction expansion, determine the inverse z-transform of the following functions:
(t 3 + t 2 + t + 1)δ (t )dt
e.
∫
∞
−∞
cos 2 (2π t + 0.1π )δ (t + 1)dt
∫ Ans. d. ∫
Ans a: Ans. e.
∞
−∞ ∞
e − tδ (t − 1)dt = e −1 (t 3 + t 2 + t + 1)δ (t )dt = 1
g. y[n] = x[n] + x[n − 1] + x[n − 2] i. y[n] = nx[n] + x[n − 1] + x 2 [n − 2] Ans. g. linear, time invariant, causal, BIBO stable. i. non-linear, time variant, causal, BIBO unstable Hints: to show y[n] is time variant, substitute x[n] with x[n − L] into the right-hand side of the equation. Can we get y[n − L] ?
Ans. b. x[n] = 2cos(ω x n + 0.1π ) , where ω x = 0.2π + 2π k and k is an integer Select k = ±1 ,
ω1 = 0.2π + 2π = 2.2π ω2 = 0.2π − 2π = −1.8π ⇒ 1.8π
The corresponding continuous time frequencies are: F1 = F2 =
a. x = […,1,2,3,…] Ans: Period: N = 3
X [k ] = DFS { x[n]} = ∑ x[n]e− jk (2π / N ) n
n =0
N −1
= 1 + 2e = 1 + 2e
− jk (2π /3) − jk 2π 3
+ 3e
− jk (2π /3)2 4π 3
H (ω ) =
0.2 1 − 0.8e − jω ⎛ 2π Y [k ] = DFS { y[n]} = H ⎜ k ⎝ 3
k = 0, Y [0] = X [0] ⎛ 2π ⎞ k = 1, Y [1] = H ⎜ ⎟ X [1] ⎝ 3 ⎠ 0.2 = X [1] 2π −j 1 − 0.8e 3 ⎛ 4π ⎞ k = 2, Y [2] = H ⎜ ⎟ X [2] ⎝ 3 ⎠ 0.2 = X [2] 4π −j 3 1 − 0.8e
•
-0.8
Ans:
x[n] = −0.5δ [n + 1] + δ [n] + 0.5δ [n − 1] + δ [n − 2] − 0.8δ [n − 3]
pp.74 Q1.2 Compute the following integrals
a. d.
∫ ∫
∞
−∞
e − tδ (t − 1)dt
∞
−∞
−∞
∞
∫
−∞
cos 2 (2π t + 0.1π )δ (t + 1)dt = cos 2 (2π − 0.1π ) = cos 2 (0.1π )
pp75 Q1.7 Given the discrete time sinusoid x[n] = 3cos(0.1π n − 0.2π ) , can you determine another sinusoid with digital frequency 0 < ω < 2π with the same samples
C0 =
X ( z) 1 z = z 2 z =0
With
C1 = = = And C2 = = = X ( z) ( z − 2) z z =2 X ( z) ( z − 1) z z =1
z +1 z −1 ( z − 1)( z − 2) z z =1 2 = −2 −1
z +1 z−2 ( z − 1)( z − 2) z z =2 3 3 = 1× 2 2
Therefore
X ( z ) = −2 3 z 1 z + + z −1 2 z − 2 2
3 x[n] = IZ { X ( z )} = −2u[n] + × 2n u[n] + δ [n] 2 n −1 = (−2 + 3 × 2 )u[n] + δ [n]
pp. 81 Q1.38 Determine the discrete Fourier series (DFS) expansion for the following periodic signals (one period is explicitly shown, beginning at n=0):
Ans: yes.
x[n] = 3cos(0.1π n − 0.2π ) = 3cos(ω0 n − 0.2π ) ,where ω0 = 0.1π
ω = ω0 + 2kπ = 0.1π + 2kπ
, where k is an integer. Select k = −1 , and
ω = 0.1π − 2π = −1.9π
c. X ( z ) = Ans: Formula
z +1 , ( z − 1)( z − 2)
z >2
x[n] = a nu[n] ⇔ X ( z ) =
z , z−a
z>a z , z−a z<a
x[n] = a nu[− n − 1] ⇔ X ( z ) = −
First, the original equation can be rewritten as:
X ( z) z +1 = z z ( z − 1)( z − 2)
Second, write the above equation in partial fraction expansion:
C X ( z ) C0 C1 = + + 2 z z z −1 z − 2
We can find the coefficients as follows: