数字信号处理(英文版)复习重点题型和答案

•
-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
∞
−∞
(t 3 + t 2 + t + 1)δ (t )dt
e.
∫
ຫໍສະໝຸດ Baidu
∞
−∞
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
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:
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
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: 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
⎞ ⎟ X [k ] ⎠
, where X [k ] are computed in the previous question.
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
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.
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
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
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:
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 ×
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
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
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
−∞
∞
∫
−∞
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
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 =
+ 3e
− jk
, k = 0,1, 2
k = 0, k = 1, k = 2,
X [0] = 1 + 2 + 3 = 6 X [1] = 1 + 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
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:
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⎝ ⎠
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):
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
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: