T5级数字信号处理(A)参考答案及评分标准

湖北汽车工业学院 数字信号处理 考试试卷 (A题)

参考答案及评分标准

命题教师：湛柏明 审题教师：张洪

1、(16’) The block diagram illustrating the process of digital processing of an analog signal is depicted in Figure A.

]Figure A: Block diagram

It is known that the sampling pulse p(t) is p(t) where T = 0.2 s. The input signal x1(t)

1,for0 t 1s

0,otherwise

n

(t nT)

Assume that the impulse response of the digital system is h[n] = μ[n]-μ[n-3] a. Determine the sequence x4[n].(6’)

b. Determine (6’) and sketch (4’) the output x5[n] of the digital system. Solution:

a. Because x4[n] is obtained by sampling x1(t) at uniform intervals T= 0.2 s, in the range 0 ≤ t < 1 s,

1s1s 5 samples, thus we can get

T0.2s

x4[n] = {1 1 1 1 1} (得6分) ↑n=0

b. The output signal x5[n] is obtained by the linear convolution x5[n]

m

x[m]h[n m] x[m]h[n m]

4

4m 0

4

1 1 1 1 1

× 1 1 1

分) -------------------------------------------------------

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 The plot of x5[n] ------------------------------------------------------- 1 2 3 3 3 2 1

Thus the output signal x5[n] = {1 2 3 3 3 2 1} (得3分) ↑n=0

2、(12’) Determine the total solution for n≥ 0 of the difference equation

1

y[n 1] x[n] 4

n

with the input sequence x[n] = 0.5μ[n] and the initial conditions y[-1] = 8. Solution:

The characteristic equation is

1

0

4

1

The characteristic roots: (得2分)

4

Then the homogeneous solution is

1

forn 0 (得2分) yh[n] C()n

4

y[n]

Assume that the particular solution is yp[n] K(0.5)n [n], substituting yp[n] and x[n] into the system difference equation we have

K

K(0.5)n (0.5)n 1 (0.5)n

4

Then K 2

The particular solution is yp[n] 2(0.5)n [n] (得2分) Thus the total solution of the difference equation is

11

forn 0 y[n] C()n 2()n

42

Write the difference equation in to the recursive form:

1

y[n] x[n] y[n 1]

41

So y[0] x[0] y[ 1] 1 2 3 (得3分)

4

Substituting y[0] = 3 into the total solution y[n], we have two equations: y[0] C 2 3

Then C = 1

The total solution of the difference equation is then given by

11

forn 0 (得3分) y[n] ()n 2()n

42

3. (8’) In IIR digital filter design, we often need to design a prototype (原型) analog filter with the transfer function Ha(s), then we can use the bilinear transformation method to obtain the transfer function H(z) of the IIR digital filter. Assume that the transfer function Ha(s) of the prototype analog filter is

1 s 1

Determine the transfer function H(z) of an IIR digital filter using the bilinear transformation method. Solution:

The mapping formula of the bilinear transformation is

Ha(s)

1 z 1

s (得3分) 1

1 z

Substituting the above formula into Ha(s) we can get the transfer function H(z) of the IIR digital filter:

11 z 1

H(z) (得5分) 1

21 z

11 z 1

4、(18’) Suppose that the impulse response h[n] of a causal FIR filter is h[n] = {-4.8804 9.5661 -4.8804} It is known that the input signal of the filter is

x[n] = sin(0.2n) + sin(0.5n) + cos (0.2n) (a) Determine the frequency response H(ejω).( 6’) (b) Determine the group delay of the filter. (4’)

(c) Determine the output signal y[n] of the filter.(8’) Solution:

a. The frequency response

H(e) h[n]e j n 4.8804 9.5661e j 4.8804e j2

j

n 02

) 9.5661 e j 4.8804(ej e j ) 9.5661e j 9.7608cos(

(得4分)

The magnitude response is H(ej ) 9.7608cos( ) 9.5661

The phase response is H(ej ) , where β is a constant (得2分)

d H(ej )

1 (得4分) b. The group delay ( )

d

c. The output signal y[n]

For 0.2, H(ej0.2) 9.7608cos(0.2) 9.5661 0 (得2分) For 0.5, H(ej0.5) 9.7608cos(0.5) 9.5661 1 (得2分) This means that the output signal is x[n] = sin(0.5(n-1)). (得4分)

5、(12’) Given two length-4 discrete-time sequences g[n] and h[n], assume that their linear convolution is yL[n] = {1 1 1 1 1 1 1}. Determine the 9-point and 6-point circular convolutions of g[n] and h[n] respectively. (2×6’) Solution:

From the relationship between the linear convolution and the circular convolution: yC[n]

m

y

L

[n mL],0 n L 1

where L is the number of points that we make the circular convolution. Then the 9-point circular convolution is

yC[n] = {1 1 1 1 1 1 1 0 0} (得6分)

The 6-point circular convolution is

yC[n] = {2 1 1 1 1 1} (得6分)

4 6z 1 2.16z 2 0.16z 3

6、(12’)Given a causal IIR transfer function: H(z)

1 0.19z 2 0.03z 3

4(1 z 1)(1 0.1z 1)(1 0.4z 1)

Its factored form is H(z)

(1 0.5z 1)(1 0.2z 1)(1 0.3z 1)

(a) Draw its structure in direct form II; (4’)

(b) Draw its one of the equivalent structures; (4’)

(c) Draw its zero-pole diagram and interpret (解释，说明) that the function is whether or not BIBO

stable. (4’) Solution:

a. The filter structure in direct form II is depicted in the following figure: (得4分)

Direct form II structure An equivalent structure

b. The equivalent structure is depicted in the following figure. (得4分)

c. The system has three zeros at z 1,z 0.1,z 0.4 The system has three poles at z 0.5,z 0.2,z 0.3 The pole-zero diagram is illustrated below: (得2分) From the pole-zero diagram, we can conclude that the system is BIBO stable as all of poles of this causal system are located inside the unit circle. (得2分)

7、(12’)Given a causal IIR discrete-time system described by the transfer function

1

.

1 0.6z 1

It is known that the input sequence is x[n]=(0.4)nμ[n]. (a) Determine the impulse response h[n] of the system. (4’)

(b) Determine the output sequence y[n] using the z-transform. (4’)

(c) Determine the magnitude response |H(ej )| and phase response ( ). (4’) Solution:

(a). Determine the impulse response h[n] of the system

Taking the inverse z-transform to H(z) we get the impulse response of the system given by

H(z)

h[n] (0.6)n [n] (得4分) (b). Determine the output sequence y[n] using the z-transform

1

The z-transform of the input sequence is X(z) (得2分) 1

1 0.4z

Then the z-transform of the output sequence is

11

Y(z) H(z)X(z)

1 0.6z 11 0.4z 1

By the use of the partial-fraction expansion technique:

32

Y(z) 1 1

1 0.6z1 0.4z

Taking the inverse z-transform to Y(z) we get the output sequence

y[n] 3(0.6)n [n] 2(0.4)n [n] (得2分) (c). Determine the expression of the frequency response H(ej ) in the form |H(ej )|ej ( ) By replacing z with ejω we get the frequency response H(ej ): H(ej )

11

(得2分)

1 0.6e j 1 0.6cos( ) j0.6sin( )

The magnitude response is H(ej

)

11

221 0.6cos( ) j0.6sin( )1 0.6cos( )] 0.36sin( )