数字信号处理答案10

Chapter 10 Solutions

10.1 (a) The impulse response is given by h[n] = –0.8h[n –1] + 0.1h[n –2] + δ[n]. The first ten samples are listed in the table.

(b) The impulse response contains an infinite number of non-zero terms.

10.3 (a)(i) Without pre-warping, the transfer function for the analog filter is

15708

s 15708)

2500(2s )2500(2s )s (H 1

p 1p +=

π+π=

ω+ω=

The bilinear transformation 1

z 1z f 2s S +-=gives

1

1

z

00921.01)

z

1(4954.015708

1

z 1z 16000

15708)z (H ---+=

++-=

(ii)

The analog frequency 2.5 kHz is converted to a digital frequency Ωp1 =

8000/)2500(2π =1.9635 rads. This frequency is pre-warped to the analog frequency

2

tan

f 21p S 1p Ω=ω= 23946 rad/sec, which makes the transfer function for the analo

g filter

23946

s 23946s )s (H 1

p 1p +=

ω+ω=

After the bilinear transformation, the digital transfer function is obtained:

1

1

z

1989.01)

z

1(6.023946

1

z 1z 16000

23946)z (H --++=

++-=

(b) The magnitude responses for both filters are shown below. The –3 dB frequency for the pre-warped filter is equal to the specified 2.5 kHz.

10.4 (a) The cut-off frequency for the analog filter is 1500/(2π) = 238.73 Hz. The digital frequency that corresponds to this analog frequency is Ωp1 = 8000/)73.238(2π = 0.1875 rads. The pre-warped analog cut-off frequency is 2

tan f 21p S 1p Ω=ω= 1504.4

rad/sec, to give the transfer function 4

.1504s 4.1504)s (H +=

.

Filter with pre-warping |H(f)| f

(b) Use the bilinear transformation to get the digital transfer function

1

1z

8281.01)

z 1(0859.04

.15041

z 1z 16000

4.1504)z (H ---+=

++-=

(c) The difference equation is y[n] = 0.8281y[n –1] + 0.0859x[n] + 0.0859x[n –1]. (d)

The frequency response is Ω

-Ω--+=

Ωj j e

8281.01)

e 1(0859.0)(H . The magnitude response

may be found by taking the magnitudes of this expression for several values of Ω. It is plotted below against frequency in Hz.

(e) The magnitude for the analog transfer function is

2

2

4

.15044.1504)(H +ω=

ω

The shape of the digital filter may be found by substituting 2

tan f 2S Ω=ω:

2

2

2

2

S 4

.15042tan 160004

.15044

.15042tan f 24

.1504)(H +⎪⎭⎫ ⎝

⎛Ω=

+⎪⎭⎫ ⎝

⎛Ω=

Ω

This function may be plotted for various values of Ω. The magnitude response is plotted below against digital frequency in rads. When digital frequencies are converted to frequencies in Hz, the plot becomes identical to the one in part (d).