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4.24 From the figure, the filter has a finite impulse response. It may be described as a sum of impulse function as: h[n] = 0.5[n] + 0.4[n–1] + 0.3[n–2] + 0.2[n–2] The difference equation has the parallel form: y[n] = 0.5x[n] + 0.4x[n–1] + 0.3x[n–2] + 0.2x[n–3] 4.25 n h[n] The impulse response is finite, with samples as listed in the table. 0 1.000 1 0.300 2 0.090 3 0.027 4 0.000 5 0.000 6 0.000
x[n]
+
delay –0.14 delay –0.38
y[n]
4.23 n h[n]
The first ten samples of the impulse response are: 0 1.0 1 –1.2 2 –0.8 3 0.4 4 0.0 5 0.0 6 0.0 7 0.0 8 0.0 9 0.0
Chapter 4 Solutions 4.1 (a) The pass band gain for this filter is unity. The gain drops to 0.707 of this value at 2400 Hz and 5200 Hz. Thus, the frequencies passed by the filter lie in the range 2400 to 5200 Hz. (b) The filter is a band pass filter. (c) The bandwidth is the range of frequencies for which the gain exceeds 0.707 of the maximum value, or 5200 –2400 = 2800 Hz. 4.2 A low pass filter passes frequencies between DC and its cut-off frequency. The bandwidth is identical to the cut-off frequency. Thus, the cut-off frequency is 2 kHz. 4.3 The maximum pass band gain of the filter is 20 dB. The bandwidth is defined as the range of frequencies for which the gain is no more than 3 dB below the pass band gain, or 17 dB. This gain occurs at the cut-off frequency of 700 Hz. For a high pass filter, the bandwidth is the range of frequencies between the cut-off frequency, 700 Hz, and the Nyquist frequency (equal to half the sampling rate), 2 kHz. The bandwidth is 1300 Hz. 4.4 The low pass filter has a cut-off frequency of 150 Hz and bandwidth 150 Hz. The band pass filter has cut-off frequencies at 250 Hz and 350 Hz for a bandwidth of 100 Hz. The high pass filter has a cut-off frequency of 400 Hz and a bandwidth of 100 Hz, which extends from its cut-off frequency to the Nyquist limit at half the sampling rate. 4.5 (a) The low pass filter output is on the left. The high pass filter output is on the right. (b) An approximation to the original vowel signal can be found by adding the high and low pass waveforms together.
The impulse response samples for a FIR filter serve directly as bk coefficients, so y[n] = x[n] + 0.3x[n–1] + 0.09x[n–2] + 0.027x[n–3] This result may also be seen by writing the impulse response in terms of impulse functions: h[n] = [n] + 0.3[n–1] + 0.09[n–2] + 0.027[n–3]
4 3.41
5 2.05
6 1.23
7 0.74
8 0.44
9 0.27
0
2
4
6
8
n
(c) n y[n] 0 1.0
y[n]
5 4 3 2 1 0 -1 -2 -3 -4 -5 -2 0 2 4 6 8
1 –1.6
2 4.44
3 –4.10
4 3.29
5 –3.66
6 3.29
7 –2.96
8 2.67
y[n]
1
0.8
0.6
0.4
0.2
0 -2
-1
0
1
2
3
4
5
6
n
4.8 (a) (b) 4.9 (b) (c) 4.10 n y[n]
y[n] = –0.25y[n–1] + 0.75x[n] – 0.25x[n–1] y[n] = y[n–1] – x[n] – 0.5x[n–1] (a) The system is non-recursive. b0 = b1 = b2 = 1/3 The system is recursive. a0 = 1, a1 = –0.2, b0 = 1 The system is recursive. a0 = 1, a1 = 0.5, b0 = 1, b1 = –0.4 (a) 0 1.0
9 –2.40
n
(d) n y[n] 0 0.5 1 0.5 2 1.5 3 2.5 4 1.5 5 0.5 6 0.0 7 0.0 8 0.0 9 0.0
y[n]
5 4 3 2 1 0 -1 -2 -3 -4 -5 -2 0 2 4 6 8
n
4.11 n x[n] n y[n] 4.12 n x[n] n y[n] 0 2.00 0 0.6
x[n]
n
4.6
(a)
linear
(b)
non-linear
(c)
non-linear
(d)
linear
4.7 Since the new input is shifted to the right by two positions from the original input, the new output is shifted to the right by two positions from the original output.
y[n]
1 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 -2 0 2 4 6 8
0 1.00 0 1.00
1 1.000 1 0.750
2 1.000 2 0.813
3 1.000 3 0.797
4 5 1.000 0.000 4 5 0.801 –0.20
6 0.000 6 0.050
7 0.000 7 –0.013
8 0.000 8 0.003
9 0.000 9 –0.001
1 0.00 1 –0.5
2 –1.00 2 –0.1
3 0.00 3 0.25
4 0.00 4 –0.1
5 0.00 5 0.0
6 0.00 6 0.0
7 0.00 7 0.0
8 0.00 8 0.0
9 0.00 9 0.0
+
delay –0.5
y[n]
+
delay
–0.2
+
0.3
4.21
The direct form 2 equations are: w[n] = x[n] + 1.2w[n–1] – 0.5w[n–2] y[n] = w[n] – 0.2 w[n–1] (a) y[n] = –0.14 y[n–1] – 0.38 y[n–2] + x[n] w[n]
n
4.13 The overall input x[n] for any sampling instant is the sum of the inputs x1[n] and x2[n]. This overall input is applied to the difference equation in the normal way to obtain outputs.
The difference equation for the second second-order section is
y[n] = y1[n] + 0.3y1[n–2] Substituting the first equation into the second gives y[n] = (–0.1x[n] + 0.2x[n–1] + 0.1x[n–2]) + 0.3(–0.1x[n–2] + 0.2x[n–3] + 0.1x[n–4]) = –0.1x[n] + 0.2x[n–1] + 0.07x[n–2] + 0.06x[n–3] + 0.03x[n–4] 4.20 x[n]
5
1 –0.1
2 3.0
3 1.7
4 0.8
பைடு நூலகம்
5 –0.1
6 0.0
7 0.0
8 0.0
9 0.0
y[n]
4 3 2 1 0 -1 -2 -3 -4 -5
-2
0
2
4
6
8
n
(b) n y[n]
y[n]
0 1.0
5 4 3 2 1 0 -1 -2 -3 -4 -5 -2
1 0.6
2 3.36
3 4.02
4.22
x[n]
+
delay –0.14 delay –0.38
y[n]
(b)
w[n] = x[n] – 0.14w[n–1] – 0.38w[n–2] y[n] = w[n]
Note that the difference equation diagram for this part is the same as that for part (a). w[n]
n x[n] n y[n] 4.14 n x[n] n y[n] 4.15
0 0.00 0 0.00
1 0.807 1 0.807
2 1.200 2 0.837
3 1.007 3 0.467
4 5 0.400 0.500 4 –0.053 5 0.32
6 0.600 6 0.375
7 0.700 7 0.430
9 0.989 9 0.568
x[n] delay delay delay x[n–3] x[n–1]
1.0
+
y[n]
–0.8
0.5
4.16 4.17 4.18 4.19
y[n] = 0.5y[n–2] + 1.2x[n] – 0.6x[n–1] + 0.3x[n–2] y[n] = 2.1x[n–1] – 1.5x[n–2] w[n] = x[n] + 0.3w[n–1] – 0.1w[n–2] y[n] = 0.8w[n] – 0.4w[n–2] The difference equation for the first second-order section is y1[n] = –0.1x[n] + 0.2x[n–1] + 0.1x[n–2]
8 0.800 8 0.485
9 0.900 9 0.54
0 0.00 0 0.00
1 0.394 1 0.394
2 0.632 2 0.317
3 0.777 3 0.523
4 5 0.865 0.918 4 0.446 5 0.561
6 0.950 6 0.502
7 0.970 7 0.569
8 0.982 8 0.527
