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7/31/2019 ECE 437(10P)
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ECE 437
DIGITAL SIGNAL PROCESSING
PROJECT 1
KISHAN NANDAKUMAR.
A20252276
FALL 2011
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1. y(n)+ay(n-1)=x(n).
where a.) a = 0.9.
b.) a = -0.9.
Solution:
Converting the following system into H(z).
We have H(z) = z/(z+a) ( 1)
a.) Substituting the value a = 0.9 in (1). We get H(z) = z/(z+0.9).
H(z) has one zero at z = 0 and one pole at p = -0.9. The pole-zero plot and impulse response of the system is
shown in Figure 1.1
Figure 1.1 Pole-zero pattern and Impulse Response when a = 0.9.
The frequency response plots for |H()| and illustrated in Figure 1.2.
Figure 1.2 Magnitude and Phase response for the system with a = 0.9.
The Discrete Fourier Transform of the impulse response is shown in Figure 1.3. The length of magnitude |H(w)|
is 93.
Figure 1.3 Magnitude and Phase spectra of the Fourier Transform for system with a = 0.9.
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Higher sampling rate and longer sampling duration help minimize the effects of aliasing and leakage. By
increasing the samples from N=10 in figure 1.4, and N= 50 in figure 1.5, I have demonstrated how the leakage
can be minimized.
Figure 1.4 Impulse response and calculated DFT (N = 10samples).
Figure 1.5 Impulse response and calculated DFT (N =50 Samples).
I have demonstrated by increasing the samples from N=10 in figure 1.6, and N= 60 in figure 1.7 how the length
of FFT changes calculated spectra in comparison to DTFT.
Figure 1.6 DFT and DTFT (N = 10samples).
Figure 1.7 DFT and DTFT (N = 60 Samples).
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DFT is a sampled version of DTFT for a finite signal. Figure 1.8 shows DTFT for the period to and the x
represents the sampled DFT for a sample value N= 60.
Figure 1.8 DTFT and sampled DFT for N = 60.
b.) Substituting the value a = -0.9 in (1). We get H(z) = z/(z-0.9).
H(z) has one zero at z = 0 and one pole at p = 0.9.
Figure 2.1 Pole-zero pattern and Impulse Response when a = 0.9.
The frequency response plots for |H()| and illustrated in Figure 2.2.
Figure 2.2 Magnitude and Phase response for the system with a = 0.9.
The Discrete Fourier Transform of the impulse response is shown in Figure 2.3. The length of magnitude |H(w)|
is 93.
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Figure 2.3 Magnitude and Phase spectra of the Fourier Transform for system with a = 0.9.
Higher sampling rate and longer sampling duration help minimize the effects of aliasing and leakage. By
increasing the samples from N=10 in figure 2.4, and N= 50 in figure 2.5, I have demonstrated how the leakage
can be minimized.
Figure 2.4 Impulse response and calculated DFT (N = 10samples).
Figure 2.5 Impulse response and calculated DFT (N =50 Samples).
I have demonstrated by increasing the samples from N=10 in figure 2.6, and N= 60 in figure 2.7 how the length
of FFT changes calculated spectra in comparison to DTFT.
Figure 2.6 DFT and DTFT (N = 10samples).
Figure 2.7 DFT and DTFT (N = 60 Samples).
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DFT is a sampled version of DTFT for a finite signal. Figure 2.8 shows DTFT for the period to and the x
represents the sampled DFT for a sample value N= 60.
Figure 2.8 DTFT and sampled DFT for N = 60.
2. H(z) = (1- rcos()z-1
))/(1- 2rcos(o)z-1
+ r2z
-2)
where a) = /8 and r = 0.95b) = /4 and r = 0.95c) = /8 and r = 0.99
Solution:
a.) Substituting = /8 and r = 0.95 in H(z).
We have H(z)= (z2-0.8777z)/(z
2-1.7554z+0.9025).
H(z) has two zeros at z1 = 0 and z2 = 0.8777 and two poles at p1 = 0.8777 + 0.3635i and p2 = 0.8777 - 0.3635i.
The pole-zero plot and Impulse Response of the system is shown in Figure 3.1.
Figure 3.1 Pole-zero Plot and Impulse Response when = /8 and r = 0.95.
The frequency response plots for |H()| and illustrated in Figure 3.2.
Figure 3.2 Magnitude and Phase response for the system with = /8 and r = 0.95.
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The Discrete Fourier Transform of the impulse response is shown in Figure 3.3. The length of magnitude |H(w)|
is 193.
Figure 3.3 Magnitude and Phase spectra of the Fourier Transform for system with = /8 and r = 0.95.
Higher sampling rate and longer sampling duration help minimize the effects of aliasing and leakage. By
increasing the samples from N=10 in figure 3.4, and N= 120 in figure 3.5, I have demonstrated how the leakage
can be minimized.
Figure 3.4 Impulse response and calculated DFT (N = 10samples).
Figure 3.5 Impulse response and calculated DFT (N =120 Samples).
I have demonstrated by increasing the samples from N=10 in figure 3.6, and N= 100 in figure 3.7 how the length
of FFT changes calculated spectra in comparison to DTFT.
Figure 3.6 DFT and DTFT (N = 10samples).
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Figure 3.7 DFT and DTFT (N = 100 Samples).
DFT is a sampled version of DTFT for a finite signal. Figure 3.8 shows DTFT for the period to and the x
represents the sampled DFT for a sample value N= 110.
Figure 3.8 DTFT and sampled DFT for N = 110.
b.) Substituting = /4 and r = 0.95 in H(z).
We have H(z)= (z2-0.6718z)/(z
2-1.3435z+0.9025).
H(z) has two zeros at z1 = 0 and z2 = 0.6718 and two poles at p1 = 0.6717 + 0.6718i and p2 = 0.6717 - 0.6718i.
The pole-zero plot and Impulse Response of the system is shown in Figure 4.1
Figure 4.1 Pole-zero Plot and Impulse Response when = /4 and r = 0.95.
The frequency response plots for |H()| and illustrated in Figure 4.2.
Figure 4.2 Magnitude and Phase response for the system with = /4 and r = 0.95.
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The Discrete Fourier Transform of the impulse response is shown in Figure 4.3. The length of magnitude |H(w)|
is 193.
Figure 4.3 Magnitude and Phase spectra of the Fourier Transform for system with = /4 and r = 0.95.
Higher sampling rate and longer sampling duration help minimize the effects of aliasing and leakage. By
increasing the samples from N=10 in figure 4.4, and N= 120 in figure 4.5, I have demonstrated how the leakage
can be minimized.
Figure 4.4 Impulse response and calculated DFT (N = 10samples).
Figure 4.5 Impulse response and calculated DFT (N =120 Samples).
I have demonstrated by increasing the samples from N=10 in figure 4.6, and N= 100 in figure 4.7 how the length
of FFT changes calculated spectra in comparison to DTFT.
Figure 4.6 DFT and DTFT (N = 10samples).
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Figure 4.7 DFT and DTFT (N = 100 Samples).
DFT is a sampled version of DTFT for a finite signal. Figure 4.8 shows DTFT for the period to and the x
represents the sampled DFT for a sample value N= 110.
Figure 4.8 DTFT and sampled DFT for N = 110.
c.) Substituting = /8 and r = 0.99 in H(z).
We have H(z)= (z2-0.9146z)/(z
2-1.8293z+0.9801).
H(z) has two zeros at z1 = 0 and z2 = 0.9146 and two poles at p1 = 0.9146 + 0.3188i and p2 = 0.9146 - 0.3188i.
The pole-zero plot and Impulse Response of the system is shown in Figure 5.1
Figure 5.1 Pole-zero Plot and Impulse Response when = /8 and r = 0.99.
The frequency response plots for |H()| and illustrated in Figure 5.2.
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Figure 5.2 Magnitude and Phase response for the system with = /8 and r = 0.99.
The Discrete Fourier Transform of the impulse response is shown in Figure 5.3. The length of magnitude
|H(w)| is 985.
Figure 5.3 Magnitude and Phase spectra of the Fourier Transform for system with = /8 and r = 0.99.
Higher sampling rate and longer sampling duration help minimize the effects of aliasing and leakage. By
increasing the samples from N=10 in figure 5.4, and N= 220 in figure 5.5, I have demonstrated how the leakage
can be minimized.
Figure 5.4 Impulse response and calculated DFT (N = 10samples).
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Figure 5.5 Impulse response and calculated DFT (N =220 Samples).
I have demonstrated by increasing the samples from N=10 in figure 5.6, and N= 190 in figure 5.7 how the length
of FFT changes calculated spectra in comparison to DTFT.
Figure 5.6 DFT and DTFT (N = 10samples).
Figure 5.7 DFT and DTFT (N = 190 Samples).
DFT is a sampled version of DTFT for a finite signal. Figure 5.8 shows DTFT for the period to and the x
represents the sampled DFT for a sample value N= 250.
Figure 5.8 DTFT and sampled DFT for N = 250.