# 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. ##### VHDL Refresher ECE 437 April 13, 2015 Motivation ECE 337 is a prerequisite But… –You may have taken 337 a few semesters previous –Breaks causes memory
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