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**ME606 Digital Signal Processing, Melbourne Institute Of Technology, Australia**

**Master of Engineering (Telecommunications) School of Information Technology and Engineering**

**z-Transforms, Filters Concepts**

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Question 1: Designing a low pass FIR filter using Windowed Fourier Series approach.

**Answer: 1. Impulse Response**

**Figure 1 Impulse response of the low pass filter in rectangular window**

**Amplitude Spectrum**

**Figure 2 Frequency Spectrum**

**2. Filter frequency response at wp = pi/8 is shown below**

**Figure 3 Filter frequency response at pi/8**

This is measured as 5.91dB which is almost -6dB

**Maximum ripple in the pass band is given below figure as 0.6821dB**

**Figure 4 Maximum ripple in pass band**

3. If the stopband starts at -20dB, the ratio of the transient band to the passband

Transient band = ws - wp

ws = -20dB dB

wp = -0.7093dB

Transient band = -20.7093dB

(Transient band)/(pass band) = (-20.7093)/(-0.7093) = 29.19dB

4. When m is 255 the impulse and amplitude response are below

**Figure 5 When the filter lenth is increased**

a. If the stopband starts at -20dB, the ratio of the transient band to the passband

Transient band = ws - wp

ws = -20dB dB

wp = -0.5711dB

Transient band = -20.5711dB

(Transient band)/(pass band) = (-20.5711)/(-0.5711) = 36.02dB

b. Ripple in pass band is 0.7dB

**Figure 6 Maximum pass band**

**c. The effect of the filter length on the ripples and stopband attenuation: **If the filter length is increased the ratio of the transition to the pass band is increased, whereas the stop band attenuation is -20dB itself.

5. Using the Hamming window the frequency response obtained is as below

**Figure 7 Hamming window**

The stop band attenuation is -60dB. In the pass band the ripples are eliminated. This is the advantage of windowing.

Now if Hanning windowing is applied , the changes are as below

**Figure 8 Hanning window**

Here also the stop band attenuation is -60dB.

6. Now the pass band is changed to π/4

**Figure 9 When the pass band is changed**

The transition band is increased whereas the stop band remains same as -20dB itself and the pass band is changed with more ripples.

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**Question 2: Designing a bandpass FIR filter using Windowed Fourier Series approach**

**Answer:** 1. The impulse response is as below

**Figure 10 Response of the system**

The amplitude spectrum is as below

**Figure 11 Frequency Spectrum**

2. The pass band edges are given by

wp1 = -26.23dB , wp2 = -26.23dB and the frequency is almost 0.5 rad

**Figure 12 Pass band edges**

The frequency at -6dB is as below

**Figure 13 Frequency at -6dB**

3. Maximum ripple in pass band is 0.7707dB

**Figure 14 Maximum pass band**

4. Now the pass band is changed to π/4

**Figure 15 Impulse response**

**Figure 16 Fequency response**

Wp1 = -2.788db and wp2 = -28.7dB

Here the pass band is more wider than the previous one which allows more frequencies to pass through.

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**Question 3: Designing a band-stop FIR filter using Frequency sampling approach. In this section, you must write your own code to design a band stop filter using the knowledge you gained from the sections 1-1 and 1-2**

**Answer: 1. Mathlab code**

wp = pi/8; % lowpass filter bandwidth

M = 121;

n = -(M-1)/2:(M-1)/2; % selection time window

h0 = (wp/pi)*sinc((wp/pi)*n); % truncated impulse response

h1 = h0.*rectwin(M)'; % windowing

h = h1.*exp(j*w0*n); % Lowpass to bandpass conversion

h2 = h1-h; % bandstop conversion

figure;

figure;plot(h2)

ylabel('Impulse response')

xlabel('Samples')

% spectrum

FFTsize = 512;

pxx = 20*log10(abs(fft(h2,FFTsize)));

fxx = (0:(FFTsize/2)-1)*(pi/(FFTsize/2));

figure;plot(fxx,pxx(1:FFTsize/2));ylabel('Amplitude [dB]');xlabel('frequency [Radian]');grid on

**2. The frequency spectrum**

**Figure 17 Frequency spectrum**

**3. Plot is as below**

**Figure 18 Spectrum**

4. The amplitude at w=[w/4, 3w/4, w/2] are -37.14dB, -48.02dB and -42.26dB

5. The plot is as below

**Figure 19 Spectrum when heading in included**

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**Question 4: Designing a low pass FIR filter using Frequency sampling approach**

**Answer: **1. The responses are as below

**Figure 20 Both the spectrums**

In the rectangular window the number of side lobes are higher. Hanning window minimizes the same.

2. The plot is

**Figure 21 Spectrum**

Section 2

3. Given that

H(Z) = ((1 - o.2Z^{-1})/((1 - o.5Z^{-1})(1 - o3Z^{-1})))

We need to find the ROC

Re arranging this we get

(Z(Z - 0.2))/((Z - 0.5)(Z - 0.3))

By partial fraction solving we get

1 + (0.175/(Z - 0.5)) - (0.075/(Z - 0.3))

Taking inverse Z transform we get

∂(n) + 0.175 (0.5)^{n} - 0.075(0.3)^{n}

ROC will be n>o

4. The system is unstable as the poles lie on the left half.

5. It is a general form of filter

We can represent y(n) = ∑_(k=0)^M (b_{k} X(n-k)) - ∑_(k=0)^N (a_{k} Y(n-k))

On taking Z transform (Y(Z))/(X(Z)) = (bo+b1Z-1+ ---±BMZ-M)/(a0+a1Z±--±aNZ-N)

6. The values can be found out if we know the a,b, M and N values

7. The plot is

**Figure 22 Various Figures**

8. It is a band pass filter

9. The block diagram is

**Figure 23 Block Diagam**

10. y(n) = ∑_(k=0)^M (b_{k} X(n-k)) - ∑_(k=0)^N (a_{k} Y(n-k))

11. The plot is

**Figure 24 Impulse response**

**Section 3 Filtering**

12. The plot os

**Figure 25 Power spectral density**

It is the power spectral density estimate.

13. The plot is

**Figure 26 Filter characteristics**

Delay by 0.099 compared to input.

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