Question

3.2 Simple Bandpass Filter Design The L-point averaging filter is a lowpass filter. Its passband width is controlled by L, being inversely proportional to L. In fact, you can use the GUI altidemo to view the frequency response for different averagers and measure the passband widths. It is also possible to create a filter whose passband is centered around some frequency other than zero. One simple way to do this is to define the impulse response of an L-point FIR as: [n] = (wen), 0<n<L Where L is the filter length, and is the center frequency that defines the frequency location of the passband For example, we would pick we = 0.45m if we want the peak of the filter's passband to be centered at 0.457. The bandwidth of the bandpass filter is controlled by L, the larger the value of L, the narrower the bandwidth. This particular filter is also discussed in the section on useful filters in Chapter 7. (a) Generate a bandpass filter that will pass a frequency component at W=0.451. Make the filter length (L) equal to 12. Since we are going to be filtering the signal defined in section 3.1(b), measure the gain of the filter at the three frequencies of interest: W=0.22W=0.45 and 6 = 0.85 (b) The passband of the BPF filter is defined by the region of the frequency response where | H (cha) is close to its maximum value. If we define the maximum to be Hmaz, then the passband width is defined as the length of the frequency region where the ratio H(c)/Hmax is greater than 1/2 = 0.707. Figure 3 shows how to define the passband and stopband. Note: you can use MATLAB's find function to locate those frequencies where the magnitude satisfies | Hle) > 0.707H max- The stopband of the BPF filter is defined by the region of the frequency response where (1) is close to zero. In this case, we will define the stopband as the region where |H() is less than 10% of the maximum Make a plot of the frequency response for the L=12 bandpass filter from part (a), and determine the passband width (at the 0.707 level). Repeat the plot for L=24 and L = 48, so you can explain how the width of the passband is related to filter length L. i.e., what happens when Lis doubled or halved. (c) Comment on the selectivity of the L = 12 bandpass filter. In other words, which frequencies are "passed by the filter." Use the frequency response to explain how the filter can pass one component at w = 0.45X, while reducing or rejecting the others at = 0.22x and = 0.857 (6) Generate a bandpass filter that will pass the frequency component at = 0.457, but now make the filter length (L) long enough so that it will also greatly reduce frequency components at (or near) @=0.22x and =0.858. Determine the smallest value of L so that 7 BANDPASS FILTER (centered at 0.40) 0.8F PASSBAND 0.6F 0.4 STOPBAND 0.2 STOPBAND 05 1 25 3 1.5 2 Frequency (radians) Figure 3. Passband and Stopband for a typical FIR bandpass filter. In this case, the maximum value is 1, the passband is the region where the frequency response is greater than 1/V2 = 0.707, and the stopband is defined as the region where the frequency response is less than 25% of the maximum. • Any frequency component satisfying | S 0.22 will be reduced by a factor of 10 or more." • Any frequency component satisfying 0.851 SS will be reduced by a factor of 10 or more. This can be done by making the passband width very small. (e) Use the filter from the previous part to filter the sum of 3 sinusoids” signal from Section 3.1. Make a plot of 100 points of the input and output signals, and explain how the filter has reduced or removed two of the three sinusoidal components. (f) Make a plot of the frequency response (magnitude only) for the filter from part (d), and explain how H() can be used to determine the relative size of each sinusoidal component in the output signal In other words, connect a mathematical description of the output signal to the values that can be obtained from the frequency response plot.

Answer 1

r Cax-wila (jfeletw) } ~ 1 hena z 2 .. Con lwen n) 7eurol jun L-1 ( همان E hh) t wel 1 jwn e L-1 jeven e to ( . 2 سے 2 du e Lt F [ C 구구 n0 j (wetw) Ymunro -1 1 e + ] @z-w) in Wetw) - С.

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%matlab

wc=0.45*pi;
w=0:0.001:pi;
L=12;
H=(1-exp(1i.*L.*(wc-w)))./(1-exp(1i.*(wc-w)));
H=H+(1-exp(-1i.*L.*(wc+w)))./(1-exp(-1i.*(wc+w)));
H=H./L;
plot(w./pi,abs(H));
xlabel('\omega/\pi');
ylabel('|H(e^{j\omega})|');

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1.2 X: 0.4504 Y: 1.08 1 X: 0.3839 Y: 0.7627 X: 0.522 Y: 0.7629 0.8 (7) HI 0 0.6 0.4 X: 0.22 Y: 0.2251 X: 0.8502 Y: 0.1892 0.2 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 WXT

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therefore

gain at

w=0.45*pi is 1.08

w=0.85*pi is 0.1894

w=0.22*pi is 0.2251

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$H|_{\max }=1.08$

Now cut-off gain =1.08/sqrt(2)=0.7637

therefore

cut-off frequencies are

w1=0.3839*pi

w2=0.522*pi

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d)

we make magnitude at w=0.22*pi less than or equal to 0.1

by matlab

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%matlab

w1=0.22*pi;
wc=0.45*pi;
L=1;
H=(1-exp(1i.*L.*(wc-w1)))./(1-exp(1i.*(wc-w1)));
while abs(H)>0.1
L=L+1;
H=(1-exp(1i.*L.*(wc-w1)))./(1-exp(1i.*(wc-w1)));
end
L

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L=26

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taking L=30

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%matlab

wc=0.45*pi;
w=0:0.001:pi;
L=30;
H=(1-exp(1i.*L.*(wc-w)))./(1-exp(1i.*(wc-w)));
H=H+(1-exp(-1i.*L.*(wc+w)))./(1-exp(-1i.*(wc+w)));
H=H./L;
plot(w./pi,abs(H));
xlabel('\omega\times \pi');
ylabel('|H(e^{j\omega})|');
---------------------------------

1.2 1 0.8 mə) 0 0.6 0.4 0.2 X: 0.2209 Y: 0.1003 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 W XT

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%matlab