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Using Pole Zero Placement Method, design a second-order notch filter with a sampling rate of 10,000...

Using Pole Zero Placement Method, design a second-order notch filter with a sampling rate of 10,000 Hz, a 3dB bandwidth of 200 Hz, and narrow stop-band centered at 3,000 Hz. From the transfer function, determine the difference equation.

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Answer #1

Given data, Sampling rate () is 10,000 Hz, 3dB bandwidth (BWidB) is 200 Hz, and Narrow stop-band centered at() is 3,000 Hz The transfer function by using Pole-zero placement for a second-order notch filter as, H(z) Now find the all required values as shown in below steps, The required magnitude of the poles as, ~ 1-(200 /10000) x π r0.9372 Find the angle of the pole location by using the center frequency as, 4 3000 10000 x360° 108°

The unit-gain scale factor formula is, (to have a unit passband gain) 2-2cos θ 1-2(0.9372)cos (108°)+(0.9372) 2 -2 cos (108°) K -0.9387 From the equation (1). H(z) H(2)- 0.9387 (z2-2z cos (108)+1) -2(0.9372)z cos(108)+P 0.9387(z0.6180z +1 +0.5792z+0.8783) 0.93872+0.5801z+0.9387) (z +0.5792z+0.8783) 0.9387 +0.5801z1 +0.9387z2 1+0,5792z+0.8783z2

The difference equation can be written as, H(z) H(z 1+0.5792z +0.8783z 0.9387+0.5801z +0.9387 y(n)+0.5792y (n-1)+0.8783y (n-2) 0.9387x (n)+0.5801x (n-1)+ 0.9387x (n-2) y(n) 0.9387x (n)+0.5801r(n-1)+0.9387x (n-2)-0.5792y (n-1)-0.8783y(n-2) 0.9387+0.5801z-1 +0.9387z2 1+0.5792z-1+0.8783z

Checking the Notch filter result, (By using MATLAB).

b = [0.9387 0.5801 0.9387];

a = [1 0.5792 0.8783];

[h,w] = freqz(b,a);

plot(w/pi,20*log10(abs(h)))

ylim([-40 10])

xlim([0 1])

xlabel('Normalized Frequency ( imespi rad per sample)')

ylabel('Magnitude in dB')

plot is,

10 5 -5 C-10 S -15 CD-20 -25 -30 -35 40 0 0.1 0.2 0.3 04 0.5 0.6 0.7 0.8 0.9 1 Normalized Frequency(xT rad per sample)

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