Question

MUST BE IN MATLAB

Design a low pass filter for this signal. Set the pass band frequency to 4.9 GHz and the stop band frequency to 5.6 GHz. Allow for 1 dB of attenuation in the pass band and require at least 20 dB of attenuation in the stop band. a. First design a Butterworth filter. Use the command buttord() to determine the order and the normalizing frequency for the filter. Use [Num,Den]=butter() to determine the numerator and denominator coefficients of your filter. b. Filter the difference scan. Use the command filter(Num,Den,data) where Num and Den are the numerator and denominator coefficients of the filter, and data is your difference scan. Save the filtered data to another vector. Plot the frequency spectrum of the filtered signal and compare your results to the original signal.

Answer 1

clear;
clc;
close all;
Ws = 10*10^9;
Wp = 4.9*10^9/Ws/2;
Ws = 5.6*10^9/Ws/2;
[n,Wn] = buttord(Wp,Ws,1,20);
[num,den] = butter(n,Wn);
disp("Coefficinets of Numerator => ");
disp(num);
disp("Coefficinets of Denominator => ");
disp(den);
[z,p,k] = butter(n,Wn);
sos = zp2sos(z,p,k);
freqz(sos,2024,10*10^9);
title(sprintf('n = %d Butterworth Lowpass Filter',n))

n = 20 Butterworth Lowpass Filter -500 Magnitude (dB) -1000 - 1500 0 0.5 1 1.5 2 3.5 4 4.5 5 2.5 3 Frequency (Hz) *10° 0 -500 Phase (degrees) -1000 -1500 -2000 0 0.5 1 1.5 2 3 3.5 4.5 5 2.5 Frequency (Hz) *10°

Coefficinets of Numerator => 1.0e-04 * Columns 1 through 10 0.0000 0.0000 0.0003 0.0017 0.0073 0.0232 0.0581 0.1161 0.1887 0.2517 Columns 11 through 20 0.2768 0.2517 0.1887 0.1161 0.0581 0.0232 0.0073 0.0017 0.0003 0.0000 Column 21 0.0000 Coefficinets of Denominator => Columns 1 through 10 — 1.0000 -9.8789 47.8796 -150.6886 344.2368 -605.1754 847.7741 -967.4110 911.9745 -716.3472 Columns 11 through 20 470.9123 -259.2916 119.2754 -45.5554 14.2958 -3.6272 0.7262 -0.1105 0.0120 -0.0008 Column 21 0.0000