A continuous-time signal is processed by a discrete-time filter using the model

Question

A continuous-time signal is processed by a discrete-time filter using the model of discrete-time processing of continuous-time signals (Lecture 6). The overall (continuous-time) system meets the following specifications: Sampling frequency F_s = 1000 Hz: Passband = [100, 200] Hz: Stopband 1 = [0, 50] Hz Stopband 2 = [250, F_s/2] Hz: Passband ripple = 3 dB: Stopband attenuation = 40 dB. Sketch the magnitude response in decibels (i.e., 20log_10 |H(J ohm)| = 20 log_10 |H(j2pi F)|) for the overall system in the form of a tolerance scheme in terms of the continuous-time frequency F in Hz for the frequency range [0, F_s/2] Hz, indicating the passband edge frequency, the stopband edge frequency, the passband ripple and the stopband attenuation.

Explanation / Answer

This is MATLAB help for you and explanation is as follow

Normalization is done by dividing with sampling frequency

fp1=100/1000; %normalized passband start frequency

fp2=200/1000; %normalized passband edge frequency

Ap1= 3;

As1 =40;

fst1=50/1000;%stopband edge normalized frequency

fst2=250/1000;

Stopband=fdesign.bandstop(1000,fp1,fp2,Ap1,Ap2,fst1,fst2);

fvtool(Stopband);%for plotting magnitude response

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