Question

[MATLAB Scriptfile task]

Bandpass filter Modulation Band-limiting Envelope detection BPF RECT LPF BPF sine wave Vo BPF 2 RECT LPF BPF 2 sine wave BPF 8 RECT LPF BPF 8 sine wave Fig. 2 The block diagram of an 8-channel voice vocoder (RECT means rectification and it can be implemented by taking the absolute value of a sound signal; LPF=lowpass filter and BPF band pass filter) design of the logarithmic filter bank- fimn: the lowest frequency t fmax: the highest frequency 8 N: number of bands or number of channels 8 The output fco contains N+1 frequency values for N bandpass filters. For example: fco (i) of the ith bandpass filter. function fco-bands_cutoff (fmin, fmax, N); xmin loq10 (fmin/165.4+1) /2.1; xmax= log10 (fmax/165.4+1) /2.1; 8relative value N; xxmin:dx : xmax; fco-zeros (1, N+1) ; for li-165.4(10 (x (i) * 2.1)-1); end t Script to be used the following codes demonstrate how you can design a band-pass filter from 1000 Hz to 3000 Hz and then plot its frequency response: Fs-16000; % sampling frequency [BB,AA]-butter(3, [1000 3000]/(Fs/2)); % The filter order is set to 3 and [1000 3000]/(Fs/2) are normalized frequencies between 0 and 1 H,F]=freqz(BB,AA,256, Fs); Fs is the sampling frequency plot(Fabs(H)); Then you can use the filter command in Matalb to filter the input signal. For example Y filter(BB,AA,X); % X is the input and Y is the filtered signal. The tone or sine carrier in Fig. 2 for each band can be generated with the following commends: t=(0:length (orig) - 1 )/Fs;% creating a time vector tone_carrier=sin(2*pi*Fc*t); The sampling frequency of the input wav file is Fs and the center frequency of the

Design N-band tone vocoder with a given figure (below) in MATLAB implementing the given script file(bands_cutoff). This program should be able to process any sound(.wav) file. Then, graph the band-passed signals and amplitude envelopes extracted(after rectification and low-pass filtering) and waveforms of the original sound and vocoded sound. Additionally, using the output of the script file, make spectrograms of the original sound and the synthesized sound.

Answer 1

close all,
clear all,
clc,

ProjectPath = pwd;
SoundPath = strcat(ProjectPath,'\BirdChirp.wav');
FilterOrder=3;
No_of_Bands=3;
fmin = 200;
fmax = 300;

[y, Fs, nbits] = wavread(SoundPath);
Freq = (bands_cutoff(fmin,fmax,No_of_Bands))/(Fs/2);
L=length(Freq);
subplot(4,3,[1 2 3]); plot(y); title('Original Signal');
c=4;
for r=1:L-1
[b a] = butter(FilterOrder,[Freq(r) Freq(r+1)],'bandpass');
[H F] = freqz(b,a,256,Fs);
FilteredSignal = filter(b,a,y);   
subplot(4,3,c); plot(F,abs(H));
str = strcat('Freq. Response of Band-Pass Butter Worth Filter (Order=',num2str(FilterOrder),') for Freq. Range - [',num2str(Freq(r)),', ',num2str(Freq(r+1)),']') ;
title(str);
subplot(4,3,c+1); plot(FilteredSignal); title('Filtered Signal');
subplot(4,3,c+2); spectrogram(FilteredSignal); title('Spectrogram');
c=c+3;
end

Original Signal 3 Freq. Response of Band-Pass Butter Worth Filter (Order- 3) for Freq. Range [0.035281,0.Q41448 Filtered Signal Spectrogram 10 000 000 2000 2000 3000 4000 5000 6000 2 0.2 0.4 0.6 0.8 Normalized Frequency ( rad/sample) 10 Freq. Response of Band-Pass Butter Worth Filter (Order-3) for Freq. Range [0.041848,0.047881] Filtered Signal Spectrogram 8000 600 4000 2000 000 2000 3000 4000 5000 6000 4. 0.2 0.4 0.8 08 Normalized Frequency (n rad/sample) 10 Freg. Response of Band-Pass Butter Worth Filter (Order - 3) for Freq. Range [0.047881,0.054422 Filtered Signal Spectrogram 6000 0.5 4000 2000 1000 2000 3000 4000 5000 6000 2 6 0.2 0.4 0.6 0.8 1 Normalized Frequency rad/sample) x10