multiplication in frequency domain equals convolution in time domain mismatch
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We know that a convolution in the time domain equals a multiplication in the frequency domain.
As per the article below,
In order to multiply one frequency signal by another, (in polar form) the magnitude components are multiplied by one another and the phase components are added.
To prove this, i created two sinusoidal signals,
Freq1 = 1000;
Freq3 = 3000;
Fs = 16000;
T = 1/Fs;
Nos = (0:128-1)*T;
Amp = 1.0;
Signal1 = Amp*sin(2*pi*Freq1*Nos);
Signal3 = Amp*sin(2*pi*Freq3*Nos);
In the time domain, i convolved these two signals,
Signal4 = conv(Signal1,Signal3);
Taking a 32 point FFT of the two signals & and a 32 point FFT of the convolved signal,
NFFT = 32;
freqdata1 = fft(Signal1,NFFT);
freqdata2 = fft(Signal3,NFFT);
freqdata4 = fft(Signal4,NFFT);
in the frequency domain, i multipled the magnitude components of the two individual signals and added the phase component of the two signals. I compared this Magnitude and phase value with the Convolved signal's phase and magnitude value. I expected the values, [Newmag' NewPhase'] & [Mag3' Phase3'] to be similar since the a convolution in time domain equals a multiplication in the frequency domain. But they are not. What am i missing here ? What have i done wrong ?
for ii = 2:((length(freqdata1)/2)+1)
sig1_cc = real(freqdata1(1,ii));
sig1_dd = imag(freqdata1(1,ii));
Mag1(ii-1) = sqrt((sig1_cc^2)+(sig1_dd^2));
Phase1(ii-1) = atan(sig1_dd/sig1_cc);
sig2_cc = real(freqdata2(1,ii));
sig2_dd = imag(freqdata2(1,ii));
Mag2(ii-1) = sqrt((sig2_cc^2)+(sig2_dd^2));
Phase2(ii-1) = atan(sig2_dd/sig2_cc);
sig3_cc = real(freqdata4(1,ii));
sig3_dd = imag(freqdata4(1,ii));
Mag3(ii-1) = sqrt((sig3_cc^2)+(sig3_dd^2));
Phase3(ii-1) = atan(sig3_dd/sig3_cc);
Newmag(ii-1) = Mag1(ii-1)*Mag2(ii-1);
NewPhase(ii-1) = Phase1(ii-1) + Phase2(ii-1);
end
[Newmag' NewPhase']
[Mag3' Phase3']
Accepted Answer
Merida
on 8 May 2019
0 votes
More Answers (2)
For discrete time domain signal:
n=0:100;
k = 0:200;
w = (pi/100)*k;
m=n'*k;
X11=(exp(-j*pi/100)).^m;
x1=cos(pi*n/2); % generate random number x1
x2=sin(pi*n/2); % generate random number x2
X1 = x1*X11;
X2 = x2*X11;
x = conv(x1,x2)
n1=0:200;
m1=n1'*k;
X12=(exp(-j*pi/100)).^m1;
X3 = x*X12
X = X1.*X2
stem(n1,X3)
stem(n1,X)
1 Comment
The imaginary components turn out to be noise level.
n=0:100;
k = 0:200;
w = (pi/100)*k;
m=n'*k;
X11=(exp(-j*pi/100)).^m;
x1=cos(pi*n/2); % generate random number x1
x2=sin(pi*n/2); % generate random number x2
X1 = x1*X11;
X2 = x2*X11;
x = conv(x1,x2)
n1=0:200;
m1=n1'*k;
X12=(exp(-j*pi/100)).^m1;
X3 = x*X12
X = X1.*X2
stem(n1, abs(X3))
stem(n1, abs(X))
stem(n1, abs(X3)-abs(X))
Narendhar
on 1 Oct 2024
0 votes
: Write MATLAB code to demonstrate that the convolution in spatial domain is equivalentto multiplication in the frequency domain.
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