How to do FT Time shift and Time scaling properties

Currently i am trying use matlab to do the FT properties -time shift and time scaling, can someone help me in the implementation? i would like to prove that Timeshift: F[x(t-τ) ]= e^(-jwτ) F[x(t)]
Time scaling: F[x(at)]=1/(|a|) X(w/a)
Thanks.

 Réponse acceptée

Wayne King
Wayne King le 28 Mar 2013
Modifié(e) : Wayne King le 28 Mar 2013
t=0:0.001:0.1-0.001;
Fs = 1e3;
freq1 = 100;
x1=cos(2*pi*freq1*t);
Delay=2;
yp = fft(x1);
yp = yp(1:length(x1)/2+1);
f = 0:Fs/length(x1):500;
yp = yp.*exp(-1i*2*pi*f*Delay*(1/Fs));
yp = [yp conj(fliplr(yp(2:end-1)))];
y = ifft(yp,'symmetric');
plot(t(1:100),x1(1:100),'b');
hold on;
plot(t(1:100),y(1:100),'r');

2 commentaires

Hi Wayne, thanks for your help. I am able to observe the shift but i couldn't get the same plot result when comparing:
D=2;
Fs=1000;
freq1=100;
t=0:0.001:1-0.001;
x=cos(2*pi*freq1*(t-D));
y1 = fft(x);
and
x1=cos(2*pi*freq1*t);
Delay=2;
yp = fft(x1);
yp = yp(1:length(x1)/2+1);
f = 0:Fs/length(x1):500;
yp = yp.*exp(-1i*2*pi*f*Delay*(1/Fs));
any reason why? Thanks.
Your code just fails when sampling frequency increases.
For f = 1e4 it gives matrix dimensions must match error. Other wise if I keep your sampling frequency at 1e6
After that point, even if I increase the delay by 2000. I could not observe any delay.
Sorry but it seems your code works for one case.

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Plus de réponses (6)

Wayne King
Wayne King le 29 Mar 2013
Modifié(e) : Wayne King le 29 Mar 2013
They agree if you get the delay right. You're not delaying the signal by 2. You're trying to delay the signal by two samples, but that has to take into account the sampling interval, so you're actually delaying the signal by 0.002 seconds.
t = 0:0.001:1-0.001;
freq1 = 100;
Fs = 1000;
x1=cos(2*pi*freq1*t);
Delay=2;
yp = fft(x1);
yp = yp(1:length(x1)/2+1);
f = 0:Fs/length(x1):500;
yp = yp.*exp(-1i*2*pi*f*Delay*(1/Fs));
yp = [yp conj(fliplr(yp(2:end-1)))];
yrec = ifft(yp,'symmetric');
Compare with
D=2;
Fs=1000;
freq1=100;
t=0:0.001:1-0.001;
x=cos(2*pi*freq1*(t-(D*(1/Fs))));
y1 = fft(x);
y1T = ifft(y1,'symmetric');
max(abs(y1T-yrec))
You can see the above are identical. Thank you for accepting my answer if I have helped you.

1 commentaire

ong
ong le 29 Mar 2013
Hi there, Thanks for the clear explanation. However, i have one more issue, does it work the same way if i m having my fft display in the angle instead of abs?
I tried having it plot in the phase domain but there is some differ in the result.

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Wayne King
Wayne King le 28 Mar 2013
Modifié(e) : Wayne King le 28 Mar 2013
n = 0:159;
x = cos(pi/4*n);
y = cos(pi/4*(n-2));
xdft = fft(x);
ydft = fft(y);
xdft(21)
ydft(21)
Note that 80+i0 has become 0-80i due to the predicted phase shift of e^{-i\pi/2}
Obviously, the only way to properly "prove" that theorem is mathematically.
The scaling one you have to be careful with in discrete-time because scaling doesn't work the same with a discrete variable as it does with continuous time.

1 commentaire

ong
ong le 28 Mar 2013
Is there anyway that i can do it in this form:
t=0:0.001:1
x1=cos(2*pi*freq1*t)
Delay=2
yp = fft(x1);
yp = yp.*exp(-j*2*pi*t*Delay);

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ong
ong le 3 Avr 2013

0 votes

To update on this question, Wayne King provided the explanation and the steps provided are accurate. However there is one problem, instead of ifft the abs function, it was to display in the phase domain, here, the phase for the time shift properties and the function:cos(2*pi*freq1*(t-(D*(1/Fs) doesnt match.
There are error to the phase shift, anyone can help please?
Thanks.

1 commentaire

visit and you will get easy made time shifting, time scaling and amplitude scaling all in one project. surely both projects will help you.

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SHREEVARSHINI R
SHREEVARSHINI R le 24 Oct 2021

0 votes

1.Write a MATLAB program to find Fourier transform of the signal Ate-btu(t)
2.Write a MATLAB program to perform amplitude scaling, time scaling and time shift on the signal x(t) = 1+t; for t=0 to 2
Sk Group
Sk Group le 25 Oct 2021

0 votes

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