Discover MakerZone

MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi

Learn more

Discover what MATLAB® can do for your career.

Opportunities for recent engineering grads.

Apply Today

How to do FT Time shift and Time scaling properties

Asked by ong on 28 Mar 2013

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.

0 Comments

ong

Products

No products are associated with this question.

4 Answers

Answer by Wayne King on 28 Mar 2013
Edited by Wayne King on 28 Mar 2013
Accepted answer
 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');

1 Comment

ong on 29 Mar 2013

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.

Wayne King
Answer by Wayne King on 28 Mar 2013
Edited by Wayne King on 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 Comment

ong on 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);

Wayne King
Answer by Wayne King on 29 Mar 2013
Edited by Wayne King on 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 Comment

ong on 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.

Wayne King
Answer by ong on 3 Apr 2013

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.

0 Comments

ong

Contact us