# What's the purpose of doing fftshift twice?

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Mark Golberg on 9 Jun 2016
Commented: Jan on 24 Feb 2017
Hello,
I have a signal in time domain (x).
Can someone please explain the difference between:
y1 = fftshift(fft(x));
y2 = fftshift(fft(fftshift(x)));
As I understood y2 is the better/more common approach. Why???
Thank you!!
y2 does not really make sense. fftshift is a frequency domain operation, applying it in the time domain does not make sense. See Walter's answer for more detail.

John BG on 9 Jun 2016
Mark
command fftshift is used to visualise the FFT within [-Fs/2 Fs/2] instead of [0 Fs] that is the interval that the FFT takes as default.
Let be the following signal:
dt=.01;
t=[0:dt:10];
x0=10*sin(t);
x=10*sin(t)+10*cos(t); % 2nd tone has phase shift equivalent to pi/4 rad
xd1=fftshift(x);
X=fft(x);
Xd=fft(xd1);
figure(1);plot([1:1:length(X)],abs(X),'b',[1:1:length(X)],fftshift(abs(X)),'g');grid on;
legend('|X|','fftshift(|X|)');
figure(2);plot([1:1:length(X)],angle(X),'b',[1:1:length(X)],fftshift(angle(X)),'g');grid on;
legend('arg(X)','fftshift(arg(X))');
If you apply a frequency operation directly on time domain, for this example you get the following:
figure(3);plot(t,x0,'b',t,x,'r');grid on
text(2.63,x0(find(t==2.63)),'\leftarrow 10*sin(t)','FontSize',14);
text(1.16,x(find(t==1.16)),'\leftarrow 10*sin(t)+10*cos(t)','FontSize',14);
figure(4);plot(t,x0,t,xd1);
text(5.38,x0(find(t==5.38)),'\leftarrow 10*sin(t)','FontSize',14);grid on;
text(2.41,xd1(find(t==2.41)),'\leftarrow fftshift(10*sin(t)+10*cos(t))','FontSize',14);
the typical phase imbalance oscilloscope also helps:
figure(5);plot(x0,x,x0,xd1);
As explained in MATLAB help, there is further phase cross-over when working with 2D signals
. If you find this answer of any help solving your question
John
Jan on 24 Feb 2017
John BG has accepted this answer by his own. It is not clear, if it helps the OP.

Walter Roberson on 9 Jun 2016
No, if your data is in the time domain, then using fftshift(x) before usng fft() would only be before the case where you had negative time represented in your data and time zero was at the middle of the data. The first version you gave is significantly more likely for time domain data: it assumes the data starts at time 0 and it shifts the result of the fft so as to center the plot in the frequency domain.
You only use fftshift() twice if the data you are using fft() or ifft() on is centered in the array.