FFT time delay, incorrect alighnment of the output wave

Hi,

I shift the FFT of a sampled signal by a certain number of samples and take the inverse FFT to get a delayed signal. I use the formula:

x[n+k] -> FFT(x[n]) * exp(-j*2*pi*k*m/N), where N = the length of the signal, k is the amount of time delay in time domain and m is the running index.

When I take the IFFT, I see that the signal starts correctly but then it gets misaligned (For example, in Figure 1 you can see there are 2 peaks in place of one peak at t~70 s). Why am I getting this result?

My code is below:

%for fs=2*10^8 Hz and f_signal=300 MHz

Fs=300000000; %fs=300MHz

T=1/Fs;  %the  sampling period

t=0:T:((0.7*10^-6)-T);

L=length(t);

f_signal=3*10^6;

x=cos(2*pi*f_signal*t);   % signal_T=1/(3*10^6); %= 3.3333e-07

X=fft(x);

delayed_samples= 126; %delay the signal by 126 samples:

Y=X.*exp(1j*2*pi*delayed_samples*[0:L-1]/L ); 

y_1=real(ifft(Y));

figure, plot(Fs.*t, x, 'g'), hold on, plot(Fs.*t, y_1, 'r'),  legend ('x','delayed x') ,   title 'signals vs. samples' 

%%%end of the script%%%%%%%

Figure 1 is

Answer 1

[Matt J]

Because it is a circulant shift. The discontinuity between x(1) and x(end) has been shifted circulantly to around frequency=83.

[Sophie]

If I need them to be aligned what do I do?

[Matt J]

Sample full periods of the sinusoid, so that there is no discontinuity between x(1) and x(end).