Unexpected behavior of FFT

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MAWE
MAWE on 22 Sep 2022
Commented: MAWE on 23 Sep 2022
I have this signal
r = x.*cos(2*pi*f1.*t) + i.*cos(2*pi*f2.*t) + n
where x and i are BPSK symbols {+1,-1} and n is additive Gaussian noise.
When I find the positive part of the PSD using fft I was expecting to get two peaks one at f1 and one at f2. However, I get the peaks towards the hight frequency fs/2 where fs is the sampling rate. Why is this happening?
EDIT: See figure
I = imread('PSD_test.png');
imshow(I)
EDIT: I calculate the PSD from fft like so
N=length(r);
Unrecognized function or variable 'r'.
rFFT = fft(r,N);
prr = 2.*abs(rFFT).^2/(fs*N);
  5 Comments
MAWE
MAWE on 22 Sep 2022
@Paul I edited the questions to explain how I calculated the PSD from fft.

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Answers (1)

Bjorn Gustavsson
Bjorn Gustavsson on 22 Sep 2022
In the field (IS-radar) I work we have typical carrier-frequencies of 224-1000 MHz and the modulation band-width is a couple of MHz. Here you try to use BPSK with baud-lengths of 16 samples - which corresponds to a really high frequency. To my understanding you don't have a large ferquency-span between the modulation-frequency and the Nyquist-frequency. Does the signal look sensible if you plot it for a time-period of a couple of bits.
  5 Comments
MAWE
MAWE on 23 Sep 2022
@Bjorn Gustavsson Also, the bandwidth of the signal is 500 MHz

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