Unexpected behavior of FFT

3 次查看（过去 30 天）

显示更早的评论

MAWE 2022-9-22

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1809915-unexpected-behavior-of-fft

评论： MAWE 2022-9-23

PSD_test.png

在 MATLAB Online 中打开

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 个评论
显示 3更早的评论隐藏 3更早的评论

MAWE 2022-9-22

@Paul The actual code is involved. I just tried to put the gest to understand why I get unexpected behavior. Do I need to shift the frequencies of the fft by fftshift, for example? I did that but didn't work

MAWE 2022-9-22

@Paul I edited the questions to explain how I calculated the PSD from fft.

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Bjorn Gustavsson 2022-9-22

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1809915-unexpected-behavior-of-fft#answer_1058730

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.