MATLAB Answers

Help with Chirp FFT

조회 수: 49(최근 30일)
Randa Qashoa
Randa Qashoa 22 May 2019
댓글: Daniel M 21 Nov 2019
I am trying to make a fft for a chirp signal with a frequency range of 15-25 MHz over a period of 10 micro seconds. When I attempt to plot the power vs. the frequency, I end up with a "square"-like wave from 15 MHz to 25 MHz and another one from 25 MHz to 35 MHz. I only need to generate the one wave that is from 15-25 MHz. Please find attached the plot and I have included the code below:
frames = 16;
nfft = SPF/frames;
f_max = 25e6;
f_min = 15e6;
d_chirp = 10e-6;
hchirp = dsp.Chirp('InitialFrequency',f_min,'TargetFrequency',f_max,...
'TargetTime', d_chirp,'SweepTime', d_chirp, 'SampleRate', SPF, 'SamplesPerFrame', nfft);
chirpData = (step(hchirp))';
freq =(0:nfft-1)*Fs/nfft;
xlabel('Frequency (Hz)');
xlim([0 50e6]);
power vs. freq..jpg

  댓글 수: 1

Hieu Nguyen
Hieu Nguyen 21 Nov 2019
I have the same problem; the bandwidth seems to double its length. Also, sweeping up and down have different results

로그인 to comment.


Daniel M
Daniel M 21 Nov 2019
You are plotting the incorrect frequency vector. fft() does not return the frequencies in order from most negative to most positive. Follow the example in the documentation for more information.
Change your freq and power vectors to:
freq2 = (0:(nfft-1)/2)*Fs/nfft;
power2 = power(1:(nfft/2));
Also note that this isn't typically what is considered the power. This is just the energy. Power is energy squared. You would also have to divide by nfft to scale properly.

  댓글 수: 1

Daniel M
Daniel M 21 Nov 2019
I also recommend increasing your sampling rate because it is not sufficient to fully represent this signal. At least 60 MHz should be OK.

로그인 to comment.

이 질문에 답변하려면 로그인을(를) 수행하십시오.


Translated by