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I'm trying to frequency shift a sine wave at 50kHz by using a complex exponential at 15kHz. I should be getting an FFT with a peak at 65kHz.

Instead I'm getting a strange looking peak at 52.4kHz. Anyone know why this is?

This is probably a simple issue but I've been stuck for too long.

enter image description here

Here is my Matlab code:

clc,clear close all srate = 300e3; npnts = srate*5; %generate number of points for 5 seconds of sampling time = (0:npnts-1)/srate; % signal settings freq1 = 50e3; ampl = 1; signal = ampl*sin(2*pi*freq1*time); % amplitude spectrum via Fourier transform signalX = fft(signal); signalAmp = 2*abs(signalX)/npnts; % need to multiply by 2 to recover amplitude from negative freqs % divide by npnts to normalize fourier coefficients % shift signal using complex exp freq2 = 15e3; shift = exp(1i*freq2*time).*signal; signalshift = fft(shift); signalAmpshift = 2*abs(signalshift)/npnts; % vector of frequencies in kHz hz = linspace(0,srate/2,floor(npnts/2)+1)/1e3; %go from 0 to nyquist, frequency resolution is defined by last term ns=1; xlimits = [(freq1-ns*freq2)/1e3 (freq1+ns*freq2)/1e3]; figure(1); stem(hz,signalAmp(1:length(hz)),'ks','linew',2,'markersize',10) hold on stem(hz,signalAmpshift(1:length(hz)),'linew',2,'markersize',10) xlim(xlimits) legend({'Signal';'Shifted'})
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  • $\begingroup$ For starters, you are missing a factor of $2\pi$ in your complex exponential. But more importantly, your input signal should be analytic before you shift it. $\endgroup$ Commented Apr 4, 2023 at 18:05
  • $\begingroup$ @Jazzmaniac what do you mean analytic? $\endgroup$ Commented Apr 4, 2023 at 18:12
  • $\begingroup$ add the $2\pi$ factor, you'll get your peak at 65Hz $\endgroup$ Commented Apr 4, 2023 at 18:31
  • $\begingroup$ @PrematureCorn analytic signals have no negative frequency components. You real-valued 50 kHz sine has a frequency component at +50 kHz, and one at -50 kHz. After shifting, it would have one at -45 kHz and one at +65 kHz. That might, or might not be (and I think that's what Jazzmaniac is understandably assuming here) what you want. $\endgroup$ Commented Apr 4, 2023 at 18:36
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    $\begingroup$ Please don't abandon your questions. Jdip put effort into providing a working answer, which you have not even upvoted. I have no interest in answering such users. $\endgroup$ Commented Apr 5, 2023 at 15:02

2 Answers 2

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Just summarizing the comments.

  1. Your complex exponential is missing a factor of $2\pi$ hence you are not shifting by 15 kHz but by $15/(2\pi) kHz = 2.837 kHz$
  2. The shifted frequency is not an integer multiple of the frequency resolution (sampler rate divided by FFT length) so you don't get a single line but "spectral leakage"
  3. Your input signal is real, i.e. it has equal (but conjugate) amounts of positive and negative frequencies. After multiplying the signal is complex. You are shifting both frequency component of the original signal the same way. Make sure you look at the "correct" components.
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  • $\begingroup$ Could you please add to your answer what the "correct" components are? Also, what would be the best practice of handling a real signal multiplied by a complex exponential? Should I filter it? $\endgroup$ Commented Apr 4, 2023 at 19:29
  • $\begingroup$ @PrematureCorn: "correct" in this context depends really on what exactly you want to do and why. There are some applications where complex signals are perfectly fine and others where they really don't make sense. $\endgroup$ Commented Apr 4, 2023 at 20:58
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    $\begingroup$ @PrematureCorn this may help: $2\cos(2\pi 50e3 t) = e^{j2\pi 50e3 t}+e^{-j2\pi 50e3 t}$, which is how we see directly the positive 50KHz and negative 50KHz frequencies (which appear as two bins in the FFT). If you multiply that by $e^{j2\pi 15e3 t}$ you get $e^{j2\pi 65e3 t}+e^{-j2\pi 35e3 t}$. If you only wanted the 65KHz, you either need to high pass filter, or use the analytic signal as explained in the comments under your post. $\endgroup$ Commented Apr 5, 2023 at 2:44
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    $\begingroup$ @PrematureCorn If Hilmar answered your primary question, please select it to close this out. Review your other questions as well if still open and let the responders know what is missing still in your understanding. Thanks! $\endgroup$ Commented Apr 5, 2023 at 21:21
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signal 1

frt = 300e3; % [Hz] sampling rate dt=1 % 1 second observation time n = frt*dt % amount time samples t = (0:n-1)/frt; % [s] time reference % signal 1 f1 = 50e3; % [Hz] A1 = 1; s1 = A1*sin(2*pi*f1*t);

Put enough frequency samples in F to avoid low resolution.

When F = frt*[0:n]/n resulting FFT amplitudes show errors > 10%.

Use more frequencies than time samples to avoid FFT amplitude low resolution.

nf=4*n % amaount frequency samples F = frt*[0:nf]/nf; % [Hz] FFT frequency reference S1 = 1/(n)*fft(s1,nf); % not centered S1_amp = abs(S1); S1c_amp=abs(fftshift(S1)); % centered S1 FFT F1c=frt*[-nf/2:nf/2]/nf; % centered frequency reference figure(1) plot(F1c(1:min(numel(F1c),numel(S1c_amp))),S1c_amp(1:min(numel(F1c),numel(S1c_amp)))); grid on xlabel('f [Hz]') title('2-side centered FFT(s1) : s1=A*sin(2*pi*f1*t)')

enter image description here

S1p_amp=2*S1_amp;F1p=F; % F1p S1p are just for 1-side |FFT| display S1p_amp(nf/2+1:end)=[]; F1p(nf/2+1:end)=[]; figure(2) plot(F1p,S1p_amp) grid on xlabel('f [Hz]') title('1-side FFT(s1)')

enter image description here

signal 2

Corrected 2*pi missing in s2 exponential

f2 = 15e3; % [Hz] A2=1; s2 = A2*exp(1j*2*pi*f2*t); S2 = 1/n*fft(s2,nf); S2c_amp = abs(fftshift(S2)); figure(3) plot(F1c(1:min(numel(F1c),numel(S2c_amp))),S2c_amp(1:min(numel(F1c),numel(S2c_amp)))); grid on xlabel('f [Hz]') title('2-side centered FFT(s2) : s2=A*exp(1j*2*pi*t)')

enter image description here

*frequency shift s3=s1.s2

s3 = s1.*s2; S3 = 1/n*fft(s3,nf); S3c_amp = abs(fftshift(S3)); figure(4) plot(F1c(1:min(numel(F1c),numel(S3c_amp))),S3c_amp(1:min(numel(F1c),numel(S3c_amp)))); % plot(F1c,S3c_amp) grid on xlabel('f [Hz]') title('2-side centered FFT(s2) : s3=s1*s2')

enter image description here

Using stem for FFT signals may clutter small peaks.

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