5
\$\begingroup\$

derrzet derrzet

Hi,

I'd like to understand the context in between an oscillogramm and the resulting FFT. My example is a full-wave rectifier and I am try to calculate some harmonics.

Because the function is symmetric I only need to calculate the an-values: $$ A_n = a_n = \frac{2}{\pi}\int_0^{2\pi} \! f(t) * cos(nt) \, \mathrm{d}t \\ a_n = \frac{2}{\pi}\int_0^{2\pi} \! |sin(t)| * cos(nt) \, \mathrm{d}t $$ dividing up for integrating $$ a_n = \frac{2}{\pi} \left( \int_0^{\pi} \! sin(t) * cos(nt) \, \mathrm{d}t + \int_{\pi}^{2\pi} \! (-sin(t)) * cos(nt) \, \mathrm{d}t \right) \\ $$ result of the integral $$ a_n = \frac{2}{\pi} \left( -\frac{cos(\pi n) + 1}{n^2 - 1} - \frac{cos(2\pi n) + cos(\pi n)}{n^2 - 1} \right) $$ My question is whether my way was right up to here and how I have to transfer this into the fft?

\$\endgroup\$
6
  • \$\begingroup\$ You can check your results with that in almost any textbook on Fourier transforms. However, your equation shown in your question blows up when n=1 so it can't be correct. \$\endgroup\$
    – Barry
    Commented Jun 12, 2016 at 11:54
  • \$\begingroup\$ ok you are right, but i still not see my failure. \$\endgroup\$
    – C-Jay
    Commented Jun 12, 2016 at 12:22
  • \$\begingroup\$ \$\pi n+1\$ in the cosinus is an strange result (adding 1 radian?), it should be something like \$(n+1)\pi\$. \$\endgroup\$
    – Roger C.
    Commented Jun 12, 2016 at 12:57
  • \$\begingroup\$ Your question is a bit confusing... Are you really trying to calculate Fast Fourier Transform based on a real oscillogram? To me it looks like you're trying to do Fourier series on a graph of a function and not on an oscillogram from a device. There are no integrals in FFT and you feed FFT with a time-series, not a function. \$\endgroup\$
    – AndrejaKo
    Commented Jun 12, 2016 at 13:55
  • \$\begingroup\$ @AndrejaKo the oscillogram and FFT are from a real rectifier and I like to calculate the amplitudes of some harmonics. \$\endgroup\$
    – C-Jay
    Commented Jun 12, 2016 at 16:18

2 Answers 2

1
\$\begingroup\$

If it helps, you have a sine wave multiplied by a square wave added to a phase shifted sine wave multiplied by another phase shifted square wave.

Addition translates to addition in fourier space, multiplication maps to convolution.

So what you're going to get is the FT of a square wave shifted along the w axis and with an imaginary component, I think.

\$\endgroup\$
2
  • \$\begingroup\$ That confuses me. Where do you see any square waves? \$\endgroup\$
    – C-Jay
    Commented Jun 29, 2016 at 9:18
  • \$\begingroup\$ @C-Jay Think of a square wave as an on-off; apply it to a sine wave to get half the rectified output. \$\endgroup\$ Commented Jul 5, 2016 at 20:35
0
\$\begingroup\$

For understanding FFT, first you need to link continuous time to discrete time (which happens in most digital oscilloscopes - take into account the sampling frequency and the cut-off frequency):

https://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem

Then try to understand a discrete time fourier transform:

https://en.wikipedia.org/wiki/Discrete-time_Fourier_transform

After understanding all this, you should be able to follow up the FFT easily.

If you are good with Matlab, you can simulate with a lot of signals (search fft).

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.