FFT for Sinusoidal Signal: 100 kHz, Peaks @ k=100, 412

[rjunior]

Suppose you have the FFT of a sinusoidal signal sampled at 100 kHz and it has energy peaks at k = 100 and k = 412. How many points is the FFT? What is the frequency of the signal?

 

[uart]

Ok it's a straight forward question, but the answer depends upon whether "k" is a zero based index or a unity based index.

So what is the first element of your FFT. Is it X(0) or is it X(1)?

 

[rjunior]

uart,

Thanks for your attention. Let suppose it is X(0).

 

[uart]

uart,

Thanks for your attention. Let suppose it is X(0).

Ok that's the easiest case. So "k" runs from 0 through to "N-1".

The frequency spacing is: df = fs/N

And the frequency represented by the kth component is k df, provided it is less than the Nyquist frequency (one half fs).

The sample at k=100 will represent the sine wave frequency, and that at k=412 it's image (above the Nyquist frequency).

So you can pretty easily solve what you require (your signal frequency f1 and number of samples N) from the following equations.

df = fs/N
f1 = 100 df
fs - f1 = 412 df

 

[rjunior]

Dear uart,

Thank you for your attention. I have developed the equation but how i do to get signal frequency. Thanks in advance

df = fs/N
f1 = 100 df
fs - f1 = 412 df

fs - 100 df = 412 df
fs/N (N - 100 ) = 412 fs/N
N - 100 = 412
N = 512

 

[MisterX]

Okay, so you know the maximum frequency in the FFT output is one half the sampling frequency (fs), right? The FFT will typically have output like this

X(0) is the 0-frequency (a.k.a DC) component

X(1) through X(N/2) are for positive frequencies up to fs/2. The distance between adjacent frequencies is fs/N.

X(N/2 + 1) to X(N - 1) are for negative frequencies. X(N/2+ 1) is for the most negative frequency. The frequency also increases by fs/N for each next sample in this region.

There are two peaks in the magnitude of the spectrum. This has to do with the conjugate symmetry of the FFT of all real-valued signals.

so
|X(1)| = |X(N - 1)|
|X(2)| = |X(N - 2)|
and so on

 

[uart]

Dear uart,

Thank you for your attention. I have developed the equation but how i do to get signal frequency. Thanks in advance

Try solving for N first, then use f1 = 100 fs / N

Rearrange the above equations to eliminate f1 and you get :

100 df = fs - 412 df

If you subst in df = fs/N you can factorize out fs and solve directly for N.