CODEWITHSUNDEEP
fftdft
Katoch
Katoch

Reputation: 2777

what is the result inside fft bin at different index?

For example if I have 10 samples in the FFT in a system sampled at 100 Hz. Then how are the results represented in FFT bin output ? The frequency resolution is going to be how many Hz each DFT bin represents. This is, as you have noted, given by fs/N. In this case :--

resolution = 100/10 = 10 hz

Definition of FFT states that frequency range [0,fs] is represented by N point.

But when the result comes :--

so my question is :--
X[6] is the 60hz signal & its value will be complex conjugate of 40 hz signal .. is it right ?
Or X[6] does not contains spectral of 60hz but it is only complex conjugate of X[4] ?

Please suggest.

Upvotes: 0

Views: 875

Answers (2)

Dan Boschen
Dan Boschen

Reputation: 302

The frequencies of each bin in the DFT go from $n = 0 to N-1$ where each frequency is given as $n F_s/N$.

Due to the sampling process, for any signals (either real and complex signals) the upper half of this frequency spectrum that extends from $F_s/2$ to 1 sample less than $F_s$ is equivalent to the negative half spectrum (the spectrum that extends from $-F_s/2% to 1 sample less than 0.). This applies to the DFT and all digital signals.

This is demonstrated in the graphics below, showing how an analog spectrum (top line), convolves with the sampling spectrum (second line) of a system sampled at 20 Hz to result in the digital spectrum of a sampled signal. Since the spectrum is unique over a 20 Hz range (and repeats everywhere else), we only need to show the frequencies over any 20 Hz range to represent the signal. This can be identically shown from -10 Hz to +10 Hz, or 0 to 20 Hz.

Representing the sampled spectrum of a real signal sampled at 20 Hz over the range of -10Hz to +10Hz:

Sampling real 1

The same signal can also be represented over the range of 0 to 20 Hz:

Sampling real 2

Representing the sampled spectrum of a complex signal sampled at 20 Hz over the range of -10Hz to +10Hz:

Sampling complex 1

The same signal can also be represented over the range of 0 to 20 Hz:

Sampling complex 2

Since the spectrum for the DFT is discrete, the samples go from $n = 0 to N-1$ where each frequency is given as $F_s/n$ as described above. There is also only N samples in the DFT of course but as you rotate the samples in the DFT, you effectively move through the expanded spectrum above. It is helpful to view the spectrum on the surface of a cylinder with a circumference that goes from 0 to $F_s$ where you see that $F_s$ is equivalent to 0, and going backwards from 0 is equivalent to going into the negative half spectrum.

So in your example specifically:

X[0] = DC

X1 = 10

X2 = 20

X3 = 30

X4 = 40

X[5] = 50

X[6] = 60

X[7] = 70

X[8] = 80

X[9] = 90

Can also be represented as

X[0] = DC

X1 = 10

X2 = 20

X3 = 30

X4 = 40

X[5] = -50

X[6] = -40

X[7] = -30

X[8] = -20

X[9] = -10

Note that the command "FFTSHIFT" in MATLAB shifts the DFT vector accordingly to produce the following order representing the range from -F_s/2 to +F_s/2 :

fftshift([X[0], X1, X2, X3, X4, X[5], X[6], X[7], X[8], X[9]]) =

[X[5], X[6], X[7], X[8], X[9], X[0], X1, X2, X3, X4]

Upvotes: 3

Paul R
Paul R

Reputation: 212959

Assuming your input signal is purely real, then X[6] is just the complex conjugate of X[4]. The top half of the spectrum is essentially redundant for real signals. See: this question and answer for further details.

Note that the input signal should be bandwidth-limited to fs / 2, otherwise aliasing will occur, so for a baseband signal there should not be any components >= fs / 2.

Upvotes: 1

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