Analogue Frequency of Band-limited signal (Discrete Fourier Transform)

1. Aug 28, 2013

zaka

Hi, I have the following question:

A signal x(t) which is band-limited to 10kHz is sampled with a sampling frequency of 20kHz. The DFT (Discrete Fourier Transform) of N= 1000 samples of x(n) is then computed. To what analogue frequency does the index k=120 respond to?

I'm trying to relate the above to a previous worked example that I encountered, the question was similar as follows:
The sampling frequency, is fs = 120 Hz. The DFT of N=6 samples. To what analogue frequency does the index
k=1 correspond to? What about index k=5 ?

For k =1
fk = k (fs / N)
= (1)*(120/6) = 20Hz

For k=5
fk = (k-N)*(fs / N)
= (5-6)*(120/6) = -20Hz

Why for k=5 was k subtracted from N. Is there a general forumla for this sort of question?

And what does this mean for the question I'm trying to resolve?
Is fk for that = 120*(2000/1000)

or fk for that = (120-1000)*(2000/1000)

or something else completely?

Any help would be much appriciated.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 29, 2013

collinsmark

For a discrete-time signal x[n], the DFT of that signal, F[k] is not only discrete, but also periodic.*

Let's step back a little, and allow me to use an analogy. Imagine a continuous function that can be expressed as a function of an angle θ, like f(θ). It doesn't change anything if you add 2π to θ; subtract 2π from θ; or add any integer multiple of 2π to θ, be it positive or negative. The location of f at θ = -10 deg is the same location as f at θ = -10 + 360 deg, which is the same as θ = -10 + 720 deg, or even θ = -10 -360 deg. It's all the same thing. 350 degrees is the same as 350 - 360 deg = -10 deg.

Similarly, F[k] is also periodic with a period of N. Add or subtract any integer multiple of N, positive or negative to k, and you get the same result. In other words,

F[k] = F[k + mN],

where m is any integer.

So, specific to your example, subtracting N from k (which is the same thing as adding -N to k) is analogous to computing that 350 deg = 350 + -360 deg = -10 deg.

*(As a matter of fact, there is a sort-of assumption that x[n] is also periodic. In practice, the original signal is not necessarily periodic, but when the DFT is taken, the mathematics assume that x[n] is periodic with a period of N times the sample time. It is for this reason why the time-based sample data is commonly windowed with a Hamming window or Hanning window [or some other type of window] prior to taking the DFT. But that's another story, and I won't get into that here.)

Yes there is!

(But I don't want to give it to you. You can figure this out yourself.)

Actually, you may have to break things up into different cases though. It's one of those, "if this, use this formula, otherwise use this other one over here" sort of thing.

In the process, it might help to break F[k] into sections. Let's restrict ourselves to k = 0 to k = N-1.

F[0] is always the DC component (corresponding to 0 Hz).

F[N/2] is the Nyquist frequency element, if N is an even number. (If N is an odd number, there is no Nyquist frequency element, and the Nyquist frequency is in between a couple of elements.

The lower half of F[k] (k: 0 < k < N/2]) represent positive frequency elements.

The upper half of F[k] (k: N/2 < k < N]) represent negative frequency elements. They start out as large, negative frequency elements and end at small negative frequency elements.

(Boldface mine.)

I assume you mean 20,000 as opposed to 2000.

Given that correction, they're technically both right!
[Edit: Well, they are both correct from a purely mathematical viewpoint. But arguably, only one of them is correct from a more practical point of view.]

But there's no need in subtracting off N from k as you did in your second choice. 120 is less than 1000/2, meaning k is below the Nyquist frequency index, and represents a positive frequency.

Last edited: Aug 29, 2013