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I came across this question recently, and I am very confused by (b)(ii).

b(i) gives $x_n$ = [0.5, 0, -0.5, 0].

My approach to (ii) was to recognise that $X_m$ represents the frequency content of the signal at $\omega = \frac{m \omega_0}{N}$, where $N$ is the number of samples (4 in this case) and $\omega_0$ is the sampling frequency (4Hz or 8$\pi$ rad/s in this case).

So, for $m=0$, the frequency of that spectral component is 0Hz, for $m=1$, the frequency of that spectral component is 1Hz, and so on. Thus, the non-zero spectral components are at 1 Hz and 3Hz.

Here, though, is as far as I have been able to get.

The mark-scheme says something about only being able to "pick out" frequencies at half of the sampling frequency, and that this is somehow related to the property of the DFT that $X_{N−m} = X_m^*$. They use this to conclude that the non-zero components in the DFT represent a frequency of 1Hz.

However, I don't really understand what all that means, and how it all fits together. I'd really appreciate it if someone could tell me in detail how to get from where I've gotten to, to the mark-scheme's conclusion.


1 Answer 1


So, insert $(0,0,0,1)$ into your IDFT. You should get a single oscillation at the frequency for $m=3$, right.

Just look at the formula you're getting (your DFT sum formula elegantly collapses to just one summand), and read the frequency from that.

Hint: $e^{jx}$ is $2\pi$-periodic in $x$.

  • $\begingroup$ Why (0,0,0,1)?? $\endgroup$
    – VJ123
    Apr 11, 2022 at 12:46
  • $\begingroup$ because your trouble seems to be with the $m=3$ component, and that is a frequency domain vector only having that $m=3$ component. $\endgroup$ Apr 11, 2022 at 12:48
  • $\begingroup$ Ok so putting (0,0,0,1) into my IDFT gives (0.25, -0.25j, -0.25, 0.25j). Where do I go from here? $\endgroup$
    – VJ123
    Apr 11, 2022 at 12:58
  • $\begingroup$ don't look at the numerical values, look at the formula (I explicitly said that, and that it elegantly collapses) $\endgroup$ Apr 11, 2022 at 12:59
  • $\begingroup$ Ah I see. Using de Moivre's theorem, the formula collapses to 1/4 (-j)^n, though I can't quite see how to read the frequency from this? $\endgroup$
    – VJ123
    Apr 11, 2022 at 13:08

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