I'm going through the derivation of the Goertzel DFT computation in the Wikipedia article, but I'm a little rusty, and I can't replicate one of the steps. Here's the background:

The filter output function is

$$y[n] = e^{j\omega_0n} \sum^n_{k=0} x[k] e^{-j\omega_0k}$$

We fix $\omega_0 = 2\pi\frac{k}{N}$, where $N$ is the number of samples in the input sequence.

Substituting that into the above equation, I get:

$$y[n] = \sum^n_{k=0} x[k] e^{j2\pi\frac{k}{N}n}e^{-j2\pi\frac{k}{N}k}$$

which simplifies to

$$y[n] = \sum^n_{k=0} x[k] e^{j\frac{2\pi k}{N}(n - k)}$$

as I see it. At this point, I'm supposed to observe that $e^{+j2\pi k}=1$, and get

$$y[N]=\sum _{n=0}^{N}x[n]e^{-j2\pi {\frac {nk}{N}}}$$

I don't see how factoring out $e^{+j2\pi k}$ helps here, or how to get the bottom term as a series in $n$, not $k$.


I see part of it now. The term $y[N]$ should be interpreted as $y[n=N]$, which then gives you

$$y[N] = \sum^N_{k=0} e^{\frac{j2\pi k}{N}(N-k)} x[k]$$

which will factor into the constant 1 value asked for.

Edit: The above is also incorrect, but I figured out what the issue was. The variable $k$ is reused in the Wikipedia article for both the series index and the chosen discretized frequency bin number, even though they mean different things. (Here's an edited version that swaps in another letter.)

I'm going to use $m$ for the frequency bin number from here on out. As stated clumsily in the Wikipedia article, $m$ is chosen from $\{ 0, 1, 2, \dots, N - 1 \}$, which is to say it can be anything from zero to one less than the length of the signal in samples. For clarity, I also throw out $k$ since it's completely useless and confusing. $n$ is the signal sample index.

$$y[N] = \sum^N_{n=0} x[n] e^{\frac{j2\pi m}{N}(N-n)}$$

This factors easily into

$$y[N] = \sum^N_{n=0} x[n] e^{j2\pi m} e^{-j2\pi \frac{mn}{N}}$$

which we can then simplify as stated in the article to

$$y[N] = \sum^N_{n=0} x[n] e^{-j2\pi \frac{mn}{N}}$$


  • $\begingroup$ no. What do you want to show exactly ? $\endgroup$ – reuns Oct 8 '16 at 20:35
  • $\begingroup$ @user1952009 I don't understand your question. Is there something unclear about my question or answer? $\endgroup$ – bright-star Oct 8 '16 at 20:56
  • $\begingroup$ yes, everything. The derivation of $y[N] = \sum_{m=0}^N x[m] e^{-2 i \pi m k/N}$ is very clear on wikipedia. Note that this method for evaluating one bin of the DFT with more numerical accuracy isn't very important in the theory of filters and discrete Fourier transform, and I don't even see why you would need it. $\endgroup$ – reuns Oct 8 '16 at 21:16

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