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Referring to this link, about Goertzel,I am confused about the final equation after N iterations Where does

real = (q1 - q2 * cosine) 

and

imag = (q2 * sine) 

come from?

And how does the transposition theorem work? (i.e. swapping input and output stages for direct form II realization)

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1 Answer 1

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They are needed in order to compute the complex-valued DFT output.

Desired realization

$ H_1(z) = \frac{1}{1-e^{-j\frac{2\pi k}{N}} z^{-1}} $

The iteration computes

$ H_2(z) = \frac{1}{1-2 \cos(\frac{2\pi k}{N})z^{-1} + z^{-2}} = \frac{1}{1-e^{-j\frac{2\pi k}{N}} z^{-1}} \frac{1}{1-e^{j\frac{2\pi k}{N}} z^{-1}} $

So that final step implements $1-e^{j\frac{2\pi k}{N}} z^{-1}$

because

$ H_1(z) = H_2(z)[1-e^{j\frac{2\pi k}{N}} z^{-1}] $

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  • $\begingroup$ Yes, But can someone derive the real and imag term? $\endgroup$
    – gari
    Commented Jun 27, 2019 at 16:58
  • $\begingroup$ @gari does referring you to Euler's formula representing the complex exponential as sum of sine and cosine help you? $\endgroup$ Commented Jun 28, 2019 at 7:21
  • $\begingroup$ That's fine. But why does the term q1 and q2 appear in real? $\endgroup$
    – gari
    Commented Jun 30, 2019 at 5:30

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