## Goertzel algorithm

Started by 7 years ago4 replieslatest reply 7 years ago425 views

Hi,

For Goertzel algorithm, does it require N+1 or N samples to generate the bin corresponding to N-point DFT?

In Rick's book I see the following statement "For the Goertzel algorithm we retain only every Nth, or (N+1)th,". I am not sure why it "N+1 or N". Could someone please clarify?

Thanks,

Krishna

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Hi,

The Goertzel difference equations are

w(n)=2cos(2\pi m/N) w(n-1) - w(n-2) + x(n)

y(n)=w(n) - e^(-j2 \pi m/N) * w(n-1)

To find X(m), we need to calculate y(N), so we should find w(N) and w(N-1). Then the first equation should be evaluated N+1 times (assuming that the samples of x(n) start from index n=0 to n=N-1). However, if we are looking for the abs(X(m))^2, we can obtain the following equation from y(n) given above

abs(X(m))^2 = w(N-1)^2 + w(N-2)^2 - w(N-1)*w(N-2)*2*cos(2* \pi *m/N)

this is equation (13-83) of the book. Now, using this equation, we need to evaluate the w(n) equation by N times.

Figure 13-44 shows an example with N=64, for X(15) and abs(X(15)) the first stage should be evaluated by 65 and 64 times, respectively.

Please verify and correct me if I am making any mistake.

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Thanks Tavassoli.

I am just wondering if DFT/FFT and Goertzel evaluates the same bin, why should Goertzel depend on additional sample?

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I couldn't tell you exactly how without wrapping myself in math for a while, but if you mind your p's and q's you should only need $$N$$ points.  Get the math right and the Goertzel has the exact same effect as doing a one-bin FFT (you can even extract both inphase and quadrature parts -- if you get the math right).