# Transfer function of resonant filter with 2 poles, peak at $f_0 = 500\text{ Hz}$, and $\Delta f = 32\text{ Hz}$

This a contest question. I'd like some help because I can't find any materials related to this topic.

https://www.qconcursos.com/questoes-de-concursos/questao/ecd6c966-51

My english translation:

Find the transfer function of a resonant filter with 2 poles, peak on $f_0 = 500\text{ Hz}$, $\Delta f = 32\text{ Hz}$, and sampling frequency $f_s = 10 \text{ kHz}$.

-> a) $H(z) = \frac{0.062}{1-1.8831z^{-1}+0.09801z^{-2}}$

b) $H(z) = \frac{0.081}{3-2.641z^{-3}+0.09801z^{-2}}$

c) $H(z) = \frac{0.082}{1-1.8831z^{-1}+0.09801z^{-2}}$

d) $H(z) = \frac{0.762}{1-2.8831z^{-1}+0.09801z^{-2}}$

e) $H(z) = \frac{0.262}{2-1.8831z^{-1}+0.09801z^{-2}}$

Any hints?

• Am I correct that "contest" here means a qualifying exam for a university? Also, does the arrow besides answer a) mean that is the correct answer? – MBaz Sep 10 '18 at 15:01
• Also please indicate if $\Delta f = f_2 - f_1$ where $f_1$, and $f_2$ are $-3$ dB cutoff frequencies of a second order filter. – Fat32 Sep 10 '18 at 15:08
• Yes. @MBaz this is a qualify to get a federal employee. That contest has been finished and this website collect some question about the several contest. And the answer is a) – miguel747 Sep 10 '18 at 15:12
• @Fat32 tbh i didnt know that $\Delta f$ means the different between $f_2$ (cutoff frequency) and $f_1$. Who is $f_1$ in this case? I saw $\Delta f$ only when I studied angular modulation (FM/PM). – miguel747 Sep 10 '18 at 15:17
• in analog filter design (of second order) the peak frequency $f_0$ is called as the resonant frequency, whereas the cutoff frequencies f1,f2 are the frequencies at which the frequency response falls to $1/\sqrt{2}$ of their peak value $H(f_0)$; i.e., $$H(f_{c1}) = H(f_{c2}) = H(f_0) / \sqrt{2}$$ Those frequencies correspond to -3dB points then. And their difference is the -3dB bandwidth of the filter as given by $\Delta f = |f_2 - f_1|$. – Fat32 Sep 10 '18 at 15:35

You mistyped a few crucial numbers (the coefficients of $z^{-2}$). With the correct coefficients as shown in the link in your question, you quickly see that choice $b)$ can be discarded because it has a third order polynomial in the denominator.

For a peaking resonant filter we need two complex conjugate poles, so we can discard choices $c)$ and $d)$ because they have real-valued poles (due to the negative coefficient of $z^{-2}$). So we're left with $a)$ and $e)$, for which we need to compute the angles of the poles, because this is (approximately) where the frequency response peak will occur. Having obtained those angles we just need to check which one is closer to the angle corresponding to a frequency of $f_0=500\textrm{ Hz}$ (with $f_s=10\textrm{ kHz}$).

So the angle we're looking for is

$$\theta=2\pi\frac{500}{10000}=\frac{\pi}{10}$$

and it turns out that the poles of transfer function $a)$ have just that angle.

• Thank you @Matt L. I wish you would send any materials about this topic? I've never computed the angle when occur the peak responser as $\theta = 2\pi\frac{f_0}{f_s}$ before. it was a new to me. And what about the $\Delta f$? dont we need use it? – miguel747 Sep 10 '18 at 15:36
• @miguel747: Try to learn about the relation between poles and zeros and the frequency response. You'll find a lot about this on this site. Poles close to the unit circle will cause the frequency response to peak, and the angle of the poles determine the location of the peak with respect to frequency. I guess you should read some basic DSP text; it's hard to explain all that stuff in a single answers. Browse this site for book and website recommendations. There quite a few good (and free) books, e.g., one by Orfanidis. – Matt L. Sep 10 '18 at 16:05
• @miguel747: And yes, we didn't need $\Delta f$ in this example, because there was only 1 filter with the correct peak frequency (assuming that at least one of the answers must be the correct one). – Matt L. Sep 10 '18 at 16:06