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As part of an assignment, I was given a filter with a specific set of poles and zeros, namely poles at $0.51 \pm 0.68i$ and zeroes at $1.5$ and $−0.7$. This part of the assignment is titled as "Design of band-pass filters" so it must be that this filter is bandpass.

What troubles me is that this is the filter response:enter image description here

Since we are talking about a band-pass filter, I'd expect the plot of the Amplitude Response to cross -3dB twice and as you can clearly see, that's not the case. One other explanation that kind off makes sense to me is that it's considered bandpass in the sense that only the frequencies from about 0.5 to 1.5 are highly amplified compared to all the others which are either less amplified or attenuated.

So, is it actually a bandpass filter? If so, why?

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    $\begingroup$ It is, for the reason you said it is! $\endgroup$
    – Jdip
    Jan 7 at 18:06
  • $\begingroup$ Oh great then! Makes sense that way. Thanks! $\endgroup$
    – Nyquist-er
    Jan 7 at 18:11
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    $\begingroup$ @AhsanYousaf I have a question regarding that, consider that the zeros remain at the same place but now the poles become $-0.68 \pm 0.51i$, that is we went from the poles $a \pm bi$ to $-b \pm ai$. Using your formula we again get 0.85, but plotting the amplitude response of the new filter, the peak is around 2.4 radians/sample and not at 0.85 as you previously correctly pointed out. Why does the formula not work this time? $\endgroup$
    – Nyquist-er
    Jan 7 at 18:23
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    $\begingroup$ @Nyquist-er Regarding the -3dB points: These may be relative to the largest amplitude of the response; that is, points 3dB below the peak. $\endgroup$
    – MBaz
    Jan 7 at 18:32
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    $\begingroup$ You shouldn't have $\omega^2$ and $j\omega$ in the numerator and denominator. Try to evaluate your $H(\omega)$ and plot its magnitude. You'll see that it's not correct. However, your plots are okay because the coefficients are correct. $\endgroup$
    – Matt L.
    Jan 7 at 20:08

2 Answers 2

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I'd expect the plot of the Amplitude Response to cross -3dB twice

When you're designing filters, you're more concerned with the relative response than with absolute -- so if the peak gain of a filter isn't unity, then you take your 3dB points relative to the actual peak.

Eyeballing that filter response, it's 3dB down from the peak twice -- at 0.75 radians/sample, and at 1.2 radians/sample.

Having said all that, I want to see a "bandpass" filter have zero response (not 0dB, but zero in absolute terms) at the frequency extremes in continuous-time, and at least zero response at DC. It's pretty easy to make a filter with a null at DC -- that filter's response isn't even 20dB down from the peak at DC, and it's attenuation at high frequency isn't stellar, either. I'd call that a "sorta bandpass", or a "crappy bandpass" -- even if I had designed it and were recommending that it be used for some reason.

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For poles at $0.51 \pm 0.68i$ and zeroes at $1.5$ and $-0.7$ the transfer function $H(z)$ is:

$$H(z) = \frac{z^2 - 0.8z - 1.05}{z^2 - 1.02z + 0.7225}$$

and $H(e^{j\omega})$ is:

$$ H(e^{j\omega}) = H(z)\lvert_{z=e^{j\omega}} = \frac{e^{2j\omega} - 0.8e^{j\omega} - 1.05}{e^{2j\omega} - 1.02e^{j\omega} + 0.7225} $$

which looks like the following: BPF

and is quite clearly a bandpass filter. This can also be seen from the pole-zero plot below:

zplane

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    $\begingroup$ -1 : this answers basically just shows the same Bode graphs that the one in the question (but in a linear scale), and to the main question « why is it a bandpass filter », answers « it's quite clear ». $\endgroup$
    – Blackhole
    Jan 8 at 11:36

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