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How can I quicly determine if a given filter's transfer function like: \$ H(s)=\frac{k}{s^2+ks} \$, or \$H(s)=\frac{1}{s+k}\$, is either a low-pass, high-pass, or band-pass ?

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If you plot the function \$|H(j \omega)|\$ over \$\omega\in[0,+\infty]\$ (\$j\$ being the imaginary unit), you obtain what is called "Bode plot" (specifically the magnitude part).

Once you have the plot, it will be easy to discern what kind of filter you have on your hands, since the plot will show a gain \$>1\$ (i.e. \$0dB\$) in the frequency region where the signal can pass:

  • a low [frequency]-pass filter will be \$>1\$ in the low frequency region, the left side of the plot

  • a high [frequency]-pass filter will be \$>1\$ in the high frequency region, the right side of the plot

  • a band-pass filter will be \$>1\$ in the central part, delimiting a band of frequencies allowed to pass.

It is important to remember that the "pass" definition is a simplification: the plot you just created tells you how damped (\$<1\$) or amplified (\$>1\$) a signal having a specified frequency is when the filter acts on it. As the plot will never be exactly zero (made exception for certain specific and limited scenarios), all signals will actually pass through the filter, only they will be damped enough not to be detectable or relevant.

The "damped enough" threshold is the \$-3dB\$ (i.e. a gain of \$0.7\$) line mentioned in the comments to the other answers.

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Yes. Evaluate the function as s approaches zero and as s approaches infinity. That will give you a very quick look at low and high pass filters. Band pass can be a little trickier, and may require some factoring first to get it to a form that makes sense to apply the aforementioned process.

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  • \$\begingroup\$ Thanks! One more question: Suppose if I end (after using L'Hopital) with a constant. i.e. not approaching infinity/zero. Does that mean it it is a band-pass filter? \$\endgroup\$ – JBee May 26 '16 at 18:11
  • \$\begingroup\$ @JBee You might be able to show that works for some cases, but I do not know of an "official" theorum that supports it. If the quick analysis of s=0 or s=inf doesn't work, you can always look at where the poles and zeros fall. \$\endgroup\$ – Brendan Simpson May 26 '16 at 18:19
  • \$\begingroup\$ @JBee: Filters are supposed to be stable; you expect a constant. The main question is whether it's a non-zero constant. \$\endgroup\$ – MSalters May 27 '16 at 8:56
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Remember that s represents frequency and the overall equation gain. Think of what happens when s is very low or even 0, and then what happens when s approaches infinity.

In your second example, at s=0 you get 1/k, and at s=∞ you get 0. This is therefore a low-pass filter. The rolloff point of the filter is when s=k.

The first example is the same thing with another s in the denominator. You still get 0 for s=∞, but the equation blows up when s=0. This is because the 1/s added from the second example represents a integrator.

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  • \$\begingroup\$ you mean s = -k? \$\endgroup\$ – njzk2 May 26 '16 at 20:37
  • \$\begingroup\$ The pole is at $$s=-k$$. The -3dB point is at $$\omega = \pm k$$, which means that $$s = j\omega = \pm k\sqrt{-1}$$. Which is as different from $$s=k$$ as it is from $$s=-k$$ \$\endgroup\$ – Ben Voigt May 27 '16 at 2:05

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