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enter image description here

If the zeros were on the imaginary axis, it sure would have been a NOTCH filter.

But, since the complex conjugate zeros are on the left of the jW axis, the transfer function has second order terms in both numerator and denominator.

I tried the Bode plot of the transfer function and ended up getting a high pass filter by assuming zeros at z1,z2= -1+i , -1 -i and poles at p1= -3 and p2= -5.

But does the asymptotic approximations end up with correct results?

How can i identify the type of filter if the transfer function is not in any of the standard forms ,for example , in case of a notch filter :

enter image description here

edit: i agree with the fact that there are many other filters apart from the 5 basic ones,but is there any way of predicting the behavior (approximating) given any pole-zero plot like the one above.

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    \$\begingroup\$ There are infinitely more filters that are undefined by a name compared to those that are named. \$\endgroup\$
    – Andy aka
    Dec 16, 2015 at 13:31
  • \$\begingroup\$ i agree, but how to predict their behaviour based on the given pole zero plot ? \$\endgroup\$
    – Ashik Anuvar
    Dec 16, 2015 at 13:33
  • \$\begingroup\$ From the pole-zero plot you know where the poles and zeros are. Then you can write down H(s) as zeros go in the nominator and poles in the denominator. Then you know H(s) and plot it over s. Simple :-) Have you seen: web.mit.edu/2.14/www/Handouts/PoleZero.pdf ? \$\endgroup\$
    – Bimpelrekkie
    Dec 16, 2015 at 13:40
  • \$\begingroup\$ yes, i did the same , used bode plot by hand and ended up with high pass ( under asymptotic approx),but i doubt that and want some intuitive explanation on this kinda problem \$\endgroup\$
    – Ashik Anuvar
    Dec 16, 2015 at 13:43
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    \$\begingroup\$ "...but i doubt that..." How can you doubt the hard mathematical approach ? Mathematics don't lie when applied correctly. pole/zero diagram, H(s), bode plot. They are essentially the same information presented in a different way. How can there be doubt ? Perhaps it is you not understanding ? Then what do you not understand ? \$\endgroup\$
    – Bimpelrekkie
    Dec 16, 2015 at 14:58

2 Answers 2

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Certainly, it is not one of the classical response types - but a mixture. To desribe the response in words:

In your pn-diagram, the two real poles have larger pole frequecies than the zero frequency of the pair of zeros. From this it can be concluded that the frequency response has, in principle, a highpass-notch behaviour. However, because the zeros have a small real part the notch depth is finite.

The corresponding transfer function contains a second-order polynominal in both, numerator and denominator. The pole Q is very low (Q<0.5) and the zero-Q is rather high.

enter image description here

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  • \$\begingroup\$ which software did u use to simulate this? any specific ones for beginners? \$\endgroup\$
    – Ashik Anuvar
    Dec 16, 2015 at 14:59
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    \$\begingroup\$ The above simulation was produced by a small Java applet called "Sysquake" (google for sysquake). A pair of poles and zeros can be arbritarily placed in the s-domain (using the mouse) - and the magnitude respone as well as the step response is displayed. \$\endgroup\$
    – LvW
    Dec 16, 2015 at 19:58
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The gain at any frequency (i.e. any point on the positive \$j\omega\$ axis), \$\small G(\omega)\$, can be found by drawing vectors from each pole and zero to that point, and then calculating $$\small G(\omega)=\frac{product\:of\:zero\:vector\:lengths}{product\:of\:pole\:vector\:lengths}$$ For example, if there is a zero very close to the jω axis, the relevant vector length will be a minimum as the frequency locus passes that point, and the overall gain will be correspondingly low.

Similarly a pole close to the \$j\omega\$ axis will give a high gain at the frequency of nearest approach.

This is a good intuitive method of estimating amplitude frequency response.

It also works well in the z-domain, where proximity to the unit circle may be interpreted similarly.

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