I am designing a 2nd order IIR digital filter :

My tf equation with coefficients is

b = [0 1.209e09]
a= [9.2175   -2.6952    1.0000]

I have couple of questions :

  1. How to get unity scaling .. as a formula... so that i can always multiply it with the 'sys' tf.
  2. I usually seen that coefficients are less than 1 , is there a way i can make them so ?
  3. My coefficients are calculated from some adaptive filter optimization algorithm, so they keep on changing, is there a way i can reduce the filter to filter coefficients changes to a a minimum.?

It's important to specify at which frequency you want unity gain. But assuming you mean DC ($\omega=0$), because that filter has a low pass characteristic, the DC gain of an IIR filter is given by


It's also common to normalize the denominator coefficients such that $a[0]=1$. In your example that would give

a = [1.00000 -0.29240 0.10849]


b = [0.00000 131163547.59967]

Finally, normalizing by the DC gain $(1)$ will give you a filter with unity gain at DC. This leaves the denominator coefficients unchanged, and the new normalized numerator coefficients are given by

b = [0.00000 0.81609]

  • $\begingroup$ Thankyou so much :) $\endgroup$ – BandW Jun 23 '20 at 9:01
b = [0 1.209e09]
a= [9.2175   -2.6952    1.0000]

% original -----------------------
sys = tf(b,a,0.1,'Variable','z^-1');

% fixes --------------------------

% scale coefficients by b(2) 

bode(sys,'-', sys1,'--')

Which results:

Transfer function 'sys' from input 'u1' to output ...

           1.209e+09 z^-2
 y1:  -------------------------
      9.217 - 2.695 z^-1 + z^-2

Transfer function 'sys1' from input 'u1' to output ...

 y1:  -------------------------------------------
      7.624e-09 - 2.229e-09 z^-1 + 8.271e-10 z^-2
  • $\begingroup$ can you please write the code how you did it... how did you get y1 ? also now we very small coefcients.. is that correct ?.. also pole (z-2) is not 1 anymore $\endgroup$ – BandW Jun 22 '20 at 15:00
  • $\begingroup$ Updated the code. Hope it helps.... . $\endgroup$ – Juha P Jun 22 '20 at 17:05
  • $\begingroup$ Thankyou so much :) $\endgroup$ – BandW Jun 23 '20 at 9:01

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