I've got the following bode plot from a black box: enter image description here

I've calculated the zero to be 1055 and pole to be 67. Thus I'm using the transfer function \$H(s) = (s-1055)/(s-67)\$ but this gives a clearly wrong bode plot:

Bode plot from WolframAlpha

WolframAlpha calculation

I'm guessing my transfer function is wrong. Can anyone see how?


3 Answers 3


The plot from Alpha looks pretty close to your "black box" measurement to me.

The only difference is a pre-factor.

Try \$\frac{67}{1055}\frac{s-1055}{s-67}\$:

enter image description here


First, note that that the transfer function you give has the pole (and zero too) in the right hand plane, i.e., the system it describes is unstable. I suspect you meant:

\$H(s) = \dfrac{s + 1055}{s+67}\$

However, even this is not in standard form. Putting this transfer function into standard form yields:

\$ H(s) = \dfrac{s + 1055}{s + 67} = \dfrac{1055}{67} \dfrac{\frac{s}{1055} + 1}{\frac{s}{67} + 1}\$

So, now you see where the undesired gain has come from. Knowing (only) the pole and zero, you should guess instead:

\$H(s) =\dfrac{\frac{s}{1055} + 1}{\frac{s}{67} + 1}\$

enter image description here


Actually, you have half of a Bode plot from a Black Box. A Bode plot shows magnitude and phase. You cannot do the fit you're trying to do without employing phase information. More accurately, you can, but it might be difficult to interpret your results. If you had your phase, you'd have more confidence in your fit.

  • \$\begingroup\$ I was moreso worried in the gain of 24dB that my transfer function had. The phase information wasn't important and I shouldn't have included it. \$\endgroup\$
    – Sticky
    Oct 31, 2012 at 9:18

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