This is my Bode plot enter image description here The solution manual says the function plotted is $$\frac{s^2+0.02s+1001}{s^2+2s+101)(s^2+20s+10100)}$$ What I know is the Bode plot for $$\frac{s^2+2ζω_ns+ω_n^2}{ω_n^2}$$ and the inverse of that.

I can see three peaks at ω=1 , ω=10 , ω=100.

Since the plot is going up after the peak we have a second order equation on the numerator. Then we have a second peak and then the plot is almost a straight line so I guess there is a second order equation in the denominator followed by another one afterwards because of the next peak. The phase diagram makes sense as well for the above.

The initial value is -120 so we must have a constant given by :$$ 20logK=-120=>K=10^{-6}$$ So my transfer function must be of the following form : $$10^{-6}\frac{s^2+2ζ_1 1s+1^2}{\frac{s^2+2ζ_2 10s+10^2}{10^2}\frac{s^2+2ζ_3 100s+100^2}{100^2}} $$

Can I find the ζs? This is a question from an exam so I should be able to , by hand . But I also get a quite different result.

This is what I'm using : enter image description here

  • \$\begingroup\$ Your graph clearly has a null at 1 rad/s yet the first formula's numerator (result manual?) does not yield a null at 1 but \$\sqrt{1001}\$. Can you correct your formula or your graph please. \$\endgroup\$
    – Andy aka
    Commented Jul 21, 2017 at 10:29
  • \$\begingroup\$ Also, in your final formula you are using \$\omega_n^2\$ in the complex part and not \$\omega_n\$ \$\endgroup\$
    – Andy aka
    Commented Jul 21, 2017 at 10:34
  • \$\begingroup\$ I added what I follow from my textbook . Is that incorrect ? \$\endgroup\$ Commented Jul 21, 2017 at 10:35
  • \$\begingroup\$ Is what incorrect? \$\endgroup\$
    – Andy aka
    Commented Jul 21, 2017 at 10:36
  • \$\begingroup\$ I uploaded a picture including what I followed to get the transfer function. I followed the formula precisely. \$\endgroup\$ Commented Jul 21, 2017 at 10:39

2 Answers 2



The solution given in the book is wrong because the numerator in the given solution would make the notch frequency at \$\sqrt{1001}\$ and clearly it is at about 1 radian per second.


how can I find the damping ratios ?

Pictorial analysis of the bode plot: -

enter image description here

I've drawn the red lines on to show what I consider to be flow of the frequency response should the peaks and nulls be subdued. This allows me to say that the resonant peak at 10 rad/s is about 12 dB and ditto at 100 rad/s.

Knowing that for a fairly undamped filter, Q (quality factor) is the peaking value as per this graph on this answer: -

enter image description here

You could use the more precise formula detailed lower down in that picture but I suspect assuming the peaking amplitude = Q is good enough.

So, we can say that Q is approximately 12 dB converted to a real number i.e. about 4. Because Q = 1/2\$\zeta\$, \$\zeta\$ = ~0.125.

We can also fairly well say that with the three resonances at factors of ten difference there is little interaction to muddy the waters too much.


The thing is to rewrite your original equation in a well-ordered form, following a low-entropy format, this is the key to unveil the various quality factors \$Q\$ and the resonant frequencies \$\omega_0\$. A second-order polynomial form obeys the following expression when the terms \$a_0\$ and \$b_0\$ are different than 0 and factored as a leading term:

\$H(s)=H_0\frac{1+\frac{s}{Q_N\omega_{0N}}+\left(\frac{s}{\omega_{0N}}\right)^2}{1+\frac{s}{Q_D\omega_{0D}}+\left(\frac{s}{\omega_{0D}}\right)^2}\$ in which \$H_0=\frac{a_0}{b_0}\$, \$Q_D=\frac{\sqrt{b_2}}{b_1}\$ and \$\omega_{0D}=\frac{1}{\sqrt{b_2}}\$. Substitute the terms \$a_1\$ and \$a_2\$ for the numerator quality factor and resonant frequency.

Now your expression is \$\frac{s^2+0.02s+1001}{(s^2+2s+101)(s^2+20s+10100)}\$. However, from the Bode plot, you have a dc gain of -120 dB, a double zero in the numerator which creates a notch at the resonant frequency of 1 rad/s then followed by 4 resonating poles located at 10 and 100 rad/s. Start by factoring 1001 in the numerator \$N(s)\$. You should find \$N(s)=1001(1+\frac{0.02}{1001}s+\left(\frac{s}{1001}\right)^2)\$. Now proceed with the denominator \$D(s)\$ by factoring 101 and 10100. You should get: \$D(s)=101\times10100(1+\frac{0.02}{101}s+(\frac{s}{101})^2)(1+\frac{20}{10100}s+(\frac{s}{10100})^2)\$

From these expressions, you should extract the dc gain \$H_0=\frac{1001}{101\times10100}=981\times10^{-6}\$ or -60 dB. So you already see that this value does not match the dc gain from the Bode plot. Looks like a 60-dB attenuation is missing. Determine the \$Q\$s and the resonant frequencies with the given formulas and then plot the result to see how it matches your chart:

enter image description here

Then plot the polynomial forms and the raw-expression:

enter image description here

As expected, the dc gain is wrong, the notch is misplaced but the poles seem to be ok (the \$x\$-axis in in Hz and not in rad/s). So it seems, at first look, that the problem lies in the numerator while the denominator looks ok. Considering the missing 60-dB attenuation (a ratio of 1000), I have updated the dc gain to \$H_0=\frac{1.001}{101\times10100}=9.8\times10^{-7}\$ which is -120 dB. I can now divide all terms in the numerator by 1001 and plot the newly-resulting Bode plot:

enter image description here

The horizontal scale is now in rad/s and the final result does not look bad : )

enter image description here

So the correct high-entropy expression is given below, please note the correct dimensions of the various coefficients to keep a unitless transfer function:

enter image description here

You have the values of the various \$Q\$s and resonant frequencies in the Mathcad shots. The damping ratio \$\zeta\$ and \$Q\$ are linked by \$\zeta=\frac{1}{2Q}\$. You will learn more about low-entropy expressions and Fast Analytical Circuits Techniques (FACTs) here.


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