How to draw the Bode diagram for a given transfer function?

Question

With this transfer function: $$G(s)=\displaystyle\frac{10(s+1)}{s(0.1s+1)}$$ I need to do operations to draw the Bode diagram manually

I have this:

$G(jw)=\displaystyle\frac{10jw+10}{-0.1w^2+jw}$

$|G|=\displaystyle\frac{\sqrt{100w^2+100}}{\sqrt{0.01w^4+w^2}}$

$|G|dB=20\log{\displaystyle\frac{\sqrt{100w^2+100}}{\sqrt{0.01w^4+w^2}}}$

$|G|dB=10\log{(100w^2+100)}-10\log{(0.01w^4+w^2)}$

Then, I have to find values for w, but in many examples I see that the expression |G|dB is a single term, like:

$|G|dB=10\log{(100w^2+100)}$

$100w^2\gg 100$

$100w\gg 10$

$w\gg0.1$

How can I calculate the value of w in this case with two terms ($|G|dB=10\log{(100w^2+100)}-10\log{(0.01w^4+w^2)}$)?

Answer 1

$|G(s)| = 10 { |s+1| \over |s| |{s \over 10} +1| } $.

Hence $\gamma(\omega) = 20 \log_{10} |G(i\omega)| = 20 (\log_{10} 10 + \log_{10} |1+ i \omega| - \log_{10} |i \omega| - \log_{10} |1+ {i \omega \over 10}|)$, or $\gamma(\omega) = 20(1 + \log_{10} \sqrt{1+ \omega^2}- \log_{10} \omega -\log_{10} \sqrt{1+ ({\omega \over 10})^2} )$.

Now draw each part separately and add them together.

The first term is constant.

The third term contributes -20 db/decade and passes through the $\gamma(10^0) = 0$ db, $\omega = 10^0$ point.

To a reasonable approximation, the second term contributes nothing for $\omega < 10^0$ and +20 db/decade for $\omega \ge 0^0$.

Similarly, the fourth term contributes nothing for $\omega < 10^1$ and -20 db/decade for $\omega \ge 10^1$.

Using Octave (excuse my rather rusty Matlab):

w = logspace(-2,2);
semilogx (w, 20*log10(abs(10*(i*w+1)./((i*w).*((i*w)./10+1)))))

gives

Answer 2

If your end goal is to just acquire the Bode plot, then typically you'd say that $G(s) = \frac{10(s+1)}{s(s+0.1)}$ has a zero at $s=-1$ and poles at $s=0$, $s=-0.1$, use this to draw magnitude and phase plots for each of the individual contributing parts separately and then combine. That's how I remember doing it in undergrad signal processing classes anyway. Here's an example with a very similar problem: http://lpsa.swarthmore.edu/Bode/Example2/index.html