0
\$\begingroup\$

How do I draw the Bode plot for:

$$ H(s) = \frac{100000(s + 10) ^3 }{s(s + 100)(s + 1000j)(s − 1000j)} $$

I searched tutorials for the Bode plot, but I can't find a solved or raw example where \$(s-j)(s+j)\$ appears. Do I just consider the imaginary \$j^2\$ as \$-1\$ and go on?

No one would explain to us how to do it, but we always get it wrong when \$j\$ appears separate and not as \$s=j\omega\$.

\$\endgroup\$
6
  • \$\begingroup\$ I really don't know what you are asking. I tend to imagine it's because you've not written enough. But it may be a language thing. Can you confirm that you understand that \$\left(s+j\,1000\right)\cdot\left(s-j\,1000\right)=s^2+1\times 10^6\$? (Note, no 'j' implied other than \$s=\sigma+j\,\omega\$.) Yes, \$j^2=-1\$. \$\endgroup\$
    – jonk
    Sep 21, 2019 at 5:51
  • \$\begingroup\$ You can plot it here \$\endgroup\$
    – G36
    Sep 21, 2019 at 6:07
  • \$\begingroup\$ \$(s+j)(s-j)\$ give a 2nd order term, \$(s^2 +1)\$, with \$\zeta =0\$, so the bode plot amplitude goes to infinity at \$\omega _n = 1\$ rad/sec, and the HF drop-off is -40 dB/dec. \$\endgroup\$
    – Chu
    Sep 21, 2019 at 6:50
  • \$\begingroup\$ @jonk yes, i understand no other j is implied. Now if i have s^2+1×10^6. How to i draw this component? And also how do i draw (s+10)^3 component ? \$\endgroup\$ Sep 21, 2019 at 10:33
  • \$\begingroup\$ @Chu Ok so for my case given the bode plot amplitude goes to infinity at 10^3 rad/sec? How does this affect my final result when i try to add all the components of this H(s) when I hit this component that goes to infinity and then drops-off? i just ignore the spike in that point and look only at -40db/dec? \$\endgroup\$ Sep 21, 2019 at 11:47

2 Answers 2

1
\$\begingroup\$

Substitute \$s = j\omega\$, and plot the value for \$|H(j\omega)|\$. For the \$ j\$ question, \$ j^2 = -1\$.

Here you can find a good lesson and some quick resource on how to plot the bode plot of signals.

So, we start with, $$ H(s) = \frac{100000(s + 10) ^3 }{s(s + 100)(s + 1000j)(s − 1000j)}, $$ and replace \$ s = j\omega\$, resulting in, $$ H(j\omega) = \frac{100000(j\omega + 10) ^3 }{j\omega(j\omega + 100)(j\omega + 1000j)(j\omega − 1000j)}, $$

that can be further simplified to get $$ H(j\omega) = \frac{100000 j(j\omega + 10) ^3 }{\omega(j\omega + 100)(\omega + 1000)(\omega − 1000)}. $$

The magnitude of the signal is: $$ \left| H(j\omega)\right| = \frac{100000 (\omega^2+100)\sqrt{\omega^2+100}}{\omega \left(\omega + 1000\right) \sqrt{\omega^{2} + 10000} \left|{\omega - 1000}\right|}$$

Now, you either put that in to some software or plot it manually, I used octave to get this get this:

f = @(w) (100000*(w.^2+100).*sqrt(w.^2+100))./(w.*(w+1000).*(w.^2+10000).*abs(w-1000))
t = 10.^(0.001:0.001:5);
loglog(t,f(t))

Resulting in, Bode plot of the transfer function H(j omega)

Which resembles a lot the approximation using the asymptotes, the greatest differences are near the zeros and poles, where it differs the most. Also, this plot does not reproduce completely the peak at \$ \omega = 1000 \text{ rad/s}\$. But as you change the resolution of the plot (the values used for \$ t\$) it becomes higher and higher (infinity gain).

\$\endgroup\$
4
  • \$\begingroup\$ Ok i figured it out that at 10^3 rad/sec my component goes to infinity and then drops to -40 db/dec but when i try to draw and add the components togheter what happens when i'm in that point where it goes to infinity and how does it influence my result? i just ignore it and look at how much i drops off (-40db/dec) and that is how it affects my diagram or ?...... I'm confused because we worked only with finite values for components \$\endgroup\$ Sep 21, 2019 at 11:50
  • \$\begingroup\$ Draw the asymptotes first, that will be an approximation. Then you should calculate the values around the poles and zero. For the poles with \$ (s + \sigma - j\omega)\$ where \$ \sigma > 0\$ the approximation will be more or less ok, but for the \$ \sigma = 0\$ the approximation will not be ok. \$\endgroup\$
    – jDAQ
    Sep 21, 2019 at 18:23
  • \$\begingroup\$ Well if i try to calculate Zeta from s^2 + 2 *Zeta * Wn * s + Wn^2 i have no zeta wich means it will not be ok since it is 0..... should i give values by myself then? \$\endgroup\$ Sep 22, 2019 at 15:36
  • \$\begingroup\$ I never mentioned Zeta. \$ Z , \zeta \$. This system is not a second order one, I have no idea why you would want to find Zeta, and of course, do not just "give values by [yourself]" \$\endgroup\$
    – jDAQ
    Sep 22, 2019 at 18:42
0
\$\begingroup\$

Let's just say you didn't have a computational package, you first find the magnitude of the transfer function. Then plug in different frequencies that would be the x scale the y scale would be the output of the magnitude function.

Similarly if you want to find the phase find an expression for the phase of the transfer function and then plug in different frequencies and the y result is the phase

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.