I want to draw the bode plot of a tranfer function:

$$ H(j\omega)=\frac{100j\omega T}{j\omega T + 10}, T=1s $$

Now I have

$$ H(j\omega)=\frac{100}{1 + \frac{10}{j\omega T}}, T=1s $$

Using a double log scale:

$$ 20*\log{H(j\omega)}=20*\log{\frac{100}{1 + \frac{10}{j\omega T}}}, T=1s $$

And can I just insert omega and compute the points for the plot?

Like for omega = 1000

$$ 20*\log{\left(\frac{100}{1 + \frac{10}{2*\pi*1000}}\right)}=40-20*\log{\left(1 + \frac{10}{2*\pi*1000}\right)}=39.9... $$

Is that correct?

  • \$\begingroup\$ Alfred is absolutely right about needing to split the complex-valued H(j*omega) into magnitude and phase before you apply the double-log scales (for magnitude Bode plot). If you're just looking for a tool that can take an arbitrary Laplace transform and give you the Bode plot, take a look at this example. \$\endgroup\$
    – compumike
    Jun 28, 2012 at 1:20
  • \$\begingroup\$ I'm trying to understand with this example how this generally works, beside any tools that I could use. \$\endgroup\$
    – Sam
    Jun 28, 2012 at 1:46
  • \$\begingroup\$ Is your question about showing why plotting the magnitude on log-log axes make the magnitude plot have approximately straight-line segments in two different regions? \$\endgroup\$
    – compumike
    Jun 28, 2012 at 2:52
  • \$\begingroup\$ The question is more general but I thought this example could show - with all steps - how to draw a bode plot so everybody can understand and draw this. I can't see anything obviously easy and good to understand on wikipedia. It's just to mathematical for beginners. \$\endgroup\$
    – Sam
    Jun 28, 2012 at 2:59

1 Answer 1


You've got to back up a step or three. The transfer function is complex valued so, to plot it, you need two plots, usually magnitude and phase. The magnitude plot is usually log-log but the phase plot is lin-log.

So, you need to find the magnitude and then take the log before plotting the Bode magnitude. To find the magnitude, multiply H by its conjugate and then take the root.

$$|H(j\omega)|^2 = \dfrac{100}{1 + \dfrac{10}{j \omega T}}\dfrac{100}{1 - \dfrac{10}{j \omega T}} = \dfrac{100^2}{1 + \dfrac{10^2}{(\omega T)^2}}$$

Also, omega is the radian frequency while f is the frequency. So, if omega = 1000, you don't multiply by 2 pi. However if f = 1000, you do.

UPDATE: fixed denominator of transfer function to match OP's original


We should try to put this transfer function in standard from so that can identify the asymptotic gain, the type, and the pole/zero frequency. Since the variable \$\omega\$ appears with highest exponent 1, it is a 1st order filter. There are only two types of 1st order filters: low-pass and high-pass. In standard form the OP's transfer function is:

\$H(j\omega) = 100 \dfrac{\frac{j\omega}{\omega_0}}{1 + \frac{j\omega}{\omega_0}}\$

\$ \omega_0 = \frac{10}{T}\$


\$ |H(j\omega)| = 100 \dfrac{\frac{\omega}{\omega_0}}{\sqrt{1 + (\frac{\omega}{\omega_0})^2}}\$

Now, if we stare at this a bit and ask it some questions, we can imagine exactly what this looks like.

When \$ \omega << \omega_0\$, the denominator is effectively "1" and so, the transfer function is decreasing by a factor of 10 as \$ \omega\$ decreases by a factor of 10. On a log-log scale, this is a line with a slope of +1.

When \$ \omega >> \omega_0\$, the denominator is effectively \$ \frac{\omega}{\omega_0} \$ and so the transfer function is effectively constant with a value of 100.

If we plot these two lines on a log-log plot and have them intersect at \$ \omega = \omega_0\$, we've created an asymptotic Bode magnitude plot. In fact, it's easy to see that when \$ \omega = \omega_0\$, the magnitude is \$ \frac{100}{\sqrt{2}}\$ so the lines we plotted are actually the asymptotes of the (magnitude) transfer function. That is, the function approaches these lines at the frequency extremes but never actually gets to them (on a log-log plot, \$ \omega = 0 \$ is at negative infinity)

  • \$\begingroup\$ I'm starting to understand: $$ \frac{H}{dB}(w)=20*\log\left(\frac{100}{\sqrt{1+\frac{1}{{(\omega T)}^2}}}\right) $$ now I need to form this as far I understand: $$ \frac{H}{dB}(w)=20*\log\left(\frac{100}{\sqrt{1+\frac{1}{{\left(\frac{\omega}{\omega_e}T\right)}^2}}}\right) $$ The question is, what is the next step? \$\endgroup\$
    – Sam
    Jun 28, 2012 at 1:30
  • \$\begingroup\$ Split the fraction, so you end up with 20log100 - 20log(denominator), and then you should be able to figure the rest out. Notice that the sqrt turns into a (1/2) in front of the second 20log(f(w)) expression. \$\endgroup\$
    – Shamtam
    Jun 28, 2012 at 2:07
  • \$\begingroup\$ What does omega_e represent? By the way, while I was eating supper, it occurred to me that the "T" in your equation is a period. If so, then you shouldn't be using omega but, rather, f. Typically, we write the transfer functions where the variable omega is divided by a constant omega_p or omega_z (for pole and zero respectively). Anyhow, the final equation in your comment above has a unit off. If you're dividing the variable omega by some fixed omega_e, the units cancel so you can't have that "T" in there. Also, this transfer function is for a 1st order, high pass filter. \$\endgroup\$ Jun 28, 2012 at 2:22
  • \$\begingroup\$ Good point @Shamtam \$\endgroup\$
    – Sam
    Jun 28, 2012 at 2:23
  • \$\begingroup\$ I saw the omega_e in somewhere in an example, I thought there must be used a ratio, but it didn't help me to draw the plot. \$\endgroup\$
    – Sam
    Jun 28, 2012 at 2:43

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