I am really struggling to see how the lecturer took this transfer function and produced the bode plot that I have in my notebook. It is not the plotting so much that is confusing, I just don't understand how he manipulated the transfer function into this form that he did and why it needed to be in that form in the first place when it looked solvable in its original form.

$\textrm{(b)}\quad T(s) = 180\displaystyle\frac{s(1+0.01s)}{(1+0.05s)(1+0.001s)}$

Looking at the transfer function shown I would have thought that all one would simply need to do is evaluate the gain at $20\log(180)$ and then plot the bode plot using the two zeros and two poles.

Supposedly to plot the asymptotic bode plot for this function correctly one needs to get it into the following form:

$$3600\times\frac{\left(\frac s{20}\right)\left(1+\frac s{100}\right)}{\left(1+\frac s{20}\right)\left(1+\frac s{1000}\right)}$$

Sorry for the bad format but basically the answer provided has it rising to its pole at $20$ on the $71\textrm{ dB}$ line and going flat until it hits $100 \textrm{ rad/s}$ and then rising up again at $20\textrm{ dB}$ per decade until it meets its pole at $1000$. I don't understand what is going on and everything I have written hitherto is just my observation of how the lecturer demonstrated.

If anyone with any knowledge on the mechanics of plotting bode plots by hand could shed some light on this for me I'd greatly appreciate it. Simon.


1 Answer 1


Quick Answer cause I'm at work and don't have time to do the plotting but basically, in equation b you already have the expression i what i like to call "The Bode Form", to only a gain and terms of the form $(1+s/A)$ with $A$ some value. I'm sure the professor explained this, that when you take logarithm you add and subtract all this different terms because the log turns multiplication into addition and division into subtraction, etc. This individual terms, When $s < A$ are approximately equal to 1 (a constant line at 0db Gain in the plot), and when$ s > A$ are approximately equal to $s/A$ ( a 20db/decade drop in the gain).

So the plot would be an increasing line of 20db/decade from up to your first pole at 20 rad/s, where the gain would be $20\log_{10}(3600)$, and then it would remain at that until 100 rad/s where the next pole appears which again makes the gain drop at 20db/decade, and then at the next pole at 1000 rad/s it drops at 40db/decade and that's it, that's your plot.

Could you maybe put up the image you have in your notebook so everyone could see what the professor did.

Attempt at illustrating below. This video may be instructive.

enter image description here

  • $\begingroup$ Feel free to delete my addition. Let me know if it's OK, or if you want changes. $\endgroup$
    – Peter K.
    Commented Apr 1, 2016 at 12:40
  • 1
    $\begingroup$ Oops, yeah it was in rads/sec not hertz. Yeah the plot looks good. Let's wait and see if the OP had the same in his notebook. $\endgroup$
    – bone
    Commented Apr 1, 2016 at 13:23
  • $\begingroup$ It doesn't match exactley but I think what is done here is correct. What I am not getting though is why when I evaluate it without multiplying the numerator by (20/20) to put it in that form it comes out completely different. Why is it necessary to make the gain 3600 and the zero in the numerator over 20? Thats the crux of my confusion and why I originally posted. $\endgroup$
    – burton01
    Commented Apr 2, 2016 at 7:07

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