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.