CDS101 answer:
Case 1: The zero is purely imaginary, i.e., s=0+Aj, (A>0):
Note that imaginary zeros (pure or otherwise) occur in pairs. So if
Aj is zero, then so is -Aj and the corresponding terms in the transfer function are (s-Aj) and (s+Aj). Now, to obtain the Bode Magnitude plot, we
substitute jw for s in the TF. After we do this, we notice that when w=A,
(jw-Aj)=(jA-jA)=0 (!). Therefore, the magnitude of the transfer function at
the zero is 0 (which is the lowest possible value, since |G(jw)| >=0 always).
Case 2: The zero is imaginary, i.e., s=B+Aj, with A>0 (this
automatically means that s=B-Aj is also a zero):
The corresponding terms in the transfer function are (s-(B+Aj)) and (s-(B-Aj)). In this case, when w=A, the contribution of these terms to |G(jw)| is
minimum, which results in a low response. In the example on slide 9 from Lecture6.1, the zeros of Hq2f were -0.02 +/- 0.6321j. Hence when w=A=0.6321,
we have a low response.
Case 3: The zero is real. i.e, s=B (B real):
In this case, the contribution of this zero to |G(jw)| is minimum only
when w=0. So this really doesn't give us much in terms of undertanding what the zero does in the Bode plot.
The above is only a qualitative explanation of the effect of zeros
on the Bode plot of a TF. Quantitatively, there is a corner frequency
associated with each zero (and each pole). [See here
for a quick description of corner frequency or break point.] This is relevant because at each corner
frequency, there is a change in the behavior of the Bode plot
(both magnitude and phase). If the zero is real (s=B), then the corner
frequency is simply w=B. If the zeros are a pair of imaginary numbers (B +/-
Aj), then the corner frequency is w=sqrt(A2+B2).
Some gory details:
A first order zero is represented in the Bode Magnitude Plot by an increase in
slope by +1 (20dB/dec) at the corresponding break point.
If 1/A is a zero (i.e, (jwA+1) is a term in the numerator of the TF),
then w=1/A is called a break point. Also, at the break point, the transition
in the slope is not sharp, but is a smooth connection between the lines (called
asymptotes) for w<<1/A and w>>1/A. This smoothness is got by making sure that
at w=1/A, the plot passes through a point that has a |G(jw)| value that is
approximately 3dB (factor of 1.4) greater than the value that we would have got at w=1/A had we just implemented a sharp transition.
In the Bode Phase Plot, there is a phase increase of 90deg between the plot for w<<1/A and w>>1/A. The transition is implemented smoothly by making sure that at
w=1/A, the phase is L+45deg, where L = phase in deg for w<<1/A. Further, we
also ensure that at w=1/A, the phase plot has a slope of 45deg.
Similar heuristics are available for second order zeros, poles, and terms of
the form Ksn (the latter determine the low frequency behavior).
See p. 345-351
(and particularly p. 351) in "Feedback Control of Dynamic Systems" 3/ed by
Franklin, Powell, and Emami-Naeini, for a summary of heuristics for drawing
Bode plots. The rationale for these heuristics is also explained.
Therefore, we determine all the zeros and poles, and at
each break point (corresponding to a zero or a pole), we follow the heuristics
and obtain a complete Bode plot.