# Pole zero plot

1. Jan 7, 2013

### Steve Collins

I am attempting the question shown in the attachment.

It can be seen that the poles are located at -2 ± 3j which expressed in terms of s is (s + 2)2 + 32.

This is the denominator, but how is the numerator of the transfer function found?

Edit:

Looking at the Laplace look-up table I would want the numerator to be 3 giving:

3/((s + 2)2 + 32) so that i could use e-atcosωt to perform the reverse Laplace transform in part b.

Is this correct?

#### Attached Files:

• ###### pole zero config.pdf
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Last edited: Jan 7, 2013
2. Jan 8, 2013

### rude man

There are no zeros in your plot, ergo there is no numerator other than a constant. The constant cannot be determined from the plot (unless it's contained in those funny numbers within the white part of the plot. I have never seen a plot like that before.) You can assume it's 3 but any other real constant is OK also. That should be obvious since L-1{cF(s)} → cf(t), c a constant.

My table says L-1{1/[(s+a)2 + b2]} → (1/b)e-atsin(bt).

3. Jan 8, 2013

### Steve Collins

Yes you are correct, I have misread the table and copied the entry from the line above.

4. Jan 8, 2013

### rude man

BTW you could also have done the inversion by partial fraction expansion, if you're comfy with manipulating complex numbers just a reminder probably ...