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I am having trouble where to start an attempt at sketching the Bode plot for the following transfer function: Transfer Function in Question I know I need to rewrite it into its proper form, making both the lowest-order term in the numerator and denominator unity... but the denominator is what's throwing me off. If I keep all of the terms separate in the denominator, how can I find the poles of the transfer function?

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You need to calculate the roots of numerator and denominator.

For the numerator you have \$(s^2+s+1)=0 \Leftrightarrow s_{n_{1,2}}=-\frac{1}{2} \pm j\sqrt{\frac{3}{4}}\$

For the denominator you have \$(s+1) =0 \Leftrightarrow s_{d_{1}}=-1\$ as well as $$(2 s^2+1) =0 \\ \Leftrightarrow s^2=- {\frac{1}{2}}\\ \Leftrightarrow s_{d_{2,3}}=\pm j\sqrt{\frac{1}{2}} $$ because a product is zero if one or more of its factors are zero.

Therefore your transfer function is $$ W(s)=\frac{(s-s_{n_{1}}) \cdot (s-s_{n_{2}})}{(s-s_{d_{1}})\cdot(s-s_{d_{2}})(s-s_{d_{3}})} $$

Here is a picture of what the bode plot should look like (sorry for just using Matlab instead of drawing by hand): Bode plot Detail of Bode plot Please note that the transfer function drops with 20dB/decade after the resonance.

As for the resonance: @Mario has explained very well that you see a resonance peak because of the complex conjugated pole pair on the imaginary axis. Because it is in the denominator, the peak points upwards. As there is no dampening (the peak is on the imaginary axis), you get an arbitrary high value.

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  • \$\begingroup\$ Matlab is not a lot of use when there's an infinite resonance. The OP says 'sketch'. \$\endgroup\$ – Chu Jun 29 '16 at 10:48
  • \$\begingroup\$ Agreed, but with the poles and zeros that close together, sketching this bode plot is also hard to get right. You are welcome to create a sketch and add it as an answer. The question also says that finding the poles gave the author difficulties. I merely added the figure because it was little extra work and might help to start the sketch. After all, it is correct for all points except the resonance frequency. I'm looking forward to see how you will sketch an infinite value. \$\endgroup\$ – cx05 Jun 29 '16 at 11:12
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The poles can be found by checking where denominator becomes zero.

In this case you have a product of two terms and if one of them is zero the whole expression is zero.

(1+s) == 0 results in s = -1

(1 + 2 s^2) == 0 results in s = \$\pm j/\sqrt 2\$

The complex conjugate pair indicates that you have resonance. It will appear as a peak in your bode plot

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Solve for numerator = 0 (for zeros) and denominator = 0 (for poles). For a pole-zero diagram, plot the imaginary values to the y-axis and the real values to the x-axis.

I use Mathcad to do so, but you could do it by hand of course.

enter image description here enter image description here

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