# Zeros and poles in the following bode plot

According to the definition, zeros are the frequencies which make transfer function to be zero, and poles are the frequencies which make transfer function to be infinity.

The definition can be seen MIT handout, and Wikibook.

But considering the following transfer function $$H(i \omega) = \frac{(i \omega)(2+i \omega ) }{(1+i \omega )^2}$$

But according to definition, $$i \omega$$ is an imaginary number, but 1 and 2 is a real number, How is it possible to make transfer function to become zero or infinity.

When plotting bode plot, it is like below: The pole is 1, the zero is 2.

Why?

Can anybody provide a mathematical explanation?

• In your "transfer function" you have set s=jw, which is a special case on the jw axis where sigma = zero. The magnitude plot you show is the magnitude of this special case, not where the general transfer function goes to zero or infinity. – John D Jun 10 at 3:20
• Because this is a slice through the 3D view that influences the response – JonRB Jun 10 at 6:32
• Read the comments to @pr871 excellent answer to this question: electronics.stackexchange.com/questions/503713/… . He explains what goes to infinity at a pole and how. – Sredni Vashtar Jun 10 at 13:53

How is it possible to make transfer function to become zero or infinity.

You will only get zeros and poles in the bode plot if those zeros and poles lie on the $$\j\omega\$$ axis. The bode plot is one part of the bigger picture: - The above diagram incorporates both the bode plot and the pole zero diagram. It's for a 2nd order low pass filter (just an example). As you can see from viewing from the right, the bode plot would look like this: - Note that the frequency at which the above amplitude is maximum is not quite the same as the frequency on the $$\j\omega\$$ axis at which the pole coincides. I mention this In case anyone thinks it is an error. This is typical of 2nd order low-pass filters.

And, if you were to look downwards from above you would see the pole zero diagram (aka the s-plane diagram): - At the pole points above, the transfer function response will be "infinity" but these poles are not aligned to the $$\j\omega\$$ axis and, in effect, they are in a realm that has no physical existence.

Hopefully this will help you realize that "s-plane real numbers" lie on a different axis to the $$\j\omega\$$ axis. It's sometimes called the $$\\sigma\$$ (sigma) axis and represents how much damping the system has. Low damping means that poles get close to the $$\j\omega\$$ axis. High damping means that poles move further from the $$\j\omega\$$ axis.

Pictures from here.

Your transfer function has two zeros and two poles. The zeros are at s = 0 and s = -2, while both of the poles are at s = -1.

Particularly you need to remember that a Bode plot is given in two parts: the magnitude and the phase of your transfer function. The graph you provided is only the magnitude plot of H(s), which is an incomplete picture of the Bode plot for H(jw). The following are the magnitude and phase Bode plots of H(s): Another way of analyzing your transfer function and see when it becomes zero, or infinity is to perform the Root Locus. Your transfer function will be zero at the zeros and infinity at the poles. In a root locus plot, we see that the function will approach the zeros from its poles, that is, H(s) will be stable if for every pole, there is a correspondent zero, amoung other rules.

Here's the Matlab code I created to plot these graphs:

num = [1,2,0];
den = [1,2,1];
H = tf(num,den)
bode(H)
rlocus(H)


Mathematically, as a commenter pointed out, the plot is the Magnitude plot of H(s) at s=jw, i.e. 20*log10(abs(H(jw))) and not the complex H(s) itself. The magnitude is real.

Have a look here

https://web.njit.edu/~levkov/classes_files/ECE232/Handouts/Frequency%20Response.pdf

and I think equations 1.14 /1.15 are a good start. Remember you can break up H(s) into constituent terms H1(s)*H2(s)/H3(s) etc... and calculate the magnitude for each term individually.