Zero & Pole Plots: Filters, Frequency Response & Impulse Inputs

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Zero-pole plots are essential for understanding filters and their frequency response. To translate a zero-pole plot into a frequency response, one must derive a transfer function from the plotted poles and zeros. This transfer function can be expressed as H(s) = ((s-z1)(s-z2)(s-z3))/((s-p1)(s-p2)(s-p3)), where z and p represent the zeros and poles, respectively. The frequency response can then be obtained by analyzing the magnitude and phase of the transfer function. The gain factor, which is a constant in front of the poles and zeros, is also crucial for complete characterization.
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Hi I am after any information regarding zero pole plots, specifically with reference to filters.

I understand the process of getting the transfer function and plotting it. But once its been plotted how do you go about translating that into a frequency response.

I am also wondering about how from the plots you get the response to an impulse input.

Many thanks!
 
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Dirac8767 said:
Hi I am after any information regarding zero pole plots, specifically with reference to filters.

I understand the process of getting the transfer function and plotting it. But once its been plotted how do you go about translating that into a frequency response.

I am also wondering about how from the plots you get the response to an impulse input.

Many thanks!

It sounds like you are a little confused with the terminology, or you are not being specific enough in your question. Can you give an example of what you're trying to figure out?

When you say that you plot the transfer function, are you referring to plotting the magnitude and phase of it, as in a Bode plot, or are you talking about plotting a pole-zero diagram of the transfer function?

If you are talking about Bode plots, then those are already your frequency response, graphically. It tells you how much an input signal at a specific frequency will be attenuated and how much its phase will shift at the output of the system. Multiplying this transfer function by your input signal in the frequency domain is equivalent to convolving your input signal in the time domain with the impulse response of the system, ether way it gives you the output of the system.

As to your other question, the frequency response of a system which is expressed by its transfer function, is itself the impulse response in the frequency domain. If you can work your way backwards: first find the poles, zeros, and gain of your bode plots, second write out the transfer function with the poles, zeros, and gain you found, and finally take the inverse laplace transform of this transfer function. That should give you the impulse response of the system.
 
DragonPetter said:
...

Sorry maybe i wasnt clear enough, i meant plotting the information from the the transfer function(i.e. the poles and zeros) onto the argand diagram. That bit is all understood.

I am just unsure about how to get from that argand diagram form to a frequency response.

Forget about the question regarding the impulse response as I've figured that bit out now.
 
Also let me explain it to you this way.

An impulse response is the output of a system that you get when you apply a delta function (also called impulse function) to its input in the time domain. Any input you apply to the system will give you an output of the system equal to the convolution of this input with the impulse response. This can be proven mathematically, but this is the fact we use when working these problems.

In the frequency domain, the convolution operation becomes a multiplication operation, and so your transfer function

H(S) = Y(S)/X(S),

and because a delta function is equal at all frequencies in the frequency domain, or in other words, it is a constant, so X(S) = 1 , while x(t) is a delta impulse. you can see that when you apply an impulse, X(S) = 1, and the output Y(S) is equal to the laplace transform of the impulse response of the system H(S), which agrees with what I've said above.

Now, reorganize as Y(S) = H(S)X(S) looks like
y(t) = h(t)*x(t) in the time domain

and with X(S) = 1, by the definition of the impulse response, h(t) must be your impulse response.

so H(S) is the laplace transform of the impulse response of the system.
 
Dirac8767 said:
Sorry maybe i wasnt clear enough, i meant plotting the information from the the transfer function(i.e. the poles and zeros) onto the argand diagram. That bit is all understood.

I am just unsure about how to get from that argand diagram form to a frequency response.

Forget about the question regarding the impulse response as I've figured that bit out now.

Alright, in that case, you take the points labeled as zeros and each one is the sum of its complex coordinates. Say you have a zero plotted at (1,5j), then the zero is 1+5j. You do the same for the poles that are plotted, say you find one at (2,4j), then a pole is 2+4j

Now you simply write out a transfer function that has all of the poles and zeros you got from the plot.

Using the example pole and zero:

H(s) = (s-(1+5j))/(s-(2+4j))

if you have multiple zeroes and poles, say z1, z2, z3 and p1, p2, p3 you'd write the equation as

H(s) = ((s-z1)(s-z2)(s-z3))/((s-p1)(s-p2)(s-p3)) since anytime s is equal to one of your zeros, your equation becomes 0 and any time your s is equal to one of your poles, you get a divide by 0.

Edit: When you work out the expression, you will get a a constant in front of of the poles and zeros, and this is your gain factor.
 
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