Facebook
Google
GitHub
Linkedin
Converting Transfer Function Into Circuits ..#2
- Thread starter Mibiss
- Start date
It would be helpful for us if you could tell us something about your knowledge of analog signal processing.
Specifically: For example, do you know which filter function the given transfer function belongs to?
Perhaps it would be helpful when you try to describe the given function with words - using poles and zeros.
Do you know something about the effects of poles/zeros on the filter function?
How would you describe the characteristics of a function like H(s)=1/[1+sT+(s/wp)²] ?
Hello there,
This is actually like a two part question.
The first is to find an implementation and the second is to use as few parts as possible.
An interesting and informative way to do this is to use cascade integrators and a set of gains that feed back or feed forward into the input of the integrators. This might get a little more complicated for this transfer function for the actual implementation though. This is a good way to look at it in theory as it emulates a flow graph of the system.
Instead, you can try using two low pass filters in cascade with the necessary forward gains, and one offset. You can get the form of the low pass filters from one of the previous suggestions in this thread, which is using a partial fraction expansion.
A partial fraction expansion may or may not work well for all systems, but if the system can actually decompose into first order functions, it could be simpler to implement the circuit in real life. For this circuit I think that would be the case, so you should try to learn how to do a partial fraction expansion. This will convert the transfer function into a sum of first order low pass filters in this form:
Hs=A/(s+B)+C/(s+D)+E
where A,B,C,D, and E are constants. You can then transform the first two terms there into low pass filters.
There is a chance you can use just resistors to form the summation but you could look into that.
To get the gains in that function you may be able to get away with using just one op amp, but you'll have to look into that also.
Doing a partial fraction expansion usually means solving a set of simultaneous algebraic equations. You'd have to know how to do that. This of course is if you are not allowed to use automatic math software that does algebra and calculus.
I don't understand the poles and zeros completely. Are they not related to the stability of the filter? The numerator are the poles and the denominator are the zeros of the transfer function? No I don't think i know about the effects on the filter function.
If you do a partial fraction expansion you'll see the solution right away.
"The numerator are the poles..."
At first - neither the numerator nor the denominator "are the poles".
Secondly, when you had some lessons in "Linear Circuit Theory" you should know that the zeros of the numerator/denominator have something to do with the zeros and the poles of the transfer function.
After reading that the TO does not know the meaning of poles or zeros of the transfer function, I am afraid that it will not help him much to transform the given function by partial fraction decomposition.
By learning and applying this mathematical manipulation, he will only lose a lot of time without it helping him to solve the problem.
I am afraid that he will not "see the solution right away".
(Remember his post#8: "I don't understand the poles and zeros completely. Are they not related to the stability of the filter? The numerator are the poles and the denominator are the zeros of the transfer function? No I don't think i know about the effects on the filter function.")
Hi there,
Yes, you are right there I believe, I figured it will take some more time on his part to ask about these issues also.
We could go into more detail if he does not understand what we mean, but he'll have to indicate that first.
I could mention that the first order functions I was talking about have the form:
A/(s+B)
and that would look like a low pass filter, and moreover, we can break 'A' down into two factors C and B so we end up with:
C*[B/(s+B)]
where the B/(s+B) part is a regular RC low pass filter with B=1/RC (RC=R*C here).
That might help.
