Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.
A transfer function can be a IIR or an FIR function in either cases you can express it as a polinomy, in case of IIR function you perform long division and take a given number of coeficients, in this case your answer will be aproximately.
Returning to the topic if you take the polinomy's coeficients of the input and multiply them with the polinomy's coeficients of the tranfer function, you will find the polinomy's coeficients of the response. This theory is according to multiplication-convolution properties of Laplace transform.
I think the terms IIR and FIR are reserved for discrete-time transfer function based on the z-transform.
For laplace transform, one of the ways to attack the inverse laplace transform is by using partial-fraction expansion. You take your transfer function and break it up into a sum of smaller terms each with 1st or 2nd order poles, then you perform a inverse laplace transform by looking at tables.
Please google "partial fraction expansion" and you will find lots of examples.
The terms IIR and FIR are not exclusive of Z transform they represent the impulse response of the transfer function, in order words, if the numbers of polinomy coeficents of the transfer function are finite or not.
The question was how to find the output whitout inverse the Laplace transform, as far as I am concerned fraction expansion is a technique used to find the inverse transform and so, it is not the answer to the given question.
Considering input and transfer function expressed as polinomials in the S domain, when you multiply the polinomy's coeficients of the input signal with the poliniomy's coeficients of the transfer function, you are indeed performing the convolution of the input signal with the transfer function in time domain , the resulting polinomy gives the coeficients of the output in the S domain.
Thats is the way to find the output without performing inverse.
To the best of my knowledge, FIR and IIR filters always refer to digital filters. I have never seen them used for s-domain filters. FIR have all zeros sometimes referred to as MA (moving average) and IIR have both poles and zeros sometimes referred to as ARMA (auto-regressive moving average). Am I wrong in these definitions?
Check out this **broken link removed** to see how FIR filters are implemented in Matlab.
To the best of my knowledge, FIR and IIR filters always refer to digital filters. I have never seen them used for s-domain filters. FIR have all zeros sometimes referred to as MA (moving average) and IIR have both poles and zeros sometimes referred to as ARMA (auto-regressive moving average). Am I wrong in these definitions?
Check out this h**p://www.mathworks.com/access/helpdesk/help/toolbox/signal/fir1.html to see how FIR filters are implemented in Matlab.
You are probably right, I got confused in my answer, The answer given does not apply for Laplace Transform , but for Z transform. Multiplying the polinomys of signal imput to the transfer function gives the output on Z transforms, since the degrees from each terms refers to the delay in time, the same does not happen in S domain where the degree from each terms refers to the differentiaition order.
Multiplying the polinomys is the same that findind the convolutions of the coeficients in time domain ( for discrete Systems).
Considering that I think the original question remains, once fraction expansions is a way to inverse the Laplace transform and in order to perform convolution in time domain, you need inverting the transfer function.
nasser,
If the transfer function is in the s-domain, and the input is in the time domain, you must get the LaPlace transform of the input. Multiplying the transfer function by the LaPlace transform of the input results in the output in the s-domain.
~
To get the output in the time domain without using the inverse LaPlace transform you must do the following:
Determine the time-domain impulse response of the network (not an easy thing to do).
Convolve the impulse respons with the input in the time domain.
Regards,
Kral
If you are asking about discrete values, you can use the "Sliding Tape Method".
Lathi signals and systems book gives a good example of this, all you need is the input function x[n] and the transfer h[n] and you can compute the output.
If you are refering to continous values - convolution will give you the output.
This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register.
By continuing to use this site, you are consenting to our use of cookies.
