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How to define the transfer function with exp

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milad karimshoushtari
milad karimshoushtari 2015년 5월 4일
답변: Meet Doshi 2017년 10월 10일
I cannot make this transfer function either in matlab or in simulink, It has a wierd delay! can anyone help me please?! exp means a delay?! T(z,s) is output temperature and Tin(s) is input temperature.
s: is the laplace variable z,v,k1,k2,m(z) are constant
if gives this error ??? Error using ==> tf.exp at 31 The input argument of the "lti/exp" command must be a transfer function of the form -M*s.

답변 (2개)

Sriram Narayanan
Sriram Narayanan 2015년 5월 6일
편집: Sriram Narayanan 2015년 5월 6일
It is not possible to specify an exponential delay in the transfer function other than the form exp(-M*s) when using the laplace variable. However, this delay could be specified by seperating the terms as exp(-(z/v)*s))*exp(m/(s + k2))*exp(-(k1*z/v)) where the last two terms are constant gains for the transfer function.
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milad karimshoushtari
milad karimshoushtari 2015년 5월 6일
thanks for your answer sriram. but the problem is with the third term exp(m/(s + k2)) , s is laplace variable! its other than the form exp(-M*s)
Sriram Narayanan
Sriram Narayanan 2015년 5월 6일
One suggestion I have is to calculate the 1st order or higher-order approximation using Taylor expansion, similar to performing a Pade approximation for exp(-M*s) which is (1 - M*s/2)/(1 + M*s/2) for 1st order.

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Meet Doshi
Meet Doshi 2017년 10월 10일
hd = exp(tf([-2 0],1))

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