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From: moonman on 14 Jul 2010 09:34
For term like exp(-1.5 s) x 3/(s^2+4s+3) We can take the Ztransform or discrete form by this code

clc
Td=1.5;
T=1;
num=[0 0 3];
den=[1 4 3];
Gs=tf(num,den,'InputDelay',Td)
Gz=c2d(Gs,T)
[numz,denz]=tfdata(Gz);
[A,B,C,D]=tf2ss(numz{1},denz{1})

But when i am having exp(1.5 s) x 3/(s^2+4s+3) (( negative sign not there how to calculate)) s is Laplace Domain

Thanks
From: moonman on 14 Jul 2010 11:45
Any Comments on this






"moonman " <moonnightingale(a)yahoo.com> wrote in message <i1kecb$im$1(a)fred.mathworks.com>...
> For term like exp(-1.5 s) x 3/(s^2+4s+3) We can take the Ztransform or discrete form by this code
>
> clc
> Td=1.5;
> T=1;
> num=[0 0 3];
> den=[1 4 3];
> Gs=tf(num,den,'InputDelay',Td)
> Gz=c2d(Gs,T)
> [numz,denz]=tfdata(Gz);
> [A,B,C,D]=tf2ss(numz{1},denz{1})
>
> But when i am having exp(1.5 s) x 3/(s^2+4s+3) (( negative sign not there how to calculate)) s is Laplace Domain
>
> Thanks
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