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Bilinear transformation: s-plane to z-plane

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naresh850

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Hi,
In Bilinear transformation s-plane is transformed in z- plane by formulla.

s = (2/T)(1-z^-1)/(1+z^-1); T = Sampling time period

in analog domain
s = d/dt

its in discrete domain

dx/dt = (x(n)-x(n-1))/T where is x(t) is analog signal.T = Sampling time period

so s = (1-z^-1)/T

But in bilinear transform, why 2/(1+z^-1) is added in to define s?

Reagards,
NP
 

mahdithdn

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hi,

1- dx/dt(t=n) = (x(n)-x(n-1))/T feedback expanssion
2- dx/dt(t=n+1) = (x(n)-x(n-1))/T feedforward expanssion

so
the addition of expression 1 & 2 is.

sX(s) + exp(sT)sX(s) = (2/T)(X(s)-exp(sT)X(s))

and z = exp(sT)

=> s = (2/T)(1-z^-1)/(1+z^-1)

regards
 

naresh850

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Could u plz explain brifly?
i cant get u.

Regards,
 

mahdithdn

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Hi,

there are many approximations for z = exp(sT). some are simple but they deviate form z = exp(sT) and some are more complex.
forward and backward difference are only two simple approximations. but the other one is bilinear. its philosophy is based on derivation at the point (t=(n+1/2)T).
we can use this derivation by two means,
dx/dt(t = (n+1/2)T) = x(n) - x(x-1)/T;
but the other method is to use average of derivation at point (t = nT) and (t = (n+1)T).
dx/dt(t = (n+1/2)T) = 0.5(dx/dt(t=nT) + dx/dt(t=(n+1)T);
now use backward difference expansion for dx/dt(t=nT) and forward difference expansion for dx/dt(t=(n+1)T).
dx/dt(t=nT) = (x(n)-x(n-1))/T backward difference expansion.
dx/dt(t=(n+1)T) = (x(n)-x(n-1))/T forward difference expansion.
the bilinear transform is obtained.
regards
 

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