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# Inverse Z-Transform with Examples

represents the discrete time fourier transform of (r^-n*x (n)).

The Fourier inverse of X (r*e^ (j*ω)) is

x (n)*(r^-n) = (1/(2*π))* X (r*e^ (j*ω))* e^ (j*ω*n) dω

Multiplying by (r^n) on both sides, we get,

x (n)= (1/(2*π))* X (r*e^ (j*ω))* (r*e^ (j*ω))^n dω

Changing the variable of integration from ω to z

r*e^ (j*ω) = z

Differentiating the above equation, we get

r*j*( e^ (j*ω)) dω = dz

dω = (dz/ r*j*( e^ (j*ω)))

That is,   dω = (dz/ j*(r*( e^ (j*ω))))

dω = (dz/(z*j))

Therefore,

x (n) = (1/(2*π))* X (z)*(z^n)/ (j*z) dz =(1/(2*π*j))* X (z)*(z^(n-1)) dz

Since the integration over a(2*π) interval in ‘ω’, in terms of ‘z’ corresponds to one transfers around a circle |z|=r

x (n) =(1/(2*π*j))* X (z)*(z^(n-1)) dz

Which gives the inverse z-transform of x(z)

X (z) =(b0+(b1*z^(-1))+(b2*z^(-2))+………………..+(bm*z^(-m)))/(1+(a1*z^(-1)) +
(a2*z^(-2))+…………………..+(an*z^(-n)));

Where (n>m).

The roots of the numerator polynomial are those values of ‘z’ for which x(z)=0 and are referred as zeroes of X (z).Zeros are represented by ‘0’ in the z-plane.

The roots of the denominator polynomial are those values of ‘z’ for which X (z) = ∞ and referred as poles of X (z).poles are represented by ‘x’ in the z-plane.