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### Inverse Z transform using long division method

While carrying out the long division method, it always converts the given expression in the simplest form. That means as far as possible, it should be in form of Z divided by some polynomial. |Z| > 1 it is causal sequence. For the causal sequence the denominator polynomial should have maximum power of Z transform on its left. Thus the given expression is in the total proper form.

Find inverse Z transform of  X(z) is = Z/Z-1   |Z|>1

X(Z) = 1 + Z-1 +  Z-2  +  Z-3+..........

∞
X(Z) = ∑ x(n) Z-n
n=0

X(Z) is = x(0) + x(1) Z-1 + x(2) Z-2  + x(3) Z-3

X(n) =  {x(0) ,  x(1), x(2), x(3) }

x(n) = { 1, 1, 1, 1 }

x(n) = u(n)

Z-1{Z-1/Z-1} = u(n)

This is a standard Z transform pair.
While carrying out the long division method, it always converts the given expression in the simplest form. That means as far as possible, it should be in form of Z divided by some polynomial. |Z| > 1 it is causal sequence. For the causal sequence the denominator polynomial should have maximum power of Z transform on its left. Thus the given expression is in the total proper form.

Find inverse Z transform of  X(z) is = Z/Z-1   |Z|>1

X(Z) = 1 + Z-1 +  Z-2  +  Z-3+..........

∞
X(Z) = ∑ x(n) Z-n
n=0

X(Z) is = x(0) + x(1) Z-1 + x(2) Z-2  + x(3) Z-3

X(n) =  {x(0) ,  x(1), x(2), x(3) }

x(n) = { 1, 1, 1, 1 }

x(n) = u(n)

Z-1{Z-1/Z-1} = u(n)

This is a standard Z transform pair.