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Z-transform properties (Summary and Simple Proofs)

All of these properties of z-transform are applicable for discrete-time signals that have a Z-transform. Meaning these properties of Z-transform apply to any generic signal x(n) for which an X(z) exists. (x(n) \underleftrightarrow { \quad \quad z\quad \quad } X(z))

Property Mathematical representation Exceptions/

ROC

Linearity a1x1(n)+a2x2(n) = a1X1(z) + a2X2(z) At least

ROC1∩ROC2

Time shifting x(n-k) \underleftrightarrow { \quad z\quad } z-kX(z) ROC of x(n-k)
Scaling anx(n) \underleftrightarrow { \quad z\quad } x(z/a) If r1 <|z|< r2,

then

|a|r1<|z|<|a|r2

Time reversal x(-n) \underleftrightarrow { \quad z\quad } x(1/z) 1/r2<|z|<1/r1
Differentiation

in Z-domain

or

Multiplication by n

nkx(n) \underleftrightarrow { \quad z\quad } [-1]kzk \frac { { d }^{ k }X(Z) }{ d{ Z }^{ k } } ROC = All R
Convolution x(n)*h(n) \underleftrightarrow { \quad z\quad } x(Z)*h(Z) At least

ROC1∩ROC2

Correlation x(n)⊗y(n) \underleftrightarrow { \quad z\quad } x(Z).y(Z-1)
Initial Value theorem

in Z-transform

x(0) = \lim _{ z\rightarrow \infty } x(Z) For a causal

signal x(n)

Final Value theorem

in Z-transform

x(\infty ) = \lim _{ z\rightarrow 1 } x(Z)(1-Z-1) For a causal

signal x(n)

Conjugation x*(n) \underleftrightarrow { \quad z\quad } x*(Z*) ROC of x(n)
Parseval’s

Theorem

\sum _{ n=-\infty }^{ \infty }{ } x(n).h*(n) = \int _{ -\pi }^{ \pi }{ } F(z)F*(z) dz

Parseval’s relation tells us that the energy of a signal is equal to the energy of its Fourier transform.

Proofs of the properties of Z-transform

Linearity

We will be proving the following property of Z-transform

a1x1(n)+a2x2(n) = a1X1(z) + a2X2(z)

Proof

According to the definition of z-transform, we have:

X(z) = \sum _{ n=-\infty }^{ \infty }{ } x(n)z-n

Here, the input signal is

x(n) = a1x1(n)+a2x2(n)

Put this equation in the formula for z-transform

X(z) = \sum _{ n=-\infty }^{ \infty }{ } [a1x1(n)+a2x2(n)]z-n

Separating the two terms

X(z) = a1 \sum _{ n=-\infty }^{ \infty }{ } x1(n)z-n + a2 \sum _{ n=-\infty }^{ \infty }{ } x2(n)z-n

We’ve taken out a1 and a2 from the summation since they are constants. Thus, we get

X(z) = a1X1(z) + a2X2(z)

Time Shifting

We will be proving the following property of Z-transform

x(n-k) \underleftrightarrow { \quad z\quad }  z-kX(z)

From the definition of z-transform we have

X(z) = \sum _{ n=-\infty }^{ \infty }{ } x(n)z-n

Let’s assume,

n-k = m

n = m+k

We can rewrite the z-transform equation as

X(z) = \sum _{ m=-\infty }^{ \infty }{ } x(m)z-(m+k)

= \sum _{ m=-\infty }^{ \infty }{ } x(n)z-mz-k

= z-kX(z)

Thus

x(n-k) \underleftrightarrow { \quad z\quad } z-kX(z)

Scaling in z-domain

We will be proving the following property of Z-transform

anx(n) \underleftrightarrow { \quad z\quad }  x(z/a)

From the definition, the z-transform of anx(n) is:

z[anx(n)] = \sum _{ n=-\infty }^{ \infty }{ } anx(n)z-n

= \sum _{ n=-\infty }^{ \infty }{ } x(n)(a-1z)-n

= \sum _{ n=-\infty }^{ \infty }{ } x(n)(z/a)-n

Hence,

anx(n) \underleftrightarrow { \quad z\quad } X(z/a)

ROC of X(z) is always r1<|z|<r2

Thus the ROC of X(z/a) is |a|r1<|z|<|a|r2

Scaling in the time domain corresponds to expanding or shrinking of the z-plane.

Time reversal

We will be proving the following property of Z-transform

x(-n) \underleftrightarrow { \quad z\quad }  x(1/z)

From the definition, the z-transform of x(-n) is:

z[x(-n)] = \sum _{ n=-\infty }^{ \infty }{ } x(-n)z-n

Put -n=m

Incorporating the corresponding change in limits:

z[x(-n)] = \sum _{ m=-\infty }^{ \infty }{ } x(m)zm = \sum _{ m=-\infty }^{ \infty }{ } x(m)z-mx-1

= \sum _{ m=-\infty }^{ \infty }{ } x(m)(z-1)-m  = X[z-1]

Thus ROC always is r1<|z|<r2

In this case, ROC is

1/r2<|z|<1/r1

Differentiation in z-domain/Multiplication by n

We will be proving the following property of Z-transform:

nkx(n) \underleftrightarrow { \quad z\quad }  [-1]kzk \frac { { d }^{ k }X(Z) }{ d{ Z }^{ k } } 

From the definition, the z-transform of x(n) is:

z[x(n)] = \sum _{ n=-\infty }^{ \infty }{ } x(n)z-n

Differentiating both sides with respect to z, we get:

dX(z)/dz = x(n) \sum _{ n=-\infty }^{ \infty }{ } x(-n)z-n-1

= x(n) \sum _{ n=-\infty }^{ \infty }{ } x(-n)z-nz-1

Rearranging LHS and RHS

-z(dX(z)/dz) = \sum _{ n=-\infty }^{ \infty }{ } nx(n)z-n

= z{nx(n)}

Hence, proved.

Convolution

We will be proving the following property of Z-transform:

x(n)*h(n) \underleftrightarrow { \quad z\quad }  x(Z)*h(Z)

The convolution of x(n) and h(n), y(n) can be represented as:

y(n) = x(n).h(n)

= \sum _{ k=-\infty }^{ \infty }{ } x(k)h(n-k)

The z-transform of the above equation is:

Y(z) = \sum _{ n=-\infty }^{ \infty }{ } \sum _{ k=-\infty }^{ \infty }{ } x(k)h(n-k)z-n

Adjusting the summation and replacing z-n by z-(n-k)z-k

\sum _{ k=-\infty }^{ \infty }{ } x(k)\sum _{ n=-\infty }^{ \infty }{ } h(n-k)z-(n-k)z-k

But \sum _{ n=-\infty }^{ \infty }{ } h(n-k)z-(n-k)= h(z)

Placing h(z) in the prior equation

Y(z) = h(z)\sum _{ k=-\infty }^{ \infty }{ } x(k)z-k

=h(z).x(z). Hence, proved.

Initial Value Theorem

We will be proving the following property of Z-transform:

x(0) = \lim _{ z\rightarrow \infty }  x(Z)

From the definition, the z-transform of x(n) is:

z[x(n)] = \sum _{ n=-\infty }^{ \infty }{ } x(n)z-n

Let’s expand the above equation:

X(z) = x(0)z0+x(1)z-1+x(2)z-2+x(3)z-3

Applying limits of \lim _{ z\rightarrow \infty } to the above series reduces the terms 1/z,1/z2… to 0.

Thus x(0) =\lim _{ z\rightarrow \infty } X(z). Hence, proved.

Final Value Theorem

We will be proving the following property of Z-transform:

x(\infty ) = \lim _{ z\rightarrow 1 }  x(Z)(1-Z-1)

The z-transform of a discrete causal signal [x(n)-x(n-1)] is given by:

z[x(n)-x(n-1)] = [x(n)-x(n-1)]z-n

which can be rewritten as:

X(z) – z-1X(z) = x(n)z-n  –  x(n-1)z-n

X(z)(1 – z-1) =\sum _{ n=0}^{ \infty }{ } x(n)z-n – z-1 \sum _{ n=0}^{ \infty }{ } x(n-1)z-(n-1)

Taking limit \lim _{ z\rightarrow 1 } on both sides and expanding the equation we get

X(z)(1 – z-1) = x(0) + x(1)z-1….-z-1[x(-1)z + x(0) + x(1)z-1….]

As\lim _{ z\rightarrow 1 } the terms of z exponents will be reduced to 1. Since x(n) is causal, x(-1) = 0.

Thus, we are finally left with x(\infty ) = \lim _{ z\rightarrow 1 } x(Z)(1-Z-1). Hence, proved.

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