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)). We will also specify the Region of Convergence of the transform for each of the properties.

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	a_nx(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	n^kx(n) $\underleftrightarrow { \quad z\quad }$ [-1]^kz^k $\frac { { d }^{ k }X(Z) }{ d{ Z }^{ k } }$	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)  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

a_nx(n)  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)  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)z^m= $\sum _{ m=-\infty }^{ \infty }{ }$ 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:

n^kx(n)  [-1]^kz^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)  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^-nby 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) =  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() =  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.

Proofs of the properties of Z-transform

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