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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)). 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 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.

## One thought on “Z-transform properties (Summary and Simple Proofs)”

1. nobody says:

Typo in time shifting proof, n instead of m at summation!

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