In this chapter, we will understand the basic properties of Z-transforms.

Linearity

It states that when two or more individual discrete signals are multiplied by constants, their respective Z-transforms will also be multiplied by the same constants.

Mathematically,

$$a_1x_1(n)+a_2x_2(n) = a_1X_1(z)+a_2X_2(z)$$

Proof − We know that,

$$X(Z) = sum_{n=-infty}^infty x(n)Z^{-n}$$

$= sum_{n=-infty}^infty (a_1x_1(n)+a_2x_2(n))Z^{-n}$

$= a_1sum_{n = -infty}^infty x_1(n)Z^{-n}+a_2sum_{n = -infty}^infty x_2(n)Z^{-n}$

$= a_1X_1(z)+a_2X_2(z)$ (Hence Proved)

Here, the ROC is $ROC_1bigcap ROC_2$.

Time Shifting

Time shifting property depicts how the change in the time domain in the discrete signal will affect the Z-domain, which can be written as;

$$x(n-n_0)longleftrightarrow X(Z)Z^{-n}$$

Or $x(n-1)longleftrightarrow Z^{-1}X(Z)$

Proof −

Let $y(P) = X(P-K)$

$Y(z) = sum_{p = -infty}^infty y(p)Z^{-p}$

$= sum_{p = -infty}^infty (x(p-k))Z^{-p}$

Let s = p-k

$= sum_{s = -infty}^infty x(s)Z^{-(s+k)}$

$= sum_{s = -infty}^infty x(s)Z^{-s}Z^{-k}$

$= Z^{-k}[sum_{s=-infty}^infty x(m)Z^{-s}]$

$= Z^{-k}X(Z)$ (Hence Proved)

Here, ROC can be written as Z = 0 (p>0) or Z = ∞(p<0)

Example

U(n) and U(n-1) can be plotted as follows

Z-transformation of U(n) cab be written as;

$sum_{n = -infty}^infty [U(n)]Z^{-n} = 1$

Z-transformation of U(n-1) can be written as;

$sum_{n = -infty}^infty [U(n-1)]Z^{-n} = Z^{-1}$

So here $x(n-n_0) = Z^{-n_0}X(Z)$ (Hence Proved)

Time Scaling

Time Scaling property tells us, what will be the Z-domain of the signal when the time is scaled in its discrete form, which can be written as;

$$a^nx(n) longleftrightarrow X(a^{-1}Z)$$

Proof −

Let $y(p) = a^{p}x(p)$

$Y(P) = sum_{p=-infty}^infty y(p)Z^{-p}$

$= sum_{p=-infty}^infty a^px(p)Z^{-p}$

$= sum_{p=-infty}^infty x(p)[a^{-1}Z]^{-p}$

$= X(a^{-1}Z)$(Hence proved)

ROC: = Mod(ar1) < Mod(Z) < Mod(ar2) where Mod = Modulus

Example

Let us determine the Z-transformation of $x(n) = a^n cos omega n$ using Time scaling property.

Solution −

We already know that the Z-transformation of the signal $cos (omega n)$ is given by −

$$sum_{n=-infty}^infty(cos omega n)Z^{-n} = (Z^2-Z cos omega)/(Z^2-2Zcos omega +1)$$

Now, applying Time scaling property, the Z-transformation of $a^n cos omega n$ can be written as;

$sum_{n=-infty}^infty(a^ncos omega n)Z^{-n} = X(a^{-1}Z)$

$= [(a^{-1}Z)^2-(a^{-1}Z cos omega n)]/((a^{-1}Z)^2-2(a^{-1}Z cos omega n)+1)$

$= Z(Z-a cos omega)/(Z^2-2az cos omega+a^2)$

Successive Differentiation

Successive Differentiation property shows that Z-transform will take place when we differentiate the discrete signal in time domain, with respect to time. This is shown as below.

$$frac{dx(n)}{dn} = (1-Z^{-1})X(Z)$$

Proof −

Consider the LHS of the equation − $frac{dx(n)}{dn}$

$$= frac{[x(n)-x(n-1)]}{[n-(n-1)]}$$

$= x(n)-X(n-1)$

$= x(Z)-Z^{-1}x(Z)$

$= (1-Z^{-1})x(Z)$ (Hence Proved)

ROC: R1< Mod (Z) <R2

Example

Let us find the Z-transform of a signal given by $x(n) = n^2u(n)$

By property we can write

$Zz[nU(n)] = -Zfrac{dZ[U(n)]}{dz}$

$= -Zfrac{d[frac{Z}{Z-1}]}{dZ}$

$= Z/((Z-1)^2$

$= y(let)$

Now, Z[n.y] can be found out by again applying the property,

$Z(n,y) = -Zfrac{dy}{dz}$

$= -Zfrac{d[Z/(Z-1)^3]}{dz}$

$= Z(Z+1)/(Z-1)^2$

Convolution

This depicts the change in Z-domain of the system when a convolution takes place in the discrete signal form, which can be written as −

$x_1(n)*x_2(n) longleftrightarrow X_1(Z).X_2(Z)$

Proof −

$X(Z) = sum_{n = -infty}^infty x(n)Z^{-n}$

$= sum_{n=-infty}^infty[sum_{k = -infty}^infty x_1(k)x_2(n-k)]Z^{-n}$

$= sum_{k = -infty}^infty x_1(k)[sum_n^infty x_2(n-k)Z^{-n}]$

$= sum_{k = -infty}^infty x_1(k)[sum_{n = -infty}^infty x_2(n-k)Z^{-(n-k)}Z^{-k}]$

Let n-k = l, then the above equation cab be written as −

$X(Z) = sum_{k = -infty}^infty x_1(k)[Z^{-k}sum_{l=-infty}^infty x_2(l)Z^{-l}]$

$= sum_{k = -infty}^infty x_1(k)X_2(Z)Z^{-k}$

$= X_2(Z)sum_{k = -infty}^infty x_1(Z)Z^{-k}$

$= X_1(Z).X_2(Z)$ (Hence Proved)

ROC:$ROCbigcap ROC2$

Example

Let us find the convolution given by two signals

$x_1(n) = lbrace 3,-2,2rbrace$ …(eq. 1)

$x_2(n) = lbrace 2,0leq 4quad andquad 0quad elsewhererbrace$ …(eq. 2)

Z-transformation of the first equation can be written as;

$sum_{n = -infty}^infty x_1(n)Z^{-n}$

$= 3-2Z^{-1}+2Z^{-2}$

Z-transformation of the second signal can be written as;

$sum_{n = -infty}^infty x_2(n)Z^{-n}$

$= 2+2Z^{-1}+2Z^{-2}+2Z^{-3}+2Z^{-4}$

So, the convolution of the above two signals is given by −

$X(Z) = [x_1(Z)^*x_2(Z)]$

$= [3-2Z^{-1}+2Z^{-2}]times [2+2Z^{-1}+2Z^{-2}+2Z^{-3}+2Z^{-4}]$

$= 6+2Z^{-1}+6Z^{-2}+6Z^{-3}+…quad…quad…$

Taking the inverse Z-transformation we get,

$x(n) = lbrace 6,2,6,6,6,0,4rbrace$

Initial Value Theorem

If x(n) is a causal sequence, which has its Z-transformation as X(z), then the initial value theorem can be written as;

$X(n)(atquad n = 0) = lim_{z to infty} X(z)$

Proof − We know that,

$X(Z) = sum_{n = 0} ^infty x(n)Z^{-n}$

Expanding the above series, we get;

$= X(0)Z^0+X(1)Z^{-1}+X(2)Z^{-2}+…quad…$

$= X(0)times 1+X(1)Z^{-1}+X(2)Z^{-2}+…quad…$

In the above case if Z → ∞ then $Z^{-n}rightarrow 0$ (Because n>0)

Therefore, we can say;

$lim_{z to infty}X(z) = X(0)$ (Hence Proved)

Final Value Theorem

Final Value Theorem states that if the Z-transform of a signal is represented as X(Z) and the poles are all inside the circle, then its final value is denoted as x(n) or X(∞) and can be written as −

$X(infty) = lim_{n to infty}X(n) = lim_{z to 1}[X(Z)(1-Z^{-1})]$

Conditions −

• It is applicable only for causal systems.
• $X(Z)(1-Z^{-1})$ should have poles inside the unit circle in Z-plane.

Proof − We know that

$Z^+[x(n+1)-x(n)] = lim_{k to infty}sum_{n=0}^kZ^{-n}[x(n+1)-x(n)]$

$Rightarrow Z^+[x(n+1)]-Z^+[x(n)] = lim_{k to infty}sum_{n=0}^kZ^{-n}[x(n+1)-x(n)]$

$Rightarrow Z[X(Z)^+-x(0)]-X(Z)^+ = lim_{k to infty}sum_{n = 0}^kZ^{-n}[x(n+1)-x(n)]$

Here, we can apply advanced property of one-sided Z-Transformation. So, the above equation can be re-written as;

$Z^+[x(n+1)] = Z[X(2)^+-x(0)Z^0] = Z[X(Z)^+-x(0)]$

Now putting z = 1 in the above equation, we can expand the above equation −

$lim_{k to infty}{[x(1)-x(0)+x(6)-x(1)+x(3)-x(2)+…quad…quad…+x(x+1)-x(k)]}$

This can be formulated as;

$X(infty) = lim_{n to infty}X(n) = lim_{z to 1}[X(Z)(1-Z^{-1})]$(Hence Proved)

Example

Let us find the Initial and Final value of x(n) whose signal is given by

$X(Z) = 2+3Z^{-1}+4Z^{-2}$

Solution − Let us first, find the initial value of the signal by applying the theorem

$x(0) = lim_{z to infty}X(Z)$

$= lim_{z to infty}[2+3Z^{-1}+4Z^{-2}]$

$= 2+(frac{3}{infty})+(frac{4}{infty}) = 2$

Now let us find the Final value of signal applying the theorem

$x(infty) = lim_{z to infty}[(1-Z^{-1})X(Z)]$

$= lim_{z to infty}[(1-Z^{-1})(2+3Z^{-1}+4Z^{-2})]$

$= lim_{z to infty}[2+Z^{-1}+Z^{-2}-4Z^{-3}]$

$= 2+1+1-4 = 0$

Some other properties of Z-transform are listed below −

Differentiation in Frequency

It gives the change in Z-domain of the signal, when its discrete signal is differentiated with respect to time.

$nx(n)longleftrightarrow -Zfrac{dX(z)}{dz}$

Its ROC can be written as;

$r_2< Mod(Z)< r_1$

Example

Let us find the value of x(n) through Differentiation in frequency, whose discrete signal in Z-domain is given by $x(n)longleftrightarrow X(Z) = log(1+aZ^{-1})$

By property, we can write that

$nx(n)longleftrightarrow -Zfrac{dx(Z)}{dz}$

$= -Z[frac{-aZ^{-2}}{1+aZ^{-1}}]$

$= (aZ^{-1})/(1+aZ^{-1})$

$= 1-1/(1+aZ^{-1})$

$nx(n) = delta(n)-(-a)^nu(n)$

$Rightarrow x(n) = 1/n[delta(n)-(-a)^nu(n)]$

Multiplication in Time

It gives the change in Z-domain of the signal when multiplication takes place at discrete signal level.

$x_1(n).x_2(n)longleftrightarrow(frac{1}{2Pi j})[X1(Z)*X2(Z)]$

Conjugation in Time

This depicts the representation of conjugated discrete signal in Z-domain.

$X^*(n)longleftrightarrow X^*(Z^*)$