3/31/2020 Properties Of Z Transform With Proof Pdf
The z-transform therefore exists (or converges) if X(z)= X. 2 Properties of the region of convergence The properties of the ROC depend on the nature of the signal. Assuming that the signal has a finite amplitude and that the z-transform is a rational function.
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 −
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.
$X^*(n)longleftrightarrow X^*(Z^*)$
Z-Transform has following properties:
Linearity Property
If $,x (n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(Z)$
and $,y(n) stackrel{mathrm{Z.T}}{longleftrightarrow} Y(Z)$
Then linearity property states that
$a, x (n) + b, y (n) stackrel{mathrm{Z.T}}{longleftrightarrow} a, X(Z) + b, Y(Z)$
Time Shifting Property
If $,x (n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(Z)$
Then Time shifting property states that
$x (n-m) stackrel{mathrm{Z.T}}{longleftrightarrow} z^{-m} X(Z)$
Multiplication by Exponential Sequence Property
If $,x (n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(Z)$
Then multiplication by an exponential sequence property states that
$a^n, . x(n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(Z/a)$
Time Reversal Property
If $, x (n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(Z)$
Then time reversal property states that
$x (-n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(1/Z)$
Differentiation in Z-Domain OR Multiplication by n Property
If $, x (n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(Z)$
Then multiplication by n or differentiation in z-domain property states that
$ n^k x (n) stackrel{mathrm{Z.T}}{longleftrightarrow} [-1]^k z^k{d^k X(Z) over dZ^K} $
Convolution Property
If $,x (n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(Z)$
and $,y(n) stackrel{mathrm{Z.T}}{longleftrightarrow} Y(Z)$
Then convolution property states that
$x(n) * y(n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(Z).Y(Z)$
Correlation Property
If $,x (n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(Z)$
and $,y(n) stackrel{mathrm{Z.T}}{longleftrightarrow} Y(Z)$
Then correlation property states that
$x(n) otimes y(n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(Z).Y(Z^{-1})$
Initial Value and Final Value Theorems
Initial value and final value theorems of z-transform are defined for causal signal.
Initial Value Theorem
For a causal signal x(n), the initial value theorem states that
$ x (0) = lim_{z to infty }X(z) $
This is used to find the initial value of the signal without taking inverse z-transform
Final Value Theorem
For a causal signal x(n), the final value theorem states that
$ x ( infty ) = lim_{z to 1} [z-1] X(z) $
This is used to find the final value of the signal without taking inverse z-transform.
Region of Convergence (ROC) of Z-Transform
The range of variation of z for which z-transform converges is called region of convergence of z-transform.
Properties of ROC of Z-Transforms
The concept of ROC can be explained by the following example:
Example 1: Find z-transform and ROC of $a^n u[n] + a^{-}nu[-n-1]$
$Z.T[a^n u[n]] + Z.T[a^{-n}u[-n-1]] = {Z over Z-a} + {Z over Z {-1 over a}}$
$$ ROC: |z| gt a quadquad ROC: |z| lt {1 over a} $$
The plot of ROC has two conditions as a > 1 and a < 1, as you do not know a.
In this case, there is no combination ROC.
Here, the combination of ROC is from $a lt |z| lt {1 over a}$
Hence for this problem, z-transform is possible when a < 1.
Causality and StabilityCausality condition for discrete time LTI systems is as follows:
A discrete time LTI system is causal when
Stability Condition for Discrete Time LTI Systems
A discrete time LTI system is stable when
Z-Transform of Basic Signals
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