1 Crore+ students have signed up on EduRev. Have you? 
In this chapter, we will understand the basic properties of Ztransforms.
Linearity
It states that when two or more individual discrete signals are multiplied by constants, their respective Ztransforms will also be multiplied by the same constants.
Mathematically,
Proof − We know that,
(Hence Proved)
Here, the ROC is
Time Shifting
Time shifting property depicts how the change in the time domain in the discrete signal will affect the Zdomain, which can be written as;
Or
Proof −
Let s = pk
(Hence Proved)
Here, ROC can be written as Z = 0 (p>0) or Z = ∞(p<0)
Example
U(n) and U(n1) can be plotted as follows
Ztransformation of U(n) cab be written as;
Ztransformation of U(n1) can be written as;
So here (Hence Proved)
Time Scaling
Time Scaling property tells us, what will be the Zdomain of the signal when the time is scaled in its discrete form, which can be written as;
Proof −
(Hence proved)
ROC: = Mod(ar1) < Mod(Z) < Mod(ar2) where Mod = Modulus
Example
Let us determine the Ztransformation of x(n) = a^{n}cosωn using Time scaling property.
Solution −
We already know that the Ztransformation of the signal cos(ωn)cos(ωn) is given by −
Now, applying Time scaling property, the Ztransformation of ancosωnancosωn can be written as;
Successive Differentiation
Successive Differentiation property shows that Ztransform will take place when we differentiate the discrete signal in time domain, with respect to time. This is shown as below.
Proof −
Consider the LHS of the equation −
(Hence Proved)
ROC: R1< Mod (Z) <R2
Example
Let us find the Ztransform of a signal given by
By property we can write
Now, Z[n.y] can be found out by again applying the property,
Convolution
This depicts the change in Zdomain of the system when a convolution takes place in the discrete signal form, which can be written as −
Proof −
Let nk = l, then the above equation cab be written as −
(Hence Proved)
ROC : ROC ⋂ ROC2
Example
Let us find the convolution given by two signals
x_{1}(n) = {3,−2,2} ...(eq. 1)
x_{2}(n) = {2,0≤4 and 0 elsewhere} ...(eq. 2)
Ztransformation of the first equation can be written as;
Ztransformation of the second signal can be written as;
So, the convolution of the above two signals is given by −
Taking the inverse Ztransformation we get,
x(n) = {6,2,6,6,6,0,4}
Initial Value Theorem
If x(n) is a causal sequence, which has its Ztransformation as X(z), then the initial value theorem can be written as;
Proof − We know that,
Expanding the above series, we get;
In the above case if Z → ∞ then Z^{−n}→0 (Because n>0)
Therefore, we can say;
(Hence Proved)
Final Value Theorem
Final Value Theorem states that if the Ztransform 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 −
Conditions −
Proof − We know that
Here, we can apply advanced property of onesided ZTransformation. So, the above equation can be rewritten as;
Now putting z = 1 in the above equation, we can expand the above equation −
This can be formulated as;
(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
Now let us find the Final value of signal applying the theorem
Some other properties of Ztransform are listed below −
Differentiation in Frequency
It gives the change in Zdomain of the signal, when its discrete signal is differentiated with respect to time.
Its ROC can be written as;
Example
Let us find the value of x(n) through Differentiation in frequency, whose discrete signal in Zdomain is given by
By property, we can write that
Multiplication in Time
It gives the change in Zdomain of the signal when multiplication takes place at discrete signal level.
Conjugation in Time
This depicts the representation of conjugated discrete signal in Zdomain.
3 videos50 docs54 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
3 videos50 docs54 tests
