1 Crore+ students have signed up on EduRev. Have you? 
Discrete Time Fourier Transform(DTFT) exists for energy and power signals. Ztransform also exists for neither energy nor Power (NENP) type signal, up to a certain extent only. The replacement z = e^{jw} is used for Ztransform to DTFT conversion only for absolutely summable signal.
So, the Ztransform of the discrete time signal x(n) in a power series can be written as −
The above equation represents a twosided Ztransform equation.
Generally, when a signal is Ztransformed, it can be represented as −
Or
If it is a continuous time signal, then Ztransforms are not needed because Laplace transformations are used. However, Discrete time signals can be analyzed through Ztransforms only.
Region of Convergence
Region of Convergence is the range of complex variable Z in the Zplane. The Z transformation of the signal is finite or convergent. So, ROC represents those set of values of Z, for which X(Z) has a finite value.
Properties of ROC
The Ztransform is uniquely characterized by −
Signals and their ROC
x(n)  X(Z)  ROC 

δ(n)  1  Entire Z plane 
U(n)  1/(1−Z^{−1})  Mod(Z)>1 
a^{n}u(n)  1/(1−aZ^{−1})  Mod(Z)>Mod(a) 
−a^{n}u(−n−1)  1/(1−aZ^{−1})  Mod(Z)<Mod(a) 
na^{n}u(n)  aZ^{−1}/(1−aZ^{−1})^{2}  Mod(Z)>Mod(a) 
−a^{n}u(−n−1)  aZ^{−1}/(1−aZ^{−1})^{2}  Mod(Z)<Mod(a) 
U(n)cosωn  (Z^{2}−Zcosω)/(Z^{2}−2Zcosω+1)  Mod(Z)>1 
U(n)sinωn  (Zsinω)/(Z^{2}−2Zcosω+1)  Mod(Z)>1 
Example
Let us find the Ztransform and the ROC of a signal given as x(n)={7,3,4,9,5}, where origin of the series is at 3.
Solution − Applying the formula we have −
ROC is the entire Zplane excluding Z = 0, ∞, ∞
3 videos50 docs54 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
3 videos50 docs54 tests
