Test: Z Transform Inversion - Electrical Engineering (EE) MCQ

Explanation: All the methods mentioned above can be used to calculate the inverse z-transform of the given signal.

Explanation: Since the ROC is the exterior circle, we expect x(n) to be a causal signal. Thus we seek a power series expansion in negative powers of ‘z’. By dividing the numerator of X(z) by its denominator, we obtain the power series

So, we obtain x(n)= {1,3/2,7/4,15/8,31/16,….}.

Explanation: Using the power series expansion for log(1+x), with |x|<1, we have

Explanation: First, we note that we should reduce the numerator so that the terms z-2 and z -3 are eliminated. Thus we should carry out the long division with these two polynomials written in the reverse order. We stop the division when the order of the remainder becomes z -1. Then we obtain
X(z)= 1+2z -1+(1/6 z^-1)/(1+5/6 z^-1+1/6 z^-2 ).

Explanation: First we eliminate the negative powers of z by multiplying both numerator and denominator by z2.
Thus we obtain X(z)= z²/(z²-1.5z+0.5)
The poles of X(z) are p1=1 and p2=0.5. Consequently, the expansion will be
(X(z))/z = z/((z-1)(z-0.5)) = 2/((z-1) ) – 1/((z-0.5) )( obtained by applying partial fractions)
=>X(z)= 2z/(z-1)-z/(z-0.5).

Explanation: To eliminate the negative powers of z, we multiply both numerator and denominator by z2. Thus,
X(z)=(z(z+1))/(z^-2-z+0.5)
The poles of X(z) are complex conjugates p1=0.5+0.5j and p2=0.5-0.5j
Consequently the expansion will be
X(z)= (z(0.5-1.5j))/(z-0.5-0.5j) + (z(0.5+1.5j))/(z-0.5+0.5j).

Explanation: First we express X(z) in terms of positive powers of z, in the form X(z)=z³/((z+1)〖(z-1)〗² )
X(z) has a simple pole at z=-1 and a double pole at z=1. In such a case the approximate partial fraction expansion is
(X(z))/z = z²/((z+1)〖(z-1)〗² ) =A/(z+1) + B/(z-1) + C/〖(z-1)〗²
On simplifying, we get the values of A, B and C as 1/4, 3/4 and 1/2 respectively.
Therefore, we get X(z)= z/(4(z+1)) + 3z/(4(z-1)) + z/(2〖(z-1)〗² ) .

Explanation: The partial fraction expansion for the given X(z) is
X(z)= 2z/(z-1)-z/(z-0.5)
In case when ROC is |z|>1, the signal x(n) is causal and both the terms in the above equation are causal terms. Thus, when we apply inverse z-transform to the above equation, we get
x(n)=2(1)ⁿu(n)-(0.5)ⁿu(n)=(2-0.5ⁿ)u(n).

Explanation: The partial fraction expansion for the given X(z) is
X(z)= 2z/(z-1)-z/(z-0.5)
In case when ROC is |z|<0.5,the signal is anti causal. Thus both the terms in the above equation are anti causal terms. So, if we apply inverse z-transform to the above equation we get x(n)= [-2+0.5ⁿ]u(-n-1).

Explanation: The partial fraction expansion of the given X(z) is
X(z)= 2z/(z-1)-z/(z-0.5)
In this case ROC is 0.5<|z|<1 is a ring, which implies that the signal is two sided. Thus one of the signal corresponds to a causal signal and the other corresponds to an anti causal signal. Obviously, the ROC given is the overlapping of the regions |z|>0.5 and |z|<1. Hence the pole p2=0.5 provides the causal part and the pole p1=1 provides the anti causal part. SO, if we apply the inverse z-transform we get x(n)= -2u(-n-1)-(0.5)ⁿu(n).