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Which of the following method is used to find the inverse ztransform of a signal?
Explanation: All the methods mentioned above can be used to calculate the inverse ztransform of the given signal.
What is the inverse ztransform of X(z)=1/(11.5z^{1}+0.5z^{2} ) if ROC is z>1?
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,….}.
What is the inverse ztransform of X(z)=log(1+az^{1}) z>a?
Explanation: Using the power series expansion for log(1+x), with x<1, we have
What is the proper fraction and polynomial form of the improper rational transform
X(z)= (1+3z^{1}+11/6 z^{2}+1/3 z^{3})/(1+5/6 z^{1}+1/6 z^{2} )?
Explanation: First, we note that we should reduce the numerator so that the terms z2 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} ).
What is the partial fraction expansion of the proper function X(z)= 1/(11.5z^{1}+0.5z^{2} )?
Explanation: First we eliminate the negative powers of z by multiplying both numerator and denominator by z2.
Thus we obtain X(z)= z^{2}/(z^{2}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/((z1)(z0.5)) = 2/((z1) ) – 1/((z0.5) )( obtained by applying partial fractions)
=>X(z)= 2z/(z1)z/(z0.5).
What is the partial fraction expansion of X(z)= (1+z^{1})/(1z^{1}+0.5z^{2} )?
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=0.50.5j
Consequently the expansion will be
X(z)= (z(0.51.5j))/(z0.50.5j) + (z(0.5+1.5j))/(z0.5+0.5j).
What is the partial fraction expansion of X(z)=1/((1+z^{1} )(1z^{1})^{2})?
Explanation: First we express X(z) in terms of positive powers of z, in the form X(z)=z^{3}/((z+1)〖(z1)〗^{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 = z^{2}/((z+1)〖(z1)〗^{2} ) =A/(z+1) + B/(z1) + C/〖(z1)〗^{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(z1)) + z/(2〖(z1)〗^{2} ) .
What is the inverse ztransform of X(z)= 1/(11.5z^{1}+0.5z^{2}2 ) if ROC is z>1?
Explanation: The partial fraction expansion for the given X(z) is
X(z)= 2z/(z1)z/(z0.5)
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 ztransform to the above equation, we get
x(n)=2(1)^{n}u(n)(0.5)^{n}u(n)=(20.5^{n})u(n).
What is the inverse ztransform of X(z)= 1/(11.5z^{1}+0.5z^{2} ) if ROC is z<0.5?
Explanation: The partial fraction expansion for the given X(z) is
X(z)= 2z/(z1)z/(z0.5)
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 ztransform to the above equation we get x(n)= [2+0.5^{n}]u(n1).
What is the inverse ztransform of X(z)= 1/(11.5z^{1}+0.5z^{2} ) if ROC is 0.5<z<1?
Explanation: The partial fraction expansion of the given X(z) is
X(z)= 2z/(z1)z/(z0.5)
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 ztransform we get x(n)= 2u(n1)(0.5)^{n}u(n).
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