12. What is the unit sample response of the system described by the difference equation y(n)=0.5y(n-1)+2x(n)?
A. 0.5(2)nu(n)
B. 2(0.5)nu(n)
C. 0.5(2)nu(-n)
D. 2(0.5)nu(-n)
Answer: B
By applying the z-transform on both sides of the difference equation given in the question we obtain,
\(\frac{Y(z)}{X(z)}=\frac{2}{1-0.5z^{-1}}\)=H(z)
By applying the inverse z-transform we get h(n)=2(0.5)nu(n).
13. Which of the following method is used to find the inverse z-transform of a signal?
A. Counter integration
B. Expansion into a series of terms
C. Partial fraction expansion
D. All of the mentioned
Answer: D
All the methods mentioned above can be used to calculate the inverse z-transform of the given signal.
14. What is the inverse z-transform of
X(z)=\(\frac{1}{1-1.5z^{-1}+0.5z^{-2}}\) if ROC is |z|>1?
A. {1,3/2,7/4,15/8,31/16,….}
B. {1,2/3,4/7,8/15,16/31,….}
C. {1/2,3/4,7/8,15/16,31/32,….}
D. None of the mentioned
Answer: A
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
15. What is the inverse z-transform of
X(z)=\(\frac{1}{1-1.5z^{-1}+0.5z^{-2}}\) if ROC is |z| < 0.5?
A. {….62,30,14,6,2}
B. {…..62,30,14,6,2,0,0}
C. {0,0,2,6,14,30,62…..}
D. {2,6,14,30,62…..}
Answer: B
In this case, the ROC is the interior of a circle. Consequently, the signal x(n) is anti causal. To obtain a power series expansion in positive powers of z, we perform the long division in the following way:
A. 1+2z-1+\(\frac{\frac{1}{6}z^{-1}}{1+\frac{5}{6} z^{-1}+\frac{1}{6} z^{-2}}\)
B. 1-2z-1+\(\frac{\frac{1}{6} z^{-1}}{1+\frac{5}{6} z^{-1}+\frac{1}{6} z^{-2}}\)
C. 1+2z-1+\(\frac{\frac{1}{3} z^{-1}}{1+\frac{5}{6} z^{-1}+\frac{1}{6} z^{-2}}\)
D. 1+2z-1–\(\frac{\frac{1}{6} z^{-1}}{1+\frac{5}{6} z^{-1}+\frac{1}{6} z^{-2}}\)
Answer: A
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