21. If W4100 = Wx200, then what is the value of x?
A. 2
B. 4
C. 8
D. 16
Answer: C
We know that according to the periodicity and symmetry property,
100/4=200/x=>x=8.
22. What are the values of z for which the value of X(z)=0?
A. Poles
B. Zeros
C. Solutions
D. None of the mentioned
Answer: B
For a rational z-transform X(z) to be zero, the numerator of X(z) is zero and the solutions of the numerator are called ‘zeros’ of X(z).
23. What are the values of z for which the value of X(z)=∞?
A. Poles
B. Zeros
C. Solutions
D. None of the mentioned
Answer: A
For a rational z-transform X(z) to be infinity, the denominator of X(z) is zero and the solutions of the denominator are called ‘poles’ of X(z).
24. If X(z) has M finite zeros and N finite poles, then which of the following condition is true?
A. |N-M| poles at origin(if N>M)
B. |N+M| zeros at origin(if N>M)
C. |N+M| poles at origin(if N>M)
D. |N-M| zeros at origin(if N>M)
Answer: D
If X(z) has M finite zeros and N finite poles, then X(z) can be rewritten as X(z)=z -M+N.X'(z).
So, if N>M then z has positive power. So, it has |N-M| zeros at origin.
25. If X(z) has M finite zeros and N finite poles, then which of the following condition is true?
A. |N-M| poles at origin(if N < M)
B. |N+M| zeros at origin(if N < M)
C. |N+M| poles at origin(if N < M)
D. |N-M| zeros at origin(if N < M)
Answer: A
If X(z) has M finite zeros and N finite poles, then X(z) can be rewritten as X(z)=z-M+N.X'(z).
So, if N < M then z has a negative power. So, it has |N-M| poles at the origin.
26. Which of the following signals have a pole-zero plot as shown below?
A. a.u(n)
B. u(an)
C. anu(n)
D. none of the mentioned
Answer: C
From the given pole-zero plot, the z-transform of the signal has one zero at z=0 and one pole at z=a.
So, we obtain X(z)=z/(z-A.
By applying inverse z-transform for X(z), we get
x(n)= anu(n).
27. Which of the following signals have a pole-zero plot as shown below?(Let M=8 in the figure)
A. x(n)=an, 0≤n≤8 =0, elsewhere
B. x(n)=an, 0≤n≤7 =0, elsewhere
C. x(n)=a-n, 0≤n≤8 =0, elsewhere
D. x(n)=a-n, 0≤n≤7 =0, elsewhere
Answer: B
From the figure given, the z-transform of the signal has 8 zeros on the circle of radius ‘a’ and 7 poles at the origin.
So, X(z) is of the form
X(z)=\(\frac{(z-z_1) (z-z_2)……(z-z_8)}{z^7}\)
By applying inverse z-transform, we get x(n)=an, 0≤n≤7
=0, elsewhere.
28. The z-transform X(z) of the signal x(n)=anu(n) has:
A. One pole at z=0 and one zero at z=a
B. One pole at z=0 and one zero at z=0
C. One pole at z=a and one zero at z=a
D. One pole at z=a and one zero at z=0
Answer: D
The z-transform of the given signal is X(z)=z/(z-A.
So, it has one pole at z=a and one zero at z=0.
29. What is the nature of the signal whose pole-zero plot is as shown?
A. Rising signal
B. Constant signal
C. Decaying signal
D. None of the mentioned
Answer: C
From the pole-zero plot, it is shown that r < 1, so the signal is a decaying signal.
30. What are the values of z for which the value of X(z)=0?
A. Poles
B. Zeros
C. Solutions
D. None of the mentioned
Answer: B
For a rational z-transform X(z) to be zero, the numerator of X(z) is zero and the solutions of the numerator are called ‘zeros’ of X(z).
31. If Y(z) is the z-transform of the output function, X(z) is the z-transform of the input function and H(z) is the z-transform of the system function of the LTI system, then H(z)=?
A. \(\frac{Y(z)}{X(z)}\)
B. \(\frac{X(z)}{Y(z)}\)
C. Y(z).X(z)
D. None of the mentioned
Answer: A
We know that for an LTI system, y(n)=h(n)*x(n)
On applying z-transform on both sides we get,
Y(z)=H(z).X(z)=>H(z)=\(\frac{Y(z)}{X(z)}\)
32. What is the system function of the system described by the difference equation y(n)=0.5y(n-1)+2x(n)?
A. \(\frac{2}{1+0.5z^{-1}}\)
B. \(\frac{0.5}{1+2z^{-1}}\)
C. \(\frac{0.5}{1-2z^{-1}}\)
D. \(\frac{2}{1-0.5z^{-1}}\)
Answer: D
Given difference equation of the system is y(n)=0.5y(n-1)+2x(n)
On applying z-transform on both sides we get, Y(z)=0.5z-1Y(z)+2X(z)
33. 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).
34. 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.
35. 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
36. 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