LTI System Analysis MCQ Quiz – Objective Question with Answer for LTI System Analysis

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?

Which of the following series has an ROC as mentioned 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?
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)

=>\(\frac{Y(z)}{X(z)}=\frac{2}{1-0.5z^{-1}}\)=H(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

X(z)=\(\frac{1}{1-1.5z^{-1}+0.5z^{-2}}=1+\frac{3}{2}z^{-1}+\frac{7}{4}z^{-2}+\frac{15}{8}z^{-3}+\frac{31}{16}z^{-4}+…\)

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

 

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:

Thus

X(z)=\(\frac{1}{1-1.5z^{-1}+0.5z^{-2}}=2z^2+6z^3+14z^4+30z^5+62z^6+…\)

In this case x(n)=0 for n≥0.Thus we obtain x(n)= {…..62,30,14,6,2,0,0}

 

37. What is the inverse z-transform of X(z)=log(1+az-1) |z|>|a|?

A. x(n)=(-1)n+1 \(\frac{a^{-n}}{n}\), n≥1; x(n)=0, n≤0

B. x(n)=(-1)n-1 \(\frac{a^{-n}}{n}\), n≥1; x(n)=0, n≤0

C. x(n)=(-1)n+1 \(\frac{a^{-n}}{n}\), n≥1; x(n)=0, n≤0

D. None of the mentioned

Answer: C

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

X(z)=\(\sum_{n=1}^∞ \frac{(-1)^{n+1} a^n z^{-n}}{n}\)

Thus

x(n)=(-1)n+1 \(\frac{a^n}{n}\), n≥1

=0, n≤0.

 

38. What is the proper fraction and polynomial form of the improper rational transform

X(z)=\(\frac{1+3z^{-1}+\frac{11}{6} z^{-2}+\frac{1}{3} z^{-3}}{1+\frac{5}{6} z^{-1}+\frac{1}{6} z^{-2}}\)?

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

X(z)=1+2z-1+\(\frac{\frac{1}{6} z^{-1}}{1+\frac{5}{6} z^{-1}+\frac{1}{6} z^{-2}}\).

 

39. What is the partial fraction expansion of the proper function
X(z)=\(\frac{1}{1-1.5z^{-1}+0.5z^{-2}}\)?

A. \(\frac{2z}{z-1}-\frac{z}{z+0.5}\)

B. \(\frac{2z}{z-1}+\frac{z}{z-0.5}\)

C. \(\frac{2z}{z-1}+\frac{z}{z+0.5}\)

D. \(\frac{2z}{z-1}-\frac{z}{z-0.5}\)

Answer: D

First we eliminate the negative powers of z by multiplying both numerator and denominator by z2.

Thus we obtain X(z)=\(\frac{z^2}{z^2-1.5z+0.5}\)
The poles of X(z) are p1=1 and p2=0.5. Consequently, the expansion will be

\(\frac{X(z)}{z} = \frac{z}{(z-1)(z-0.5)} = \frac{2}{(z-1)} – \frac{1}{(z-0.5)}\)

(obtained by applying partial fractions)

=>X(z)=\(\frac{2z}{(z-1)}-\frac{z}{(z-0.5)}\).

 

40. What is the partial fraction expansion of
X(z)=\(\frac{1+z^{-1}}{1-z^{-1}+0.5z^{-2}}\)?

A. \(\frac{z(0.5-1.5j)}{z-0.5-0.5j} – \frac{z(0.5+1.5j)}{z-0.5+0.5j}\)

B. \(\frac{z(0.5-1.5j)}{z-0.5-0.5j} + \frac{z(0.5+1.5j)}{z-0.5+0.5j}\)

C. \(\frac{z(0.5+1.5j)}{z-0.5-0.5j} – \frac{z(0.5-1.5j)}{z-0.5+0.5j}\)

D. \(\frac{z(0.5+1.5j)}{z-0.5-0.5j} + \frac{z(0.5-1.5j)}{z-0.5+0.5j}\)

Answer: B

To eliminate the negative powers of z, we multiply both numerator and denominator by z2. Thus,

X(z)=\(\frac{z(z+1)}{z^{-2}-z+0.5}\)

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)= \(\frac{z(0.5-1.5j)}{z-0.5-0.5j} + \frac{z(0.5+1.5j)}{z-0.5+0.5j}\).

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