Discrete Time System Analysis MCQ Quiz – Objective Question with Answer for Discrete Time System Analysis

1. Resolve the sequence Resolve the sequence into a sum of weighted impulse sequences.

A. 2δ(n)+4δ(n-1)+3δ(n-3)
B. 2δ(n+1)+4δ(n)+3δ(n-2)
C. 2δ(n)+4δ(n-1)+3δ(n-2)
D. None of the mentioned

Answer: B

We know that, x(n)δ(n-k)=x(k)δ(n-k)
x(-1)=2=2δ(n+1)
x(0)=4=4δ(n)
x(2)=3=3δ(n-2)

Therefore, x(n)= 2δ(n+1)+4δ(n)+3δ(n-2).

2. The formula y(n)=\(\sum_{k=-\infty}^{\infty}x(k)h(n-k)\) that gives the response y(n) of the LTI system as the function of the input signal x(n) and the unit sample response h(n) is known as ______________

A. Convolution sum
B. Convolution product
C. Convolution Difference
D. None of the mentioned

Answer: A

The input x(n) is convoluted with the impulse response h(n) to yield the output y(n). As we are summing the different values, we call it a Convolution sum.

 

3. What is the order of the four operations that are needed to be done on h(k) in order to convolute x(k) and h(k)?

A. Step-1:Folding
B. Step-2:Multiplication with x(k)
C. Step-3:Shifting
D. All of the above

Answer: D

The four operations that are needed to be done on h(k) in order to convolute x(k) and h(k) is

Step-1:Folding
Step-2:Multiplication with x(k)
Step-3:Shifting
Step-4:Summation

First, the signal h(k) is folded to get h(-k). Then it is shifted by n to get h(n-k). Then it is multiplied by x(k) and then summed over -∞ to ∞.

4. The impulse response of a LTI system is h(n)={1,1,1}. What is the response of the signal to the input x(n)={1,2,3}?

A. {1,3,6,3,1}
B. {1,2,3,2,1}
C. {1,3,6,5,3}
D. {1,1,1,0,0}

Answer: C

Let y(n)=x(n)*h(n)(‘*’ symbol indicates convolution symbol)

From the formula of convolution we get,
y(0)=x(0)h(0)=1.1=1
y(1)=x(0)h(1)+x(1)h(0)=1.1+2.1=3
y(2)=x(0)h(2)+x(1)h(1)+x(2)h(0)=1.1+2.1+3.1=6
y(3)=x(1)h(2)+x(2)h(1)=2.1+3.1=5
y(4)=x(2)h(2)=3.1=3

Therefore, y(n)=x(n)*h(n)={1,3,6,5,3}.

 

5. Determine the output y(n) of a LTI system with impulse response h(n)=anu(n), |a|<1with the input sequence x(n)=u(n).

A. \(\frac{1-a^{n+1}}{1-a}\)

B. \(\frac{1-a^{n-1}}{1-a}\)

C. \(\frac{1+a^{n+1}}{1+a}\)

D. None of the mentioned

Answer: A

Now fold the signal x(n) and shift it by one unit at a time and sum as follows
y(0)=x(0)h(0)=1

y(1)=h(0)x(1)+h(1)x(0)=1.1+a.1=1+a

y(2)=h(0)x(2)+h(1)x(1)+h(2)x(0)=1.1+a.1+a2.1=1+a+a2

Similarly, y(n)=1+a+a2+….an=\(\frac{1-a^{n+1}}{1-a}\).

 

6. x(n)*(h1(n)*h2(n))=(x(n)*h1(n))*h2(n).

A. True
B. False

Answer: A

According to the properties of convolution, the Convolution of three signals obeys the Associative property.

 

7. Determine the impulse response for the cascade of two LTI systems having impulse responses h1(n)=\((\frac{1}{2})^2\) u(n) and h2(n)=\((\frac{1}{4})^2\) u(n).

A. \((\frac{1}{2})^n[2-(\frac{1}{2})^n]\), n<0

B. \((\frac{1}{2})^n[2-(\frac{1}{2})^n]\), n>0

C. \((\frac{1}{2})^n[2+(\frac{1}{2})^n]\), n<0

D. \((\frac{1}{2})^n[2+(\frac{1}{2})^n]\), n>0

Answer: B

Let h2(n) be shifted and folded.

so, h(k)=h1(n)*h2(n)=\(\sum_{k=-\infty}^{\infty} h_1 (k)h_2 (n-k)\)

For k<0, h1(n)= h2(n)=0 since the unit step function is defined only on the right hand side.

Therefore, h(k)=\((\frac{1}{2})^k (\frac{1}{4})^{n-k}\)

=>h(n)=\(\sum_{k=0}^n (\frac{1}{2})^k (\frac{1}{4})^{n-k}\)

=\((\frac{1}{4})^n \sum_{k=0}^n(2)^k\)

=\((\frac{1}{4})^n.(2^{n+1}-1)\)

=\((\frac{1}{2})^n[2-(\frac{1}{2})^n], n>0\)

8. x(n)*[h1(n)+h2(n)]=x(n)*h1(n)+x(n)*h2(n).

A. True
B. False

Answer: A

According to the properties of the convolution, convolution exhibits the distributive property.

 

9. An LTI system is said to be causal if and only if?

A. Impulse response is non-zero for positive values of n
B. Impulse response is zero for positive values of n
C. Impulse response is non-zero for negative values of n
D. Impulse response is zero for negative values of n

Answer: D

Let us consider a LTI system having an output at time n=n0 given by the convolution formula
y(n)=\(\sum_{k=-{\infty}}^{\infty}h(k)x(n_0-k)\)

We split the summation into two intervals.

=>y(n)=\(\sum_{k=-{\infty}}^{-1}h(k)x(n_0-k)+\sum_{k=0}^{\infty}h(k)x(n_0-k)\)

=(h(0)x(n0)+h(1)x(n0-1)+h(2)x(n0-2)+….)+(h(-1)x(n0+1)+h(-2)x(n0+2)+…)

As per the definition of the causality, the output should depend only on the present and past values of the input. So, the coefficients of the terms x(n0+1), x(n0+2)…. should be equal to zero.

i.e h(n)=0 for n<0 .

 

10. x(n)*δ(n-n0)=?

A. x(n+n0)
B. x(n-n0)
C. x(-n-n0)
D. x(-n+n0)

Answer: B

x(n)*δ(n-n0)=\(\sum_{k=-{\infty}}^{\infty} x(k)\delta(n-k-n_0)\)
=x(k)|k=n-n0
=x(n-n0)

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