250+ TOP MCQs on Discrete Time Convolution – 2 and Answers

Signals & Systems Questions and Answers for Entrance exams on “Discrete Time Convolution – 2”.

1. What is the distributive property of a discrete time convolution?
A. [x₁(n) + x₂(n)]*h(n) = x₁(n)* [x₂(n) + h(n)]
B. [x₁(n) + x₂(n)] = x₁(n)* [x₂(n) + h(n)]
C. [x₁(n) + x₂(n)]*h(n) = x₁(n)* h(n)+ x₂(n) * h(n)
D. [x₁(n) + x₂(n)]*h(n) = x₁(n)* h(n)* x₂(n) * h(n)

Answer: C
Clarification: x₁(n) + x₂(n)]*h(n) = x₁(n)* h(n) + x₂(n)* h(n),)], x₁(n) and x₂(n) are inputs and h(n) is the impulse response of discrete time system.

2. What is this property of discrete time convolution?
x[n]*h[n]=y[n], then x[n]*h[n-n₀] = x[n-n₀]*h[n] = y[n-n₀]
A. Distributive
B. Commutative
C. Sym property
D. Shifting property

Answer: D
Clarification: x[n]*h[n]=y[n], then x[n]*h[n-n₀]= x[n-n₀]*h[n] = y[n-n₀]
This gives x[n-n₁]*h[n-n₀] = y[n-n₀-n₁]
Is the shifting property of discrete time convolution.

3. What is the sum of impulses in a convolution sum of two discrete time sequences?
A. S_y = S_xS_h, S_x=∑x(k) and S_h = ∑h(n-k)
B. S_y = S_x+S_h, S_x=∑x(k-1) and S_h = ∑h(n-k)
C. S_y = S_x-S_h, S_x=∑x(k) and S_h = ∑h(n-k)
D. S_y = S_x*S_h, S_x=∑x(n) and S_h = ∑h(n-k)

Answer: A
Clarification: S_y=S_x+S_h, , S_x = ∑x(k) and S_h = ∑h(n-k), the sum of impulses in a convolution sum of two discrete time sequences is the product of the sums of the impulses in the two individual sequences. Here, y(n)=x(n)*h(n).

4. How can a cascade connected discrete time system respresented?
A. y[n] = x[n] + t[n] + r[n]
B. y[n] = x[n] * t[n] * r[n]
C. y[n] = x[n] * t[n] + r[n]
D. y[n] = x[n] + t[n] * r[n]

Answer: B
Clarification: y[n] = x[n]*t[n]*r[n], is how we can represent a cascade connected discrete time system.
Proof:
If Y1[n]=x[n]*t[n]
y[n]=Y1*r[n], using properties.

5. How can a parallel connected discrete time system respresented?
A. y[n] = x[n] + t[n] + r[n]
B. y[n] = x[n] * t[n] * r[n]
C. y[n] = x[n] * (t[n] + r[n])
D. y[n] = x[n] + t[n] * r[n]

Answer: B
Clarification: y[n] = x[n]*(t[n]+r[n]) is how we can represent a parallel connected discrete time system.
Proof:
If Y1[n]=t[n]+r[n]
y[n]=Y1*r[n], using properties.

6. How can we solve discrete time convolution problems?
A. The graphical method only
B. Graphical method and tabular method
C. Graphical method, tabular method and matrix method
D. Graphical method, tabular method, matrix method and summation method

Answer: C
Clarification: Discrete time convolution problems are mostly solved by a graphical method, tabular method and matrix method. Even if the graphical method is very popular, the tabular and matrix method is more easy to calculate.

7. Which method uses sum of diagonal elements for discrete time convolution?
A. Matrix method only
B. Graphical method and tabular method
C. Graphical method, tabular method and matrix method
D. Graphical method, tabular method, matrix method and summation method

Answer: A
Clarification: Even if the graphical method is very popular, the tabular and matrix method is more easy to calculate. And matrix method uses the sum of diagonal elements for discrete time convolution.

8. Which method is close to a graphical method for discrete time convolution?
A. Matrix method only
B. Tabular method
C. Tabular method and matrix method
D. Summation method

Answer: B
Clarification: Tabular method is close to graphical method for discrete time convolution except that tabular representation of sequences is employed instead of graphical representation. Here every input is folded and shifted ad represented by a row.

9. The sample of x(n)={1,2,3,1} and h(n)={1,2,1,-1}, origin at 2, is 7.
A. True
B. False

Answer: A
Clarification: The input starts at n=0 and impulse at n=-1. So, output starts at n=0+(-1)=-1.
Output at is =4+4=1=7 samples. So, its true.

10. The convolution of x(n)={1,2,3,1} and h(n)={1,2,1,-1}, origin at 2, is?
A. {1,4,8,8,3,-2,-1}, origin at 4
B. {1,4,8,8,3,-2,1}, origin at 4
C. {1,3,8,8,3,-2,-1}, origin at 4
D. {1,4,8,3,-2,-1}, origin at 4