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

1. What is the distributive property of a discrete time convolution?

a) [x_{1}(n) + x_{2}(n)]*h(n) = x_{1}(n)* [x_{2}(n) + h(n)]

b) [x_{1}(n) + x_{2}(n)] = x_{1}(n)* [x_{2}(n) + h(n)]

c) [x_{1}(n) + x_{2}(n)]*h(n) = x_{1}(n)* h(n)+ x_{2}(n) * h(n)

d) [x_{1}(n) + x_{2}(n)]*h(n) = x_{1}(n)* h(n)* x_{2}(n) * h(n)

View Answer

Explanation: x

_{1}(n) + x

_{2}(n)]*h(n) = x

_{1}(n)* h(n) + x

_{2}(n)* h(n),)], x

_{1}(n) and x

_{2}(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-n_{0}] = x[n-n_{0}]*h[n] = y[n-n_{0}]

a) Distributive

b) Commutative

c) Sym property

d) Shifting property

View Answer

Explanation: x[n]*h[n]=y[n], then x[n]*h[n-n

_{0}]= x[n-n

_{0}]*h[n] = y[n-n

_{0}] This gives x[n-n

_{1}]*h[n-n

_{0}] = y[n-n

_{0}-n

_{1}] 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_{x}S_{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)

View Answer

Explanation: 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]

View Answer

Explanation: 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]

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

Explanation: 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

View Answer

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