Signals & Systems Questions and Answers – Discrete Time Convolution – 2

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) [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)
View Answer

Answer: c
Explanation: 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
View Answer

Answer: d
Explanation: 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) Sy = SxSh, Sx=∑x(k) and Sh = ∑h(n-k)
b) Sy = Sx+Sh, Sx=∑x(k-1) and Sh = ∑h(n-k)
c) Sy = Sx-Sh, Sx=∑x(k) and Sh = ∑h(n-k)
d) Sy = Sx*Sh, Sx=∑x(n) and Sh = ∑h(n-k)
View Answer

Answer: a
Explanation: Sy=Sx+Sh, , Sx = ∑x(k) and Sh = ∑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).
advertisement
advertisement

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

Answer: b
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

Answer: b
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.
Note: Join free Sanfoundry classes at Telegram or Youtube

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

Answer: c
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

Answer: a
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.
advertisement

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

Answer: b
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

Answer: a
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.
advertisement

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

Answer: a
Explanation: Using tabular method:
The convolution of x(n)={1,2,3,1} & h(n)={1,2,1,-1} is {1,4,8,8,3,-2,-1}, origin at 4
Hence, result is {1,4,8,8,3,-2,-1}, origin at 4.

Sanfoundry Global Education & Learning Series – Signals & Systems.

To practice all areas of Signals & Systems for Entrance exams, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.