This set of Digital Signal Processing Multiple Choice Questions & Answers (MCQs) focuses on “Analysis of Discrete time LTI Systems”.

1. 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

View Answer

Explanation: 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 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

View Answer

Explanation: 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 as 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)?

Step-1:Folding

Step-2:Multiplicaton with x(k)

Step-3:Shifting

Step-4:Summation

a) 1-2-3-4

b) 1-2-4-3

c) 2-1-3-4

d) 1-3-2-4

View Answer

Explanation: 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 ∞.

a) {1,3,6,3,1}

b) {1,2,3,2,1}

c) {1,3,6,5,3}

d) {1,1,1,0,0}

View Answer

Explanation: 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)=a^{n}u(n),|a|<1with the input sequence x(n)=u(n).

a) (1-a^{(n+1)})/(1-a)

b) (1-a^{(n-1)})/(1-a)

c) (1+a^{(n+1)})/(1+a)

d) None of the mentioned

View Answer

Explanation: 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+a

^{2}.1=1+a+a

^{2}

Similarly, y(n)=1+a+a

^{2}+….a

^{n}= (1-a

^{(n+1)})/(1-a)

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

a) True

b) False

View Answer

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

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

a) (1/2)^{n}[2-(1/2)^{n}], n<0

b) (1/2)^{n}[2-(1/2)^{n}], n>0

c) (1/2)^{n}[2+(1/2)^{n}], n<0

d) (1/2)^{n}[2+(1/2)^{n}], n>0

View Answer

Explanation: Let h2(n) be shifted and folded.

so, h(k)=h1(n)*h2(n)=

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

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

a) True

b) False

View Answer

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

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

View Answer

Explanation: Let us consider a LTI system having an output at time n=n0given by the convolution formula

=(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.

that is, h(n)=0 for n<0 .

10. x(n)*δ(n-n_{0})=

a) x(n+n_{0})

b) x(n-n_{0})

c) x(-n-n_{0})

d) x(-n+n_{0})

View Answer

11. Is the system with impulse response h(n)=2^{n}u(n-1) stable?

a) True

b) False

View Answer

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