Test: Discrete Time System Analysis - Electrical Engineering (EE) MCQ

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).

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.

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 ∞.

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}.

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+a².1=1+a+a²
Similarly, y(n)=1+a+a²+….aⁿ= (1-a⁽ⁿ⁺¹⁾)/(1-a).

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

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.

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

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 .

Explanation: