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Signals & Systems Questions on “Convolution : Impulse Response Representation for LTI Systems – 2”.
1. The convolution sum is given by _____ equation.
Answer: A
Clarification: By the definition of convolution sum we can write the equation as
x[n]*h[n] = ∑∞k=-∞ x[k]h[n-k].
2. When the sequences x1 [n] = u [n] and x2 [n] = u [n-3], the output of LTI system is given as _____
A. y[n] = n-2, n>3
B. y[n] = n-2, n≥3
C. y[n] = n+2, n>3
D. y[n] = n-2, n≤3
Clarification: The output y[n] =∑∞k=-∞u(k)u(n-k-3), by solving the above summation either by graphically or by direct summation we get .
3. The impulse response h (t) of an LTI system is given by e-2t.u(t) . What is the step response?
A. y(t) = 1⁄2 (1 – e-2t) u (t)
B. y(t) = 1⁄2 (1 – e-2t)
C. y(t) = (1- e-2t) u (t)
D. y(t) = 1⁄2 (e-2t) u (t)
Clarification: Given x (t) = u (t) and h (t) = e-2t.u(t). By using convolution integral
We get output y (t) as y(t) = 1⁄2 (1 – e-2t) u (t).
4. Is (t)*h(t) = h(t)*x(t)?
A. True
B. False
Clarification: By the properties of convolution we say that x(t)*h(t) = h(t)*x(t)
It can be proved using the convolution integral
5. Compute u (t) convolved with itself?
A. y(t)=t.u(t)
B. y(t)=u(t)
C. y(t)=t2.u(t)
D. y(t)=t.u(t-1)
Clarification: By taking x (t) = u (t) and h (t) = u (t) and substituting in the integral
On solving the given integral we get y (t) = t u (t).
6. Convolve the signals e-2t u(t), e-3t u(t). Determine the output?
A. y(t) = (e-2t – e-3t)u(t)
B. y(t) = (e-2t – e-3t)
C. y(t) = (e-3t – e-2t)u(t)
D. y(t) = (e-t – e-3t)u(t)
8. Convolve graphically the below given signals, and determine the correct sequence?
A. Y (-1) = 0, y (1) = 2, y (3) = 2
B. Y (-1) = 2, y (1) = 2, y (3) = 2
C. Y (-1) = 0, y (1) = 0, y (3) = 2
D. Y (-1) = 0, y (1) = 3, y (3) = 2
Clarification: Step 1: sketch x (τ) and h (-τ)
Step 2: Obtain the product x (τ) h (t-τ) and the area under this product will give y (0)
Step 3: sketch h (1-τ) and compute y (1) and so on
Step 4: similarly sketch h (-1-τ) and compute y (-1) and so on.
By following above steps we get the output as
10. Find the convolution of x1[n] = {1, 2, 3, 4} and x2[n] = {2, 1, 2, 1}.
A. Y[n] = {14, 10, 14, 10}
B. Y [n] = {14, 16, 14, 16}
C. Y [n] = {14, 16,-14,-16}
D. Y [n] = {14,-16,-14, 16}
Answer: B
Clarification: By using convolution sum we get x1[n]*x2[n] = {14, 16, 14, 16}. This can be verified using tabular method of convolution.