Two continuous time system is described as x1 (t)=te-t u(t) and x2 (t)=te-2t u(t), then the convolution of x1(t) & x2(t) will be?a)(t-1)e-2tu(t)+(t+1)e-tu(t)b)(t-2)e-2tu(t)+(t+2)e-tu(t)c)(t+2)e-2tu(t)+(t-2)e-tu(t)d)(t+1)e-2tu(t)+(t-1)e-tu(t)Correct answer is option 'C'. Can you explain this answer?

Question

Two continuous time system is described as  x1 (t)=te-t u(t) and x2 (t)=te-2t  u(t), then the convolution of x1(t) & x2(t) will be?a)(t-1)e-2tu(t)+(t+1)e-tu(t)b)(t-2)e-2tu(t)+(t+2)e-tu(t)c)(t+2)e-2tu(t)+(t-2)e-tu(t)d)(t+1)e-2tu(t)+(t-1)e-tu(t)Correct answer is option 'C'. Can you explain this answer?

Answer

To find the convolution of x1(t) and x2(t), we need to perform the integral:

y(t) = ∫[x1(τ)x2(t-τ)]dτ

Substituting the given expressions for x1(t) and x2(t), we have:

y(t) = ∫[(τ)(e^-τ)(t-τ)(e^-2(t-τ))]dτ

Expanding the expression inside the integral:

y(t) = ∫[τ(e^-τ)(t)e^-2(t)+τ(e^-τ)(-τ)e^-2(t)+τ(e^-τ)(t)e^-2τ]dτ

Simplifying:

y(t) = ∫[τ^2(e^-τ)(t)e^-2(t)-τ^2(e^-τ)(t)e^-2τ]dτ

Using the properties of integration, we can split the integral into two parts:

y(t) = ∫[τ^2(e^-τ)(t)e^-2(t)]dτ - ∫[τ^2(e^-τ)(t)e^-2τ]dτ

For the first integral, we can use integration by parts. Let u = τ^2(e^-τ)(t) and dv = e^-2(t)dτ. Then du = 2τ(e^-τ)(t)dτ and v = -1/2e^-2(t).

Applying integration by parts:

∫[τ^2(e^-τ)(t)e^-2(t)]dτ = (-1/2τ^2(e^-τ)(t)e^-2(t)) + ∫[τ(2τ(e^-τ)(t)e^-2(t))]dτ

Simplifying:

∫[τ^2(e^-τ)(t)e^-2(t)]dτ = (-1/2τ^2(e^-τ)(t)e^-2(t)) + 2∫[τ^2(e^-τ)(t)e^-2(t)]dτ

Moving the second integral to the left side:

(1 - 2)∫[τ^2(e^-τ)(t)e^-2(t)]dτ = (-1/2τ^2(e^-τ)(t)e^-2(t))

Simplifying:

-∫[τ^2(e^-τ)(t)e^-2(t)]dτ = (-1/2τ^2(e^-τ)(t)e^-2(t))

Multiplying both sides by -1:

∫[τ^2(e^-τ)(t)e^-2(t)]dτ = (1/2τ^2(e^-τ)(t)e^-2(t))

Now, let's evaluate the second integral:

∫[τ^2(e^-τ)(t)e^-2τ]dτ

Using integration by parts again, let u = τ^2(e^-τ)(t) and dv = e^-2τdτ. Then du = 2τ(e^-τ)(t)dτ and v = -1/2e^-2τ.

Applying integration by parts:

∫[τ^2(e^-τ)(t)e^-2τ]dτ = (-1/2τ^2(e^-τ)(t)e^-2τ) + ∫[τ(2τ(e^-τ)(t)e^-2τ)]dτ

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