For the signal x1(t) = e– k1t u(t) and x2(t) = e

For the signal x₁(t) = e^{– k₁t} u(t) and x₂(t) = e^{– k₂} u(t). their convolution will be—

Given -
x₁(t) = e^k1^t u(t)
x₂(t) = e^–k2^t u(t)
x(t) = x₁(t)^* x₂(t) = x₁(s).x₂(s)

x₁(s) =	1
	s – k₁

x₂(s) =	1
	s – k₂

x₁(s).x₂(s) =	1	= x(s)
	(s – k₁) (s + k₂)

By using partial fraction

x(s) =	A	+	B
	(s – k₁)		(s + k₂)

x(s) =	As + Ak₂ + Bs – Bk₂
	(s – k₁) (s + k₂)

A + B = 0 or A = – B …(i)
Ak₂ – Bk₁ = 1 …(ii)
From equations (i) and (ii), we get
or
Ak₂ + Ak₁ = 1
or

A =	1
	k₁ + k₂

and

B =	- 1
	k₁ + k₂

Now,

x(s) =	1		1	-	1
	k₁ + k₂		s – k₁		s + k₁

By taking inverse Laplace transform

x(t) =	1	[e^k1^t– e– ^k2^t]
	k₁ + k₂

Hence, alternative (A) is the correct choice.