What would be the convolution of these 2 signals? :
$$x_1(t)=A\cos(2\pi ft), -\infty<t<\infty$$ $$x_2(t)=\cos(2\pi ft), -\infty<t<\infty$$
I know the formula of the convolution : $$(x_1 * x_2)(t) = \int_{-\infty}^{+\infty} x_2(\tau) \cdot x_1(t-\tau) \, d\tau$$ but I don't know how to use it for this problem
$$ x_1(t)\circledast x_2(t) = x(t)$$
Use the property of Convolution
$$ x(t) = x_1(t)\circledast x_2(t)$$
$$ \frac{dx(t)}{dt} = x_1(t)\circledast \frac{dx_2(t)}{dt}$$
$$Or$$
$$ \frac{dx(t)}{dt} = x_2(t)\circledast \frac{dx_1(t)}{dt}$$
$$ \frac{d^2x(t)}{dt^2} = x_1(t)\circledast \frac{d^2x_2(t)}{dt^2}$$
$$ \frac{d^2x(t)}{dt^2} = - \omega^2 Acos(\omega t)\circledast cos(\omega t) $$
$$x(t) = Acos(\omega t)\circledast cos(\omega t) $$
$$ \frac{d^2x(t)}{dt^2} = -\omega^2x(t) $$
$$ \frac{d^2x(t)}{dt^2} + \omega^2x(t) = 0$$
Now the whole problem has reduced into an 2nd order linear differential equation
$$m^2+ \omega^2= 0$$
$$m = +i\omega ; -i\omega$$
$$x(t) = C_1 e^{i\omega t} + C_2 e^{-i\omega t}$$
Now how to find the Constants
Use Laplace Transformation
$$X(s) = \frac{C_1s + i\omega C_1+ sC_2 - i\omega C_2}{s^2+\omega^2}$$
From question
$$X(s) = \frac{As^2}{(s^2+\omega^2)^2}$$
$$(C_1+ C_2)s +i\omega(C_1-C_2)= \frac{As^2}{s^2+\omega^2}$$