Can somebody explain how the convolution is calculated? I have two functions:
$$ x(t) = (\frac{t}{2}-1) [u(t-2) - u(t-4)] $$
$$ v(t) = u(t-1)-u(t-3)+\delta(t-4) $$
Can somebody explain how can I calculate the convolution of these two functions analytically?
I always get stuck while calculating the limit of the integrals.
(Note: $u(t)$ is the unit step function and $\delta(x)$ is the Dirac-Delta function or Impulse Function.)
 A: Convolution is given by: $$\int_{\infty}^{\infty} x(\tau)v(\tau-t)d\tau$$
The first signal is defined i.e. $x(t)$ from 2 to 4 but the second signal is from 1 to 3 with an additional delta impulse at 4.(I assume that n was t).
$1<t<3 :$ $$\int_1^t ((\frac{\tau -t}{2})-1)d\tau = \int_1^t (\frac {\tau^2}{4}- \frac {\tau t}{2}-\tau)d\tau=-\frac{t^2}{2}-t+\frac{5}{4}$$
As the signal enters it's lower limit is set at 1 as there is no signal on the part of $v(t)$ before 1 but the upper limit goes to t as the $x(t)$ is moving. At the end when $t=3$ both signals overlap.
$3<t<4 :$ $$\int_{t-1}^3 ((\frac{\tau -t}{2})-1)d\tau = \frac{t^2-2t-8}{4}$$
Then as the $x(t)$ is moving t>1 hence $t-1$ to $3$, as $3$ the endpoint for v(t).
$4<t<5: $ $$\int_{t-1}^3 ((\frac{\tau -t}{2})-1)d\tau + (\frac{t-4}{2}-1) = \frac{t^2-2t-8}{4}+\frac{t}{2}-3=\frac{t^2-20}{4}$$
The signal from 3 to 5 is split into because when t>4 the delta function is also to be included. If there was no delta signal the 3<t<4 section would have lasted till 5.
$5<t<6: $ $$\frac{t}{2}-3=\frac{t-6}{2}$$
This is plainly the delta signal. x(t) convolved with $\delta(t-k)$ is x(t-k)
