Question

A signal f(t) = 6t and a second signal g(t) = t are convolved. Using the fundamental definition of convolutionA signal f(t) = 6t and a second signal g(t) = t ar find the value of the convolution at t = 4 seconds.

Hint: Because convolution is a linear operator and obeys the principle of superposition, the convolution of k2are constant multipliers of . (Points : 5) 24
64
48
96

Answer 1

64

Step-by-step explanation:

The convolution of two functions is given by:

`(f*g)(t)=\int\limits^t_0 f(\tau)g(t-\tau) \, d\tau` (1)

In this case:

`f(t)=6t\\g(t)=t`

So, let's replace the functions in the equation (1):

`(f*g)(t)=\int\limits^t_0 6\tau*(t-\tau) \, d\tau`

`(f*g)(t)=\int\limits^t_0 6\tau t \, d\tau - \int\limits^t_0 6\tau^(2)   \, d\tau`

Integrating with respect to τ

`6t((\tau^(2) )/(2) )\left \{ {{t=t} \atop {t=0}} \right. - 6((\tau^(3) )/(3) ) \left \{ {{t=t} \atop {t=0}} \right.`

Evaluating the integrals:

`(f*g)(t)=3t^(3) -2t^(3) =t^(3)`

Finally evaluating f*g at t=4:

`(f*g)(4)=4^(3) =64`