1. ## Convolution integral

Hey guys. just wondering if anyone can help me with this piece of convolution integral.

i am asked to convolute f(t) and g(t) where

f(t) = 2r(t-1) - 2r(t-2) - u(t-2) - u(t-3) where r is a ramp function and
u is a unit step function

and

g(t) = d(t-2) - d(t-3) where d is the unit impulse
function or delta dirac function

i use the graphical approach and make use of the sifting property of the impulse function. does that sound right? i have an answer but i'm not sure if rite. any help or input would be much appreciated.

2. By definition:

$(f*g)(t)=\int_{-\infty}^{\infty} f(\tau)g(t-\tau)\;d\tau$

and so for scalars $a$ and $b$ :

$(f*(a\times g+b \times h)(t)=a(f*g)(t)+b(f*h)(t)$

Again by definition:

$(f*\delta_k)(t)=\int_{-\infty}^{\infty} f(\tau)\delta((t-\tau)-k)\;d\tau=f(t-k)$

Which should allow you to write down the result you require (there may be a problem if $f$ is discontinuous at $k$).

CB

3. thanks for the quick response CaptainBlack. this is my answer

for t <3, y(t) = 0
for 3<t<4, y(t) = 2(t-1)
for 4<t<5, y(t) = 2(t-1) + 1
for 5<t<6,y(t) = 1

where y(t) is the convolution of f(t) and g(t).

there is a discontinuity of f(t) at t=2 but i have taken that into consideration in the answer above. can you confirm if this is correct?

4. Originally Posted by squirby
thanks for the quick response CaptainBlack. this is my answer

for t <3, y(t) = 0
for 3<t<4, y(t) = 2(t-1)
for 4<t<5, y(t) = 2(t-1) + 1
for 5<t<6,y(t) = 1

where y(t) is the convolution of f(t) and g(t).

there is a discontinuity of f(t) at t=2 but i have taken that into consideration in the answer above. can you confirm if this is correct?
Well other than any exceptional points, if my algebra is right I have:

$(f*g)(t)=2r(t-3)-4r(t-4)+2r(t-5)-u(t-4)+u(t-6)$

CB