# Convolution of delta function

for linear time invariant system,
y(t)=h(t)*x(t) where y(t) is the output , x(t) is the input and h(t) is the impulse response.(* is the convolution)

The definition of convolution is
y(t)=integration from -infinity to +infinity (h(tau)x(t-tau)d(tau)

p/s: i don't know how to use mathematical equation inhere. just joining the group

i know that if h(t) =delta(t) then y(t)=x(t).

now here comes my confusion, if h(t)=f(t)delta(t), then what is y(t)?
can anybody give me any ideas?

thanks for the help.

zai

Use the convolution property. If you have functions 1, 2, and 3:

1 * (2 * 3) = (1 * 2) * 3

Thank you for the reply. For this particular property i knew it already. my question is if
f(t)=[x(t).y(t)]*z(t) and y(t) is a delta function. How do i solve this problem?

Zai

Thank you for the reply. For this particular property i knew it already. my question is if
f(t)=[x(t).y(t)]*z(t) and y(t) is a delta function. How do i solve this problem?

Zai

What happens when you multiply a function by a delta function?

What is $$x(t)\delta(t)$$ equal to? Thinking about this graphically may help.

if delta(t) = 1 for t = 0,1,2,3 and 0 for else
then,
you get x(t) for t = 0,1,2,3 which is also a delta(t) function.

What happens when you multiply a function by a delta function?

What is $$x(t)\delta(t)$$ equal to? Thinking about this graphically may help.

thank you for the reply. As far as i know if $$x(t)\delta(t)$$, the answer is x(t) at t=0, i.e x(0). isn't it? So, do we need to consider x(0) as a constant convolve with z(t) then? it doesn't seem right to me.

zai

Hi Zai,

I'm a bit late with my answer.

"i know that if h(t) =delta(t) then y(t)=x(t)."
Actually it is x(0) because of one of the delta function properties:

integration from -infinity to +infinity (delta(t-tau)f(t)dt) = f(tau).

Convolution of a function f(t) with a delta distribution moves the function to the point (in time) where the delta impulse takes place.

I guess your hole integration becomes:
y(t)=integration from -infinity to +infinity (f(tau)delta(tau)x(t-tau)d(tau)

You could use the product rule of integration and use the convention that the integral over the delta distribution is the Heaviside Step function Theta.

Now you have:
Theta(tau) f(tau) x(t-tau) - int(Theta(tau) [f(tau)x(t-tau)]' d tau
where ' is the derivative. Usually the derivative of f(tau)x(t-tau) should be easily to obtain by product rule again.

You could also use the Laplace transform and your convolution becomes a multiplication. There are tables with often used formulas and possibly you can find one with f(t) delta(t). Afterward you have to transform it back in the time domain again.