Convolution with Impulse Function

AI Thread Summary
The discussion centers around the validity of the convolution operation involving the Heaviside step function and the impulse function. It is clarified that the equation h(t)*Heaviside(t-t0) = h(t0) is not true, while the relationship h[n]*IMPULSE[n-t0] = h[n-t0] holds true for multiplication. Participants emphasize that convolution with the Dirac delta function yields the original function, reinforcing that x[n]*δ[n] = x[n] and x[n]*δ[n + a] = x[n + a]. The conversation highlights a misunderstanding in applying these properties correctly in a specific example, leading to incorrect results. Ultimately, the correct application of convolution properties is affirmed, pointing out errors in the initial calculations.
ha9981
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Is this true:

h(t)* Heaviside(t-t0) = h(t0)

If this is true saves my work a lot. It appears not to be true I tried proving it...
 
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ha9981 said:
Is this true:

h(t)* Heaviside(t-t0) = h(t0)

If this is true saves my work a lot. It appears not to be true I tried proving it...

I don't think that's true. It's true if you replace the Heaviside step function with a Dirac delta function.
 
Sorry I meant:

h[n]*IMPULSE[n-t0] = h[n-t0]

I tried it with a question that wanted to do a convolution that would expand out as follows.

x[n] = δ[n] + 2δ[n-1] - δ[n-3]
h[n] = 2δ[n+1] + 2δ[n-1]

x[n] * h[n] = 2δ[n]*δ[n+1] + 2δ[n]*δ[n-1] + 4δ[n-1]δ[n+1] + 4δ[n-1] - 2δ[n-3]δ[n+1] - 2δ[n-3]δ[n-1]

x[n] * h[n] = 2δ[n+1] + 2δ[n-1] + 4δ[n+1] + 4δ[n-1] - 2δ[n+1] - 2δ[n-1]

x[n] * h[n] = 4δ[n+1] + 4δ[n-1]

But it turns out answer is wrong. So I crapped out.
 
cepheid said:
I don't think that's true. It's true if you replace the Heaviside step function with a Dirac delta function.

It is only true for h(n) multiplied by δ[n-t0]

Not for Convolution.

AM I CORRECT?
 
ha9981 said:
It is only true for h(n) multiplied by δ[n-t0]

Not for Convolution.

AM I CORRECT?


Yeah ok. It is true for multiplication: x[n]δ[n - a] = x[a]. The property I was thinking of for convolution was that the convolution of any function with the Dirac delta function (or unit impulse function in discrete time) is just equal to the function itself:

x[n]*δ[n] = x[n]

which follows if you just think of the definition of a convolution sum. A function in discrete time can be represented as a sequence of shifted and scaled unit impulses. What also follows fairly intuitively from that is just that:

x[n]*δ[n + a] = x[n + a]
 
cepheid said:
which follows if you just think of the definition of a convolution sum. A function in discrete time can be represented as a sequence of shifted and scaled unit impulses. What also follows fairly intuitively from that is just that:

x[n]*δ[n + a] = x[n + a]

I have a hard time finding that to be correct only because it fails to yield the right answer in the example above.
 
ha9981 said:
I have a hard time finding that to be correct only because it fails to yield the right answer in the example above.

Oh okay. It must be wrong then. :rolleyes:

How about if you correct the problem in red below:

ha9981 said:
Sorry I meant:

h[n]*IMPULSE[n-t0] = h[n-t0]

I tried it with a question that wanted to do a convolution that would expand out as follows.

x[n] = δ[n] + 2δ[n-1] - δ[n-3]
h[n] = 2δ[n+1] + 2δ[n-1]

x[n] * h[n] = 2δ[n]*δ[n+1] + 2δ[n]*δ[n-1] + 4δ[n-1]δ[n+1] + 4δ[n-1] - 2δ[n-3]δ[n+1] - 2δ[n-3]δ[n-1]

The term in red should be 4δ[n-1]*δ[n-1]. Also, how about if you show the steps in your latest attempt?
 
These theorems are right in all sense..
x(n)*delta(n)=x(n)
x(n)*delta(n-n0)=x(n-n0)
x(n).delta(n)=x(0)
x(n).delta(n-n0)=x(n0)
and u have done it in a wrong way and so u got it wrong. That just meant that u r wrong not the theorom...
 

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