Hi guys ,,,

I'm studying the Convolution from Signals & systems course, usually i have X(t): input signal and h(t): response to find y(t)utput signal of a system
as y(t) = x(t)* h(t) = ∫x(τ) h(t-τ) dτ = ∫h(τ) x(t-τ) dτ

But How we can find h(t) if we have x(t) & y(t)??

really i tried to find it but i have confused,, i attached photo with two examples A & B and assume that y(t) is an arbitrary function .

In example A ,, x(t) represents u(t):unit-step function so the Y(t) represents S(t)
==> H(t) = d/dt( S(t) ) = d/dt( exp(-t) ) = -exp(-t) Is it correct ??

In example B ,, x(t) represents the rect function u(t)-u(t-1) so the Y(t) represents S(t)-S(t-1)
==> H(t) = d/dt( S(t)-S(t-1) ) = ?? Is it correct ??

and Is there a General method to find H(t)??

 

Hi,
Do you know Laplace transform?

\(\mathcal{L}\{f(t)*g(t)\} = F(s).G(s)\)

Where: *: convolution operator
And F(s), G(s) are the Laplace transform of f(t), g(t), respectively.

Therefore, in your case:

y(t) = x(t)* h(t)

Lapace transform:

\(\mathcal{L}\{y(t)\} = \mathcal{L}\{x(t)* h(t)\} = X(s).H(s)\)

From which:

\(H(s) = \frac{\mathcal{L}\{(y(t)\}}{X(s)} \)

\(h(t)= \mathcal{L}^{-1} \{H(s)\}\)

 

Here is an example for case A:

x(t) = u(t)
In your picture, y(t) is not e^(-t ). Let's assume that y(t) have this form:

y(t) = 0 for t< a
y(t) = e^(-t ) for t>= a

And therefore we can rewrite it as follows:

y(t) = e^(-t ).u(t-a)

\(\mathcal{L}\{x(t)\} = \mathcal{L}\{u(t)\} = \frac{1}{s} \)

\(\mathcal{L}\{y(t)\} = \mathcal{L}\{ e^{-t} u(t-a)\} = \frac{ e^{-as} }{s+1} \)

\(H(s) = \frac{\mathcal{L}\{(y(t)\}}{X(s)} = \frac{ s.e^{-as} }{s+1} = e^{-as}(1 - \frac{1}{s+1} )\)

\(h(t)= \mathcal{L}^{-1} \{H(s)\}= \mathcal{L}^{-1} \{e^{-as}(1 - \frac{1}{s+1} )\} = \mathcal{L}^{-1} \{ \ e^{-as} \} - \mathcal{L}^{-1} \{ \frac{e^{-as}}{s + 1} \} \)

\(h(t)= \delta (t-a) - e^{-t}.u(t-a) \)

PS. Hope I didn't make any mistake.

 

Here is an example for case A:

x(t) = u(t)
In your picture, y(t) is not e^(-t ). Let's assume that y(t) have this form:

y(t) = 0 for t< a
y(t) = e^(-t ) for t>= a

And therefore we can rewrite it as follows:

y(t) = e^(-t ).u(t-a)

\(\mathcal{L}\{x(t)\} = \mathcal{L}\{u(t)\} = \frac{1}{s} \)

\(\mathcal{L}\{y(t)\} = \mathcal{L}\{ e^{-t} u(t-a)\} = \frac{ e^{-as} }{s+1} \)

If F(s)=L{f(t)} then L{u(t−a)⋅g(t−a)}=e^(−as). G(s)
so L{y(t)} = L{e^(-t)u(t−a)}= e^(-a)*e^(−as)/(1+s)

\(H(s) = \frac{\mathcal{L}\{(y(t)\}}{X(s)} = \frac{ s.e^{-as} }{s+1} = e^{-as}(1 - \frac{1}{s+1} )\)

\(h(t)= \mathcal{L}^{-1} \{H(s)\}= \mathcal{L}^{-1} \{e^{-as}(1 - \frac{1}{s+1} )\} = \mathcal{L}^{-1} \{ \ e^{-as} \} - \mathcal{L}^{-1} \{ \frac{e^{-as}}{s + 1} \} \)

\(h(t)= \delta (t-a) - e^{-t}.u(t-a) \)

PS. Hope I didn't make any mistake.

thanx v v very much Mr. anhnha

 

If F(s)=L{f(t)} then L{u(t−a)⋅g(t−a)}=e^(−as). G(s)
so L{y(t)} = L{e^(-t)u(t−a)}= e^(-a)*e^(−as)/(1+s)

Yes, you are correct.

\(\mathcal{L}\{u(t)\} = \frac{1}{s} \)


\(\mathcal{L}\{u(t -a)\} = \frac{ e^{-as} }{s}\)

\(\mathcal{L}\{(y(t)\}= \mathcal{L}\{\ e^{-t}u(t -a)\} = \frac{e^{-a(s+1)}}{s + 1} \)


And then you need to multiply the result above with e^-a.

 

Yes, you are correct.

\(\mathcal{L}\{u(t)\} = \frac{1}{s} \)


\(\mathcal{L}\{u(t -a)\} = \frac{ e^{-as} }{s}\)

\(\mathcal{L}\{(y(t)\}= \mathcal{L}\{\ e^{-t}u(t -a)\} = \frac{e^{-a(s+1)}}{s + 1} \)


And then you need to multiply the result above with e^-a.

yes, i solved it as attached photo(A) and verified it.
and i tried to solve example B as attached photo(B). Is it correct?? if correct how we can complete it
thanx 4 your help Mr. Anhnha

 

For part A.

Here is my opinion:

There are two ways to write your function:
1.

\(y(t) =\begin{cases}0 & t<a \\ e^{-t} & t \geq a\end{cases} \)

2.

\(y(t) = e^{-t}.u(t-a) \)

#1 and #2 are the same.

You wrote it like this:

\(y(t) =\begin{cases}0 & t<a \\ e^{-t}u(t-a) & t \geq a\end{cases} \)

I don't think it is incorrect but it is redundant.

\(e^{-t}.u(t-a) \) already says that y(t) = 0 for t<a.

And your solution for part A is correct. Now I recognize that I did make a few mistakes.

I will check part B now.

 

For part B, you need to look at the definition of S(t) again.

 

For part B, here is my view. You have to check that if it is correct or not.

x(t) = u(t) - u(t-1)
y(t) = s(t) - s(t-1)

\(X(s) = U(s)- e^{-s}U(s) = U(s)(1 - e^{-s})\)

\(Y(s) = S(s)- e^{-s}S(s) = S(s)(1 - e^{-s})\)

\(H(s) = \frac{Y(s)}{X(s)} = \frac{S(s)}{U(s)} \)

And therefore H(s) is same as in case A.

 

For part B, you need to look at the definition of S(t) again.

Ok,, see attached photo

 

I think you can't use the formula h(t) = dS(t)/dt in this case.
You can check my post #9.

 

I think you can't use the formula h(t) = dS(t)/dt in this case.
You can check my post #9.

i see it and i hope which is the correct answer
thanx 4 your help

 

I have another question as attached photo ,, Given the impulse response as a rect(t-2/2) and required the unit-step response
i tried to get it by S(t)=∫h(t) dt but i faced the problem with integration of the unit-step function...

 

I have another question as attached photo ,, Given the impulse response as a rect(t-2/2) and required the unit-step response
i tried to get it by S(t)=∫h(t) dt but i faced the problem with integration of the unit-step function...

Well, you know that u(t-1) - u(t-3) is equal to 1 for t= [1; 3] and zero otherwise, right?

\(\int_0^t (u(t-1) -u(t-3))dt = \int_1^3 1.dt = 2\)

BTW, I think you should research to know why your method h(t) = ds(t)/dt doesn't work. I don't remember it now. Hope someone can help.

 

Well, you know that u(t-1) - u(t-3) is equal to 1 for t= [1; 3] and zero otherwise, right?

\(\int_0^t (u(t-1) -u(t-3))dt = \int_1^3 1.dt = 2\) <=== Incorrect i don't find any answer equals to 2 in multiple choices of this question .

BTW, I think you should research to know why your method h(t) = ds(t)/dt doesn't work. I don't remember it now. Hope someone can help.

Hope someone can help

 

Hope someone can help

I think your paper may be misprinted.
Maybe, the exercise is like this:
Given unit step response rect((t-2)/2)
Find: unit impulse response

 

I think your paper may be misprinted.
Maybe, the exercise is like this:
Given unit step response rect((t-2)/2)
Find: unit impulse response

ok anyone tell me about the main question
Our solution is correct on not??

 

I'm not following some of your notation.

What is the difference between y(t) and Y(t)?

What is S(t)?

You don't need others to tell you if your proposed answer is correct or not. Once you have found what you believe is h(t), then convolve it with x(t) and see if you get the correct y(t).

You need to start developing the ability to check your own work. The people paying you will be paying you precisely because they need to solve problems that they don't know the answer to.