# Thread: Solving the transfer function

1. ## Solving the transfer function

Hi all,

I have been trying to solve the follow question but with no success. Anyone has idea on finding H(s)?

In one special case, the equation can be written as follow but again I have no clue on solving it.

Anyone has any idea?

Thanks,
cfy30

2. If you know that x(t) is differentiable, then I would differentiate the whole thing, take the Laplace Transform, and solve for Y/X. Is anything preventing you from doing that?

3. I am not sure I get your idea. I still can't figure Y(s)/X(s) after differentiation.

cfy30

4. Yeah, I see what you mean. This is a doozy of a problem. Is this a textbook problem? If so, there's probably some trick you're supposed to see in order to solve it. I tried doing the Laplace Transform directly, and then fiddling around with interchanging the order of integration. You can get some interesting equations that way, but not towards getting the final ratio desired. And you can try tricks with integration by parts, but that ends up doing the same thing as interchanging the order of integration. One thought that did occur to me was this: convolution. If you were to focus only on the LT of the integral term with the x(t) multiplying it, I wonder if you couldn't use the convolution theorem to help you out there. It'd be worth trying, because then you might get something like this:

$Y(s)=X(s)-A\,X(s)Y(s)\times\,\text{something}.$

Perhaps you could work with pulling the x(t) inside the integral, and doing a change of variable.

5. The figure attached shows the system I am trying to model with the equation.

It is a feedback system to suppress x(t) at the node y(t). The system itself is quite simple but the transfer function is giving me headache. What I want to do is formulating H(s) and then determine the optimum A if the optimum A exists......

cfy30

6. Are you sure you've written down your integral equation correctly? Where's the z(t) in it?

7. I am sure the equation I have is correct. z(t) is the "wanted" signal and x(t) is the interference that needs to be canceled. When the system starts to run, x(t) will be suppressed, leaving z(t) as the output y(t). Imagine z(t) is cos(2*pi*200*t) and x(t) is cos(2*pi*50*t). y(t) contains cos(2*pi*200*t) only and is free of cos(2*pi*50*t).

cfy30

8. Your first system is not linear, so does not have a transfer function.

CB

9. Why the system is not linear?

I think the system is linear but time variant.

cfy30

10. Let me take time variant back. I believe the system is also time invariant...

11. Originally Posted by cfy30
Why the system is not linear?

I think the system is linear but time variant.

cfy30
Consider two signals $x_1(t)$ and $x_2(t)$ and corresponding outputs $y_1(t)$, $y_2(t)$. Now what is the output when the input is $z(t)= x_1(t)+x_2(t)$? (note I am here working with a necessary condition for linearity not the full condition, to show that linearity fails it is sufficient to show that a necessary condition fails)

CB

12. Did you mean "corresponding outputs $y_{1}(t)$ and $y_{2}(t).$"?

13. My thinking is, let x(t) = x1(t) and z(t) = x2(t), x1(t) and x2(t) have different frequencies.
Output of the system, y(t) is always equal to z(t) or x2(t). The system behaves as a notch filter. That is what I say the system is LTI.

cfy30

14. Originally Posted by Ackbeet
Did you mean "corresponding outputs $y_{1}(t)$ and $y_{2}(t).$"?
Yes, its my fault for not reading the spell-checker suggestions carefully before accepting

15. Originally Posted by cfy30
My thinking is, let x(t) = x1(t) and z(t) = x2(t), x1(t) and x2(t) have different frequencies.
Output of the system, y(t) is always equal to z(t) or x2(t). The system behaves as a notch filter. That is what I say the system is LTI.

cfy30
An LTI system has to be LINEAR as well as time invariant, see the Wikipedia article. What you have written above is gobbledygook.

CB