# DSP - Differential Equation for a linear system H

Discussion in 'Homework Help' started by O'Fithcheallaigh, Jan 4, 2012.

1. ### O'Fithcheallaigh Thread Starter Member

Sep 15, 2010
20
0
Hello,

I am hoping this is in the right place. Sorry if it is not.

I need to find a differential equation for a linear system, H, which is characterised by the impulse response h(t) = e-2t, where t denotes time.

I know I need to change the impluse into the Laplace equivalent, which is 1/(s+a)

But past this I am not sure. Would anyone be able to point me in the correct direction?

Seán

Last edited: Jan 4, 2012
2. ### Zazoo Member

Jul 27, 2011
114
43
You can start with the fact that H = Y/X, (where Y is output, X is input)

You are essentially reversing the process used to find the transfer function, H, from a differential equation.

3. ### O'Fithcheallaigh Thread Starter Member

Sep 15, 2010
20
0
Hello,

I know the input to the system -- it is the product of two cosines -- but I don't know what the output is.

So, if I understand you correctly, I need to relate the output to the input to get the differential equation?

I am sorry if I am not getting it, the heavy maths in all a bit new to me.

Seán

4. ### Zazoo Member

Jul 27, 2011
114
43
If you just need a differential equation, then you don't need to know the output.

When you solve H=Y/X for Y you get:

YH = X
Y(s+2) = X
Ys+2Y = X

Applying the inverse Lapalce to the above will give you a differential equation in y(t) and x(t) (known).

If you actually need to solve for the output of the system with impulse response h(t) and input x(t), you can convolve the two functions, or you can find H(s) and X(s) and multiply the two together and then perform the inverse transform on the resulting product.
I would guess convolution would probably be easier since you can use Euler's identity to rewrite the cosines as complex exponentials.

Last edited: Jan 5, 2012
5. ### O'Fithcheallaigh Thread Starter Member

Sep 15, 2010
20
0
Ahhhhh ok, see, that's kinda what I was thinking, but when I was talking to some people about it they were confusing me a bit ...although no one mentioned the inverse transform.

That is new to me ...now, is the inverse Laplace the same as going from the frequency domain to the time domain? I doubr it is that easy ...since inverse is usually 1 over something.

Thanks for your help, I do see where I am going now ...or, at least a bit further down the road until I get the inverse thing sorted haha.

Many, many thanks.

Seán

6. ### Zazoo Member

Jul 27, 2011
114
43
Yes, exactly.

O'Fithcheallaigh likes this.
7. ### O'Fithcheallaigh Thread Starter Member

Sep 15, 2010
20
0
Hey,

That's great, thanks a lot for your help.

Seán

8. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,349
731
This may be of some help.

Common Fourier Transform pairs, handy to save, somebody here (I forgot!) posted it and I saved it.

File size:
432.5 KB
Views:
19
9. ### O'Fithcheallaigh Thread Starter Member

Sep 15, 2010
20
0
Hello,

That's fantastic, thats for the file. I saved a copy myself!

Can I ask another question relating to what has been discussed already? The next thing I need to do is "Derive a formula for the output signal y as a function of t". Is that not what the differential equation is??

EDIT: Or would I need an equation which is specific to this system, and not a general solution?

Seán

Last edited: Jan 6, 2012
10. ### Zazoo Member

Jul 27, 2011
114
43
The differential equation also includes a derivative of y, which is unknown since y is unknown.

In other words, In order to get an explicit formula for y in terms of t, you need to solve the differential equation for y.
The convolution approach will also work to solve for y.

11. ### O'Fithcheallaigh Thread Starter Member

Sep 15, 2010
20
0
Hello.

Okay, thanks for that! I will see how I get on.

I could be back to ask questions though!

Hope that is cool.

Seán

12. ### O'Fithcheallaigh Thread Starter Member

Sep 15, 2010
20
0
Hello.

I think I might have got what I need here ...I have been going through examples and idfferent things, and I was wondering if someone would mind having a quick look over what I have done, just to see if I am right.

$\frac{dy(t)}{dt}$ + 2y(t) = x(t)

If I let u(t) = $e^{\int 2dt}$ = $e^{2t}$, and multiply both sides by u(t)

$e^{2t}$ $\frac{dy(t)}{dt}$ + 2$e^{2t}$ y(t) = $e^{2t}$ x(t)

Because 2$e^{2t}$ = $\frac{d}{dt} e^{2t}$ I get

$e^{2t} \frac{dy(t)}{dt} + \frac{d}{dt}e^{2t} y(t) = e^{2t} x(t)$

Applying the reverse product rule gets me

$\frac{d}{dt}$ ($e^{2t}$ y(t)) = $e^{2t}$ x(t)

Integrating both sides...

$\frac{d}{dt}$ ($e^{2t}$ y(t)) dt = ∫ $e^{2t}$ x(t) dt

-> $e^{2t}$ y(t) = ∫ $e^{2t}$ x(t) dt + c

Now, dividing both sides by u(t) gives

y(t) = $e^{-2t}$ (∫ $e^{2t}$ x(t) dt + c)

I think I typed all that out correctly anyway!

Seán

Last edited: Jan 6, 2012
13. ### Zazoo Member

Jul 27, 2011
114
43
Looks good to me.

14. ### O'Fithcheallaigh Thread Starter Member

Sep 15, 2010
20
0
Hello.

That's great, thanks!

Seán