DSP - Differential Equation for linear system H

[SMOF]

Hello.

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.

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

Thanks in advance.

Seán

 

[rude man]

1. Transform your impulse equation into the frequency (Laplace) domain: h(t) --> H(s)

2. Inverse-transform the resultant H(s) transfer function back to the time domain for your diff. eq.

Hint: H(s) = Y(s)/X(s) where X is transform of time input x(t) and Y is transform of output y(t).

 

[SMOF]

Hello,

Thanks for your reply.

I will have a go at this in the morning, and I will let you know how I get on.

Thanks a lot. I just wasn't sure how to tackle it.

Yours
Seán

 

[SMOF]

1. Transform your impulse equation into the frequency (Laplace) domain: h(t) --> H(s)

2. Inverse-transform the resultant H(s) transfer function back to the time domain for your diff. eq.

Hint: H(s) = Y(s)/X(s) where X is transform of time input x(t) and Y is transform of output y(t).

Hello again.

Can I just make sure I have this right?

I got the Laplace transfer of h(t), which is H(s) = 1/(s+a).

So, now I need to get the transform of the input, which is

0.5 cos(ω₂ - ω₁)t) + 0.5 cos(ω₂ + ω₁)t)

But I am really not sure about the Laplace transform of this. I was thinking it was ...0.5 s/(s² - ω²) ...but I don't think that is right because in the time domain it is ω₂ - ω₁, and I am not too sure how to get this in the frequency domain.

Seán

 

[MisterX]

So, now I need to get the transform of the input, which is

0.5 cos(ω₂ - ω₁)t) + 0.5 cos(ω₂ + ω₁)t)

You're supposed to describe system behavior for any x(t) input. There should be no specific expression for "transform of the input" for this problem besides "X(s)".

The goal is to have and equation for the system relating the output y and its derivatives to the input x and its derivatives.

 

[SMOF]

Hello,

And thanks fot the reply.

I am afriad I am getting lost on this one.

So, if I can ask, once I have the transform of the impulse, where do I go from there?

I thought I needed to get the transform of the input to the system, and from that, get an equation for the output by relating the input to the system H.

Seán

 

[MisterX]

Hello,

And thanks fot the reply.

I am afriad I am getting lost on this one.

So, if I can ask, once I have the transform of the impulse, where do I go from there?

I thought I needed to get the transform of the input to the system, and from that, get an equation for the output by relating the input to the system H.

Seán

Do you know what is the transform of the derivative of a function in terms of that function's transform?

 

[SMOF]

I'm sorry, I don't.

Seán

 

[MisterX]

equation 11 is the general rule for the nth derivative. For the first derivative, this would be as follows.

$x(t) \stackrel{\mathcal{L}}{\leftrightarrow} X(s)$

$\frac{dx(t)}{dt} \stackrel{\mathcal{L}}{\leftrightarrow} sX(s) - x(0)$

 

[SMOF]

Hey,

Okay, yes ...I get that ...but (and I am very sorry if this sounds rude, I don't mean it to be) I don't understand how this helps me.

I am more used to the hands on stuff ...building different things ...so, the maths behind it all is very new to me.

But from what you said earlier, that I need to relate the input to the output ...I don't actually know what the output of the system is ...or, at least I don't know if I know it.

If I knew specific steps to take, I might be able to figure it out. Kind of what like Rude Man suggested, except it seem to me like the inverse transform from step one would just take me back to where I was at the start.

Again, sorry for not getting this ...maybe I have just been trying to figure it out for a bit, I just have a bit of a mental block.

Seán

 

[SMOF]

Oh, I think I may have got a bit further on ...I will try it and get back either way :)

Seán

 

[rude man]

Hello again.

Can I just make sure I have this right?

I got the Laplace transfer of h(t), which is H(s) = 1/(s+a).

So, now I need to get the transform of the input, which is

0.5 cos(ω₂ - ω₁)t) + 0.5 cos(ω₂ + ω₁)t)

But I am really not sure about the Laplace transform of this. I was thinking it was ...0.5 s/(s² - ω²) ...but I don't think that is right because in the time domain it is ω₂ - ω₁, and I am not too sure how to get this in the frequency domain.

Seán

The input in your case is an impulse = δ(t). That's what's meant by 'impulse response'.
y(t) is your output = exp(-2t), and Y(s) = 1/(s+2) as you have correctly ascertained.

So the question now is, I know what my response is if the input is δ(t), so what is the response to a general input x(t)? If the input is x(t), what is the output y(t)?

Well, you got the transform of the output right, but you blew it big-time on the input.
What does your Laplace table say for X(s) if x(t) = δ(t)?

Once you have the transform of δ(t) right = X(s), you can then say Y(s)/X(s) = transfer function of your system = H(s).

Do that next.

 

[SMOF]

Hello,

Well, for the table δ(t) = 1. So X(s) = 1.

If 1/(s+2) = Y(s), then H(s) = (1/(s+2))/1 ...?

 

[rude man]

Hello,

Well, for the table δ(t) = 1. So X(s) = 1.

If 1/(s+2) = Y(s), then H(s) = (1/(s+2))/1 ...?

δ(t) is not 1, δ(t) = δ(t). But L{δ(t)} = 1, that's right. And H(s) is exactly what you said. So write an equation with X(s), Y(s) and H(s)?

 

[SMOF]

Do you mean something like Y(s) = H(s)X(s)?

 

[rude man]

Do you mean something like Y(s) = H(s)X(s)?

Yes I do. You know what H(s) is now, so substitute in your equation.

 

[SMOF]

Okay, well, if

X(s) = 1
Y(s) = 1/(s+2)
H(s) = (1/(s+2))/1

I get

$\frac{1}{s+2}$ = $\frac{1/s+2}{1}$$\times$1

Which is the same as

$e^{-2t}$ = $\frac{e^{-2t}}{1}$$\times$1

I think.

 

[MisterX]

Okay, well, if

X(s) = 1
Y(s) = 1/(s+2)

No, X(s) and Y(s) represent arbitrary functions, constrained only by the transfer function Y(s)/X(s). You don't get to decide this X(s) is equal to 1!

Think of Y(s)/X(s), as describing a relationship between the output in the input in general. The impulse response h(t) and corresponding Y(s)/X(s) characterize the response of a linear system to any input.

When X(s) = 1 (a transformed impulse), Y(s) becomes the transformed impulse response, of course. But the solution you seek is a differential equation for the system with any input, and so it is not proper to have X(s) equal to 1.

 

[rude man]

You have determined what H(s) is. You did that by knowing that if x(t) = δ(t), so that
X(s) = 1, so since H(s) = Y(s)/X(s), and Y(s) you have determined to be 1/(s+2) if X(s) = 1, now you know Y(s)/X(s) for any x(t) which, when transformed is X(s). So, stay with your equation: Y(s) = H(s)X(s). That is the key to coming up with your sought-after differential equation.

The next step is transforming that algebraic equation in the frequency domain (never mind for now what I mean by that) into a differential equation in the time domain.

Clue: use cross-multiplication in your equation!

 

[SMOF]

Hello.

I get what you mean about needing to find a solution for any input. It is finally sinking in.

So, since I need to find a silution for any x(t), which is X(s), so would a better form of the equation to work with be

X(s) = Y(s)/H(s)?

The next step is transforming that algebraic equation in the frequency domain (never mind for now what I mean by that) into a differential equation in the time domain.

Clue: use cross-multiplication in your equation!

I thought to go from the frequency domain (Laplace) to the time domain, I just needed to use the Laplace tables in reverse basically. I take it that is not the case?

And if I stick with Y(s) = H(s)X(s) ...how do I cross multiply this? Anytime I have cross multiplied something it has been in the form of (a/b) x (c/d).

Seán

 

[rude man]

Hello.

I get what you mean about needing to find a solution for any input. It is finally sinking in.

So, since I need to find a silution for any x(t), which is X(s), so would a better form of the equation to work with be

X(s) = Y(s)/H(s)?



I thought to go from the frequency domain (Laplace) to the time domain, I just needed to use the Laplace tables in reverse basically. I take it that is not the case?

And if I stick with Y(s) = H(s)X(s) ...how do I cross multiply this? Anytime I have cross multiplied something it has been in the form of (a/b) x (c/d).

Seán

Substitute for H(s) what you determined H(s) to be in the equation Y(s) = H(s)X(s).

 

[SMOF]

Hello,

Sorry for my late reply, I was in work this evening.

Also, sorry for not getting what you meant ...so, (1/(s+2))/1

Y(s) = H(s)X(s) → Y(s) = $\frac{1/(s+2)}{1}$$\times$X(s)

 

[rude man]

Well, first make life easier for yourself: (1/(s+2))/1 = 1/(s+2).

Now, perform the cross-multipklication!

 

[SMOF]

Well, first make life easier for yourself: (1/(s+2))/1 = 1/(s+2)

So this is just Y(s) = H(s)?

I think I am doing the cross multiplication wrong. This is what I am getting

$\frac{1/s+2}{1}$ = $\frac{1}{s+2}$ → $\frac{1}{s+2}$ x s+2 = 1 x 1 → 1 = 1

And that can't be right.

 

[rude man]

Cross-multiply Y(s) = X(s)/(s+2)

 

[SMOF]

Where did that come from?

That would get Y(s)(s+2) = X(s)

 

[rude man]

Where did that come from?

That would get Y(s)(s+2) = X(s)

Sure would.
Now expand the left-hand-side to 2 terms instead of one.

 

[SMOF]

That should be

sY(s) + 2Y(s) = X(s)

 

[rude man]

That should be

sY(s) + 2Y(s) = X(s)

We are getting close.
We are now going to inverse-transform this equation back into the time domain.

So the 2Y(s) term is obviously transformed to 2y(t).
What are we going to do with the sY(s) term?
Hint: another poster has previously given you a hint as to what to do with this term.

(I have to leave for several hours, will get back about 5 hrs from now.)

 

[SMOF]

The sY(s) is the is the first derivative y'(t)

No worries, it is nearly 3am here, and I need some sleep!

Thanks.

Seán

 

[rude man]

The sY(s) is the is the first derivative y'(t)

No worries, it is nearly 3am here, and I need some sleep!

Thanks.

Seán

Spot-on! Now, write out the whole equation.

If you're interested, your equation has of course implicit in it an initial condition, although it's immaterial as far as deriving the equation itself, so you don't have to be concerned about it, but if you're interested I can show you how elegantly Laplace takes care of that too.

I'm on MST which is 3 hrs behind you. Unless you get back by around 5 a.m. I'm afraid you won't be hearing from me again until around 1 pm your time. I stay up late & get up late, privilege of being retired!

 

[SMOF]

So, if I have it right, it should be

y'(t) + 2y(t) = x(t)?

If you're interested, your equation has of course implicit in it an initial condition, although it's immaterial as far as deriving the equation itself, so you don't have to be concerned about it, but if you're interested I can show you how elegantly Laplace takes care of that too.

Yea, that would be great! I really want to learn as much as I can!

And again, thanks so much for all your help. Hopefully (if I finally have it right) I will be able to do it myself.

Seán

 

[rude man]

Good!

Question: have you had differential equations? Like, could you solve this one, for example if x(t) = 0 or x(t) = sin(wt)?

Also: have you had a course in electrical circuits?

 

[SMOF]

Hello,

Well, differentiating a constant is zero, but I am not sure of that holds for 0. And differentiating sin(wt) would be cos(wt).

I have done a course in electrical circuits, but it is all very rusty, and very much from a hands on point of view.

Seán

 

[rude man]

Hello,

Well, differentiating a constant is zero, but I am not sure of that holds for 0. And differentiating sin(wt) would be cos(wt).

I have done a course in electrical circuits, but it is all very rusty, and very much from a hands on point of view.

Seán

d/dt{sin(wt)} = wcos(wt), not just cos(wt).

Well, that's irrelevant. You don't seem to have the background for me to go into this any further, except to say that your H(s) is that of an electrical R-C network, with R going from x to y, and C going from y to "ground" ()zero volts).

If you applied an impule to this network, the capacitor voltage would respond with a voltage change of exp(-2t), adding that voltage to whatever was on C before the impulse came along, So the initial condition is just the voltage on the capacitor before the impulse came along.

It seems to me you're taking a course out of sequence. To fully appreciate the whole frequency-domain business you need to have taken at least a couple of weeks of differential equations. The widely used textbook, Calculus by Thomas, has a good section for that. I would not try to cover all that's in there; concentrate on linear equations with constant coefficients.

Good luck to you!