DSP - Differential Equation for linear system H

[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!

 

[SMOF]

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

Well, the online reference I used to make sure said d/dt sin x = cos x.

So, I was taking x = wt.

Anyway, that is neither here nor there. Again, thank you for all your help, and I will check out that book you mentioned.

Seán

 

[rude man]

Well, the online reference I used to make sure said d/dt sin x = cos x.

So, I was taking x = wt.

Anyway, that is neither here nor there. Again, thank you for all your help, and I will check out that book you mentioned.

Seán

d/dt{sinwt) = wcos(wt)
d/d(wt){sin(wt)} = cos(wt)

Your equation differentiates wrt to t, not wt.

 

[rude man]

Well, the online reference I used to make sure said d/dt sin x = cos x.

So, I was taking x = wt.

Anyway, that is neither here nor there. Again, thank you for all your help, and I will check out that book you mentioned.

Seán[/QUOTE

d(dt{sin(wt)} = wcos(wt)
d/d(wt){sin(wt)} = cos(wt)

You're differentiating with respect to t, not wt.
Cheers!

 

[SMOF]

Grand, I'm with you now.

Seán

 

[SMOF]

Hello,

Rather than start another topic, since this question is related to this tread, I thought I would just tack it on here.

We were asked to "Derive a formula for the output signal y as a function of t". I think I have done this correctly, but would be very grateful if someone could cast an eye over my work, in case I have done something silly ...or, just completely wrong. I have attached a pdf.

Many thanks.

 

[MisterX]

I wouldn't use t as both the function argument of y and the variable of integration. Your answer looks otherwise correct.

sY(s) -y0 + 2Y = X(s)

Y(s) = X(s)/(s +2) + y_{0}/(s +2)

The Laplace transform is linear, so we may treat the terms separately:

Y_{1}(s) + Y_{2}(s) \leftrightarrow y_{1}(t) + y_{2}(t)

Y_{1}(s) = X(s)/(s +2)

h_{1}(t) = e^{-2t}

The output of the system with impulse response h₁(t) can be expressed by convolution.

y_{1}(t) = (x * h_{1})(t) = \int _{0} ^{\infty} x(\tau)e^{-2(t- \tau)} d\tau = e^{-2t} \int _{0} ^{\infty} x(\tau)e^{2 \tau} d\tau

the second term:

Y_{2}(s) = y_{0}/(s +2)
y_{0}/(s +2) \leftrightarrow y_{0}e^{-2t}

solution:

y(t) = y_{1}(t) + y_{2}(t) = e^{-2t} \int _{0} ^{\infty} x(\tau)e^{2 \tau} d\tau + y_{0}e^{-2t} = e^{-2t} \left( \int _{0} ^{\infty} x(\tau)e^{2 \tau} d\tau + y_{0}\right)

 

[rude man]

I wouldn't use t as both the function argument of y and the variable of integration. Your answer looks otherwise correct.

sY(s) -y0 + 2Y = X(s)

Y(s) = X(s)/(s +2) + y_{0}/(s +2)

The Laplace transform is linear, so we may treat the terms separately:

Y_{1}(s) + Y_{2}(s) \leftrightarrow y_{1}(t) + y_{2}(t)

Y_{1}(s) = X(s)/(s +2)

h_{1}(t) = e^{-2t}

The output of the system with impulse response h₁(t) can be expressed by convolution.

y_{1}(t) = (x * h_{1})(t) = \int _{0} ^{\infty} x(\tau)e^{-2(t- \tau)} d\tau = e^{-2t} \int _{0} ^{\infty} x(\tau)e^{2 \tau} d\tau

the second term:

Y_{2}(s) = y_{0}/(s +2)
y_{0}/(s +2) \leftrightarrow y_{0}e^{-2t}

solution:

y(t) = y_{1}(t) + y_{2}(t) = e^{-2t} \int _{0} ^{\infty} x(\tau)e^{2 \tau} d\tau + y_{0}e^{-2t} = e^{-2t} \left( \int _{0} ^{\infty} x(\tau)e^{2 \tau} d\tau + y_{0}\right)

You know the old song "If you want your boomerang to come back, well first you've got to throw it"?

If you want y(t) you must first have an x(t). What is your input time function? Is it an impulse δ(t), a step function U(t), a ramp kt, or ?

 

[rude man]

If you want a general expression for any x(t), then

y(t) = L^-1{(X(s)/(s+2) + y(0+)/(s+2)}
= L^-1{X(s)/(s+2)} + y(0+)e^-2t

For example, if x(t) = U(t), X(s) = 1/s and
y(t) = L^-1{1/s(s+2)} + y(0+)e^-2t
= (1/2)(1 - e^-2t) + y(0+)e^-2t_.

 

[rude man]

Oh, also - if you're going to use convolution, why bring in the frequency domain at all?

The nice thing about Laplace is it avoids the screwy convolution integrals. Of course, a good table of transforms is required.

 

[MisterX]

Oh, also - if you're going to use convolution, why bring in the frequency domain at all?

The nice thing about Laplace is it avoids the screwy convolution integrals. Of course, a good table of transforms is required.

The frequency domain was used to obtain impulse response function h1(t), which was necessary for the convolution to be used. If you are able to convert time domain differential equations directly into time domain impulse response functions, please show me how.

Anyway. I understood the task to be finding a general time domain expression for y(t). I chose a different method so the result would be checked by a different method.

 

[SMOF]

Hello.

This is cool, thanks for all the replies. It is cool to see how things can be dealt with from different positions.

Seán