DSP - Differential Equation for linear system H

[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 seperately:

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

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

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

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 seperately:

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

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

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

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