Engineering Finding the system output by convolution

[jisbon]

Homework Statement

As attached below.

Relevant Equations

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Since there are initial conditions stated, I would have to craft the s equation in mind, in order to find the impulse by laplace inverse; which is this:
$(s^2Y(s)-sy(0)-y'(0))+8(sY(s)-y(0))+16Y(s)=x(s)$
$(s^2Y(s)+\frac{1}{2}s-1)+8(sY(s)-1)+16Y(s)=x(s)$
$Y(s)(s^2+8s+16)+\frac{1}{2}s-1-8=x(s)$
$\frac{Y(S)}{X(S)}=\frac{1}{(s^2+8s+16)+\frac{1}{2}s-9)}$

Now is this even correct(before I move on to the next step of finding the output)? It seems pretty messed up to me, but I can't figure out what is wrong here :( Please help, thanks.

 

[cnh1995]

You are correct so far. You might want to further simplify the denominator e.g. +16-9=7.

 

[vela]

It looks like you plugged in the wrong value for $y(0)$ in the Laplace transform of the $y'$ term.

 

[jisbon]

It looks like you plugged in the wrong value for $y(0)$ in the Laplace transform of the $y'$ term.

$Y(s)(s^2+8s+16)+\frac{1}{2}s-1+4=x(s)$
$\frac{Y(S)}{X(S)}=\frac{1}{(s^2+8s+16)+\frac{1}{2}s-3)}$
hence, inversing it will give me:

I then have to find the output, which is calculated by:

Hence the equation will be:
$\int_{-\infty}^{t} (16u(t))(\frac{2}{9}e^{-2(t-\iota)}-\frac{2}{9}e^{\frac{-13}{2}(t-\iota)})$

Does this look correct now?

 

[vela]

No, you didn't do the algebra correctly. You should get
$$Y(s) = \frac{X(s)}{s^2+8s+16} - \frac 12 \frac{s + 6}{s^2+8s+16}.$$

 

[jisbon]

Hi there :)
Thanks for your reply. After some figuring out, I realized that the steps that I did beforehand was unnecessary as to find the transfer function, my initial conditions has to be set to 0 in the first place.

Hence setting the initial conditions to 0,
$\frac{Y(S)}{X(S)}=\frac{1}{(s^2+8s+16)}$
Spilitting into partial fractions (in order to perform a Laplace universe to findimpulse function, I get):
$\frac{Y(S)}{X(S)}=\frac{1}{(s+4)^2}$
Inversing it will give me : $e^{-4t}t$

So following the formula,
the output should be:

whereby it should look like:
$y(t)=\int_{\infty}^{\infty} [u(\tau)]e^{-4t}t dt$

is this correct?

ps: if it is correct, how should I proceed next? unsure of the next steps I should take.

 

[vela]

No. You need to be more careful. The limits are wrong. The integration variable is wrong. $t-\tau$ doesn't appear in your integral.

You will still have to deal somehow with the initial conditions.

 

[jisbon]

No. You need to be more careful. The limits are wrong. The integration variable is wrong. $t-\tau$ doesn't appear in your integral.

You will still have to deal somehow with the initial conditions.

Sorry, I meant this: $y(t)=\int_{-\infty}^{t} [u(\tau)]e^{-4(t-\tau)}t dt$

Also for initial conditions, what do I have to do with it? From my notes, it appears that one have to simply integrate the above function to find out the output.

 

[jisbon]

Hi there again.
Dug some stuff up and attempted to solve this question.

Since X(t) = 16u(t), X(s) is simply 16/s
By subbing that in the equation above, I got Y(s) =

And I inverse it to find the output Y(t) =

Is this correct?
If so, how do I determine if the system is stable? Cheers

 

[vela]

You can check your answer by plugging your solution back into the differential equation.

 

[rude man]

I have never seen anyone solve a differetial equation with FINITE initial conditions by convolution. Until I'm shown to be wrong unequivocally I will state that that is impossible. Nothing like livig dangerously!

I googled two instances of solving ODE's via convolution "with given initial conditions". Those initial conditions were all zero!

PS with laplace it is of course straight-forward.

 

[vela]

I have never seen anyone solve a differetial equation with FINITE initial conditions by convolution. Until I'm shown to be wrong unequivocally I will state that that is impossible. Nothing like living dangerously!

Here's a trick to solve this particular problem using convolution. Let $w(t)=−\frac 12(1+2t)e^{−4t}$ and $u(t)=y(t)−w(t)$. Then $u(0)=u'(0)=0$, and $u$ satisfies the same differential equation $y$ does. The change of variables eliminates the non-homogeneous initial conditions. You can them solve for $u$ using convolution and add $w$ to the result to find $y$.

You probably noticed $w$ is simply a solution to the homogeneous differential equation that satisfies the initial conditions $y$ does. In other words, this is just the usual way we solve a differential equation. Find the homogeneous and particular solutions to obtain the general solution and then set the arbitrary constants in the homogeneous solution appropriately to satisfy the initial conditions.

 

[rude man]

Great post but aren't you "cheating" by coming up with the general solution in the classical (or laplace or fourier) way in the first place? You're not solving directly via convolution?

Whereas if the i.c.'s were zero the o.d.e. would simply be expressed as a laplace or Fourier transform, then inversed to get the impulse response, then convoluted. I would use graphical techniques; I have trouible with the limits of integration otherwise.

An interesting post no matter what!

 

[vela]

Great post but aren't you "cheating" by coming up with the general solution in the classical (or laplace or fourier) way in the first place? You're not solving directly via convolution?

Not really. You can use any function $w$ that satisfies the initial conditions. For example, if you use $w(t) = t-\frac 12$, the differential equation $u$ satisfies is $u'' + 8 u' + 16 u = 16 u(t) - 16 t$. The change of variables allows you to get rid of the non-homogeneous initial conditions at the cost of a more complicated convolution to find $u(t)$.

Using the homogeneous solution with appropriate constants for $w$ is convenient because it doesn't make the differential equation any more complicated. Plus you have to solve the homogeneous differential equation to find the impulse response anyway.

 

[rude man]

P.S. cute cat! I'm an ailurophile myself, currently with 1 boy.

 

[rude man]

OK @vela, I just deleted a post that I hope you or anyone else never saw!
Perhaps you could work a simple example for us:
$dy/dt + y = U(t)$
$U(t)=0, t<0$
$U(t)=1, t>0$
$y(t=0+) = y_0$

 

[vela]

Let $y(t) = p(t) + w(t)$ where $w(t)=y_0$. Then $p(t) = y(t) - y_0$ so that $p(0) = 0$. The differential equation becomes
$$(p + w)' + (p+w) = p' + p + y_0 = U(t) \quad \Rightarrow \quad p' + p = U(t)-y_0.$$ The impulse response is $h(t)=e^{-t} U(t)$, so we get
$$p(t) = \int_{-\infty}^\infty [U(\tau) - y_0][e^{-(t-\tau)} U(t-\tau)]\,d\tau = [1-e^{-t} + y_0(1- e^{-t})]U(t).$$ Therefore, we get $y(t) = p(t) + y_0 = y_0 e^{-t} + 1-e^{-t}$ for $t>0$.

Another way you could look at the general problem is that when you have non-zero initial conditions, in the s-domain you get
$$\frac{Y(s)}{H(s)} - F(s) = X(s) \quad \Rightarrow \quad Y(s) = H(s)[X(s) + F(s)]$$ where $F(s)$ is the polynomial terms from the initial conditions. Then
$$y(t) = \int_{-\infty}^\infty [x(\tau) + f(\tau)]h(t-\tau)\,d\tau.$$

For this particular problem, we'd get $F(s) = y_0$ so $f(t) = y_0 \delta(t)$, where $\delta(t)$ is the Dirac delta function. Then
$$y(t) = \int_{-\infty}^{\infty} [U(\tau) + y_0 \delta(\tau)][e^{-(t-\tau)}U(t-\tau)\,d\tau = (1-e^{-t} + y_0 e^{-t})U(t).$$

 

[rude man]

Thanks for the trouble you took. Very illumiating.
I should have thought of using $x(t)*h(t) \leftrightarrow X(s)H(s)$ myself. Been too long ..

Using strictly laplace I got a slightly different result for y(t):
$y(t) = 1 - e^{-t} + y_0(1-e^{-t})$. Big deal ...
Thx again.

 

[rude man]

Another way you could look at the general problem is that when you have non-zero initial conditions, in the s-domain you get
$$\frac{Y(s)}{H(s)} - F(s) = X(s) \quad \Rightarrow \quad Y(s) = H(s)[X(s) + F(s)]$$

I am still stuck on this part.
The correct equation surely is Y(s) = X(s)H(s) + F(s)

where $F(s)$ is the polynomial terms from the initial conditions.

For this particular problem, we'd get $F(s) = y_0$

I think $F(s) = y_0/s$ and no dirac delta is involved. Could this be why we got different results?

 

[rude man]

I wish now I had given you
$dy/dt + ay = kU(t)$,
$a$ in $sec^{-1}$. Of course, $a$ can = 1.0 sec^{-1} to preserve your previous work;

$k = 1.0 sec^{-1}$
y dimensionless,
to maintain dimensional integrity.with associated checking facility.

 

[vela]

I got
$$sY(s) - y_0 + Y(s) = \underbrace{(s+1)}_{1/H(s)} Y(s) - y_0 = \frac 1s$$
so
$$Y(s) = H(s) \left[\frac 1s + y_0\right].$$

 

[rude man]

I got
$$sY(s) - y_0 + Y(s) = \underbrace{(s+1)}_{1/H(s)} Y(s) - y_0 = \frac 1s$$
so
$$Y(s) = H(s) \left[\frac 1s + y_0\right].$$

Right. My dumb mistake. Time to go to bed early ...
Thx.
PS is your LaTex primer available to anyone? You have a lot of \codes I've never seen before.