First order Laplace transform

1. Mar 27, 2016

topcat123

1. The problem statement, all variables and given/known data
A process can be represented by the first order equation

(4δy(t)/δt) + y(t) = 3u(t)

Assume the initial state is steady (y = 0 at t = –0).

(a) Determine the transfer function of this process in the s domain.
(b) If the input is a ramp change in u(t) = 4t, determine the value of y(t)
when t = 10 s.

2. Relevant equations
Ramp change 1/s2

3. The attempt at a solution
(a)
Transfer function

{sY(s) + y(-0)} + Y(s)/4 = 3U(s)/4

sY(s) + Y(s)/4 = 3U(s)/4

G(s) = Y(s)/U(s) = ¾/(s + ¼) = 4/4(s + ¼)

(b)
This is where i am stuck i realise this is where the ramp change 1/s2 comes in.

Y(s) =¾/(s + ¼) × 1/s2 = ¾/s(s + ¼)
I think this is correct but I don't know where to go from here.

any help would be appreciated.

2. Mar 27, 2016

Staff: Mentor

Check your algebra. $\frac{3/4}{s + 1/4} \times \frac 1 {s^2} \ne \frac{3/4}{s(s + 1/4)}$

I haven't verified the rest of your work...

3. Mar 27, 2016

topcat123

sorry copied it in wrong
Y(s) =¾/(s + ¼) × 1/s2 = 3/s(s + ¼)

4. Mar 27, 2016

Staff: Mentor

That is still incorrect.

Also, when you write 3/s(s + 1/4), the usual interpretation is $\frac 3 s (s + 1/4)$, which is not what you meant.

5. Mar 28, 2016

topcat123

$$Y(s) =\frac{\frac{3}{4}}{(s+\frac{1}{4})}\frac{1}{s^2} = \frac{\frac{3}{4}}{s^2(s+\frac{1}{4})}$$

Is this better?

6. Mar 28, 2016

Staff: Mentor

Yes. To find y(t), split up the fraction using partial fraction decomposition, and then take the inverse Laplace transform of each of the resulting terms.

7. Mar 28, 2016

topcat123

OK so partial fraction. $$\frac{a}{s^2 (s + a)} = \frac{(As + B)}{s^2} + \frac{C}{(s + a)}$$
Multiply through by $$s^2 (s + a)$$
Giving $$a = (As + B) (s + a) + Cs^2$$
Removing the brackets$$As^2 + Bs + Aas + Ba + C^2$$
$$s^0..... a = Ba.... B = 1$$ $$s^1....0 = B + Aa$$ substituting in for B $$A = \frac{-1}{a}$$ $$s^2.... 0 = A + C$$ substituting in for A$$C = \frac{1}{a}$$
gives $$= \frac{(-\frac{s}{a} + 1)}{s^2} + \frac{\frac{1}{a}}{(s + a)}$$
simplifying $$= \frac{(-\frac{s}{a} + 1)}{s^2}$$ to $$-\frac{\frac{s}{a}}{s^2} + \frac{1}{s^2} = \frac{\frac{1}{a}}{s} + \frac{1}{s^2}$$
Final result$$\frac{a}{s^2(s + a)} = -\frac{\frac{1}{a}}{s} + \frac{1}{s^2} + \frac{\frac{1}{a}}{(s + a)}$$
$$-\frac{\frac{1}{a}}{s}\text { inverse to } -\frac{1}{a}$$
$$\frac{1}{s^2}\text{ inverse to } t$$
$$\frac{\frac{1}{a}}{(s + a)}\text{ inverse to }\frac{1}{a}e^{-at}$$
$$\left[-\frac{1}{a} + t + \frac{1}{a}e^{-at}\right]$$ as a = ¼ ad t = 10
$$3\left[-\frac{1}{\frac{1}{4}} + 10 + \frac{1}{\frac{1}{4}}e^{\frac{1}{4}}\right]$$
So $$y(t) = 3\left[-4 + 10 +4e^{-\frac{1}{4}}\right] = 27.35$$

Is that any where near?

8. Mar 29, 2016

Staff: Mentor

By my calculation, you're low by about a factor of 3.

What you show as y(t) above is actually y(10). What do you have for y(t)? You need to check that it is a solution of the diff. equation y' + (1/4)y = 3t, and that y(0) = 0. The solution I found satisifies this diff. equation and initial condition, and I get y(10) ≈ 75.94.

9. Mar 29, 2016

Staff: Mentor

I think you have a mistake in the last line above.
3u(t) = 12t, so after dividing by 4 you should get 3t. The Laplace transform of this is $\frac 3 {s^2}$, not $\frac {3/4}{s^2}$

10. Mar 29, 2016

topcat123

Yes I see the mistake there is also another.
$$Y(s) = \frac{\frac{12}{4}}{s^2(s + \frac{1}{4})}$$
also the second error.
$$\frac{1}{a}e^{at} = \frac{1}{\frac{1}{4}}e^{-\frac{10}{4}}$$
giving $$y(10) =12 \left[ -4 + 10 + 4e^{-\frac{10}{4}}\right ] = 75.94$$

Is this better?

Last edited by a moderator: Mar 29, 2016
11. Mar 29, 2016

Staff: Mentor

One thing though that you're a little sloppy on: You wrote $\frac{1}{a}e^{at} = \frac{1}{\frac{1}{4}}e^{-\frac{10}{4}}$. The left side is a function of t, but the right side is a number. You should have $y(t) = \frac{1}{a}e^{-at} = 4e^{-\frac{t}{4}}$. From this you get y(10) = 75.94 (approx.).