# 1st order linear DE with step function input

the 1st order linear equation is:

$y'(t) + \frac D M y(t) = f(t)$

with constants:

$D = 100kg/s$

$M = 1000kg$

$f(t) = Fu(t)$ <-- that's Force x the unit step function

an initial condition:

$y(0) = 20.8m/s$

the input is a step function scaled by the Force $F$ ($Fu(t)$) we need to solve the DE and then find the Force needed to make the final velocity $27.8m/s$.

also a block diagram with the Laplace transform:

$f(t) \longrightarrow {\frac 1M \over (s + \frac DM)}$

thank you!

here's what i have so far...

first i integrated the linear function.

$y'(t) + .1 y(t) = .001f(t)$

using $mu$ in the linear DE and the initial condition y(0) = 20.8

$y(t) = .01 + 20.79 e^(-.1t)$ that's e to the power of -.1t

the problem is i can't figure out what to do with the right side of the equation. the step function scaled by force. i need help integrating the right side. $Fu(t)$

i need to solve the equation to a point where i can input a constant value for the force in order to aim for the target velocity of 27.8m/s.

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thank you for helping with the edit Mahdi. i'll put in where i'm at so far. – Eli Jul 20 '13 at 17:48
$f(t) = Fu(t)$ yes. the Laplace of the step being the integral of $u(t)e^(-st)$ ? i don't know why the -st won't superscript. – Eli Jul 20 '13 at 18:05
$u(t)e^{-st}$ Yes, the input is a step function scaled by the force F. $f(t) = Fu(t)$. Once solved i need to find a Force that will yield a new given final velocity. – Eli Jul 20 '13 at 18:09
F is constant insofar as once it is applied to the object it will remain the same, in order to achieve the final velocity. Also, thank you for helping me Azmoti. – Eli Jul 20 '13 at 18:23
It just says "choose F such that the final velocity is 27.8m/s". The only time given is when the velocity is 20.8 at t=0. But the next part of the problem says "plot the velocity y(t) versus time. Your axis should go from 0 to 100 sec." So I guess we will use 100s as the final time. – Eli Jul 20 '13 at 18:33

We are given:

$$\tag 1 y'(t) + \frac D M y(t) = \dfrac{1}{M}f(t)$$

where:

• $D = 100kg/s$
• $M = 1000kg$
• $f(t) = Fu(t)$, Force $\times$ Heaviside unit step function
• Initial Condition (IC): $y(0) = 20.8m/s$

Rewriting $(1)$ yields:

$$\tag 2 y'(t) + \dfrac{1}{10} y = \dfrac{F}{1000} u(t)$$

Taking the Laplace Transform of $(2)$ yields:

$$\mathcal{L}\left(y'(t) + \dfrac{1}{10} y = \dfrac{F}{1000} u(t)\right) = s y(s) - y(0) + \dfrac{1}{10} y(s) = \dfrac{F}{1000 s}$$

We want to group the $y(s)$ term on the LHS side and everything else on the RHS, so we have:

$$y(s)\left(s + \dfrac{1}{10}\right) = y(0) + \dfrac{F}{1000 s} = 20.8 + \dfrac{F}{1000 s}$$

So we have (that last part is a partial fraction expansion):

$$\tag 3 y(s) = \dfrac{20.8 + \dfrac{F}{1000 s}}{s + \dfrac{1}{10}} = \dfrac{0.01 (F+20800 s)}{s (10 s+1)} = \left(\dfrac{20.8-0.01F}{s+0.1} + \dfrac{0.01 F}{s}\right)$$

Now, we need to find the Inverse Laplace Transform of $(3)$, so we have:

$$\mathcal{L}^{-1}~(y(s)) = y(t) = \mathcal{L}^{-1}~\left(\dfrac{20.8-0.01F}{s+0.1} + \dfrac{0.01 F}{s}\right) = 0.01 \left(F-(F-2080) e^{-t/10}\right)$$

So, we have:

$$y(t) = 0.01 \left(F-(F-2080) e^{-t/10}\right)$$

Now, we need to find $F$ such that the final velocity is $27.8~m/s$. We are given a final time for this velocity at $t = 100$, so we would have:

$$y(100) = 0.01 \left(F-(F-2080) e^{-10}\right) = 27.8 \rightarrow F = 2780.03$$

Thus, we have:

$$y(t) = 27.8003-7.0003 e^{-t/10}$$

A plot of this is:

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Azmoti thank you so much. Right before you posted that I was figuring out for myself about Laplacing the differentials on the left side but I was still stuck. THANK YOU!!! – Eli Jul 20 '13 at 19:05
on the original equation it is $y'(t) + (D/M) y(t) = (1/M) f(t)$ – Eli Jul 20 '13 at 19:08
so would we then divide $F/s$ by .001? – Eli Jul 20 '13 at 19:10
Oops yes I did I apologize. Little mistakes add up though, don't they. – Eli Jul 20 '13 at 19:14
I'm working it through with your solution by just multiplying $1/1000$ on the RHS. Is it as simple as that? – Eli Jul 20 '13 at 19:18