1st order linear DE with step function input

Question

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.

Answer 1

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: