1st order linear DE with step function input

[elijah78]

hi guys, this is my first post, but I've read many. so my problem is actually for an engineering class, but it's more math/physics related.
i took DE last semester and i know how to solve linear eqs but the step function/force is throwing me for a loop.
1. Homework Statement we are given a block diagram:

{applied force} f(t) --> [ (1/m) / (s+D/m) ] --> y(t) {velocity} (Laplace stuff)

D = drag = 100kg/s
m = 1,000kg

2. Homework Equations

the 1st order linear eq is:

y'(t) + (D/m)y(t) = (1/m)f(t)
Q: solve the DE for y(t) if the input is a step function scaled by the force F, f(t)=Fu(t).
the initial velocity is y(0)=20.8m/s. choose F such that the final velocity is 27.8m/s. (using time domain)

3. The Attempt at a Solution

D/m = .1
1/m = .001

so i solved the linear equation:

μ=e^∫.1dt = e^.1t

e^.1t [ y'(t) + .1y(t) = .001 ]

(e^(.1t)y(t))' = .001e^(.1t)

e^(.1t)y(t) = ∫.001e^(.1t)

e^(.1t)y(t) = [.001e^(.1t)]/.1 + c

y(t) = .01 + ce^(-.1t)now solving for c: y(0) = 20.8 t=0

20.8 = .01 + ce^(-.1x0)

c = 20.79

so now i have:

y(t) = .01 + 20.79e^(-.1t)i don't know how Force figures into this. it seems so simple but I'm stuck here. and i don't believe this is right because the plot of y(t) doesn't show acceleration. thank you so much in advance for any assistance!

elijah

 

[BruceW]

so i solved the linear equation:

μ=e^∫.1dt = e^.1t

e^.1t [ y'(t) + .1y(t) = .001 ]

Hi there elijah, welcome to physicsforums :)
In the second equation here, the function f(t) is missing. You've got the 1/m (i.e. .001), but missed out the f(t).

 

[elijah78]

thanks man :) i enjoy the integrity of this forum.i've also tried keeping the f(t) and when I've solved the linear eq I'm left with:

y(t) = f(t)/.1 + ce^(-.1t)

or

y(t) = Fu(t)/.1 + ce^(-.1t)

with no way to solve for c because i now have another variable, Fu(t) or f(t).

the way i learned linear DE's was:

y'(t) + p(t)y(t) = g(t)

where p(t) and g(t) are just numbers.
like in my equation here:

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

that's where I'm confused, i don't know how to use the f(t) so that i can set up an equation where i can throw in different force values to get the final velocity.

thank you so much for the quick reply. :)

 

[hilbert2]

the 1st order linear eq is:

y'(t) + (D/m)y(t) = (1/m)f(t)

I think you are expected to Fourier transform that equation, solve it in frequency space and then inverse transform to get the result. Have you studied integral transforms in your class?

 

[elijah78]

yes Fourier and laplace. he told us to use the time domain. the block diagram is in laplace format i.e. S instead of jw. I've also tried:

(1/m) / (s + D/m)

D/m = .1

.001 / (s + .1)

1 / (s + .1) ---> e^(-.1t)u(t)

i'm not sure if I'm doing the right thing with the m in the numerator's denominator here. but i still came out with the same e^(-.1t) value. he gave us a few in class notes on the project and didn't mention anything about using transforms on the project. he just showed us very simple steps regarding the forces (applied and drag) in the form of the linear equation. nothing new. the project sheet says nothing about using transforms besides the inclusion of the block diagram.

 

[hilbert2]

I tried to solve the equation with Fourier transform, but doing the final inverse transform caused too much trouble so I gave up.

Mathematica says that the solution of equation $y'(x)+ay(x)=b\theta(x)$ ($\theta(x)$ is the unit step function) is $y(x)=Cexp(-ax)+\frac{bexp(-ax)(exp(ax)-1)\theta(x)}{a}$. Here C is a constant determined by the initial conditions.

EDIT: A property of the solution is that $y(t)$ can approach some "terminal velocity" when $t \rightarrow \infty$

 

[BruceW]

i've also tried keeping the f(t) and when I've solved the linear eq I'm left with:

y(t) = f(t)/.1 + ce^(-.1t)

That's close, but not quite right. I'm guessing that at some point, you would have integrated 0.001*f(t)*e^(.1t) right? You need to use the rule for integrating a step function. But anyway, yeah, it looks like you are meant to use Laplace transform method instead. You can do it this way (i.e. standard integration), but I guess it is probably best to do it the recommended way (i.e. Laplace transform).

 

[elijah78]

Alright thanks so much guys, it's 5am so I'm going to go sleep on it. Laplace, laplace, laplace... As soon as I get over this speed bump the rest is a breeze, MATLAB plots, etc.

Thanks again!

 

[Chestermiller]

The Laplace transform for the differential equation and initial condition is:

$$sy(s)-20.8+\frac{D}{m}y(s)=\frac{F}{ms}$$

Step 1 is to solve this for y(s)

 

[rude man]

we are given a block diagram:

{applied force} f(t) --> [ (1/m) / (s+D/m) ] --> y(t) {velocity} (Laplace stuff)

D = drag = 100kg/s
m = 1,000kg

2. Homework Equations

the 1st order linear eq is:

y'(t) + (D/m)y(t) = (1/m)f(t)

You're supposed to stay in the time domain, so solve the diff. eq. conventionally.

Since f(t) = FU(t):

Use separation of variables: dy/{F/m - Dy/m} = dt etc.

 

[elijah78]

@rude man: can you explain that a little further? i know what separation of variables is but I'm having a hard time seeing it here.

edit: i contacted my professor and he said Laplace was acceptable. thank god, i wouldn't have figured conventional int out!

 

[rude man]

@rude man: can you explain that a little further? i know what separation of variables is but I'm having a hard time seeing it here.

For future reference, you just integrate both sides of the equation, in this case you'd get
-[ln(F/m - Dy/m)]/(D/m) = t + constant
etc.

edit: i contacted my professor and he said Laplace was acceptable. thank god, i wouldn't have figured conventional int out!

OK, keep us up to date on your progress! Yes, laplace is the only way to go.

 

[elijah78]

this place is so cool. great, intelligent people. I'm glad i came here.

 

[hilbert2]

You're supposed to stay in the time domain, so solve the diff. eq. conventionally.

Since f(t) = FU(t):

Use separation of variables: dy/{F/m - Dy/m} = dt etc.

The equation is not separable. You have to multiply both sides of the eq with integrating factor $e^{\frac{D}{m}t}$ and use the derivative of product rule to get

$\frac{d}{dt}\left(e^{\frac{D}{m}t}y(t)\right)=\frac{F}{m}e^{\frac{D}{m}t}U(t)$

And now you are able to integrate both sides of the equation.

 

[rude man]

The equation is not separable.

You're not easy to convince, are you? Even when I showed you how?

 

[BruceW]

Since f(t) = FU(t):

Use separation of variables: dy/{F/m - Dy/m} = dt etc.

I don't understand why you replace f(t) with simply F. I think this is why hilbert2 is saying the equation is not separable. Are you just calculating the solution for t>0 ? If we specify that t>0 then I guess it is fine to say f(t)=F, so then the equation is separable in that case.

 

[Chestermiller]

The Laplace transform for the differential equation and initial condition is:

$$sy(s)-20.8+\frac{D}{m}y(s)=\frac{F}{ms}$$

Step 1 is to solve this for y(s)

$$y(s)=\frac{\frac{F}{ms}+20.8}{s+D/m}=\frac{20.8}{s}\frac{(s+\frac{F}{20.8m})}{(s+D/m)}=\frac{20.8}{s}\frac{(s+D/m+\frac{F}{20.8m}-D/m)}{(s+D/m)}=\frac{20.8}{s}(1+\frac{(\frac{F}{20.8m}-D/m)}{(s+D/m)})$$
$$y(s)=\frac{20.8}{s}+\frac{(\frac{F}{m}-20.8(D/m))}{s(s+D/m)}=\frac{20.8}{s}+(F/D-20.8)(\frac{1}{s}-\frac{1}{s+D/m})=\frac{F/D}{s}-\frac{(F/D-20.8)}{s+D/m}$$
So,
$$y(t)=\frac{F}{D}-(\frac{F}{D}-20.8)\exp{(-\frac{D}{m}t)}$$

 

[rude man]

I don't understand why you replace f(t) with simply F. I think this is why hilbert2 is saying the equation is not separable. Are you just calculating the solution for t>0 ? If we specify that t>0 then I guess it is fine to say f(t)=F, so then the equation is separable in that case.

You got the geneal idea.

FU(t) means there is no input to the system for t < 0, ergo no output.

In Laplace, same thing. In fact, Laplace as most of us know it is the so-called "single-sided Laplace transform" which implies no output for t < 0. The so-called "double-sided Laplace transform" does handle systems for t < 0 but is very rarely used and then only for stochastic signals. Most treatments of the latter prefer the Fourier transform.

 

[elijah78]

chester that too is what i got, only i had plugged in some ICs along the way, making it harder for me to investigate different values of drag and what influence they have on other parameters. this makes it a little easier.

 

[elijah78]

@everybody

yeah the whole problem is about a vehicle moving from 0 =< t =< 100

 

[hilbert2]

When working out this kind of problems, it is useful to check that the final result is physically reasonable. If the function $y(t)$ is interpreted as the velocity of an object, then the equation $y'(t)+\frac{D}{m}y(t)=\frac{F}{m}U(t)$ describes the motion of an object that is subject to both an accelerating force and a damping force (friction or air resistance). From basic physics we know that is such a situation the object eventually reaches a state of terminal velocity, where the accelerating and decelerating forces are equal and opposite. Setting $y'(t)=0$ in the equation, we can easily solve that at that point the velocity of the object must be $y(t)=\frac{F}{D}$. We can see that in Chestermiller's result, y(t) approaches this value as $t\rightarrow \infty$, so the result is physically reasonable.

 

[elijah78]

after all was said and done, I've figured out that it would take 2780N to get the vehicle up to 27.8m/s.

 

[BruceW]

In Laplace, same thing. In fact, Laplace as most of us know it is the so-called "single-sided Laplace transform" which implies no output for t < 0. The so-called "double-sided Laplace transform" does handle systems for t < 0 but is very rarely used and then only for stochastic signals. Most treatments of the latter prefer the Fourier transform.

Ah right. For the single-sided Laplace transform, it doesn't matter what f(t) is for t<0, since it does not get integrated anyway. I am not used to Laplace transforms, so I forgot that usually the single-sided Laplace transform is implied (rather than the double-sided Laplace transform).

 

[rude man]

Ah right. For the single-sided Laplace transform, it doesn't matter what f(t) is for t<0, since it does not get integrated anyway. I am not used to Laplace transforms, so I forgot that usually the single-sided Laplace transform is implied (rather than the double-sided Laplace transform).

Actually, to the extent of any finite initial conditions at t = 0+, it does matter what f(t) was for t < 0. But forcing functions for t < 0 are not allowed in the one-sided Laplace transform.

 

[BruceW]

after all was said and done, I've figured out that it would take 2780N to get the vehicle up to 27.8m/s.

I agree. Nice work man :)

 

[Chestermiller]

chester that too is what i got, only i had plugged in some ICs along the way, making it harder for me to investigate different values of drag and what influence they have on other parameters. this makes it a little easier.

Here is something else that would have made life much easier for you. If you are only interested in the asymptotic velocity at very long times (as in the present case), all you have to do is multiply the Laplace Transform by s, and then take the limit of the resulting expression at s = 0. So you didn't even need to invert the transform.

Chet

 

[rude man]

Here is something else that would have made life much easier for you. If you are only interested in the asymptotic velocity at very long times (as in the present case), all you have to do is multiply the Laplace Transform by s, and then take the limit of the resulting expression at s = 0. So you didn't even need to invert the transform.

Chet

Chet, you are a true polymath!

rudy

 

[vanhees71]

As a simple minded physicist I don't understand why you use the Laplace transformation here. Isn't this problem much simpler solved directly. After all it's a first-order inhomogeneous ODE which can be solved with the standard techniques in the time domain without much difficulty, or am I missing something?

 

[rude man]

As a simple minded physicist I don't understand why you use the Laplace transformation here. Isn't this problem much simpler solved directly. After all it's a first-order inhomogeneous ODE which can be solved with the standard techniques in the time domain without much difficulty, or am I missing something?

As an electrical engineer I'm aware that most physicists don't know or use the Laplace. I am not suggesting you don't know it, you evidently do. But I worked in a dept. with mostly physicists and a few EE's & that is my experience over my lo these many years.

Here's the thing: with Laplace, a diff. eq. is immediately transformed into an algebraic equation. As a bonus, all initial conditions are automatically included, as are forcing functions.

Now the problem is immensely easy. Chestermiller has pointed this out admirably I feel, and has even shown how the solutions at t = 0+ and t = ∞ are mightily facilitated without the need to find the inverse transform at all.

The main objection to the Laplace, I have found, is that you need that table of inverse transforms. Unlike the Fourier transform, inverting a Laplace transform involves integration in the complex plane as opposed to just a single variable of integration ( s = σ + jω etc). Not a pleasant business. But that's why tables were created over the centuries for dumb-dumbs like myself .

The other big thing is that Laplace is not applicable to d.e.'s with non-constant coefficients, at least I have never had occasion to explore this possibility, or a need to.

For me, solving d.e.'s in the time domain is a jungle of approaches and "guesses", especially where forcing functions (inhomogeneous d.e.'s) are concerned. I can look up a forcing function of almost any complexity without needing to "guess".

I think in the end it's just what you are accustomed to. EE's are so steeped in Laplace that we wouldn't be able to solve de's by standard methods if our lives depended on it! I usually look for separation of variables if time-domain is mandated by the instructor, otherwise I apologize to the OP and quit!

 

[Chestermiller]

Chet, you are a true polymath!

rudy

Thanks rude man, but you are much too kind.

Your description of the applicability of Laplace Transforms, together with their advantages and disadvantages, was very complete and balanced. There is great appeal to a simple table look-up method like this. As a chemical engineer, I used Laplace Transforms only occasionally during my professional career. However, I'm sure that chemical engineers involved with process control used them much more extensively. It is just a matter of choosing the tool that is most appropriate for the kinds of problems that you encounter most frequently in your profession.

Chet

 

[elijah78]

A big big big thank you to you guys for your help. i ended up with an A on the project and a B for the semester. Thank you!

 

[rude man]

A big big big thank you to you guys for your help. i ended up with an A on the project and a B for the semester. Thank you!

Congrats and come again any old time!