Linear Systems Control

WBahn

Joined Mar 31, 2012
29,978
Also, the method of solution that you used is the long way around. While valid, there is really no need to resort to Laplace transforms for something this basic.

\(
F(t) = I_o \delta (t)
\)

We know that

\(
\int_{-\infty}^{+\infty} \delta (t) dt = \int_{0^-}^{0^+} \delta (t) dt = 1 \;\; \text{(units?)}
\)

Notice that we have a units problem since if δ(t) is dimensionless then when we integrate over all time we end up with a result that has to have units of time (since the units of an integral are equal to the units of the integrand and dt carries units of time. Thus δ(t) is NOT dimensionless and carries units that are the reciprocal of the independent variable (in our case that is inverse-time). But is the unit 1/seconds? 1/hours? 1/millennia? The only way this can be resolved is to require the use of a scaling coefficient that carries the necessary units, namely whatever the units of the equation need to be multiplied by a unit of time.

Since we need F(t) to have units of force, we need Io to have units of force multiplied by units of time, so something like newton·seconds.

\(
\int_{-\infty}^{+\infty} I_0 \delta (t) dx = I_0 \int_{0^-}^{0^+} \delta (t) dt = I_0
\)

Note that this scaling coefficient is NOT optional (except in a purely dimensionless world), no matter how much textbook authors love to be sloppy and ignore it (keep in mind that most textbook authors, particularly math textbook authors, have never worked in the real world).

If we want the result as a function of time, then we need to do this slightly differently and only integrate from negative infinity up to the time of interest.

\(
\int_{-\infty}^{t} I_0 \delta (\tau) d\tau = I_0 \int_{-\infty}^{t} \delta (\tau) d\tau = I_0 u(t)
\)

where u(t) is the unit step function.

The change of variable in the integral from t to τ is merely to avoid confusion with the variable t in the integration limit. Since the variable of integral is a dummy variable, we can replace it with any variable we want.

We can apply this to our example through direct application of Newton's Second Law:

\(
F(t) \; = \; m \cdot a(t) \; = \; m \cdot \frac{dv(t)}{dt}
\;
\frac{dv(t)}{dt} \; = \; \frac{F(t)}{m} \; = \; \frac{I_0}{m} \delta (t)
\;
dv(t) \; = \; \frac{I_0}{m} \delta (t) dt
\;
v(t) \; = \; \frac{I_0}{m} \int_{-\infty}^{t} \delta (\tau) d\tau
\;
v(t) \; = \; \frac{I_0}{m} u(t)
\)

Notice how the units work out. Io has units of newton·seconds and a newton has units of mass over acceleration which work out to be (mass·distance)/(time²). Thus I0 divided by mass has units of distance/time, which is a unit of velocity.

So given this form for v(t), what are the expressions for x(t), p(t), and E(t)?
 

MrAl

Joined Jun 17, 2014
11,389
If so,I've made a mistake in my math...Can you tell me which part of the solution in post #3 is not right?
Hi,

I'm sorry i may have misled you a little. It looked at first as though you were solving for the velocity, when really you are solving for the displacement. In that case it could be the sine function not the cosine function, as part of the solution anyway.

I can tell you the electrical circuit seems so much easier :)
 

Thread Starter

arman19940326

Joined Jul 31, 2014
43
Also, the method of solution that you used is the long way around. While valid, there is really no need to resort to Laplace transforms for something this basic.

\(
F(t) = I_o \delta (t)
\)

We know that

\(
\int_{-\infty}^{+\infty} \delta (t) dt = \int_{0^-}^{0^+} \delta (t) dt = 1 \;\; \text{(units?)}
\)

Notice that we have a units problem since if δ(t) is dimensionless then when we integrate over all time we end up with a result that has to have units of time (since the units of an integral are equal to the units of the integrand and dt carries units of time. Thus δ(t) is NOT dimensionless and carries units that are the reciprocal of the independent variable (in our case that is inverse-time). But is the unit 1/seconds? 1/hours? 1/millennia? The only way this can be resolved is to require the use of a scaling coefficient that carries the necessary units, namely whatever the units of the equation need to be multiplied by a unit of time.

Since we need F(t) to have units of force, we need Io to have units of force multiplied by units of time, so something like newton·seconds.

\(
\int_{-\infty}^{+\infty} I_0 \delta (t) dx = I_0 \int_{0^-}^{0^+} \delta (t) dt = I_0
\)

Note that this scaling coefficient is NOT optional (except in a purely dimensionless world), no matter how much textbook authors love to be sloppy and ignore it (keep in mind that most textbook authors, particularly math textbook authors, have never worked in the real world).

If we want the result as a function of time, then we need to do this slightly differently and only integrate from negative infinity up to the time of interest.

\(
\int_{-\infty}^{t} I_0 \delta (\tau) d\tau = I_0 \int_{-\infty}^{t} \delta (\tau) d\tau = I_0 u(t)
\)

where u(t) is the unit step function.

The change of variable in the integral from t to τ is merely to avoid confusion with the variable t in the integration limit. Since the variable of integral is a dummy variable, we can replace it with any variable we want.

We can apply this to our example through direct application of Newton's Second Law:

\(
F(t) \; = \; m \cdot a(t) \; = \; m \cdot \frac{dv(t)}{dt}
\;
\frac{dv(t)}{dt} \; = \; \frac{F(t)}{m} \; = \; \frac{I_0}{m} \delta (t)
\;
dv(t) \; = \; \frac{I_0}{m} \delta (t) dt
\;
v(t) \; = \; \frac{I_0}{m} \int_{-\infty}^{t} \delta (\tau) d\tau
\;
v(t) \; = \; \frac{I_0}{m} u(t)
\)

Notice how the units work out. Io has units of newton·seconds and a newton has units of mass over acceleration which work out to be (mass·distance)/(time²). Thus I0 divided by mass has units of distance/time, which is a unit of velocity.

So given this form for v(t), what are the expressions for x(t), p(t), and E(t)?
Give me some time to think... I never cared about units before...It's a serious topic to consider though...
 

MrAl

Joined Jun 17, 2014
11,389
Hello again,

I've taken another look at post #3, and i see it looks like you got the right result for the original response (not the counter force) but you did not show how you got this result from the method you set out to use. For example, it looks like for the second derivative you used M*s^2*F instead of M*(s^2*F-s*f0-df0), where f0 is the value fo the function at t=0 and df0 is the value of the derivative at t=0, and F is the function. For this problem, going about it in this way, the derivative of the function can not be zero but must be 1/M. You have to be careful because you can end up with the right result doing it the wrong way.
 

WBahn

Joined Mar 31, 2012
29,978
Give me some time to think... I never cared about units before...It's a serious topic to consider though...
Definitely take the time to care about units -- it might save your life someday. While a bit extreme, I actually do mean that literally. Many people have been killed because they, or someone else, couldn't be bothered to track their units and therefore didn't catch a simple, glaring mistake. Tracking your units -- really tracking them and using them as a fundamental part of the problem and not just something that has to be tacked onto the end -- is perhaps the single most effective error detection tool that the engineer has at their disposal. Personally, I think it's inexcusable negligence not to use that tool -- and on occasion juries have come to that same conclusion in wrongful death and injury cases.
 

MrAl

Joined Jun 17, 2014
11,389
Hi,

Good advice, and dont forget to check the base units as well, taking the Mars Climate Orbiter as example where we lost something like 500 million dollars (USD) because of a mix up of base units between different engineering parties. It should come out of their paychecks :)
 

WBahn

Joined Mar 31, 2012
29,978
Hi,

Good advice, and dont forget to check the base units as well, taking the Mars Climate Orbiter as example where we lost something like 500 million dollars (USD) because of a mix up of base units between different engineering parties. It should come out of their paychecks :)
In general I agree, although the circumstances surrounding that mission are such that catching the error would not have been simple. But I think that if the overall culture in the science and engineering communities had been such that an engrained awareness of and attention to units were the norm then this error would not have been made or gone uncaught. But given culture that led up to the point where the error was made, it was all-but-impossible for the error to get caught by then except in post-accident analysis.
 

MrAl

Joined Jun 17, 2014
11,389
In general I agree, although the circumstances surrounding that mission are such that catching the error would not have been simple. But I think that if the overall culture in the science and engineering communities had been such that an engrained awareness of and attention to units were the norm then this error would not have been made or gone uncaught. But given culture that led up to the point where the error was made, it was all-but-impossible for the error to get caught by then except in post-accident analysis.
Hi,

Well, in a 100 dollar hobby project i could easily see this happening by accident, but in a 500 million dollar project, i just cant see how they could have been so stupid. This kind of thing is taught in first physics classes. Then again, it's not their money.

I think it is the perfect example of irresponsibility because of the lack of touch with reality.

To be fair, i think they have gotten better since then, but they often dont mind taking the risks when it's not their money or their own lives at stake. We have to pray that they get the asteroid deflection program going without errors or this Earth is doomed.

Back on track here, there is an easier way to calculate the original problem but i'll wait until we've seen some more thinking and reflection.
 

WBahn

Joined Mar 31, 2012
29,978
Hi,

Well, in a 100 dollar hobby project i could easily see this happening by accident, but in a 500 million dollar project, i just cant see how they could have been so stupid. This kind of thing is taught in first physics classes. Then again, it's not their money.

I think it is the perfect example of irresponsibility because of the lack of touch with reality.

To be fair, i think they have gotten better since then, but they often dont mind taking the risks when it's not their money or their own lives at stake. We have to pray that they get the asteroid deflection program going without errors or this Earth is doomed.

Back on track here, there is an easier way to calculate the original problem but i'll wait until we've seen some more thinking and reflection.
It basically came down to parameters deep inside one piece of software interacting with parameters that were deep inside another piece of software and both pieces of software were inherited from previous projects and had worked fine and were considered vetted and flight-tested. I've heard a detailed description of just where the problem was and how the problem was masked because of the interfaces to the various pieces of code involved and I have to admit that it would have been hard to catch. You don't have the time or money to completely revalidate every line of code and every bolt on every project. But I still believe that if the culture had been such that all of the documentation and all of the code variable names and all of the interfaces reflected the units involved and if the culture were such that people intrinsically balked whenever they saw any code or any documentation or any interface that did not reflect the units involved, that this and similar things would almost never happen.
 
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