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initial and final value theorems

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PG1995

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Hi

Could you please help me with these queries related to final value theorem? Thank you.

Regards
PG
 

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I dont' follow your question for Q1.

For Q2, you forgot the derivative rule you just learned. The LT of the derivative of a function is s times the LT of that function minus the initial value of the function.

For Q3, it is understood that the lower limit is 0- for the one sided transform.
 
Thank you, Steve.

Final value theorem:
I dont' follow your question for Q1.

For Q2, you forgot the derivative rule you just learned. The LT of the derivative of a function is s times the LT of that function minus the initial value of the function.

I have updated Q1 and Q2. Kindly have a look.

Initial value theorem:
Could you please also help me with these queries related to initial value theorem? Thanks a lot.

Best wishes
PG
 

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I'll look at the new questions related to initial value theorem later.

For the final value theorem, I didn't see an update to Q1, so I'm still confused. For Q2, I'll conclude that indeed you are too sleepy. You wrote out an incorrect integral and claimed they are wrong to do that. But, they did not do that. You did. What you wrote is indeed incorrect. What they did do was apply the derivative rule, as I said. That statement is a direct application of the derivative rule - plain and simple. I expect that if you look at it after getting some sleep, you'll see that.
 
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Initial value theorem:
Could you please also help me with these queries related to initial value theorem? Thanks a lot.

Best wishes
PG

I'm finding it difficult to answer some of these questions because they reveal many misconceptions and needless probing. I don't mean that to say you shouldn't ask such questions when you feel the need, but sometimes the person trying to answer needs to point out that you may be trying to run before you can walk. This is a subject that takes time to learn. It's not overly difficult, but there are many many details that take more than a few days to absorb, and you may be trying to absorb too much, too fast.

So, I can't answer it all, but I can try to make a few points that might help.

Q1: Here my comment is that you need to be careful to ask "why" about definitions. The initial value theorem is stated in a particular way, and they tell you about that particular way. Asking "why is it so" is somewhat silly (we all do it, of course) because the answer is "that's how it is defined" . So, then we might ask, "why is it defined this way, and not another way". Usually the answer is either that it is defined in the most useful way and alternative definitions are less useful or are useless. Or, sometimes, the definition is chosen because that is the form we can actually say something about. Perhaps we would like to define something different, but the different way is unsolvable and the defined way is solvable. So, we either define the most useful way, or the way that allows us to say something useful.

In this case, I think the latter is a likely answer. We can say something useful if we define the initial value as t=0+, and we run into theoretical issues if we try other ways. The clear problem with t=0- is that the initial value at t=0- is always zero. So, how useful can that be?

Q2: Basically, just realize that short time scales match up with high frequencies and long time scales match up with low frequencies. But here we are just exploring mathematical relations. Follow the math to get the correct conclusions, and then just mentally note that t=∞ says something about s=0 and t=0 says something about s=∞. The detailed understanding will come with time. So, be patient.

Q3: This is really hard to respond to. Poles are not functions, they are locations where functions blow up. Poles and zeros can be related to functions in the sense that if we know the function is a ratio of two polynomials in s, then the poles and zeros essentially define the function. The text is merely stating that you don't need to worry about where the LT of the function blows up, which is defined by the pole locations. It simply isn't relevant to the initial value of the function, or the final value of sF(s).

Before we were talking about functions that the Laplace transform deals with, which is something different than poles. exp(st) can be thought of as a test function that we can apply to a system, and each value of s in the complex plane creates a different test function. Don't confuse this function with the function we are transforming, whether it be the input signal or the system's impulse response function. If these seems like a lot of detail and seems confusing, then I would say it is because you have not solved enough practical problems yet. Doing practical problems can more quickly help you get these concepts down.

When you are thinking about these things, use examples, and even use Matlab to explore.

For example, a unit step function u(t) has a value of 1 at t=0+. The transform is 1/s. Hence, the initial value theorem works because f(0+)=lim(s->∞) sF(s) becomes 1=s/s=1

For example, a unit ramp function has a value of 0 at t=0+. The transform is 1/s^2. Hence, the initial value theorem works because f(0+)=lim(s->∞)sF(s) becomes 0=lim(s->∞) 1/s=0

For example, an impulse function has a value of infinity at t=0+. The transform is 1. Hence, the initial value theorem works because f(0+)=lim(s->∞)sF(s) becomes ∞=lim(s->∞) s=∞
 
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Thank you very much, Steve.

Re: Final value theorem

For the final value theorem, I didn't see an update to Q1, so I'm still confused. For Q2, I'll conclude that indeed you are too sleepy. You wrote out an incorrect integral and claimed they are wrong to do that. But, they did not do that. You did. What you wrote is indeed incorrect. What they did do was apply the derivative rule, as I said. That statement is a direct application of the derivative rule - plain and simple. I expect that if you look at it after getting some sleep, you'll see that.

Q1: I did update the question but I do confess my rephrasing is still not good because I don't really know how to frame my confusion into words. If you still can't understand it then just skip it.

Q2: I think my sleeplessness wasn't the only thing to be blamed! :) Because it's still not clear to me. Please have a look here. Given below are links which might relate to the query.

1: https://math.stackexchange.com/ques...t-of-an-integral-be-moved-inside-the-integral
2: https://answers.yahoo.com/question/index?qid=20080817125341AAyOFku

Thanks a lot for the help.

Regards
PG
 

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I'm sorry, but I do not understand Q1 at all.

For Q2, I see that you are really questioning the rigor of the proof given. This is a valid point and valid concern, of course. Reading through your references, I see mention of the term "uniform convergence", which came up previously in your questions. This is getting into detailed mathematical issues that I would need some time to sort through and understand myself. Unfortunately for us both, I don't have that time available. But, honestly, this is beyond what you need to be worrying about. The limit of the integral is correct, even if their reasoning may not be perfectly rigorous. In this case, the operation they use is valid, and they skipped the rigorous steps that would prove that fact.

This gets back to my earlier point. Typically, engineers worry about the gist of a proof and not the mathematical rigor. Hence, engineering text books give proofs that meet this need. Very rarely would a mathematician accept any proof given in an engineering or a physics book.

Anyway, it's a good question and sorry I don't have the best answer off the top of my head.
 
Hi,


For Q1, this comes up from time to time. The short answer is that t is related to s (or more simply, f) as:
f=1/t

and of course:
t=1/f

so when you make f go to infinity you get what happens at t=0, and when you make t go to infinity you get what happens at f=0, and vice versa.

This is what makes the theories work so nicely, because when one is zero the other is infinity. So if you look in one domain at zero you're seeing what is happening in the other domain at t=infinity, and vice versa. This makes calculating what happens at the opposite end more simple in many cases when it is easier mathematically to work in the opposite domain at the opposite point (sometimes you dont want to have to bother to calculate the other domain response when you can use these theories instead).

Examples for:
Fs=1/(s*(s+1))
ft=1-e^-t

ft(0+)=1-1=0 [initial value]
ft(0+)=lim(s-->inf) s*Fs = lim(s-->inf) 1/(s+1) = 0 [initial value]

Also,
ft(inf)=1-0=1 [final vlaue]
ft(inf)=lim(s-->0) s*Fs=lim(s-->0) 1/(s+1) ==> 1/1 = 1 [final value] condition satisfied: all poles of s*Fs in left half plane

More complex example:
Fs=(s^2+s+1)/(s*(s^3+6*s^2+11*s+6))
ft= ??

Without calculating ft, we have:
ft(0+)=lim(s-->inf) s*((s^2+s+1)/(s*(s^3+6*s^2+11*s+6)))=0 [initial time value]
ft(inf)=lim(s-->0) s*((s^2+s+1)/(s*(s^3+6*s^2+11*s+6)))=1/6 [final time value]

Strictly speaking, the condition for the final value theorm to work is that all poles of s*Fs lie in the left half plane, but there are many cases where it will also work even when this is not so. It's very tricky when it works and when it doesnt though. An example is when we have a sinusoidal part that does not tend to zero. We still have a final value, but it is not a single final value but a set of values that include the entire sine wave. This would simply look like a bounded sine wave. There's no single 'final' value (like 1, 2, 3.256, 5.3, etc.) but the sine wave amplitude becomes stable and this is also interesting information sometimes. Without this key point there would be no such thing as an oscillator :)
 
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