Final Value Theorem Rule Clarification

In summary, the Final Value Theorem Rule is a mathematical concept used to calculate the final value of a system based on its transfer function. This theorem states that the final value of a system is equal to the limit of the system's output as time approaches infinity. It is commonly used in control systems to determine the steady-state behavior of a system and to ensure its stability. However, it is important to note that this theorem only applies to linear systems and may not be accurate for non-linear systems.
  • #1
chrispypatt
1
0
My homework problem is as follows:
Consider the Laplace transform shown below.
(4s3+15s2+s+30)/(s2+5s+6)

a. What is the value of f(t=0) and f(t=∞)? Use the initial and final value theorems.
b. Find the inverse transform f(t). Use this expression to find f(t=0) and f(t=∞) and compare with the result of part a).I know to find the final value, using FVT, limt->inff(t)=lims->0sF(s), but I am given the stipulation that the poles must be in the left side of the domain. My book words the definition of IVT and FVT by saying IVT is only valid if f(t) has no impulse functions (the function must be rational) and that for FVT, we must add the rule about the poles. I am confused because everywhere online, including how my professor explained it, is that the impulse rule applies to ONLY IVT and the poles ONLY applies to FVT.

If I went with how my prof taught, Neither theorem can be applied to the problem but if I go by the book's wording, IVT does not apply but FVT shows the final value of f(t) will go to 0.

If someone more knowledgeable with these two theorems could clarify this, that would be helpful! Thanks in advance.
 
Last edited by a moderator:
  • #3
In future posts, please don't delete the homework template -- its use is required.
chrispypatt said:
My homework problem is as follows:
Consider the Laplace transform shown below.
(4s3+15s2+s+30)/(s2+5s+6)
Presumably, the above is F(s); i.e., ##\mathcal{L}[f(t)]##.
chrispypatt said:
a. What is the value of f(t=0) and f(t=∞)? Use the initial and final value theorems.
b. Find the inverse transform f(t). Use this expression to find f(t=0) and f(t=∞) and compare with the result of part a).I know to find the final value, using FVT, limt->inff(t)=lims->0sF(s), but I am given the stipulation that the poles must be in the left side of the domain.
I'm not familiar with either of these theorems, and I don't understand how this stipulation fits in.
What I do know is that ##\mathcal{L}[f'(t)] = sF(s) - f(0)##, where F(s) is as above.
For the b) part, I would carry out long division, and for the remainder, use partial fractions to write it as a sum of two fractions.
chrispypatt said:
My book words the definition of IVT and FVT by saying IVT is only valid if f(t) has no impulse functions (the function must be rational) and that for FVT, we must add the rule about the poles. I am confused because everywhere online, including how my professor explained it, is that the impulse rule applies to ONLY IVT and the poles ONLY applies to FVT.

If I went with how my prof taught, Neither theorem can be applied to the problem but if I go by the book's wording, IVT does not apply but FVT shows the final value of f(t) will go to 0.

If someone more knowledgeable with these two theorems could clarify this, that would be helpful! Thanks in advance.
 
  • #4
chrispypatt said:
My homework problem is as follows:
Consider the Laplace transform shown below.
(4s3+15s2+s+30)/(s2+5s+6)

a. What is the value of f(t=0) and f(t=∞)? Use the initial and final value theorems.
b. Find the inverse transform f(t). Use this expression to find f(t=0) and f(t=∞) and compare with the result of part a).I know to find the final value, using FVT, limt->inff(t)=lims->0sF(s), but I am given the stipulation that the poles must be in the left side of the domain. My book words the definition of IVT and FVT by saying IVT is only valid if f(t) has no impulse functions (the function must be rational) and that for FVT, we must add the rule about the poles. I am confused because everywhere online, including how my professor explained it, is that the impulse rule applies to ONLY IVT and the poles ONLY applies to FVT.

If I went with how my prof taught, Neither theorem can be applied to the problem but if I go by the book's wording, IVT does not apply but FVT shows the final value of f(t) will go to 0.

If someone more knowledgeable with these two theorems could clarify this, that would be helpful! Thanks in advance.

A lot of the on-line literature on these matters is confusing, and sometimes even self-contradictory (proving the result in one section, then showing a couner-example in another section). For the FVT, a nice article that sets it out properly and clearly is
http://www.me.umn.edu/courses/me3281/notes/TransformSolutionsToLTISystems_Part4.pdf .
 

What is the Final Value Theorem Rule?

The Final Value Theorem Rule is a mathematical concept used to determine the steady-state or final value of a system as time approaches infinity. It is commonly used in control systems and signal analysis.

What is the purpose of the Final Value Theorem Rule?

The Final Value Theorem Rule helps to determine the long-term behavior of a system, which is important in understanding and predicting its overall performance. It is also used to assess the stability of a system.

How is the Final Value Theorem Rule calculated?

The Final Value Theorem Rule is calculated by taking the limit of the system's response as time approaches infinity. This can be done using mathematical equations or by visualizing the system's behavior graphically.

What are the assumptions of the Final Value Theorem Rule?

The Final Value Theorem Rule assumes that the system under consideration is stable and that all initial conditions have decayed over time. It also assumes that the system is linear and time-invariant.

What are the limitations of the Final Value Theorem Rule?

The Final Value Theorem Rule is only applicable to systems that are stable and have reached their steady-state behavior. It also does not take into account any noise or disturbances in the system, which can affect the final value. Additionally, it may not work for highly nonlinear systems.

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