Final value theorems to each transform pair

In summary: But the initial delta function is infinitely strong.That's what my professor said too. So it seems like the problem solution was incorrect in saying the value theorems can't be applied because of the impulse function. The initial value theorem just gives an incorrect result, but the final value theorem can still be applied to get a correct result.In summary, the initial value theorem is not applicable to the given function because it contains an impulse function and the theorem does not work if the order of the numerator is greater than or equal to the order of the denominator. However, the final value theorem can still be applied to get a correct result for the function.
  • #1
bl4ke360
20
0

Homework Statement


Find f(t) for the function F(s)=(10s^2+85s+95)/(s^2+6s+5) and apply the initial and final value theorems to each transform pair

Homework Equations


Initial value theorem: f(0)=lim s->∞ s(F(s))
Final value theorem: f(∞) = lim s->0 s(F(s))

The Attempt at a Solution


After dividing due to improper fraction:
F(s)= 10 + (25s+45)/(s^2+6s+5)

F(s)= 10+5/(s+1)+20/(s+5)
f(t)= 10δ(t)+[5e^(-t)+20e^(-5t)]u(t)

Where I'm confused is how I would apply the value theorems since there's an impulse function. When my professor did a similar problem and applied the theorems, I couldn't follow what she did, but the answer solution to this problem says the value theorems can't be applied to the function because the function is improper and the corresponding f(t) function contains an impulse.
How was my professor able to do it if it supposedly can't be done? Can someone please clarify this for me?
 
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  • #2
Both limits work, sort of.
lim s → 0 sF(s) = lim t → ∞ f(t) = 25 which is correct. No problem here.

lim s → ∞ sF(s) = lim t → 0 f(t) = ∞ which is also correct: f(0) is indeed infinite. But the latter limit just tells you f(0) = ∞ but not the "strength" of the infinity. In other words, the latter limit in the s domain can't tell if it's δ(t) or 10δ(t) or whatever.

I would agree that using the final-value theorem for t = 0+ is invalid if the order of the numerator = order of the denominator.
 
  • #3
I don't see how you got those results though, because my calculator gives lim s → ∞ sF(s) = 25, whereas that's what you got for lim s → 0 sF(s). And for s → 0 sF(s) it gave me 0 as the answer.
 
  • #4
bl4ke360 said:
I don't see how you got those results though, because my calculator gives lim s → ∞ sF(s) = 25, whereas that's what you got for lim s → 0 sF(s). And for s → 0 sF(s) it gave me 0 as the answer.

You got lim s → 0 sF(s) right & I got it wrong. Sorry. It's zero.

As for lim s → ∞ of sF(s), it's obviously ∞. Look at sF(s) = 10s + (25s^2 + 45s)/(s^2 + 6s + 5). The first term is ∞. The second term does approach 25. But ∞ + 25 = ∞.
 
  • #5
I just talked to my professor about this and she said you don't multiply the 10 by s, only the fractional part.
 
  • #6
I have second thoughts.

The initial value theorem just does not work if the order of the numerator >= order of denominator.

In your case, sF(s) = 10 + (25s^2 + 45s)/(s^2 +6s + 5). If you ignore the 10s term you get f(0+) = 25, whereas the actual f(0+) = 10δ(t) + 25. So, put simply, the initial-value theorem gives an incorrect result if the 10s term is ignored.

If the 10s term isn't ignored you get f(0+) = ∞ which is a lot closer to the truth than 25. The real f(0+) = 10δ(t) + 25 is a pulse of infinite height occurring at t = 0+ plus a constant of 25. Of course, as t increases the delta function disappears immediately after t = 0+ whereas the 25 dies down exponentially with time constants 1 and 5.
 
Last edited:

1. What is the final value theorem?

The final value theorem is a mathematical concept that relates the final value of a function to its Laplace transform. It states that the limit of the function as time approaches infinity is equal to the limit of the Laplace transform as the variable s approaches zero.

2. What is the significance of the final value theorem?

The final value theorem is important because it allows us to determine the steady-state behavior of a dynamic system without having to solve for the entire time-domain function. This makes it a useful tool in control systems and signal processing.

3. What is a transform pair?

A transform pair is a pair of mathematical functions that are related by a transformation. In the context of the final value theorem, it refers to a pair of functions that are related by a Laplace transform.

4. How is the final value theorem derived?

The final value theorem is derived using the properties of Laplace transforms, specifically the sifting property and the initial value theorem. It can also be derived from the definition of the Laplace transform and the use of contour integration.

5. What are some applications of the final value theorem?

The final value theorem has many applications in engineering and science, particularly in the analysis and design of control systems. It is also used in signal processing, circuit analysis, and differential equations. Additionally, it has applications in economics, finance, and other fields that involve modeling dynamic systems.

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