Initial value theorem
Conditions for initial value theorem are
- If t approaches to 0+, function f(t) should exist
i.e., lim t→0+ f(t) should exist
Function f(t) and its derivative f’(t) should be laplace transformable.
Statement
Initial value theorem is given by
lim t→0+ f(t) = lim s→∞ sF(s) = f(0+)
Where F(s) is laplace transform of f(t).
Proof
We know that,
𝐿[𝑓 ′ (𝑡)] = 𝑠 𝐿[𝑓(𝑡)] − 𝑓(0) = 𝑠𝐹(𝑠) − 𝑓(0)
∴ 𝑠𝐹(𝑠)
= 𝐿[𝑓 ′ (𝑡)] + 𝑓(0)
= ∫0 ∞ e −𝑠𝑡𝑓 ′ (𝑡)𝑑𝑡 + 𝑓(0)
Taking limit as 𝑠 → ∞ on both sides,
we have
lim𝑠→∞ 𝑠𝐹(𝑠) = lim 𝑠→∞ [∫0 ∞ e −𝑠𝑡𝑓 ′ (𝑡)𝑑𝑡 + 𝑓(0) ]
= lim𝑠→∞ [∫0 ∞ e −𝑠𝑡𝑓 ′ (𝑡)𝑑𝑡 ] + 𝑓(0)
= ∫0 ∞ lim [ e −𝑠𝑡𝑓 ′ (𝑡)]𝑑𝑡 + 𝑓(0)
= 0 + 𝑓(0) ∵ e−∞ = 0
= 𝑓(0)
= lim 𝑓(𝑡)
𝑡→0
∴ lim𝑠→∞ 𝑠𝐹(𝑠) = lim 𝑡→0 𝑓(𝑡) ( proved)
Final value theorem
Conditions for final value theorem are
-
Function f(t) and its derivative f’(t) should be laplace transformable.
- 𝑠𝐹(𝑠) has no pole on j-w axis and right half plane where 𝐹(𝑠) is the laplace transform of 𝑓(𝑡).
Statement
If Laplace transforms of 𝑓(𝑡) and 𝑓 ′(𝑡) exist and 𝐹(𝑠) is laplace transform of 𝑓(𝑡) i.e. 𝐿[𝑓(𝑡)] = 𝐹(𝑠),then
lim t→∞ f(t) = lims→ 0 sF(s)
Proof
We know that
𝐿{𝑓 ′ (𝑡)}
= 𝑠 𝐿[𝑓(𝑡)] − 𝑓(0)
= 𝑠𝐹(𝑠) − 𝑓(0)
So, 𝑠𝐹(𝑠) = 𝐿[𝑓 ′ (𝑡)] + 𝑓(0)
= ∫0 ∞ e- st f' (t) dt + f(0)
Taking limit s →0 on both sides of the above relation,
lims→0 sF(s) = lim s→0 [∫0 ∞ e-st f'(t)dt + f(0)]
= lims→0 [∫0 ∞ e-st f'(t) dt] + f(0)
= ∫0 ∞ lim s→0 [e-st f'(t)] dt + f(0)
= ∫0 ∞ f'(t)dt + f(0)
= [f(t)]0∞ + f(0)
= f(∞) - f(0) + f(0)
= f(∞)
= lim t→∞ f(t)
Therefore, limt→∞ f(t) = lims→ 0 sF(s)
Application of Initial value theorem
These are some instances of initial value theory applications:
-
The initial value theorem allows us to derive a signal's initial value, or x(0), directly from its Z-transform, or X(z), without first having to determine X's inverse Z-transform (z).
-
The initial value theorem is a theorem used in mathematical analysis to connect frequency domain statements to time domain behaviour as time approaches zero.
- The Initial Value Theorem can be used to calculate transient state gain.
Numerical Example
Let's consider a simple numerical example to illustrate the Laplace transform. Suppose we have a function f(t) = e-2t, and we want to find its Laplace transform.
The Laplace transform of f(t) is given by:
L{e−2t}=∫0∞e−ste−2tdt
Simplifying this integral:
L{e−2t}=∫0∞e−(s+2)tdt
Now, we can solve the integral:
L{e−2t}=-1/s+2e−(s+2)t|0∞
L{e−2t}=limt→∞(-1/s+2e−(s+2)t)−(-1/s+2)
Since limt→∞e−(s+2)t=0(because s is assumed to have a positive real part), the result is:
L{e−2t}=1/s+2
So, the Laplace transform of f(t)=e−2t is 1/s+2.
Frequently Asked Questions
What is initial and Final Value Theorem?
The Initial Value Theorem (IVT) and Final Value Theorem (FVT) are Laplace transform theorems used to determine a function's behavior as time approaches zero or infinity.
What is initial value theorem expression?
The Initial Value Theorem expression is given by lim s→∞ sF(s), where F(s) is the Laplace transform.
What are the rules for initial value theorem?
The rules for intial value theroem are that function must have a finite number of poles and the limit of sF(s) as s approaches each pole must exist.
What is the initial value theorem in PDC?
In Pole-Zero Analysis for Phase-locked Loops (PDC), the Initial Value Theorem evaluates the transient response, providing insights into the system's stability and initial conditions.
Conclusion
This article covered Initial and final value theorem of Laplace transform. We have also discussed the applications of intial value theorem. We also explained a numerical example based on the Laplace transform.
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