15.5. Examples

Let f(t)=1−e−2t

Step 1: Find F(s) F(s)=L{1−e−2t }= − s s+2

Step 2: Multiply by s sF(s)=s − =1− s s+2 s+2

Step 3: Take the limit as s→0 lim 1− =1−0=1 s+2 s→0

✅ Therefore, lim ¿ t→∞f(t)=1¿

In this example, we are given the function f(t)=1−e−2t, which represents a simple exponential decay. We start by finding its Laplace transform, denoted as F(s). The Laplace transform of the function is calculated as F(s)=1/(s+2). Next, we multiply this transform by 's', resulting in sF(s)=s/(s+2). The final step requires us to take the limit of this expression as s approaches 0. By performing this calculation, we find that the limit equals 1, which means the steady-state value of f(t) as time approaches infinity is 1.

Imagine a cup of hot coffee cooling down in a room. Initially, the coffee is very hot (f(t)=1) and slowly drops to a temperature that matches the room's temperature over time. The Final Value Theorem allows us to find out what that temperature will eventually stabilize at, which is 1 in this example. Just like the coffee cools down, the system's response approaches a final value.

Let f(t)=sin(t)

F(s)= s2 +1

sF(s)= s2 +1

lim s lim sF(s)=s→0 =0 s→0 s2 +1

But in reality, lim ¿ t→∞sin(t)¿ does not exist. ❌ Hence, FVT fails here due to oscillatory behavior.

Here, we examine a case where the Final Value Theorem does not apply. The given function f(t)=sin(t) is inherently oscillatory, meaning it continues to oscillate between -1 and 1 as time goes on. The Laplace transform for this function is F(s)=1/(s^2+1). When we multiply by s, we get sF(s)=s/(s^2+1). Taking the limit as s approaches 0 gives us 0, but this is misleading because the limit of sin(t) as t approaches infinity does not converge to any finite value—it oscillates indefinitely. Therefore, the Final Value Theorem fails here.

Consider a swing moving back and forth. Even though we can calculate an average position, the swing does not settle at any one point; it just keeps swinging forever. Similarly, the sin(t) function does not have a final value because it oscillates. The FVT fails in this case because it’s meant for functions that stabilize over time, just like a swing finally stops if no one pushes it.

Let 5s+3 F(s)= s(s+2)

Then, 5s+3
sF(s)=

s+2

lim5s+3 3 s→0

s+2 2 result: ⚙️ Final value is 2

In this example, we have a rational function with F(s)=5s+3/(s(s+2)). The first step is to multiply the Laplace transform by 's', resulting in sF(s)=(5s+3)/(s+2). We then take the limit of this expression as s approaches 0. The outcome of this limit calculation reveals that the final value equals 2, indicating that the system will stabilize at this value as time goes to infinity.

Think of a tank filling with water where the inflow balances against the outflow at one point. If we know how much water goes in compared to how much is lost, we can determine how full the tank will be after a long time. In this example, the final value of 2 can be compared to the height of the water in the tank once it is filled to its stable level. The Final Value Theorem helps us calculate this without waiting to see how it fills up.

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