Final Value Theorem

The final value theorem lets you find the steady-state value that a signal reaches as time goes to infinity, using only its Laplace transform. Control engineers use it to predict the DC gain and steady-state error of a feedback system without simulating the full time response.

Core Concept

The final value theorem (FVT) links long-time behavior of a signal to the low-frequency content of its Laplace transform. Small values of s correspond to slow, long-duration behavior. The s=0 point of X(s) tells you the total DC content of the signal.

The theorem has a strict stability condition. It applies only when x(t) actually settles to a finite constant as t goes to infinity. If X(s) has poles on the imaginary axis or in the right half-plane, x(t) either oscillates or grows without bound and the FVT gives a meaningless result. You must verify all poles of sX(s) are in the open left half-plane before using the theorem.

A common scenario is finding steady-state error in a unity-feedback control system. The error transform E(s) is computed, and then lim_{s->0} sE(s) gives the steady-state error directly. This avoids solving the full closed-loop differential equation.

Key Formula

The final value theorem states: x(infinity) = lim_{s->0} s*X(s). This holds when all poles of sX(s) lie strictly in the left half-plane, meaning Re(s) < 0 for every pole. A single pole at s=0 is acceptable in X(s) itself (it cancels when multiplied by s), but poles at s = j*omega or in the right half-plane invalidate the theorem. For a rational X(s), check the denominator roots before applying.

Given: X(s) = 5 / (s(s+2)(s+3))
Find the final value of x(t).

Step 1: Check poles of sX(s)
  sX(s) = 5 / ((s+2)(s+3))
  Poles at s = -2 and s = -3  (both in left half-plane)
  FVT is valid.

Step 2: Apply FVT
  x(inf) = lim_{s->0} sX(s)
         = lim_{s->0} 5 / ((s+2)(s+3))

Step 3: Substitute s=0
         = 5 / (2 * 3)
         = 5/6

Final Answer: x(infinity) = 5/6

Exam Tip: The FVT fails silently when poles of sX(s) are on the imaginary axis. For example, if X(s) = omega_0^2/(s(s^2+omega_0^2)), then sX(s) has poles at s = +/- j*omega_0. The signal x(t) oscillates forever and has no final value. Plugging into the formula gives omega_0^2/omega_0^2 = 1, which is completely wrong. Always factor the denominator of sX(s) before applying the theorem.

Properties Summary

• Statement: x(inf) = lim_{s->0} sX(s); extracts DC steady-state value
• Validity condition: all poles of sX(s) must be in the open left half-plane
• A pole of X(s) at s=0 is allowed; it cancels in sX(s)
• Poles on imaginary axis or right half-plane invalidate the theorem
• Used in control to find steady-state error: e_ss = lim_{s->0} sE(s)
• Dual of Initial Value Theorem which uses s approaching infinity instead

Quick Revision

• FVT: x(infinity) = lim_{s->0} sX(s)
• Gives steady-state DC value without computing inverse transform
• Always verify poles of sX(s) lie in left half-plane before applying
• A single pole at origin in X(s) is fine; it cancels with the s factor
• Imaginary axis poles in sX(s) mean the signal oscillates; FVT breaks down
• Control engineers use FVT to compute position, velocity, and acceleration error constants
• Exam trap: checking poles of X(s) instead of poles of sX(s); the extra factor of s can cancel a pole at origin, making the theorem valid even when X(s) itself has a pole at zero