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Initial and Final Value Theorems

A right sided signal's initial value $x(0)\stackrel{\triangle}{=}
\lim_{t \rightarrow 0} x(t)$ and final value $x(\infty)\stackrel{\triangle}{=}
\lim_{t \rightarrow \infty} x(t)$ (if finite) can be found from its Laplace transform $X(s)$ by the following theorems:

Proof: As $x(t)=x(t)u(t)=0$ for $t<0$, we have

$\displaystyle {\cal UL}\left[\frac{d}{dt}x(t)\right]$ $\textstyle =$ $\displaystyle \int_0^\infty \left[\frac{d}{dt}x(t)\right] e^{-st} dt=\int_0^\infty dx(t)\; e^{-st}$  
  $\textstyle =$ $\displaystyle x(t)e^{-st}\bigg\vert _0^\infty+s\int_0^\infty x(t) e^{-st}dt=sX(s)-x(0)$  

However, whether a given function $x(t)$ has a final value or not depends on the locations of the poles of its transform $X(s)$. Consider the following cases: Based on the above observation, the final value theorem can also be obtained by taking the partial fraction expansion of the given transform $X(s)$:

\begin{displaymath}X(s)=\sum_{i=0}^n \frac{C_i}{s-p_i}=\frac{C_0}{s}+\sum_{i=1}^n \frac{C_i}{s-p_i} \end{displaymath}

where $p_i's$ are the poles, and $p_0=0$ by assumption. The corresponding signal in time domain:

\begin{displaymath}x(t)=\sum_{i=0}^n C_ie^{p_it}=C_0+\sum_{i=1}^n C_ie^{p_it},\;\;\;\;\;\;\;\;t>0 \end{displaymath}

All terms except the first one represent exponentially decaying/growing or sinusoidal components of the signal. Multiplying both sides of the equation for $X(s)$ by $s$ and letting $s\rightarrow 0$, we get:

\begin{displaymath}\lim_{s\rightarrow 0} s X(s)=\lim_{s\rightarrow 0} \left[C_0+\sum_{i=1}^n \frac{s C_i}{s-p_i}\right]
=C_0 \end{displaymath}

We see that all terms become zero, except the first term $C_0$. If all poles $p_i,\;i=1,2,\cdots,n$ are on the left side of the S-plane, their corresponding signal components in time domain will decay to zero, leaving only the first term $C_0$, the final value $x(\infty)$.

Example 1:

\begin{displaymath}X(s)=\frac{1}{s(s+2)} \end{displaymath}

First find $x(t)$:

\begin{displaymath}x(t)={\cal L}[X(s)]={\cal L}\left[\frac{1}{2}\left(\frac{1}{s}+\frac{1}{s+2}\right)\right]
=\frac{1}{2}( 1+e^{-2t} )u(t) \end{displaymath}

When $t \rightarrow \infty$, we get $x(\infty)=1/2$. Next we apply the final value theorem:

\begin{displaymath}x(\infty)=s X(s)\big\vert _{s=0}=\frac{1}{s+2}\bigg\vert _{s=0}=\frac{1}{2} \end{displaymath}

Example 2:

\begin{displaymath}X(s)=\frac{1}{s(s-2)} \end{displaymath}

According to the final value theorem, we have

\begin{displaymath}x(\infty)=s X(s)\big\vert _{s=0}=-\frac{1}{2} \end{displaymath}

However, as the inverse Laplace transform

\begin{displaymath}x(t)={\cal L}[ X(s) ]={\cal L}\left[ \frac{1}{2}\left(\frac{1}{s-2}-\frac{1}{s}\right)\right]=
\frac{1}{2}[e^{2t}-1]u(t) \end{displaymath}

is unbounded (the first term grows exponentially), final value does not exist.

The final value theorem can also be used to find the DC gain of the system, the ratio between the output and input in steady state when all transient components have decayed. We assume the input is a unit step function $x(t)=u(t)$, and find the final value, the steady state of the output, as the DC gain of the system:

\begin{displaymath}\mbox{DC gain}=\lim_{s\rightarrow 0} \left[s H(s) \frac{1}{s}\right]=\lim_{s\rightarrow 0}H(s) \end{displaymath}

Example 3:

\begin{displaymath}H(s)=\frac{s+2}{s^2+2s+10} \end{displaymath}

The DC gain at the steady state when $t \rightarrow \infty$ can be found as

\begin{displaymath}\lim_{s\rightarrow 0} H(s)=0.2 \end{displaymath}


next up previous
Next: Solving LCCDEs by Unilateral Up: Laplace_Transform Previous: Unilateral Laplace Transform
Ruye Wang 2012-01-28