Dissipative Systems

Q1: What is a dissipative system?
A1: A dissipative system is a system for which the zero-input response $\to 0$ as $t \to \infty$
where $t$ is time

Q2: Why shoud I be on the lookout for dissipative systems?
A2: By definition, the zero-input response goes zero as $t \to \infty$
Hence, we know that the complete solution as $t \to \infty$ is the same as the zero-state response.

Q3: How do I determine if a system is dissipative?
A3: Suppose the system in question is a LTI system that relates an input $x(t)$ to an output $y(t)$ by a linear differential equation with constant coefficients. Then the homogeneous counterpart to that differential equation is:
$\sum\limits_{i=0}^na_i\frac{d^i}{dt^i}y_h(t)=0$
Since the equation is linear, you can guess a form of $y_h(t)$ and if the guess works, you've found the solution since it's unique.
Guess: $y_h(t)=Ae^{st}$
$\sum\limits_{i=0}^na_i\frac{d^i}{dt^i}Ae^{st}=0$
$\sum\limits_{i=0}^na_iAs^ne^{st}=0$
$\sum\limits_{i=0}^na_is^n=0$
Solving this equation gives the roots of the homogeneous equation (i.e. $s_1, s_2, ..., s_n$) and is called the characteristic equation or characteristic polynomial.

Suppose you've solved for the n roots for the nth order characteristic polynomial.
Then, $y_h(t)=\sum\limits_{i=0}^nA_ie^{s_it}$
In general, $s_i$ is complex (i.e. $s_i=a_i+jb_i$).
For sake of ease, let's look at the contribution of a specific $A_ie^{s_it}$ to $y(t)$:
$A_ie^{s_it}=A_ie^{(a_i+jb_i)t}$
$A_ie^{s_it}=A_ie^{a_it}e^{jb_it}$
So the specific contribution is the product of three parts:

1. $A_i$
2. $e^{a_it}$
3. $e^{jb_it}$

Since $A_i$ is a constant, it doesn't diminish as $t\to\infty$. Since $e^{jb_it}$ is a complex exponential, it doesn't diminish as $t\to\infty$. So $A_i$ and $e^{jb_it}$ are not dissipative.

This leaves us with $e^{a_it}$. This is the term that you care about: the only way you can diminish the effects of $A_i$ and $e^{jb_it}$ is if $e^{a_it}\to 0$ as $t\to\infty$. And $e^{a_it}$ is dissipative only if $a_i$ is negative. This in turn makes the whole contribution dissipative. Note that if $a_i$ is positive, then $e^{a_it}$ shoots up to infinity as $t\to\infty$. On the other hand, if $a_i=0$, then $e^{a_it}=1$. This outcome does nothing to counter the effects of $A_i$ or $e^{jb_it}$.

As a summary: 

$\begin{cases} a_i <0,&\mbox{Contribution is dissipative} \\ a_i\geq0,&\mbox{Contribution is not dissipative} \end{cases}$

This gives us an easy method for determining whether $y_h(t)$ is dissipative:

1. Solve for each of the roots of the characteristic polynomial corresponding to the original homogeneous equation. 
2. Examine the real part of each root. If all real parts are positive, then you know that each contribution is dissipative. The sum of dissipative terms is also dissipative, so you can then conclude that $y_h(t)$ is dissipative.