User:DukeEgr93/Responses

From PrattWiki
< User:DukeEgr93
Revision as of 02:38, 2 December 2013 by DukeEgr93 (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to navigation Jump to search

The following is a brief description of the different ways a complete solution to a differential equation might be broken up. The three possibilities are:

  1. Homogeneous and Particular (aka natural and forced)
  2. Zero-input and Zero-state
  3. Transient and Steady-state

To give an example, consider the following equation (from Haykin/Van Veen, p. 503, with a different input):

\( \begin{align} \frac{d^2}{dt^2}y(t)+5\frac{d}{dt}y(t)+6y(t)&=\frac{d}{dt}x(t)+6x(t)\\ x(t)&=u(t)-e^{-t}u(t)\\ y(0^-)&=1\\ \dot{y}(0^-)&=2 \end{align} \)

Homogeneous and Particular

The homogeneous solution is the solution to the differential equation when the forcing function \(x(t)=0\). In other words,

\( \begin{align} \frac{d^2}{dt^2}y_h(t)+5\frac{d}{dt}y_h(t)+6y_h(t)&=0\\ \end{align} \)

Using a general complex exponential as an example:

\( \begin{align} y_h(t)&=Ke^{st} \end{align} \)

yields

\( \begin{align} s^2Ke^{st}+5sKe^{st}+6Ke^st=0\\ \end{align} \)

which is only true if \(K=0\) (trivial) or

\( \begin{align} s^2+5s+6&=0 s&=-2, -3 \end{align} \)

meaning

\( \begin{align} y_h(t)=K_1e^{-2t}+K_2e^{-3t} \end{align} \)

It is impossible to find the constants without also knowing what the particular solution is.

The particular solution is going to resemble the forcing function - in this case, a constant plus an exponential with a decay of -1. That is,

\( \begin{align} y_p(t)&=C_1+C_2e^{-t}\\ \dot{y}_p(t)&=-C_2e^{-t}\\ \ddot{y}_p(t)&=C_2e^{-t} \end{align} \)

and

\( \begin{align} \frac{d^2}{dt^2}y_p(t)+5\frac{d}{dt}y_p(t)+6y_p(t)&=\frac{d}{dt}x(t)+6x(t)\\ \left(C_2e^{-t} \right) + 5\left(-C_2e^{-t} \right) + 6\left(C_1+C_2e^{-t} \right) &=0+e^{-t}+6-6e^{-t} \end{align} \)

from which, using harmonic balance, the following equations may be found:

\( \begin{align} 6C_1=6 &\longrightarrow C_1=1\\ 2C_2=-5 & \longrightarrow C_2=-\frac{5}{2} \end{align} \)

such that

\( \begin{align} y(t)=y_h(t)+y_p(t)&=K_1e^{-2t}+K_2e^{-3t}+1-\frac{5}{2}e^{-t}\\ \dot{y}&=-2K_1e^{-2t}-3K_2e^{-3t}+\frac{5}{2}e^{-t} \end{align} \)

Since the values for \(y\) and \(\dot{y}\) are known at time 0, solve:

\( \begin{align} y(0)=K_1+K_2+1-\frac{5}{2}=1&\longrightarrow K_1+K_2=\frac{5}{2}\\ \dot{y})(0)=-2K_1-3K_2+\frac{5}{2}=2&\longrightarrow 2K_1+3K_2=\frac{1}{2} \end{align} \)

such that

\( \begin{align} K_1&=7 & K_2&=-\frac{9}{2} \end{align} \)

meaning:

\( \begin{align} y_h(t)&=7e^{-2t}-\frac{9}{2}e^{-3t}\\ y_p(t)&=1-\frac{5}{2}e^{-t}\\ y(t)&=7e^{-2t}-\frac{9}{2}e^{-3t}+1-\frac{5}{2}e^{-t} \end{align} \)