User:DukeEgr93/Responses

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):

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

Note - unilateral Laplace transforms and signals starting at time 0 are assumed; \(u(t)\) is assumed everywhere.

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

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

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:

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

Zero-input and Zero-state

The complete solution is also the sum of the zero-input and the zero-state solutions. The zero-input response is how the system responds solely to the initial state, neglecting any forcing function. The form of the zero-input response looks the same as the homogeneous response above, but the coefficients can be determined without knowledge of the forcing function (the "input"). That is,

\( \begin{align} y_{zi}(t)&=K_1e^{-2t}+K_2e^{-3t}\\ \dot{y}_{zi}(t)&=-2K_1e^{-2t}-3K_2e^{-3t} \end{align} \)

Given the state at time 0 - not as a result of any input, but rather as a result of the initial conditions on \(y\), we can find:

\( \begin{align} y_{zi}(0)&= K_1+K_2&=1\\ \dot{y}_{zi}(0)&=-2K_1-3K_2&=2 \end{align} \)

from which we can determine

\( \begin{align} K_1= 5 & K_2= -4 \end{align} \)

such that

\( \begin{align} y_{zi}(t)&=5e^{-2t}-4e^{-3t} \end{align} \)

The zero-state response is a bit more complicated, as it necessarily includes the particular solution from above as well as components of the homogeneous. Among the easier ways to solve for the zero-state response is to take the Laplace transform for the differential equation and set all the initial conditions to 0. This basically means using

\( \begin{align} \frac{d^n}{dt^n}y(t)&\rightarrow s^nY(s) \end{align} \)

which in this case yields:

\( \begin{align} s^2Y_{zs}(s)+5sY_{zs}(s)+6Y_{zs}(s)&=(s+6)\left(\frac{1}{s}-\frac{1}{s+1}\right)=\frac{s+6}{(s)(s+1)}\\ Y_{zs}(s)&=\frac{s+6}{(s)(s+1)(s+2)(s+3)}=\frac{1}{s}-\frac{5}{2}\frac{1}{s+1}+\frac{2}{s+2}-\frac{1}{2}\frac{1}{s+3} \end{align} \)

meaning

\( \begin{align} y_{zs}(t)&=1-\frac{5}{2}e^{-t}+2e^{-2t}-\frac{1}{2}e^{-3t} \end{align} \)

In summary:

\( \begin{align} y_{zi}(t)&=5e^{-2t}-4e^{-3t}\\ y_{zs}(t)&=1-\frac{5}{2}e^{-t}+2e^{-2t}-\frac{1}{2}e^{-3t}\\ y(t)&=7e^{-2t}-\frac{9}{2}e^{-3t}+1-\frac{5}{2}e^{-t} \end{align} \)

which...should look familiar.