Nonhomogeneous DE System

Now we consider the general solution of a nonhomogeneous ODE system with non-zero input ${\bf x}(t)$ on the right-hand side:

$$\dot{\bf y}(t)={\bf Ay}(t)+{\bf x}(t),\;\;\;\;\mbox{or}\;\;\;\; \dot{\bf y}(t)-{\bf Ay}(t)={\bf x}(t)$$

Multiply both sides by $e^{-{\bf A}t}$:

$$e^{-{\bf A}t}\dot{\bf y}(t)-e^{-{\bf A}t}{\bf Ay}(t) =\frac{... ...eft[ e^{-{\bf A}t} {\bf y}(t) \right] =e^{-{\bf A}t}{\bf x}(t)$$

Integrate both sides

$$\int_0^t\frac{d}{d\tau}\left[ e^{-{\bf A}\tau} {\bf y}(\tau)... ...(t)-{\bf y}(0) =\int_0^t e^{-{\bf A}\tau}{\bf x}(\tau) \;d\tau$$

Multiply $e^{{\bf A}t}$ on both side:

$${\bf y}(t)=e^{{\bf A}t}{\bf y}(0)+e^{{\bf A}t}\int_0^t e^{-{... ...t}{\bf y}(0)+\int_0^t e^{{\bf A}(t-\tau)}{\bf x}(\tau) \;d\tau$$

where the first term $e^{{\bf A}t}{\bf y}(0)={\bf V}e^{{\Lambda}t}{\bf V}^{-1}{\bf y}(0)$ is the homogeneous solution of the ODE system $\dot{\bf y}={\bf Ay}$, and the second term is the particular solution.

Example 1:

$$\ddot{\bf y}+2\zeta\omega_n\dot{\bf y}+\omega_n^2{\bf y}=x(t)$$

Define $v=\dot{y}$,

$$\frac{d}{dt}\left[\begin{array}{c}y v\end{array}\right] \l... ...rray}\right]+ \left[\begin{array}{c}0 x(t)\end{array}\right]$$

or in matrix form:

$$\dot{\bf y}={\bf Ay}+{\bf x}$$

Given $\omega_n=10$ and $\zeta=0$, we have

$${\bf A}=\left[\begin{array}{cc}0 & 1 100 & 0\end{array}\right]$$

with eigenvalue and eigenvector matrices:

$${\bf D}=\left[\begin{array}{cc}10 j & 0 0 & -10\;j\end{ar... ...left[\begin{array}{cc}1 & 1 10 j & -10\;j\end{array}\right]$$

$$e^{{\bf A}t}={\bf V}e^{{\bf D}t}{\bf V}^{-1} =\left[\begin{a... ...10 t)/10\\ -10 \sin(10 t) & \cos(10 t) \end{array}\right]$$

Given $y(0)=0$ and $\dot{y}(0)=1$, we get

$$y(t)=[1\;0]e^{{\bf A}t}\left[\begin{array}{c}0\ 1\end{array... ...f A}t} \left[\begin{array}{c}0\ x(t)\end{array}\right]\;d\tau$$

Example 2:

The motion of a tuned mass damper system shown below:

can be described by the following two ODEs:

$$\left\{ \begin{array}{l} M\ddot{y}=-Ky-B\dot{y}-K_d(y-y_d)-... ...dot{y}_d=K_d(y-y_d)+B_d(\dot{y}-\dot{d}_d) \end{array}\right.$$

If we define $v=\dot{y}$ and $v_d=\dot{y}_d$, then we get

$$\left\{ \begin{array}{l} \dot{v}=\ddot{y}=[-Ky-B\dot{y}-K_d(... ...}_d=[K_d(y-y_d)+B_d(\dot{y}-\dot{d}_d)]/M_d \end{array}\right.$$

and we have

$$\frac{d}{dt}\left[\begin{array}{d}y v y_d v_d\end{arra... ...t[\begin{array}{d}0 \frac{F(t)}{M} 0 0\end{array}\right]$$

which can be written in matrix form as

$$\dot{\bf y}(t)={\bf Ay}(t)+{\bf x}(t)$$

where ${\bf x}=[0,\;\delta(t)/M,\;0,\;0]^T$.

We assume zero initial condition $y(t)=\dot{y}=v(t)=0$ and $y_d(t)=\dot{y}_d=v_d(t)=0$, and $F(t)=\delta(t)$, then we get the solution

$${\bf y}={e^{\bf A}t}{\bf y}(0) {\bf Ay}+\int_0^te^{{\bf A}\t... ...A}t}\left[\begin{array}{c} 0 1/M 0 0 \end{array} \right]$$

The first component of ${\bf y}(t)$ is:

$$y(t)=[1\;0\;0\;0]\;{\bf y}(t)=[1\;0\;0\;0] e^{{\bf A}t}\left[\begin{array}{c} 0 1/M 0 0 \end{array} \right]$$

which is the 1st row and 2nd column of $e^{{\bf A}t}/M$; and the third component of ${\bf y}(t)$ is:

$$y_d(t)=[0\;0\;1\;0]\;{\bf y}(t)=[0\;0\;1\;0] e^{{\bf A}t}\left[\begin{array}{c} 0 1/M 0 0 \end{array} \right]$$

which is the 3rd row and 2nd column of $e^{{\bf A}t}/M$;

The problem can also be solved using the Laplace transform method. Let $Y(s)={\cal L}[\;y(t)\;]$ and $Y_d(s)={\cal L}[\;y_d(t)\;]$, then ${\cal L}[\;\dot{y}(t)\;]=s Y(s)-y(0)=s Y(s)$ and ${\cal L}[\;\ddot{y}(t)\;]=s^2 Y(s)$ ${\cal L}[\;\dot{y}_d(t)\;]=s Y_d(s)$ ${\cal L}[\;\ddot{y}_d(t)\;]=s^2 Y_d(s)$.

Now the two DEs can be expressed in s-domain as:

$$\left\{\begin{array}{l} Ms^2Y=-KY-BsY-K_d(Y-Y_d)-B_ds(Y-Y_d)+1\\ M_ds^2Y_d=K_d(Y-Y_d)+B_d s(Y-Y_d)\end{array}\right.$$

Find the sum of the two equations:

$$Ms^2Y+M_ds^2Y_d=-(K-Bs)Y+1,\;\;\;\;\mbox{i.e.}\;\;\;\; (Ms^2+Bs+K)Y+M_ds^2Y_d=1$$

and rewrite the second equation as:

$$(M_ds^2+B_d s+K_d) Y_d=(K_d+B_d s) Y\;\;\;\;\;\mbox{i.e.}\;\;\;\; Y_d=\frac{K_d+B_d s}{M_ds^2+B_d s+K_d}Y$$

Substituting into the other equation we get

$$(Ms^2+Bs+K)Y+\frac{M_ds^2(K_d+B_d s)}{M_ds^2+B_d s+K_d}Y=1$$

Solving for $Y$ we get

$$Y=\frac{M_ds^2+B_d s+K_d}{ (M_ds^2+B_d s+K_d) (Ms^2+Bs+K)+M_ds^2(K_d+B_ds)}$$

Note that when $M_d=0$, the system becomes a regular damped harmonic oscillator:

$$Y=\frac{B_d s+K_d}{ (B_d s+K_d) (Ms^2+Bs+K)}=\frac{1}{Ms^2+Bs+K}$$

The motion of mass $M_d$ is:

$$Y_d=\frac{K_d+B_d s}{M_ds^2+B_d s+K_d}Y =\frac{K_d+B_d s}{ (M_ds^2+B_d s+K_d)(Ms^2+Bs+K)+M_ds^2(K_d+B_ds)}$$