Laplace Analysis of Linear Systems
The differentiation theorem can be used to convert differential equations into algebraic equations, which are easier to solve. We will now show this by means of two examples.
Moving Mass
Figure D.1 depicts a free mass driven by an external force along
an ideal frictionless surface in one dimension. Figure D.2
shows the electrical equivalent circuit for this scenario in
which the external force is represented by a voltage source emitting
volts, and the mass is modeled by an inductor
having the value
Henrys.
From Newton's second law of motion ``'', we have
![$\displaystyle f(t) = m\,a(t) \isdef m\,{\dot v}(t) \isdef m\,{\ddot x}(t).
$](http://www.dsprelated.com/josimages_new/filters/img1737.png)
![\begin{eqnarray*}
F(s) &=& m\,{\cal L}_s\{{\ddot x}\}\\
&=& m\left[\,s {\cal L...
...right\}\\
&=& m\left[s^2\,X(s) - s\,x(0) - {\dot x}(0)\right].
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/filters/img1738.png)
Thus, given
Laplace transform of the driving force
,
initial mass position, and
-
initial mass velocity,
![$ X(s)$](http://www.dsprelated.com/josimages_new/filters/img1681.png)
![$ t\ge 0$](http://www.dsprelated.com/josimages_new/filters/img1742.png)
![$ x(t)$](http://www.dsprelated.com/josimages_new/filters/img1659.png)
![$ t\ge 0$](http://www.dsprelated.com/josimages_new/filters/img1742.png)
If the applied external force is zero, then, by linearity of
the Laplace transform, so is
, and we readily obtain
![$\displaystyle X(s)
= \frac{x(0)}{s} + \frac{{\dot x}(0)}{s^2}
= \frac{x(0)}{s} + \frac{v(0)}{s^2}.
$](http://www.dsprelated.com/josimages_new/filters/img1744.png)
![$ 1/s$](http://www.dsprelated.com/josimages_new/filters/img1731.png)
![$\displaystyle u(t)\isdef \left\{\begin{array}{ll}
0, & t<0 \\ [5pt]
1, & t\ge 0 \\
\end{array}\right.,
$](http://www.dsprelated.com/josimages_new/filters/img1745.png)
![$ x(t)$](http://www.dsprelated.com/josimages_new/filters/img1659.png)
![$\displaystyle x(t) = x(0)\,u(t) + v(0)\,t\,u(t).
$](http://www.dsprelated.com/josimages_new/filters/img1746.png)
![$ x(0)=x_0$](http://www.dsprelated.com/josimages_new/filters/img1747.png)
![$ v(0)=0$](http://www.dsprelated.com/josimages_new/filters/img1748.png)
![$ x(t)=x_0$](http://www.dsprelated.com/josimages_new/filters/img1749.png)
![$ t\ge 0$](http://www.dsprelated.com/josimages_new/filters/img1742.png)
![$ v(0)$](http://www.dsprelated.com/josimages_new/filters/img1754.png)
To summarize, this simple example illustrated use the Laplace transform to solve for the motion of a simple physical system (an ideal mass) in response to initial conditions (no external driving forces). The system was described by a differential equation which was converted to an algebraic equation by the Laplace transform.
Mass-Spring Oscillator Analysis
Consider now the mass-spring oscillator depicted physically in Fig.D.3, and in equivalent-circuit form in Fig.D.4.
![]() |
By Newton's second law of motion, the force applied to a mass
equals its mass times its acceleration:
![$\displaystyle f_m(t)=m{\ddot x}(t).
$](http://www.dsprelated.com/josimages_new/filters/img1758.png)
![$ f_k(t)$](http://www.dsprelated.com/josimages_new/filters/img1759.png)
![$ k$](http://www.dsprelated.com/josimages_new/filters/img79.png)
![$ x(t)$](http://www.dsprelated.com/josimages_new/filters/img1659.png)
![$\displaystyle f_k(t)=kx(t)
$](http://www.dsprelated.com/josimages_new/filters/img1760.png)
![$ f_k = -f_m$](http://www.dsprelated.com/josimages_new/filters/img1761.png)
![$ f_k$](http://www.dsprelated.com/josimages_new/filters/img1762.png)
![$ f_m$](http://www.dsprelated.com/josimages_new/filters/img1763.png)
![$ x$](http://www.dsprelated.com/josimages_new/filters/img101.png)
![\begin{eqnarray*}
f_m(t) + f_k(t) &=& 0\\
\Rightarrow\; m {\ddot x}(t) + k x(t) &=& 0
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/filters/img1764.png)
We have thus derived a second-order differential equation governing
the motion of the mass and spring. (Note that in
Fig.D.3 is both the position of the mass and compression
of the spring at time
.)
Taking the Laplace transform of both sides of this differential equation gives
![\begin{eqnarray*}
0 &=& {\cal L}_s\{m{\ddot x}+ k x\} \\
&=& m{\cal L}_s\{{\ddo...
...orem again)} \\
&=& ms^2 X(s) - msx(0) - m{\dot x}(0) + k X(s).
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/filters/img1765.png)
To simplify notation, denote the initial position and velocity by
and
, respectively. Solving for
gives
![\begin{eqnarray*}
X(s) &=& \frac{sx_0 + v_0}{s^2 + \frac{k}{m}}
\;\isdef \; \fr...
...ta_r \;\isdef \; \tan^{-1}\left(\frac{v_0}{{\omega_0}x_0}\right)
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/filters/img1767.png)
denoting the modulus and angle of the pole residue , respectively.
From §D.1, the inverse Laplace transform of
is
, where
is the Heaviside unit step function at time 0.
Then by linearity, the solution for
the motion of the mass is
![\begin{eqnarray*}
x(t) &=& re^{-j{\omega_0}t} + \overline{r}e^{j{\omega_0}t}
= ...
...ga_0}t - \tan^{-1}\left(\frac{v_0}{{\omega_0}x_0}\right)\right].
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/filters/img1771.png)
If the initial velocity is zero (), the above formula
reduces to
and the mass simply oscillates sinusoidally at frequency
, starting from its initial position
.
If instead the initial position is
, we obtain
![\begin{eqnarray*}
x(t) &=& \frac{v_0}{{\omega_0}}\sin({\omega_0}t)\\
\;\Rightarrow\; v(t) &=& v_0\cos({\omega_0}t).
\end{eqnarray*}](http://www.dsprelated.com/josimages_new/filters/img1777.png)
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Example Analog Filter
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Laplace Transform Theorems