next up previous
Next: Superposition principle Up: Summary of Electrical and Previous: Summary of Electrical and

Series and Parallel Combinations

The through and across variables

acrossthrough0.png

Specifically, typical elements in electrical and mechanical systems include:

Constitutive Laws

  Mechanical Systems Electrical Systems
At a node Newton's 2nd Law (conservation of momentum): $\sum_k f_k=ma$ (or $\sum_k f_k=0$) Kirchhoff's Current Law (conservation of charge): $\sum_k i_k=0$
Around a loop Spatial consistency (conservation of space): $\sum_k (x_1-x_2)_k=0$ Kirchhoff's Voltage Law (conservation of energy): $\sum_k (u_1-u_2)_k=0$

  Mechanical Systems Electrical Systems
Inductive storage spring: $f(t)=k x(t)=k \int v(t) dt$ inductor: $i(t)=\frac{1}{L}\int u(t) dt$, or $u(t)=L \frac{d}{dt} i(t)$
Energy dissipater damper: $f(t)=c \dot{x}(t)=c v(t)$ resistor: $i(t)=\frac{1}{R} u(t)$, or $u(t)=R\;i(t)$
Capacitive storage mass: $f(t)=m\ddot{x}(t)=m \frac{d}{dt}{v}(t)$ capacitor: $i(t)=C\frac{d}{dt}u(t)$, or $u(t)=\frac{1}{C}\int i(t) dt$

Generic Mechanical Electrical
Conserved quantity momentum $p=mv=m\dot{x}$ charge $q$
Through variable $T$ force (momentum transfer rate): $f=dp/dt=d(mv)/dt=m d^2 x/dt^2$ current (charge transfer rate): $i=dq/dt$
$\sum_{\mbox{into node}} T_k=0$ Newton's 2nd law: $\sum_k f_k=0$ KCL: $\sum_k i_k=0$
Across variable $A_1-A_2$ relative displacement between end points voltage difference across element
$\sum_{\mbox{around loop}} (A_1-A_2)=0$ spatial consistency: $\sum_k (x_1-x_2)_k=0$ KVL: $\sum_k (u_1-u_2)_k=0$
$T=C \frac{d^{-1}}{dt^{-1}}(A_1-A_2)$   $i=\frac{1}{L}\int_{-\infty}^t (u_1-u_2) d\tau$
$T=C \frac{d^0}{dt^0}(A_1-A_2)$ $f_s=k(x_1-x_2)$ $i_R=\frac{1}{R}(u_1-u_2)$
$T=C \frac{d}{dt}(A_1-A_2)$ $ f_d=c\frac{d}{dt}(x_1-x_2)$ $i_C=C\frac{d}{dt}(u_1-u_2)$
$T=C \frac{d^2}{dt^2}(A_1-A_2)$ $ f_m=m\frac{d^2}{dt^2}(x_1)$  

Potential energy stored in spring ($f=kx$): $ w=\int f(t) dx=k \int x\; dx=\frac{1}{2}kx^2(t)$
Kinetic energy stored in mass ( $f=m\ddot{x}=m\dot{v}$): $ w=\int f(t) dx=\int m \dot{v}\; dx=m\int \frac{dv}{dt} v\; dt=m\int v\; dv=\frac{1}{2}mv^2(t) $
Magnetic energy stored in inductor ($v=L\dot{i}$): $ w=\int v(t) i(t) dt=L\int \frac{di(t)}{dt}\; i(t)\; dt=L\int i\; di=\frac{1}{2}L i^2(t)$
Electrical energy stored in capacitor ($i=C\dot{v}$): $w=\int v(t) i(t) dt=C\int \frac{dv(t)}{dt}\; v(t)\; dt=C\int v\; dv=\frac{1}{2}C v^2(t)$

Combination and Division Rules:

Note:

Masses are always combined in parallel, because they share the same across variable displacement, but not necessarily same through variable force (unless they have the same mass). This is because the relative displacement between the two ends of a mass is always zero (rigid body), therefore the displacement of a mass can only be measured absolutely with respect to the reference, i.e., the ``ground''. In comparison, the voltage across a capacitor does not have to be zero. For this reason, if a mechanical system is to be modeled by an electrical system, the mass has to be modeled by a capacitor with one end grounded, so that its voltage is always measured absolutely with respect to zero.

Example 1: RLC parallel circuit (KCL)

RCLDMK0A.png

The three elements are in parallel as they share the same across variable, the voltage $u(t)$. According to KCL, we have:


\begin{displaymath}i_C(t)+i_R(t)+i_L(t)=C\frac{du(t)}{dt}+\frac{u(t)}{R}+\frac{1}{L} \int_0^t u(\tau) d\tau=i(t) \end{displaymath}

Taking derivative on both sides, we get:

\begin{displaymath}C\frac{d^2u(t)}{dt^2}+\frac{1}{R}\frac{du(t)}{dt}+\frac{1}{L} u(t) =\frac{di(t)}{dt} \end{displaymath}

Example 2: Spring-damper-mass system

RCLDMK0.png

The three elements are in parallel as they share the same across variable, the displacement $x(t)$. According to Newton's second law, we have:


\begin{displaymath}f(t)-kx(t)-c\frac{dx(t)}{dt}=m\frac{d^2x(t)}{dt^2},\;\;\;\;\m...
...e.,}\;\;\;\;
m\frac{d^2x(t)}{dt^2}+c\frac{dx(t)}{dt}+kx(t)=f(t)\end{displaymath}

As $dx(t)/dt=v(t)$ is the velocity, the above can also be written as:

\begin{displaymath}m\frac{dv(t)}{dt}+c v(t)+k \int_0^t v(\tau) d\tau=f(t) \end{displaymath}


next up previous
Next: Superposition principle Up: Summary of Electrical and Previous: Summary of Electrical and
Ruye Wang 2012-12-12