Electromechanical Systems

Magnetic Coupling

Definition

For a conductor with length $L$, and current $i$ pass through it is in a magnetic field with a flux density $B$, a force $F=BLi$ is exerted on it.

See DC Machines

$f=BLi$ is valid for $\{\begin{array}{l}1-\text{Straight conductors perp to a magnetic field}\\ 2-\text{Circular conductors in a radial field}\end{array}$

Reminder: If the conductor moves $\perp$ to the plane of a field-conductor then from DC Machines we know that $V_b=vBL$

DC Motors

Some Definitions

$v_a=\text{The voltage that is input to the armature}$ and $v_b=\text{The back e.m.f generated by the motion of the armature opposing the current}$

Mechanically…
The rotor rotates with angular velocity $\omega$ from generated torque $T$ which is opposed by damping $c$. There is also an opposing torque of the load being applied $T_L$.

$f=Bl{i}_a$ and therefore… $T=fr=nBl{i}_a=\left(nBlr\right)i_a$

Where we can group the following $K_T=nBlr$ such that we have $T=K_T{i}_a$

We also have $v_b=nBlv=nBlr\omega =K_B\omega$

Using these relationships, we can draw the conclusion that $K_B=K_T$

Armature Circuit Relationships

Definitions

Note that here $\Omega$ is the Laplace variable for $\omega$, not resistance…

We can set up the following ODE’s for an armature circuit…
$v_a=R_a{i}_a-L_a{\frac{d{i}_a}{dt}}_{}-v_b=0$
$v_a-R_a{i}_a-L_a\frac{d{i}_a}{dt}-k_b\omega =0$

$I\frac{d\omega }{dt}=K_T{i}_a-c\omega -T_L$

Our outputs in this case are $\omega \left(t\right)$ and $i_a\left(t\right)$

$$\{\begin{array}{l}{V}_a-R_a{I}_a-L_{a}s{I}_a-k_b\Omega (s)=0\\ Is\Omega (s)=K_T{I}_a(s)-c\Omega (s)-T_L(s)\end{array}$$ $$\{\begin{array}{l}{I}_a(R_a+s{L}_a)=V_b-K_b\Omega \\ \Omega (c+Is)=K_t{I}_a-T_L\end{array}$$ $$\{\begin{array}{l}{I}_a=\frac{1}{R_a+s{L}_a}(V_a-k_b\Omega )\\ \Omega =\frac{1}{c+sI}(K_T{I}_a-T_L)\end{array}$$

Where the term in front of the voltages and currents are our mechanical or electrical impedances.

We can now create a block diagram using these relationships knowing that out input is $V_a$ and that our output is $I_a$ and $\Omega$.
Using the block diagram to derive the transfer function, you can put the denominator in a nice form to see the impedances where the denominator is the characteristic equation of the system and its roots are the poles of the system. You can find the rest of the transfer functions in a similar way.