Develop a model of the electromagnetic speaker shown in Figure 6.7.2, and obtain the transfer function relating the diaphragm displacement x to the applied voltage v.
Figure 6.7.3a shows a simplified model of the mechanical subsystem, along with its free body diagram. The mass m represents the combined mass of the diaphragm and the coil. The spring constant k and damping constant c depend on the material properties of the diaphragm. The force f is the magnetic force, which is related to the coil current i by (6.5.1), f = n B Li, where n is the number of turns in the coil. Let K_{f} = n B L. From Newton’s law
f = BLi (6.5.1)
m{\frac{d^{2}x}{d t^{2}}}=-c{\frac{d x}{d t}}-k x+K_{f}i (1)
Figure 6.7.3b shows the electrical subsystem. The coil’s inductance and resistance are L and R. The coil experiences a back emf because it is a current conductor moving in a magnetic field. This back emf is given by K_{b}{\dot{x}}. The voltage v is the signal from the amplifier. From Kirchhoff’s voltage law,
v=L{\frac{d i}{d t}}+R i+K_{b}{\frac{d x}{d t}} (2)
The speaker model consists of equations (1) and (2).
Transforming equation (1) and solving for X(s) gives
X(s)={\frac{K_{f}}{m s^{2}+c s+k}}I(s)Transforming equation (2) and solving for I(s) gives
I(s)={\frac{1}{L s+R}}\left[V(s)-K_{b}s X(s)\right]Eliminating I(s) from the previous two equations, we obtain the desired transfer function.
\frac{X(s)}{V(s)}=\frac{{K}_{f}}{m L s^{3}+(c L+m R)s^{2}+(k L+c R+K_{f}K_{b})s+k R} (3)