Model a radio station with only two broadcasting frequencies (or channels) by a circuit with

L=1, R=0, and C>0 arbitrary. We regard changing C as corresponding to turning your tuning dial. The two channels are represented by the e.m.f. E(t)= B₁ * sin(F₁ t) + B₂ *sin(F₂ t) , where F₁ is not equal to F₂ and B₁, B₂ are all given.

The solution to q" + (1/c)q = B₁ *sin(F₁t) + B₂ *sin(F₂t) is q = q_h + q_p where

q h (t)=c 1 *cos(t/sqrt(c)) + c 2 *sin(t/sqrt(c)) and

qp(t)={A 1 cos(F 1 t) + A 2 sin(F 1 t) + A 3 cos(F 2 t) + A 4 sin(F 2 t), if 1/sqrt(c) does not equal F 1 , F 2 ,

A 1 tcos(F 1 t) + A 2 tsin(F 1 t) + A 3 cos(F 2 t) + A 4 sin(F 2 t), if 1/sqrt(c)=F 1 ,

A 1 cos(F 1 t) + A 2 sin(F 1t ) + A 3 tcos(F 2 t) + A 4 tsin(F 2 t), if 1/sqrt(c)=F 2 .

Notice how the "F 1 channel" dominates if 1/sqrt(c)=F 1 and how the "F 2 channel" dominates if 1/sqrt(c)=F 2 .

To start solving this problem we must first remember what each term means; L is the inductance in henreys and becomes the q" coefficent in the DE, R is the resistance in ohms and in the DE becomes the coefficent of the q' term, C is the capacitance measured in ferrads and 1/C is the q coefficent in the DE, and E(t) is the battery or the external force of the equation. It has already been given that L=1 and R=0 which means that the DE Lq" + Rq' + (1/c)q = E(t) can be simplified to

q" + (1/c)q = E(t), and that E(t)= B 1 *sin(F 1 t) +B 2 *sin(F 2 t). So,

> restart:

> de:=diff(q(t),t$2)+(1/c)*q(t)=B1*sin(F1*t)+B2*sin(F2*t);

To make things easier to solve, we will first choose a number for c that has a whole number square.

> c:=25;

> de:=diff(q(t),t$2)+(1/c)*q(t)=B1*sin((F1)*t)+B2*sin(F2*t);

Now we need to put in an initial condition such that the charge and the current are both 0 at time 0, and call it ic.

> ic:=q(0)=0,D(q)(0)=0;

> dsolve({de,ic},q(t)): (big huge mess) There is no need to see this way maple solves the equation, however a better way which will be learned later on in the semester is to use Laplace transforms which maple can also do.

> soln:=dsolve({de,ic},q(t),method=laplace);

And simplify again as well as to turn the mess above into a function for the first case.

> q0:=s->subs({t=s,F1=2,F2=3,B1=1,B2=1},rhs(simplify(soln)));

> q0(t);

> with(plots):

> plot(q0(t),t=0..10*Pi);

The first coefficient is the amplitude=4.199 as the graph shows. Since neither the F 1 or F 2 channel is 1/sqrt(c), neither of the terms dominate the graph.

For the second case where F1=1/sqrt(c):

> de:=diff(q(t),t$2)+(1/c)*q(t)=B1*sin(sqrt(1/c)*t)+B2*sin(F2*t);

> soln:=dsolve({de,ic},q(t),method=laplace);

> q0:=s->subs({t=s,F1=1/5,F2=3,B1=1,B2=1},rhs(simplify(soln)));

> q0(t);

And the graph is:

> plot(q0(t),t=0..50*Pi);

Note the factor of t , leading to the dominate B 1 term, arising from the fact that F 1 =1/sqrt(c). This indicates you've "tuned into" the F1 channel's frequency.

For the third case where F 2 =1/sqrt(c):

> de:=diff(q(t),t$2)+(1/c)*q(t)=B2*sin(sqrt(1/c)*t)+B1*sin(F1*t);

> soln:=dsolve({de,ic},q(t),method=laplace);

> q0:=s->subs({t=s,F1=2,B1=1,B2=1},rhs(simplify(soln)));
q0(t);

And the graph is:

> plot(q0(t),t=0..50*Pi);

Note the factor of t , leading to the dominate B2 term, arising from the fact that F 2 =1/sqrt(c).

sm212_radio.mws,Midn 3/c Andrew Benjamin Nozik, 11-6-98