One can build an r-c circuit with two loops that has the property that there is a local minimum in its response to oscillatory forcing. This local minimum can be used to filter out that frequency.

Figure 38.1: A Two-Loop r-c Circuit

We derive the equations of the system using Kirchhoff's laws for voltage and current, Ohm's law for resistors, and Coulomb's law for capacitors. Kirchhoff's voltage law says that the signed sum of voltages around a closed loop is zero. Kirchhoff's current law says the signed sum of all currents into a node is zero.

The currents are shown in Figure 38.1. There is no current through the open output voltage location e_out. At the central node, i₁ and i₂ come in, while i₃ goes out, so we must have i₃=i₁+i₂, eliminating i₃.

38.1 The Laws of Kirchhoff, Ohm, and Coulomb

Ohm's law says that the voltage across a resistor of resistance r carrying a current i is , so the voltages across the resistors are

with the sign in the direction of the currents.

Coulomb's law says that the voltage across a capacitor of capacitance c with a charge q is . Charge is the integral of the current passing through it, so

with the sign in the direction of the currents.

Current is a flow of charge, or charge is an accumulation of current. By the second half of the Fundamental Theorem of Calculus (see Chapter 12 of the core text),

so we may express the resistor voltages as

The sum of the voltages around the upper loop makes

The sum of voltages around the lower loop makes

We substitute the first equation above for on the right side to obtain the pair of equations,

This is a nonautonomous linear system of the form

where the forcing term is f[t]=e_in[t]/r, , , and .

Transients
Use Theorem 24.3 of the core text to show that all solutions of

tend to zero as , when a_x, a_y, and b_y are as above. (Note: b_y>a_y.)

We are interested in the way the steady-state solution for e_out[t] depends on in the case where . Applications of Kirchhoff's voltage law around different loops give two different expressions for the output voltage:

38.2 Steady-State Solution

Now we turn our attention to the steady-state solution when , or .

Steady-State Solution
We will assume that the solution of the forced equations

may be written in the form

By substituting these forms into the equations, we will be able to solve for the constants h₁, h₂, k₁, k₂ and verify our assumption.

• 1. Substitute q₁ and q₂ into the first equation and show that the coefficients satisfy:
  
  
• 2. Substitute q₁ and q₂ into the second equation and show that the coefficients satisfy:
  
  
• 3. Write the coefficient equations from parts (1) and (2) in the form:
  
  
• 4. For example, you could use Mathematicato solve these equations as follows: aX = 1/(n r c)

  aY = 1/(m n r c)

  bX = 1/(n r c)

  bY = (1/(m n r c) + 1/(m r c))

  mat = {{-aX, aY,-w,0},     {w,0,-aX,aY},     {bX,-bY,0,-w},     {0,w,bX,-bY}}

  rhs = {0,0,0,-1/r} {h1,h2,k1,k2} = LinearSolve[mat,rhs]

  We know from above that
  
  

• Show that
  
  where and .

  Now we may use the trig trick of Hint 37.2.1 to write
  

  
  where .

• Continue your computer program, defining the square of the output amplitude, : h3 = Simplify[h1/c - h2/(m c)]

  k3 = Simplify[k1/c - k2/(m c) + 1]

  a2 = Simplify[ (h3)2 + (k3)2 ]

  38.3 A Check on a[t]²

  We can also use
  

  
  

• Show that
  
  where and .

  Show that we should also have
  

  
  and verify that this agrees with your other expression for with help from the computer.

  38.4 Where's the Min?

  Recall that we seek a minimum of the output amplitude, , because we want to design the circuit to minimize its response to certain frequencies, . How do we minimize a function defined for ?

  Although the output expression from the computer looks pretty complicated, we know that we need to look for critical points in order to minimize. We also need some additional information, since we do not have a compact interval minimization problem. The computer can differentiate for us.

• Use the computer to differentiate , da = Simplify[D[a2,w]]

• Show that the numerator of is a positive multiple of
  
  

• Prove that has a minimum at . Be complete in your reasoning. Why can't the minimum occur between 0 and ? Why can't the minimum occur between and ?

  What is the maximum of ?

  Round out your project with some specific choices of the parameters and a plot of the amplitude as a function of frequency.

  
  Figure 38.2: A Minimal Response to Forcing

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