Resonance and Fourier Theory

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Applied Differential Equations

Aleksei Beltukov

Fall 2010



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Resonance

In this section we consider an RLC -circuit driven by a sum of harmonics.The circuit is described by the following second order ODE:

Ld2q

dt2+ R

dq

dt+

1

C q =

n

An cos(ωnt) + Bn sin(ωnt) (1)

Our first order of business is to prove that the complimentary function of (1)decays exponentially regardless of the values of the parameters R, L and C .

The characteristic equation

L λ2 + R λ +1

C

= 0

has the roots

λ1,2 = − R

2 L±

R2

4 L− 1

L C .

Depending on the value of R2/(4 L)− (L C )−1 the following three cases arepossible.

Overdamped case

If R2/(4 L) > (L C )−1 then the characteristic roots λ1,2 are distinct realnumbers. Since R, L and C are positive, the root

λ2 = − R

2 L−

R2

4 L− 1

L C

is clearly a negative number. To see that the other root

λ2 = − R

2 L+

R2

4 L− 1

L C

is also negative, recall that two positive numbers compare in the same wayas their squares, i.e., a < b implies and is implied by a2 < b2. Since

R2

4 L− 1

L C < R

2

4 L,

we have R2

4 L− 1

L C <

R

2 L,



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and, consequently,

λ2 =

R2

4 L− 1

L C − R

2 L< 0.

Now since both λ1 and λ2 are negative, the complimentary function

qc = C 1 eλ1 t + C 2 eλ2 t

decays exponentially with t.

Critically damped case

If R2/(4 L) = (L C )−1 the characteristic roots λ1,2 are the same:

λ1 = λ2 = − R

2 L.

The complimentary function is then the product of a linear polynomial anda decaying exponential:

qc = (C 1 + C 2 t) e−Rt

2L .

As t →∞ the linear polynomial approaches infinity but the exponential termgoes to zero at a much faster exponential rate. Therefore qc approaches zero

exponentially as before.

Underdamped case

It remains to consider the case where R2/(4 L) < (L C )−1 and the character-istic roots are complex:

λ1,2 = − R

2 L± ω0 i, ω0 =

1

L C − R2

4 L.

Using Euler’s formula, we can write complimentary function as

qc = C 1 e(−R

2L+ω0 i) t + C 2 e(−

R

2L−ω0 i) t = e−

Rt

2L (A cos(ω0t) + B sin(ω0t)).

This is a harmonic with an exponentially decaying amplitude: so, yet again,qc decays exponentially.



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Particular solution

We have shown that regardless of the values of R, L and C that define Equa-tion (1) the characteristic function qc is exponentially decaying, or transient ,as an engineer would say. Therefore the operation of the circuit in “steady”mode is defined by q p which we now proceed to construct.

If we did not have the summation on the right-hand side of (1), our guessfor q p would be a simple harmonic. It stands to reason that for a sum of harmonics with frequencies ωn the guess should be a similar sum, i.e.:

q p =n

an cos(ωnt) + bn sin(ωnt).

We now substitute our guess into the left-hand side of Equation (1) andcompare the like terms:

Ld2q

dt2+ R

dq

dt+

1

C q =

n

(−L ω2n) (an cos(ωnt) + bn sin(ωnt))

+n

(R ωn) (−an sin(ωnt) + bn cos(ωnt))

+

n1

C (an cos(ωnt) + bn sin(ωnt))

= (♥) =n

An cos(ωnt) + Bn sin(ωnt).

Comparing the coefficients in front of cos(ωnt) and sin(ωnt) on both sides,we arrive at the following system of equations

1

C − L ω2

n

an + R ωn bn = An,

−R ωn an +

1

C − L ω2

n

bn = Bn,

whose solution is given by

an =

1C − L ω2

n

An −R ωn Bn

1C − L ω2

n

2+ R2 ω2

n

bn =

1C − L ω2

n

Bn + R ωn An

1C − L ω2

n

2+ R2 ω2

n

.



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We conclude that q p is given by

q p =n

1C − L ω2n

An −R ωn Bn

cos(ωnt) +

1C − L ω2n

Bn + R ωn an

sin(ωnt)

1C − L ω2

n

2+ R2 ω2

n

(2)

Amplitude amplification

Admittedly, the formula (2) that we obtained for the particular solution of (1) is not very revealing. However it does provide important insight intothe structure of the output q p: every harmonic component of the input ismapped into a harmonic component of the same frequency according to the

rule:An cos(ωnt) + Bn sin(ωnt) →1C − L ω2

n

An − R ωn Bn

cos(ωnt) +

1C − L ω2

n

Bn + R ωn an

sin(ωnt)

1C − L ω2

n

2+ R2 ω2

n

(3)

Equation (3) says that harmonic signals change their amplitudes and phasesas they pass through an RLC -circuit. Let us compute the amplification factorfor the amplitude of the harmonic signal as a function of the frequency ωn.Recall that the amplitude of a linear combination of sine and cosine

An cos(ωn t) + Bn sin(ωn t)is the number

A2

n + B2n. Therefore the square of the amplitude of the

right-hand side of (3) is given by1C − L ω2

n

An −R ωn Bn

1C − L ω2

n

2+ R2 ω2

n

2

+

1C − L ω2

n

Bn + R ωn An

1C − L ω2

n

2+ R2 ω2

n

2

Bringing the two fractions to a common denominator and simplifying, we get

A2n + B2

n

1C − L ω2

n2

+ R2 ω2n

which means that a signal with amplitude

A2n + B

2n and frequency ωn is

mapped into the signal with amplitude A2

n + B2n

1C − L ω2

n

2+ R2 ω2

n



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and hence the amplitude magnification factor F is given by

F (ωn) = 1 1C − L ω2

n

2+ R2 ω2

n

= 1 L2 (ω2

0 − ω2n)2 + R2 ω2

n

.

Notice that F goes to zero quadratically as ωn goes to infinity: thus anyRLC -circuit acts as a low-pass filter. Let us now consider a special casewhere the natural frequency of the circuit ω0 = 1/

√L C is one and the

resistance parameter R is small compared to one, say, R = 0.01. Let us alsoset the inductance L to one, for simplicity. Thus

F (ωn) =1

(1

−ω2n)2 + 10−4 ω2

n

.

We notice that if ωn is close to zero, the amplification factor is close to one:low-frequency waves pass without significant changes in amplitude. However,when ωn is very close to the natural frequency 1, the amplification factor isclose to 100! Figure 1 shows the graph of F (ωn) :

0

20

40

60

80

100

F(w_n)

0.5 1 1.5 2 2.5 3w_n

Figure 1: Resonance



7

As Figure 1 shows, the component with frequency 1 will dominate all

other harmonics in q p. This is the working principle of the radio: the sumof harmonics received by the antenna is fed through an RLC -circuit tunedto the frequency of interest ω0. The output of the circuit, as we now see, iscompletely dominated by the components with frequencies closest to ω0. Of course, there is much more to radio broadcasting than a simple RLC -circuitbut, in principle, a radio is just an electromagnetic resonator.

Fourier series

In the previous section we considered an RLC -circuit with the forcing term:n

An cos(ωnt) + Bn sin(ωnt).

We found that the steady solution could be obtained by shifting and scalingthe harmonics of the input. Now that we know how to find the steadysolution with a trigonometric right-hand side, we can extend that result toall periodic functions using Fourier theory. The extension is made possibleby the following theorem.

Theorem 1 (Fourier Series). Let f (t) be 2π-periodic integrable function.

Then one can express f (t) as an infinite trigonometric sum of the form:

f (t) = a0 +n=1

(an cos(n t) + bn sin(n t)) , (4)

where the coefficients are given by

a0 =1

2π

2π0

f (t) dt

an =1

π

2π0

f (t) cos(n t) dt, n > 0

bn = 1π

2π

0

f (t) sin(n t) dt, n > 0.

Proof. Proving that a periodic integrable function can be expressed as atrigonometric series is outside the scope of this handout. We will instead



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limit ourselves to proving the formulas for the coefficients. Henceforth we

assume that (4) is valid. Integrating both sides of (4) from 0 to 2π, we get 2π0

f (t) dt = 2π a0 +n=1

an

2π0

cos(n t) dt + bn

2π0

sin(n t) dt

= 2π a0,

since cosines and sines integrate to zero over their periods. This shows that

a0 =1

2π

2π0

f (t) dt,

as required. Let us now integrate both sides of (4) against cos(m t): 2π0

f (t) cos(m t) dt =

2π0

cos(m t)n=1

(an cos(n t) dt + bn sin(n t)) dt

=n=1

an

2π0

cos(m t) cos(n t) dt + bn

2π0

cos(m t) sin(n t) dt

.

Now for m > 0: 2π0

cos(m t) cos(n t) dt =

0, m = n

π, m = n

and 2π0

cos(m t) sin(n t) dt = 0.

Therefore in the above summation only the term

an

2π0

cos(m t) cos(n t) dt

with n = m is nonzero and, consequently,

2π

0

f (t) cos(m t) dt = π am, m > 0.

This is equivalent to

an =1

π

2π0

f (t) cos(n t) dt, n > 0.



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The remaining identity

bn =1

π

2π

0

f (t) sin(n t) dt, n > 0,

is proven similarly.

As an illustration of Theorem 1 let us expand the “square wave” function

f (t) =sin(t)√

1− cos2 t

into a Fourier series. The name of the function is apparent from the graph

shown in Figure 2 below.

–1

–0.5

0

0.5

1

2 4 6 8 10 12 14 16 18t

Figure 2: Square wave function



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It is straightforward to show that for the square wave all coefficients

an = 0; thus the Fourier series has sine terms only. The coefficients bn aregiven by:

bn =1

π

π0

sin(n t) dt− 2ππ

sin(n t) dt

=1

π

−cos(n t)

n

π

0

+cos(n t)

n

2π

π

=

2 (1− cos(π n))

π n

=2

π n×

0, n is even,

2, n is odd.

Thus f (t) is the sum of odd sine terms:

f (t) =4

π

∞n=0

sin((2 n + 1) t)

2 n + 1. (5)

Figures 3–4 below show the approximations of f with 3, and 7 terms, respec-tively.

–1

–0.5

0

0.5

1

2 4 6 8 10 12

Figure 3: Square wave approximated with 3 Fourier terms



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–1

–0.5

0

0.5

1

2 4 6 8 10 12

Figure 4: Square wave approximated with 7 Fourier terms

Now that we know that the square wave function has (5) as its Fourierrepresentation we can easily compute the steady output of an RLC circuitexcited by the square wave and investigate resonance in that system. Choos-ing the same parameters 1/√L C = 1 and R = 0.01 as before, we see thatthe amplitudes of the first three harmonics get magnified by the factors:

G(1) = 100, G(3) = 0.124 . . . , G(5) = 0.041 . . .

Thus the circuit will resonate with the fundamental harmonic and the restof the harmonics will be practically invisible in comparison. The phase shiftof the first harmonic is Π/2. Therefore, for this case

q p −400

πcos(t)

Exercises

1. Use Equation (3) to find the phase shift induced by an RLC -circuit.(Hint: write the left-hand side of (3) in the form a cos(ω t−φ) and the



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right-hand side in the form A cos(ω t−Φ). The phase shift is then the

difference Φ− φ.)2. Prove the following orthogonality relation of cosines:

2π0

cos(n t) cos(m t) dt =

⎧⎪⎨⎪⎩

2π, n = m = 0,

π, n = m > 0,

0, n = m.

3. Let

f (t) =

∞

n=1sin(2 n t)

2 n.

Plot the sum of the fist several terms (the more the better) of the serieson the interval [−4π, 4π]. What simple graph needs to be periodicallytranslated to generate the graph of f ? Confirm your findings with acomputation.

Resonance and Fourier Theory

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