BACKGROUND AND THEORY

The phenomenon of oscillation, or vibration, is one of the most common in nature, and everyone is acquainted with dozens of everyday examples of oscillating objects, which might range from a child on a swing to the head of a drum.

The reason for the widespread occurrence of oscillatory phenomena is, of course, that situations giving rise to vibration happen so often; for example, an object which is constrained by some force to stay in a certain position may be momentarily removed from its position of equilibrium, and then released. The restoring forces accelerates it back toward its normal position, but when it gets there, it has acquired enough velocity to "overshoot" and so to proceed out to the other side of equilibrium. Then the process repeats, and unless there is some frictional viscous "drag" on the object, it will swing back and forth indefinitely.

In the case of a series LRC circuit this oscillatory behavior is due to stored energy going back and forth from capacitor to inductor. The resistor consumes electrical energy and converts it to heat, thereby providing damping or the equivalent of viscous "drag" in a mechanical system. The description of an LRC circuit is done in the same way as the RC was done in Exp. 1. We have the series circuit and a square wave driver as in Figure 1.

Figure 1

First we use energy conservation

V(t) = V_L + V_C + V_R

in which the sum of voltages across the components must equal the signal generator voltage.

Next, we substitute in the voltage-current relations for the components

V_R = IR, V_C = , V_L = L

V = IR + + L

Then differentiate to eliminate the integral

= + + ( = 0 when V constant)

and look for a solution only when the square wave is constant (dv/dt = 0). The square wave puts energy into the circuit during the transition of the square wave from V₀ to 0 and vise versa, but we will only try to explain behavior after the transition, while the circuit oscillates on its own. Hence, we need to solve

I + RC + LC = 0 .

(1)

Assuming a solution of the form I = I₀ e^t we substitute into eqn(1) and obtain

1 + Rc + Lc² = 0

= -

This simple approach has given us an answer, but we note that R/L is often much less that so that the square root is imaginary. This immediately puts us back to oscillatory solutions as worked out in Exp. 1.

If the complex nature of the solution is written explicitly,

= - i

(2)

and

I(t) = I₀

(3)

This solution is the product of the two types of functions which you worked on in Exp. 1: a decaying exponential and an oscillatory (complex argument) part. Since we are not going to worry about "boundary conditions" of what happens when the square wave makes its transition, we are allowed to pick some convenient piece of , for example sin(t) as the "physical" solution:

I(t) = I₀sin(t)

(4)

obviously this solution could be put back into eqn (1) and verified using the result in eqn (2) for .

Notice that the resistance appears explicitly in the real exponent, and the ratio 2 L/R is the effective time constant of the system. It is traditional to define the time constant for the LR circuit, so that

(5)

and

I(t) = I₀sin(t)

(6)

describe the damped oscillation. It is also customary to write Eqn (2) in the form

² = = ₀² -

(7)

where

₀² =

(8)

which suggests that ₀for large .

Observe that R has two effects on the circuit. First, it causes the amplitude of the oscillation (i.e., the maximum excursion during a cycle) to decrease steadily from one cycle to the next. The factor is responsible for this; it is commonly called the envelope of the oscillation, for a reason evident in Figure 2.

Figure 2

Secondly, the frequency of the oscillation is altered, since we see that ² and ₀² now differ by . The natural oscillation frequency and period T for the damped oscillator are related to (as ₀ and T₀ are related to ₀), namely,

f = =

(9)

This system is said to be a dampedoscillator, and its decaytime is the time when the exponential factor has become 1/e, i.e., when the exponent is minus one. Thus, the decay time t_D is

t_D = 2.

(10)

We will now show that for cases where the decay is not too strong, or more specifically, when several oscillations occur in a decay time, the difference between the damped and undamped frequencies ( and ₀) is really very slight. First suppose there is some number n of oscillations in one decay time (n is not necessarily an integer). Then, since there is one oscillation every seconds,

n = = = ,

(11)

The ratio of ₀ to can be obtained from Eqn(7) by first dividing through by :

= 1 +

(12)

We note that ₀/ = f₀/f and use Eqn(11) to obtain

()² = ()² = 1+

(13)

Thus, for example, if there are only two oscillations in a decay time t_D, the ratio is

= = 1.003,

and the undamped frequency f₀ is only 0.3% greater than f.

The most widely used measure of oscillator decay is the "quality factor" Q, which can be defined by

Q = ₀.

(14)

It is only well-defined (or, rather, there is only a well-agreed definition of Q) when it is somewhat greater than one. In this case it is also true that ₀ = . Starting with Eqn(12) we obtain

()² = 1 + = 1 + ,

which can be solved for to give

= = .

For example, if Q = 2, = 1.033 and ₀ is just 3.3% larger than .

Thus, for Q not too small, we may approximate by ₀. Using this approximation, we can combine Eqns (14 ) and (11) to give the convenient expression

Q = n,

(15)

where n is the number of cycles per decay time.

A very important situation occurs when the damping of an oscillator becomes large. If the decay time becomes so short that the angular frequency is reduced to zero,

² = ₀² - = 0,

i.e., if

= _C = ,

(16)

then there is no oscillation at all, and the system is said to be criticallydamped. For < _C the system is overdamped. In either case, we say that the oscillator is aperiodic. Figure 3 illustrates the behavior of an oscillator released at t = 0 from rest with an initial displacement of I₀ for various values of the quantity .

Underdamped (>_c) Overdamped (<_c) Critical (=_c)

Figure 3 The time axis begins at -1 to show behavior more clearly.

The quality factor, Q, is often a source of confusion so it may help to show how its definition in Eqn(14) can lead to an alternative description in terms of energy loss. First, we must review the energy situation in the LRC circuit. At any instant the capacitor and inductor store energy according to

E_C = 1/2 CV²

E_L = 1/2 LI²

(17)

If we take R = 0 for a moment, then we can show energy is conserved and is moving back and forth between L and C. For a sinusoidally varying current

I = I₀sin(t) (R = 0),

and the energy stored in the inductor becomes

E_L = 1/2 LI₀² sin²(t)

(18)

Then for capacitance

V_C = = sin(t)

and the energy stored in the capacitor is

E_C = 1/2cos²(t)

(19)

The total energy in the circuit is conserved since the sum

E_C + E_L = 1/2 I₀²

(20)

is a constant. Here we have used the fact that since R = 0, ² = ₀² = 1/LC.

You may see from the preceding steps that the energy in the inductor is a maximum when the energy in the capacitor is zero, and vise versa, because one is a sin and the other a cos function. A consequence of this is that you only need to keep track of peak energy in the inductor to follow the cycle-to-cycle energy situation in the circuit.

Now, back to the resistor. If R 0, the stored energy is gradually lost because the current is flowing back and forth from inductor to capacitor and the power loss (I²R) in the resistor equals the rate of energy decrease. The amount of energy which the resistor removes each cycle of oscillation is

E_loss/cycle = = = I₀²