In this experiment we will move into the frequency domain for the LRC
circuit. In Exp. 2 we worked in the time domain and found that the circuit's
response to a transient was a damped sine wave of frequency 
, close to
0 which we will find is the resonant frequency of the circuit
when driven by a sine wave. One way to change from time to frequency is to
take a Fourier Transform of the time domain behavior of Exp. 2 and look at the
resulting power spectrum; it is a function of frequency which looks very much
like Fig. 2 of this experiment. Here, in Exp. 3, we will drive the circuit
with a sine wave of frequency and observe its response.
In the time domain case we have a valuable tool, the oscilloscope, to assist in
observing the circuit behavior. The equivalent too in the frequency domain is
the "sweep generator", a modified signal generator that can rapidly vary its
output frequency. In order to take data, you will have to turn the frequency
knob on the signal generator and record sine wave amplitudes. A graph of your
data will give you a plot of the response vs as shown in Fig. 2, though
you can use the sweep functions in order to get a quick visual confirmation of
the resonance curve shape on the oscilloscope.
Let us begin the analysis of the circuit of Fig. 1 by assuming that the voltage generator inside the signal generator box has the following functional form:
VS(t) = V0 eit [= V0 (cos
t + i sin t)]
We will use complex voltages, currents and impedances because the algebra is much simpler. You will need to become familiar with complex numbers and functions for a variety of applications in the future.
In linear circuits, such as the LRC, the concept of impedance is very useful. Namely, we assume that current and voltage are proportional; for an inductor, the impedance, ZL, is defined in a way which is similar to resistance:
ZL =
or I =
.
Then, take the time dependent behavior of VL:
L
= VL with VL = V0 eit ,
and plug in for the derivative
[ (i) V0 eit ] = VL =
V0 eit ,
which leads to the result
ZL = iL.
Likewise for capacitors, C
= I
so, ZC =
= -
for the impedance of the capacitor.
The beauty of the complex notation is that the algebra is much simpler than if you continued to use trigonometric functions. The major complication is in the realm of "phases" and "phase shifts". On the other hand, differential equations disappear, and you only have to add up impedances as
though they were complex resistances.
For the LRC series circuit (Fig. 1) we add up the voltages
VL + VC + VR = V0
eit ,
and replace voltages with the product of current and impedances
[ R + i ( L -
)
] I = V0 eit ,
where V0 is the amplitude of applied voltage (signal generator).
Therefore,
I(t) =
is the "solution" to the problem. However, we would rather have this in the polar form
I = I0 ei(t + 
)
in which I0 and are real numbers, the amplitude and phase of
the sinusoidal current. This is accomplished by treating the complex terms in
the denominator as the x and y components of a vector; then find the magnitude
and angle of the vector with respect to the x-axis. Usually these two
components are called the real and imaginary parts:
Re[Z] = R
Im[Z] = L -
Then I0 =
and tan(-) =
=
If we switch to more relevant parameters,
0 =
, Q =
=
then
=
=
.
And, finally (after some algebra), the voltage across Vr is
Vr = RrI =
where tan () =
Q,
for the phase, .
Also, we could have written the square root as
=
for between -
/2 and /2,
so that the voltage across the resistance is also given by
|Vr| =
This is a very useful (and simple!) relation between phase and amplitude. If we go back ot the frequency dependence, the shape of the response function (magnitude of Eq. (1))
|Vr| =
is sketched in Figure 2 as a function of frequency.
Figure 2 The Response Function of the LRC series-resonant
circuit. A constant amplitude signal (V0) drives the circuit, and
changes in frequency lead to changes in the current that flows, which you
measure as the voltage drop across the resistor. The half-power points,
1 and 2, are shown along with their
relation to the quality factor Q and the resonant frequency,
0, which corresponds to the peak of the response function.
As the Q increases, the width of the function decreases.
The maximum of Eq. 3 occurs when = 0 and the square
root is equal to unity. The "half power points", 1 and
2, are also shown in Fig. 2. Since P V2, the half
power points occur when the voltage decreases to 1/2 of its peak value, or
Q
= 1
which happens twice, = 1, below resonance, and
= 2, above resonance. If you solve the two quadratic
equations, you find that
= Q
this is the quick way to measure Q, and it is exact, having no approximations in its derivation.
When you use this technique in the lab, set the circuit at resonance, observe the amplitude, and change the frequency until the amplitude decreases to 0.707 times the maximum (resonance) value. This new frequency is one of the half-power points.
Figure 3 Phase shifts in the LRC circuit. The driving
voltage in (a) is V0 sin(t + ). A positive phase shift
(>0) is shown in (b) where 
t, the time from the current's zero
crossing to the origin, is positive. The situation for negative is shown
in (c). Notice that the time scales are different when you compare (a) with
(b) that when you compare (a) with (c); the frequency is different for the two
cases.
Now, back to the phase problem. In Fig. 3 we show how the sine waves appear on
an oscilloscope if you have selected the signal generator voltage to trigger
the sweep. The traces in (b) and (c) of Fig. 3 are two examples of how the
current in the circuit changes with time. When the frequency is below
0, the phase shift is positive as shown in (b); when the
frequency is set above resonance, the current wave form is shifted so that the
negative phase shift, (c) is observed.
In Fig. 4 the relationships between amplitude, frequency and phase shifts are
plotted. At low frequencies the current is dominated by the capacitor in the
circuit; the amplitude of the current is low and its phase is shifted by
+/2 (+90). At high frequencies the inductor controls the current so that
the amplitude is once again low, but the phase shift is now -/2 (-90). At
resonance the phase shift is zero and the amplitude is at a maximum.
Figure 4 Amplitude and phase () as a function of frequency.
Note that the phase shift is positive at low frequencies and negative for high
frequencies. Resonance is at zero phase shift, and the half-lower points are
at plus and minus 45 as predicted by Eq. 2.
The Q-Multiplier
At resonance the electromagnetic stored energy flows back and forth between the capacitor (electrostatic energy) and the inductor (magnetic energy). A cancellation of voltages across the two elements takes place because of the 180 phase shift between the capacitor and inductor voltage at resonance because of a high Q leading to small energy losses. The stored energy in the circuit can then be quite large for a modest driving voltage from the signal generator. When the capacitor has reached its peak energy the inductor has none; current is zero at that instant. We will see that this fact allows us to make a "voltage multiplier" if Q is large. On the other hand, we will be able to use this effect to measure Q by finding the ratio of capacitor voltage to circuit drive voltage.
If we define a circuit by saying that a voltage, V0, is "driving" the circuit, then at resonance
I0 = 
because cos() = 1 in Eq. 2, and I =
.
The maximum energy stored in the magnetic field of the inductor is
when VC = 0 ,
but a half cycle later the stored energy has gone to zero with the current. At this time the maximum energy stored in the electrostatic field of the capacitor is
when I = 0,
where VCO is the maximum amplitude of the capacitor's voltage. For reasonably high valuse of Q (small energy losses) the maximum stored energies are equal, but the times when the maxima occur are always 180 out of phase with each other.
So, we have, using the above expression for current,
= 
or 
= 
which is Q. The result is:
= Q.
This is a convenient way of measuring Q, provided that you are careful in how you measure the driving voltage, V0. Notice that measuring V0 at different places in the circuit is equivalent to defining a new circuit with, perhaps, fewer resistors. One of the complications here is associated with the internal resistance of the signal generator; you can't find an accurate value for V0 of the signal generator (the inside voltage) unless you remove the circuit.