2/11: Reactance of a Capacitor; Low and High-Pass Filters

in which Julia revisits the integrator and differentiator circuits as frequency filters

(A note on reading capacitors:
"1 0 4" means:
10 x 10^4 pF
so 1 x 10^5 pF
so 1 x 10^5 x 10^-12 F
so 1 x 10^-7 F
so 0.1 μF)

H&H, Lab 2-2: Differentiator circuit

As previously discussed, we began to investigate this circuit last class.  Now we're thinking about the circuit's impedance:
At f = 0, the capacitor's impedance is infinite, so it effectively acts like an open circuit--none of the signal gets through (Vout = 0).  As f approaches infinity, the capacitor's impedance approaches zero, so it effectively acts as a wire--almost all of the signal gets through (Vout = Vin).
("Input impedances" and "output impedances" are fancy names for Rth, in or out.  Remember: for a good circuit, Rout<<Rin!)

Measuring Vout across the resistor, driven with a square wave:

The square wave output--voltage over the resistor (yellow)--compared to input (purple).

Beginning from the highest peak of the output voltage:
The peak of the output voltage occurs when the voltage through the resistor (Vr) is largest, so the voltage through the capacitor (Vc) is smallest, so the capacitor is fully uncharged.
Vr then decreases because the capacitor is charging (Vc is increasing, so Vr is decreasing).
When the input voltage falls rapidly, the capacitor rapidly discharges, so the output voltage follows the input voltage in this dramatic fall.  Switching voltage causes the capacitor to charge in the opposite direction (so the voltage across the resistor slowly increases while the input plateaus at a negative voltage)--Vr increases after the peak because the capacitor is charging (Vc increasing negatively, so Vr is decreasing negatively, so Vr gets more positive).

Next we tried driving the circuit with a triangle wave:

The triangle wave output--voltage over the resistor (yellow)--compared to input (purple).

Triangle wave up: capacitor gives resistor a little extra voltage (discharging)
Triangle wave down: capacitor takes a little voltage from resistor (charging)
We still see the triangle shape in output because almost all the input voltage is going through, since the capacitor can't charge very much because the frequency is fast.

Lab 2-3, part II: differential circuit where ω is not << 1/RC

For ω < RC: For low frequencies, the capacitor's impedance is high, so the square wave doesn't pass--only the high frequencies (in the Fourier spectrum that make the square wave) associated with the peak pass through the capacitor, so the circuit effectively "differentiates" the square wave to show peaks at each rise and fall of the input wave.

A differentiator peak at low frequencies.

For omega >> RC: The high-pass filter passes a square wave, because all frequencies are high and thus pass through the "differentiator", so it no longer differentiates.

Thinking about this graphically--in the first case, omega is small so Vout/Vin ~ 0 (except where the slope of the input wave is large).  In the second case, omega is large so Vout/Vin ~ 1.
Driving with increasingly high frequencies makes the ouput square wave closer to the input--presumably because more input voltage is getting through the circuit because the capacitor has less time to charge (its impedance is lower).
Similarly, lower frequencies make the output square wave less close to the input--because less input voltage is getting through, as the capacitor's impedance is higher as it can charge for a little while.

Notes from the RC circuit reading:

The cutoff frequency, ω(0) = 1/(RC), roughly represents the boundary between frequencies that will pass and frequencies that will be blocked by an RC circuit.
When ω << 1/(RC) (at low frequencies), the capacitor's impedance is large compared to the resistor, so Vout will be almost in phase with Vin (it is difficult to see the phase change effect of the resistor at low frequencies).
When ω >> 1/(RC) (at high frequencies), the capacitor's impedance is small compared to the resistor, so Vout will approach a maximum of 90 degrees out of phase with Vin.
The ratio of two voltages is usually given in decibels:

dB = 20*log(A2/A1)

where "log" is log base 10, A2 is the amplitude of Vout, and A1 is the amplitude of Vin.

 Lab 2-4: Low-pass filter, "integrator" circuit

What does the phrase "beyond cutoff, response of low pass filter drops at 6 dB/octave" mean?
"Response" corresponds to the ratio of the amplitudes of Vout to Vin (A2/A1).
An "octave" means doubling the frequency of the input voltage signal, ω (or f).
Use the decibels formula:

dB = 20*log10(A2/A1)

According to this statement, if we double ω then:

-6 = 20*log10(Aout/Ain)

log10(Aout/Ain) = -6/20

Aout/Ain = 10^(-6/20) = 10^-.3 = 0.5

So one would expect doubling ω to double the attenuation (halve the output amplitude).

To find ω for -3 dB:

-3 = 20*log10(Aout/Ain)

Aout/Ain = 1/sqrt(2)

ω = 1/RC = 2*π*f = 1/(sqrt(2)*RC)

f = 0.71/(2*π*RC)

where R and C are the resistance and capacitance values for the particular RC circuit.

Using our capacitance and resistance values, the driving frequency for the 3dB point should be 1.06 kHz:

R = 15 kΩ, C= 0.01 μF

RC = (15 x 10^3)*(0.01 x 10^-6) = (1.5 x 10^4)*(1 x 10^-8) = 1.5 x 10^-4

1/RC = 6666.7

1/(2*π*RC) = f = 1.06 kHz

When driving the circuit with 1.06 kHz, we observed an output voltage of 2.69 V for an input voltage of 4.00 V.  This corresponds to an attenuation of (2.69/4 = ) 0.67, or 67%, which is close to the expected 70.7% amplitude decrease.

Theoretically, the phase shift at 1.06 kHz should be π/4 (45 degrees):

2*π*dt/T = π/4

dt/T = 0.125

Thus we expected a dt value of 125 μs.

Measuring the phase shift, we observed a dt of 112 μs, which is close to our expected value of 125 μs.

For a low frequency signal (10 Hz), we expected a phase shift of 0, because the capacitance's impedance would be much larger than that of the resistor.  For a high frequency signal (10 kHz), we expected a phase shift of π/2 (90 degrees), because the capacitance's impedance would be much smaller than that of the resistor.  As discussed above, experiments were in reasonable agreement with our predictions: we found no observable offset for a 10 Hz input signal and an offset of approximately 90 degrees with a 10 kHz input signal.

The 45-degree offset between 1.06 kHz input and output waves.

In general, increasing the frequency decreased the voltage (causing greater attenuation): doubling the frequency halved the output voltage (doubled the attenuation).
When driven at a frequency of 10 kHz (10*f(3dB)), the output voltage was 0.52 V for an attenuation of (4.00/0.52 = ) 7.70.
A 20 kHz frequency (20*f(3dB)) made Vout = 0.28 V, for an attenuation of (4.00/0.28 = ) 14.3.
A 40 kHz frequency (40*f(3dB)) made Vout = 0.14 V, for an attenuation of (4.00/0.14 = ) 28.6.
In the decibels formula:

dB(10 kHz) = 20*log(7.70) = 17.7

dB(20 kHz) = 20*log(14.3) = 23.1

dB(40 kHz) = 20*log(28.6) = 29.1

The decibel difference between octaves was thus (23.1 - 17.7 = ) 5.4 for the first octave and (29.1 - 23.1 = ) 6 dB for the second octave.  This agrees fairly well with the "6 dB decrease per octave" suggestion interpreted previously.

When driven at a frequency of 2*f(3dB), or 2.1 kHz, this circuit's output voltage was 1.94 V; at 4*f(3dB), or 4.2 kHz, Vout = 1.1 V; at 10*f(3dB), or 10.6 kHz, Vout = 0.52 V.

Lab 2-5: High-pass filter

A high-pass filter (which passes only high-frequency signals) was constructed with a 0.01 μF capacitor and 15 kΩ resistor to form an RC circuit.  Because the values for R and C were the same as for our low-pass filter circuit, the 3dB point is the same for this circuit: at f = 1.06 kHz.
At low frequencies, the output amplitude was proportional to the frequency
Like the low-pass filter, the limiting phase shifts for the high-pass filter were π/2 and 0 radians; however, a phase shift of π/2 occurred with a low input frequency, and a phase shift of 0 occurred with a high input frequency--the opposite of the low-pass limits.