Lab 3 RS

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RC Low-pass filter

1-50 kHz filter (20 pnts)

1. Design a low-pass RC filter with a break point between 1-50 kHz. The break point is the frequency at which the filter starts to attenuate the AC signal. For a Low pass filter, AC signals with a frequency above 1-50 kHz will start to be attenuated (not passed)

To design low-pass RC filter I had:

[math]R=10.5\ \Omega[/math]  
[math]C=1.250\ \mu F[/math]

So

[math]\omega_b = \frac{1}{RC} = 76.19\cdot 10^3\ \frac{rad}{s}[/math]
[math]f_b = \frac{\omega_b}{2\pi} = 12.13\ \mbox{kHz}[/math]

2. Now construct the circuit using a non-polar capacitor

3. Use a sinusoidal variable frequency oscillator to provide an input voltage to your filter

4. Measure the input $(V_{in})$ and output $(V_{out})$ voltages for at least 8 different frequencies$(\nu)$ which span the frequency range from 1 Hz to 1 MHz

Table1. Voltage gain vs. frequency measurements

$\nu\ [\mbox{kHz}]$	$V_{in}\ [V]$	$V_{out}\ [V]$	$\frac{V_{out}}{V_{in}}$
0.1	5.0	5.0	1.0
1.0	4.2	4.2	1.0
2.0	3.2	3.1	0.97
5.0	1.8	1.6	0.89
10.0	1.14	0.88	0.77
16.7	0.90	0.54	0.60
20.0	0.88	0.48	0.54
25.0	0.82	0.38	0.46
33.3	0.78	0.28	0.36
50.0	0.76	0.18	0.24
100.0	0.75	0.09	0.12
125.0	0.74	0.07	0.095
200.0	0.75	0.04	0.053
333.3	0.76	0.03	0.039
200.0	0.76	0.03	0.039
1000.0	0.78	0.06	0.077

5. Graph the $\log \left(\frac{V_{out}}{V_{in}} \right)$ -vs- $\log (\nu)$

phase shift (10 pnts)

1. measure the phase shift between $V_{in}$ and $V_{out}$ as a function of frequency $\nu$. Hint: you could use $V_{in}$ as an external trigger and measure the time until $V_{out}$ reaches a max on the scope $(\sin(\omega t + \phi) = \sin\left ( \omega\left [t + \frac{\phi}{\omega}\right]\right )= \sin\left ( \omega\left [t + \delta t \right] \right ))$.

See question 4 about my phase shift measurements

Questions

1. Compare the theoretical and experimentally measured break frequencies. (5 pnts)

method 1. Using fitting line

Theoretical break frequency: $12.13\ \mbox{kHz}$

The fit line equation from the plot above is $\ y=1.071-1.005\cdot x$. From intersection point of line with x-axis we find:

$\mbox{log}(f_{exper})=\frac{1.071}{1.005} = 1.066$

$f_{exp} = 10^{1.066} = 11.64\ \mbox{kHz}$

The error is:

[math]Error = \left| \frac{f_{exp} - f_{theor}}{f_{theor}} \right| = \left| \frac{11.64 - 12.13}{12.13} \right|= 4.04\ %[/math]

method 2. Using the -3 dB point

At the break point the voltage gain is down by 3 dB relative to the gain of unity at zero frequency. So the value of $\mbox{log}(V_{out}/V_{in}) = (3/20) = 0.15$. Using this value I found from plot above $\mbox{log}(f_b) = 1.1\ \mbox{kHz}$. So $f_b = (10^{1.1}) = 12.59\ \mbox{kHz}$. The error in this case is $3.79\ %$

2. Calculate and expression for $\frac{V_{out}}{ V_{in}}$ as a function of $\nu$, $R$, and $C$. The Gain is defined as the ratio of $V_{out}$ to $V_{in}$.(5 pnts)

We have:

$1)\ V_{in} = I\left(R+X_C\right) = I\left(R+\frac{1}{i\omega C}\right)$

$2)\ V_{out} = I \left(\frac{1}{i\omega C}\right)$

Dividing second equation into first one we get the voltage gain:

$\ \frac{V_{out}}{V_{in}} = \frac{I \left(\frac{1}{i\omega C}\right)}{I\left(R+\frac{1}{i\omega C}\right)} = \frac{\left(\frac{1}{i\omega C}\right)}{\left(R+\frac{1}{i\omega C}\right)} = \frac{1}{1+i\omega RC}$

And we are need the real part:

$\left |\frac{V_{out}}{V_{in}} \right | = \sqrt{ \left( \frac{V_{out}}{V_{in}} \right)^* \left( \frac{V_{out}}{V_{in}} \right)} = \sqrt{\left ( \frac{1}{1+i\omega RC}\right ) \left ( \frac{1}{1-i\omega RC}\right )} = \frac{1}{\sqrt{(1 + (\omega RC)^2}} = \frac{1}{\sqrt{(1 + (2\pi \nu RC)^2}}$

3. Sketch the phasor diagram for $V_{in}$,$V_{out}$, $V_{R}$, and $V_{C}$. Put the current $I$ along the real voltage axis. (30 pnts)

4. Compare the theoretical and experimental value for the phase shift $\theta$. (5 pnts)

The experimental phase shift is [math]\ \Theta_{exper} = (-\omega\ \delta T)_{exper}[/math]

The theoretical phase shift is [math]\ \Theta_{theory}=\arctan \ (-\omega R C)[/math]

5. What is the phase shift $\theta$ for a DC input and a very-high frequency input?(5 pnts)

Because a DC circuit doesn't have any oscillation there are no any phase shift.

6. Calculate and expression for the phase shift $\theta$ as a function of $\nu$, $R$, $C$ and graph $\theta$ -vs $\nu$. (20 pnts)

From the phasor diagram above (question 3) the angle between vectors $V_{in}$ and $V_{out}$ given by

[math]\Phi = \arctan \ (V_R/V_C) =\arctan \left( \frac{IR}{I \left(-\frac{1}{\omega C}\right)} \right) = \arctan\ (-\omega RC)[/math]

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