Lab 4 RS

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RC High-pass filter

1-50 kHz filter (20 pnts)

1. Design a high-pass RC filter with a break point between 1-50 kHz. The break point is the frequency at which the filter's attenuation of the AC signal goes to 0(not passed). For a High pass filter, AC signals with a frequency below the 1-50 kHz range will be attenuated .

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 and output voltages for at least 8 different frequencies 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
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
11.0
12.0
15.0
20.0
30.0
40.0
50.0
100.0
200.0

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 $V_{out}$

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

Experimentally measured break frequency calculation:

The fit line equation from the plot above is $\ y=-1.14+0.9101\cdot x$.

From intersection point of line with x-axis we find:

$\mbox{log}(f_{exper})=\frac{1.14}{0.9101} = 1.252$

$f_{exp} = 10^{1.252} = 17.86\ \mbox{kHz}$

The error is:

[math]Error = \left| \frac{f_{exp} - f_{theor}}{f_{theor}} \right| = \left| \frac{17.86 - 12.13}{12.13} \right|= 32.08\ %[/math]

Error pretty big. Probably is something wrong with RC measurements.

Questions

Calculate and expression for $\frac{V_{out}}{ V_{in}}$ as a function of $\nu$, $R$, and $C$.(5 pnts)

We have:

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

$2)\ V_{out} = I R$

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

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

And we are need the real part:

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

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\ \left (\frac{1}{\omega R C}\right )[/math]

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)

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.

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 4) the angle between vectors $V_{in}$ and $V_{out}$ given by

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

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