Lab 4 RS

Go Back to All Lab Reports

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

[math]\nu\ [\mbox{kHz}][/math]	[math]V_{in}\ [V][/math]	[math]V_{out}\ [V][/math]	[math]\frac{V_{out}}{V_{in}}[/math]
0.5	12.4	0.48	0.039
1.0	11.0	0.88	0.080
2.0	8.0	1.3	0.162
3.0	6.0	1.5	0.250
4.0	4.9	1.6	0.326
5.0	4.2	1.7	0.405
6.0	3.7	1.7	0.459
7.0	3.2	1.7	0.531
8.0	3.0	1.75	0.583
9.0	2.85	1.70	0.596
10.0	2.60	1.70	0.654
11.0	2.50	1.70	0.680
12.0	2.35	1.70	0.723
13.0	2.25	1.70	0.755
14.0	2.20	1.70	0.772
15.0	2.10	1.70	0.809
20.0	2.00	1.75	0.875
30.0	1.85	1.75	0.946
50.0	1.82	1.78	0.978
100.0	1.80	1.80	1.00

5. Graph the [math]\log \left(\frac{V_{out}}{V_{in}} \right)[/math] -vs- [math]\log (\nu)[/math]

phase shift (10 pnts)

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

See question 3 about my phase shift measurements

Questions

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

Theoretical break frequency: [math]12.13\ \mbox{kHz}[/math]

The fit line equation from the plot above is [math]\ y=-1.097+0.9857\cdot x[/math]. From intersection point of line with x-axis we find:

[math]\mbox{log}(f_{exper})=\frac{1.097}{0.9857} = 1.113[/math]

[math]f_{exp} = 10^{1.113} = 12.97\ \mbox{kHz} [/math]

The error is:

[math]Error = \left| \frac{f_{exp} - f_{theor}}{f_{theor}} \right| = \left| \frac{12.97 - 12.13}{12.13} \right|= 6.92\ %[/math]

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

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

We have:

[math]1)\ V_{in} = I \left(R+X_C\right) = I\left(R+\frac{1}{i\omega C}\right)[/math]

[math]2)\ V_{out} = I R [/math]

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

[math]\ \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}[/math]

And we are need the real part:

[math]\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}}[/math]

3. Compare the theoretical and experimental value for the phase shift [math]\theta[/math]. (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]

4. Sketch the phasor diagram for [math]V_{in}[/math],[math] V_{out}[/math], [math]V_{R}[/math], and [math]V_{C}[/math]. Put the current [math]I[/math] along the real voltage axis. (30 pnts)

5. What is the phase shift [math]\theta[/math] 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.

For a very high frequency input the phase shift is 0 (see plot in question 6)

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

From the phasor diagram above (question 4) the angle between vectors [math]V_{in}[/math] and [math]V_{out}[/math] 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]

Forest_Electronic_Instrumentation_and_Measurement Go Back to All Lab Reports