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 [math](V_{in})[/math] and output [math](V_{out})[/math] voltages for at least 8 different frequencies[math] (\nu)[/math] 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]
0.1	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 [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 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: [math]12.13\ \mbox{kHz}[/math]

Experimentally measured break frequency: [math]9.59\ \mbox{kHz}[/math]

 Q: The above was read off the graph?  Why not use fit results?
 A: The fit was made by using GIMP Image Editor. I do not have so much experience with ROOT. But I will try to do it. Thank you for comment.
 A1: The fit was done by ROOT. The graph above was replaced.

The fit line equation from the plot above is [math]\ y=1.071-1.005\cdot x[/math].

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

[math]\mbox{log}(f_{exper})=\frac{1.071}{1.005} = 1.066[/math]

[math]f_{exp} = 10^{1.066} = 11.64\ \mbox{kHz} [/math]

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 [math]\mbox{log}(V_{out}/V_{in}) = (3/20) = 0.15 [/math]. Using this value I found from plot above [math]\mbox{log}(f_b) = 1.1\ \mbox{kHz}[/math]. So [math]f_b = (10^{1.1}) = 12.59\ \mbox{kHz}[/math]. The error in this case is [math]3.79\ %[/math]

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]. The Gain is defined as the ratio of [math]V_{out}[/math] to [math]V_{in}[/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 \left(\frac{1}{i\omega C}\right) [/math]

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

[math]\ \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}[/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{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}}[/math]

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)

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 \ (-\omega R C)[/math]

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.

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 3) the angle between vectors [math]V_{in}[/math] and [math]V_{out}[/math] 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]

Forest_Electronic_Instrumentation_and_Measurement

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