Lab 3 RS
- 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]\displaystyle{ R=10.5\ \Omega }[/math] [math]\displaystyle{ C=1.250\ \mu F }[/math]
So
[math]\displaystyle{ \omega_b = \frac{1}{RC} = 76.2\cdot 10^3\ \frac{rad}{s} }[/math] [math]\displaystyle{ f_b = \frac{\omega_b}{2\pi} = 12.1\ \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]\displaystyle{ (V_{in}) }[/math] and output [math]\displaystyle{ (V_{out}) }[/math] voltages for at least 8 different frequencies[math]\displaystyle{ (\nu) }[/math] which span the frequency range from 1 Hz to 1 MHz
| [math]\displaystyle{ \nu\ [\mbox{kHz}] }[/math] | [math]\displaystyle{ V_{in}\ [V] }[/math] | [math]\displaystyle{ V_{out}\ [V] }[/math] | [math]\displaystyle{ \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]\displaystyle{ \log \left(\frac{V_{out}}{V_{in}} \right) }[/math] -vs- [math]\displaystyle{ \log (\nu) }[/math]
phase shift (10 pnts)
- measure the phase shift between [math]\displaystyle{ V_{in} }[/math] and [math]\displaystyle{ V_{out} }[/math] as a function of frequency [math]\displaystyle{ \nu }[/math]. Hint: you could use [math]\displaystyle{ V_{in} }[/math] as an external trigger and measure the time until [math]\displaystyle{ V_{out} }[/math] reaches a max on the scope [math]\displaystyle{ (\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: 12.1 kHz
- Experimentally measured break frequency: 9.59 kHz
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 fit line equation from the plot above is [math]\displaystyle{ \ y=0.8989-0.915\cdot x }[/math].
- From intersection point of line with x-axis we find:
- [math]\displaystyle{ \mbox{log}(f_{exper})=\frac{0.8989}{0.915} = 0.982 }[/math]
- [math]\displaystyle{ f_{exp} = 10^{0.982} = 9.59\ \mbox{kHz} }[/math]
- The error is:
[math]\displaystyle{ Error = \left| \frac{f_{exp} - f_{theor}}{f_{theor}} \right| = \left| \frac{9.59 - 12.1}{12.1} \right|= 20.7\ % }[/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]\displaystyle{ \mbox{log}(V_{out}/V_{in}) = (3/20) = 0.15 }[/math]. Using this value I found from plot above [math]\displaystyle{ \mbox{log}(f_b) = 1.1\ \mbox{kHz} }[/math]. So [math]\displaystyle{ f_b = (10^{1.1}) = 12.6\ \mbox{kHz} }[/math]. The error in this case is 4.1 %.
2. Calculate and expression for [math]\displaystyle{ \frac{V_{out}}{ V_{in}} }[/math] as a function of [math]\displaystyle{ \nu }[/math], [math]\displaystyle{ R }[/math], and [math]\displaystyle{ C }[/math]. The Gain is defined as the ratio of [math]\displaystyle{ V_{out} }[/math] to [math]\displaystyle{ V_{in} }[/math].(5 pnts)
We have:
- [math]\displaystyle{ 1)\ V_{in} = I\left(R+R_C\right) = I\left(R+\frac{1}{i\omega C}\right) }[/math]
- [math]\displaystyle{ 2)\ V_{out} = I \left(\frac{1}{i\omega C}\right) }[/math]
Dividing second equation into first one we get the voltage gain:
- [math]\displaystyle{ \ \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]\displaystyle{ \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 fRC)^2}} }[/math]
3. Sketch the phasor diagram for [math]\displaystyle{ V_{in} }[/math],[math]\displaystyle{ V_{out} }[/math], [math]\displaystyle{ V_{R} }[/math], and [math]\displaystyle{ V_{C} }[/math]. Put the current [math]\displaystyle{ I }[/math] along the real voltage axis. (30 pnts)
4. Compare the theoretical and experimental value for the phase shift [math]\displaystyle{ \theta }[/math]. (5 pnts)
The experimental phase shift is [math]\displaystyle{ \ \Theta_{\mbox{exper}} = (\omega\ \delta T)_{\mbox{exper}} }[/math]
The theoretical phase shift is [math]\displaystyle{ \ \Theta_{\mbox{theory}}=\mbox{arctan}(\omega RC) }[/math]
