Difference between revisions of "Lab 3 RS"
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<math>C=1.250\ \mu F</math> | <math>C=1.250\ \mu F</math> | ||
− | <math>\omega_b = \frac{1}{RC} = 76. | + | So |
− | <math>f_b = \frac{\omega_b}{2\pi} = 12. | + | |
+ | <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 | 2. Now construct the circuit using a non-polar capacitor | ||
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3. Use a sinusoidal variable frequency oscillator to provide an input voltage to your filter | 3. Use a sinusoidal variable frequency oscillator to provide an input voltage to your filter | ||
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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 | 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 | ||
− | + | {| border="1" cellspacing="0" style="text-align: center; width: 500px; height: 500px;" | |
− | {| border="1" | ||
|+ '''Table1. Voltage gain vs. frequency measurements''' | |+ '''Table1. Voltage gain vs. frequency measurements''' | ||
|- | |- | ||
− | ! scope="col" width="50" | + | ! scope="col" width="50" | <math>\nu\ [\mbox{kHz}]</math> |
! scope="col" width="50" | <math>V_{in}\ [V]</math> | ! scope="col" width="50" | <math>V_{in}\ [V]</math> | ||
! scope="col" width="50" | <math>V_{out}\ [V]</math> | ! scope="col" width="50" | <math>V_{out}\ [V]</math> | ||
! scope="col" width="50" | <math>\frac{V_{out}}{V_{in}}</math> | ! scope="col" width="50" | <math>\frac{V_{out}}{V_{in}}</math> | ||
− | + | |- | |
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|0.1 ||5.0 ||5.0 ||1.0 | |0.1 ||5.0 ||5.0 ||1.0 | ||
|- | |- | ||
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− | [[File:RS lab3 voltage gain.png | 800 px]] | + | [[File:RS lab3 voltage gain m2.png | 800 px]] |
=phase shift (10 pnts)= | =phase shift (10 pnts)= | ||
#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>. | #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 | + | ''See question 4 about my phase shift measurements'' |
=Questions= | =Questions= | ||
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===method 1. Using fitting line=== | ===method 1. Using fitting line=== | ||
− | + | Theoretical break frequency: <math>12.13\ \mbox{kHz}</math> | |
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− | + | 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>log(f_{ | + | :<math>\mbox{log}(f_{exper})=\frac{1.071}{1.005} = 1.066</math> |
− | :<math>f_{exp} = 10^{ | + | :<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{ | + | <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=== | ===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. | + | 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> |
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− | + | ==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>. The Gain is defined as the ratio of <math>V_{out}</math> to <math>V_{in}</math>.(5 pnts)== | |
We have: | We have: | ||
− | :<math>1)\ V_{in} = I\left(R+ | + | :<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> | :<math>2)\ V_{out} = I \left(\frac{1}{i\omega C}\right) </math> | ||
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:<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> | :<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> | ||
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==3. 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)== | ==3. 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)== | ||
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==4. Compare the theoretical and experimental value for the phase shift <math>\theta</math>. (5 pnts)== | ==4. 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 experimental phase shift is <math>\ \Theta_{exper} = (-\omega\ \delta T)_{exper}</math> |
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+ | The theoretical phase shift is <math>\ \Theta_{theory}=\arctan \ (-\omega R C)</math> | ||
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+ | [[File:Phase_table_m2.png | 600 px]] | ||
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+ | ==5. What is the phase shift <math>\theta</math> for a DC input and a very-high frequency input?(5 pnts)== | ||
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+ | Because a DC circuit doesn't have any oscillation there are no any phase shift. | ||
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+ | For a very high frequency input the phase shift is -90 degree (see plot in question 6) | ||
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+ | ==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)== | ||
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+ | 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> | |
− | [[File: | + | [[File:RS phase shift m1.png | 600 px]] |
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Latest revision as of 05:23, 28 January 2011
- 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:
So
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 MHz0.1 | 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
-vs-phase shift (10 pnts)
- measure the phase shift between and as a function of frequency . Hint: you could use as an external trigger and measure the time until reaches a max on the scope .
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:
The fit line equation from the plot above is
. From intersection point of line with x-axis we find:
The error is:
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
. Using this value I found from plot above . So . The error in this case is2. Calculate and expression for as a function of , , and . The Gain is defined as the ratio of to .(5 pnts)
We have:
Dividing second equation into first one we get the voltage gain:
And we are need the real part:
3. Sketch the phasor diagram for , , , and . Put the current along the real voltage axis. (30 pnts)
4. Compare the theoretical and experimental value for the phase shift . (5 pnts)
The experimental phase shift is
The theoretical phase shift is
5. What is the phase shift 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 -90 degree (see plot in question 6)
6. Calculate and expression for the phase shift as a function of , , and graph -vs . (20 pnts)
From the phasor diagram above (question 3) the angle between vectors
and given by