Difference between revisions of "Lab 3 RS"

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2. Now construct the circuit using a non-polar capacitor
 
2. Now construct the circuit using a non-polar capacitor
 
[[File:TF_EIM_Lab3.png | 400 px]]
 
  
 
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" cellspacing="0" style="text-align: center; width: 400px; height: 200px;"
 
 
|+ '''Table1. Voltage gain vs. frequency measurements'''
 
|+ '''Table1. Voltage gain vs. frequency measurements'''
 
|-
 
|-
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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)=
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===method 1. Using fitting line===
 
===method 1. Using fitting line===
  
:Theoretical break frequency: <math>12.13\ \mbox{kHz}</math>
+
Theoretical break frequency: <math>12.13\ \mbox{kHz}</math>
  
:Experimentally measured break frequency: <math>9.59\ \mbox{kHz}</math>
+
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:
  
  Q: The above was read off the graph?  Why not use fit results?
+
:<math>\mbox{log}(f_{exper})=\frac{1.071}{1.005} = 1.066</math>
  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>\ y=0.8989-0.915\cdot x</math>.
+
:<math>f_{exp} = 10^{1.066} = 11.64\ \mbox{kHz} </math>
:From intersection point of line with x-axis we find:
 
  
:<math>\mbox{log}(f_{exper})=\frac{0.8989}{0.915} = 0.982</math>
 
  
:<math>f_{exp} = 10^{0.982} = 9.59\ \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>
:The error is:
 
 
 
  <math>Error = \left| \frac{f_{exp} - f_{theor}}{f_{theor}} \right| = \left| \frac{9.59 - 12.13}{12.13} \right|= 20.9\ %</math>
 
  
 
===method 2.  Using the -3 dB point===
 
===method 2.  Using the -3 dB point===
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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>
 
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>
  
 
+
==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)==
'''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+R_C\right) = I\left(R+\frac{1}{i\omega C}\right)</math>
+
:<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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==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>
  
  The theoretical phase shift is <math>\ \Theta_{theory}=\arctan \ (\omega R C)</math>
+
  The theoretical phase shift is <math>\ \Theta_{theory}=\arctan \ (-\omega R C)</math>
  
  
[[File:Phase_table_m1.png | 600 px]]
+
[[File:Phase_table_m2.png | 600 px]]
  
==5. what is the phase shift <math>\theta</math> for a DC input and a very-high frequency input?(5 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.
 
  Because a DC circuit doesn't have any oscillation there are no any phase shift.
  
==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)==
+
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 <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
 
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}{i\omega C}\right|} \right] = \arctan \ (\omega RC)</math>
+
  <math>\Phi = \arctan \ (V_R/V_C) =\arctan \left( \frac{IR}{I \left(-\frac{1}{\omega C}\right)} \right) = \arctan\ (-\omega RC)</math>
 +
 
 +
 
 +
[[File:RS phase shift m1.png | 600 px]]
 +
 
  
  

Latest revision as of 05:23, 28 January 2011

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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]


RS lab3 voltage gain m2.png

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]

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]

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:

[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]

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)

Phase diagram m.png

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 theoretical phase shift is [math]\ \Theta_{theory}=\arctan \ (-\omega R C)[/math]


Phase table m2.png

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 -90 degree (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 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]


RS phase shift m1.png


Forest_Electronic_Instrumentation_and_Measurement

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