Difference between revisions of "Lab 3 RS"

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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>
  
==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:
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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>
  
==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)==
  
 
[[File:Phase diagram m.png | 600 px]]
 
[[File:Phase diagram m.png | 600 px]]
  
==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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[[File:Phase_table_m2.png | 600 px]]
 
[[File:Phase_table_m2.png | 600 px]]
  
==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.
  
==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)==
+
==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

Revision as of 07:01, 27 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 (Vin) and output (Vout) voltages for at least 8 different frequencies(ν) which span the frequency range from 1 Hz to 1 MHz

Table1. Voltage gain vs. frequency measurements
ν [kHz] Vin [V] Vout [V] VoutVin
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 log(VoutVin) -vs- log(ν)


RS lab3 voltage gain m1.png

phase shift (10 pnts)

  1. measure the phase shift between Vin and Vout as a function of frequency ν. Hint: you could use Vin as an external trigger and measure the time until Vout reaches a max on the scope (sin(ωt+ϕ)=sin(ω[t+ϕω])=sin(ω[t+δt])).
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: 12.13 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 graph above was replaced.
The fit line equation from the plot above is  y=1.0711.005x.
From intersection point of line with x-axis we find:
log(fexper)=1.0711.005=1.066
fexp=101.066=11.64 kHz


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 log(Vout/Vin)=(3/20)=0.15. Using this value I found from plot above log(fb)=1.1 kHz. So fb=(101.1)=12.59 kHz. The error in this case is 3.79 

2. Calculate and expression for VoutVin as a function of ν, R, and C. The Gain is defined as the ratio of Vout to Vin.(5 pnts)

We have:

1) Vin=I(R+XC)=I(R+1iωC)
2) Vout=I(1iωC)


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

 VoutVin=I(1iωC)I(R+1iωC)=(1iωC)(R+1iωC)=11+iωRC


And we are need the real part:

|VoutVin|=(VoutVin)(VoutVin)=(11+iωRC)(11iωRC)=1(1+(ωRC)2=1(1+(2πνRC)2

3. Sketch the phasor diagram for Vin,Vout, VR, and VC. Put the current I along the real voltage axis. (30 pnts)

Phase diagram m.png

4. Compare the theoretical and experimental value for the phase shift θ. (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 θ 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.

6. Calculate and expression for the phase shift θ as a function of ν, R, C and graph θ -vs ν. (20 pnts)

From the phasor diagram above (question 3) the angle between vectors Vin and Vout 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


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