Difference between revisions of "Lab 3 RS"

From New IAC Wiki
Jump to navigation Jump to search
 
(One intermediate revision by the same user not shown)
Line 134: Line 134:
 
  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.
  
  For a very0high frequency input the phase shift is -90 degree
+
  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)==
 
==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)==

Latest revision as of 05:23, 28 January 2011

Go Back to All Lab Reports


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

Go Back to All Lab Reports