TF EIMLab3 Writeup

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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).

[math]\displaystyle{ \omega_{break} = \frac{1}{RC} }[/math]
[math]\displaystyle{ \Rightarrow R = \frac{1}{\omega_{break} C } = \frac{1}{25 \times 10^{3} \times 9.45 \times 10^{-9}} = 4,233 \Omega }[/math]
R C [math]\displaystyle{ \omega_B }[/math] [math]\displaystyle{ \nu_B }[/math]
Ohms Farads rad/s Hz
[math]\displaystyle{ 1 \times 10^{5} }[/math] [math]\displaystyle{ 561 \times 10^{-12} }[/math] 17825 2837
[math]\displaystyle{ 96.4 \times 10^{3} }[/math] [math]\displaystyle{ 561 \times 10^{-12} }[/math] 18490 2943
[math]\displaystyle{ 10.5 }[/math] [math]\displaystyle{ 1.25 \times 10^{-6} }[/math] 76190 12126
[math]\displaystyle{ 31.3 }[/math] [math]\displaystyle{ 10.3 \times 10^{-6} }[/math] 3102 494
[math]\displaystyle{ 2.058 \times 10^5 }[/math] [math]\displaystyle{ 7.73 \times 10^{-10} }[/math] 6310 1004


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 }[/math] [math]\displaystyle{ V_{in} }[/math] [math]\displaystyle{ V_{out} }[/math] [math]\displaystyle{ \frac{V_{out}}{V_{in}} }[/math]
Hz Volts Volts
50 0.6 0.3
100 0.5 0.18
250 0.5 0.075
500 0.45 0.04
1000 0.4 0.017
2500 0.28 0.005
5056 0.16 0.005
  1. Graph the [math]\displaystyle{ \log \left(\frac{V_{out}}{V_{in}} \right) }[/math] -vs- [math]\displaystyle{ \log (\nu) }[/math]


phase shift (10 pnts)

  1. measure the phase shift between [math]\displaystyle{ V_{in} }[/math] and [math]\displaystyle{ V_{out} }[/math]

Questions

1.)compare the theoretical and experimentally measured break frequencies. (5 pnts)

[math]\displaystyle{ \omega_{break} = \frac{1}{RC} = \frac{1}{400 \times 10^{3} \times 9.45 \times 10^{-9}} = = 2.6 \times 10^{2} }[/math]

Theory Exp %diff

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)


[math]\displaystyle{ V - IR -X_CI = =V -(X_R +X_C) I = 0 }[/math]

[math]\displaystyle{ \Rightarrow }[/math] the capacitor can be added in series with the other resistor
[math]\displaystyle{ X_{tot} = X_R + X_C }[/math]

It looks like the voltage divider from the resistance section

[math]\displaystyle{ V_{out}= Re\left [\frac{X_C}{R+X_C} V_{in} \right ]= Re \left [\frac{X_C}{X_R+X_C} V_{in} \right ] }[/math]

To evaluate [math]\displaystyle{ Re[Z] = \sqrt{Z Z^*} }[/math] where[math]\displaystyle{ Z = x+iy }[/math] and [math]\displaystyle{ Z^* = x-iy }[/math]


[math]\displaystyle{ \frac{V_{out}}{V_{in}} = Re \left [ \frac{\frac{1}{i \omega C}}{R+\frac{1}{i \omega C}} \right ] = \sqrt{\frac{\frac{1}{i \omega C}}{R+\frac{1}{i \omega C}} \frac{\frac{1}{-i \omega C}}{R+\frac{1}{-i \omega C}} } = \frac{1}{\sqrt{1 + \omega^2 R^2 C^2}} }[/math]

Let

[math]\displaystyle{ \omega_b = \frac{1}{RC} = }[/math] break point (cut off ) frequency

then

[math]\displaystyle{ \frac{V_{out}}{V_{in}} = \frac{1}{\sqrt{1 + \left ( \frac{\omega}{\omega_b}\right )^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) 5.) what is the phase shift [math]\displaystyle{ \theta }[/math] for a DC input and a very-high frequency input?(5 pnts) 6.) calculate and expression for the phase shift [math]\displaystyle{ \theta }[/math] as a function of [math]\displaystyle{ \nu }[/math], [math]\displaystyle{ R }[/math], [math]\displaystyle{ C }[/math] and graph [math]\displaystyle{ \theta }[/math] -vs [math]\displaystyle{ \nu }[/math]. (20 pnts)


Forest_Electronic_Instrumentation_and_Measurement