TF EIMLab3 Writeup - Revision history

Foretony: /* 1-50 kHz filter (20 pnts) */

2011-02-10T19:49:20Z

1-50 kHz filter (20 pnts)

2011-02-10T19:47:01Z

1-50 kHz filter (20 pnts)

2011-02-10T03:59:04Z

1-50 kHz filter (20 pnts)

2011-02-10T03:25:29Z

1-50 kHz filter (20 pnts)

2011-02-10T02:30:19Z

1-50 kHz filter (20 pnts)

2011-02-10T02:25:25Z

1-50 kHz filter (20 pnts)

2011-01-22T02:33:30Z

Questions

2011-01-22T02:33:06Z

Questions

2011-01-21T06:11:33Z

Questions

2011-01-21T06:06:50Z

1-50 kHz filter (20 pnts)

@@ Line 20: / Line 20: @@
 |  <math>31.3 </math>|| <math>10.3 \times 10^{-6}</math>|| 3102||494
 |-
-|  <math>2.0 \times 10^5 </math>|| <math>7.73 \times 10^{-10}</math>|| 6468||1029
+|  <math>2.058 \times 10^5 </math>|| <math>7.73 \times 10^{-10}</math>|| 6310||1004
 |}

@@ Line 19: / Line 19: @@
 |-
 |  <math>31.3 </math>|| <math>10.3 \times 10^{-6}</math>|| 3102||494
 |}

@@ Line 17: / Line 17: @@
 |-
 |  <math>10.5 </math>|| <math>1.25 \times 10^{-6}</math>|| 76190||12126
 |}

@@ Line 16: / Line 16: @@
 |  <math>96.4 \times 10^{3}</math>|| <math>561 \times 10^{-12}</math>|| 18490||2943
 |-
-|  || || ||
+|  <math>10.5 </math>|| <math>1.25 \times 10^{-6}</math>|| 76190||12126
 |}

@@ Line 13: / Line 13: @@
 |-
 |  <math>1 \times 10^{5}</math>|| <math>561 \times 10^{-12}</math>|| 17825||2837
 |-
 |  || || ||

@@ Line 6: / Line 6: @@
 :<math>\omega_{break} = \frac{1}{RC}</math>
 : <math>\Rightarrow R = \frac{1}{\omega_{break} C } = \frac{1}{25 \times 10^{3} \times 9.45 \times 10^{-9}} =  4,233 \Omega</math>
 .)Now construct the circuit using a non-polar capacitor.

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-|Theory|| Exp || %diff||
+|Theory|| Exp || %diff
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+|  || ||
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 .)compare the theoretical and experimentally measured break frequencies. (5 pnts)
-\omega_{break} = \frac{1}{RC} = \frac{1}{400 \times 10^{3} \times 9.45 \times 10^{-9}} = = 2.6 \times 10^{2}
+<math>\omega_{break} = \frac{1}{RC} = \frac{1}{400 \times 10^{3} \times 9.45 \times 10^{-9}} = = 2.6 \times 10^{2}</math>
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 |}
-#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)
+.) 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)
-#Compare the theoretical and experimental value for the phase shift <math>\theta</math>. (5 pnts)
-#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)
-# what is the phase shift <math>\theta</math> for a DC input and a very-high frequency input?(5 pnts)
+<math>V - IR -X_CI = =V -(X_R +X_C) I = 0</math>
-# 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)
 [[Forest_Electronic_Instrumentation_and_Measurement]]

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 = 1-50 kHz filter (20 pnts)=
-#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).
+.)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).
-#Now construct the circuit using a non-polar capacitor.
-#use a sinusoidal variable frequency oscillator to provide an input voltage to your filter.
+:<math>\omega_{break} = \frac{1}{RC}</math>
-#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.
+: <math>\Rightarrow R = \frac{1}{\omega_{break} C } = \frac{1}{25 \times 10^{3} \times 9.45 \times 10^{-9}} =  4,233 \Omega</math>
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