TF EIMLab6 Writeup - Revision history

Foretony: /* Questions */

2011-03-27T17:38:39Z

Questions

2011-03-27T17:35:43Z

Pulse Sharpener

2011-03-27T17:34:29Z

Pulse Sharpener

2011-03-27T17:03:50Z

Questions

2011-03-04T04:03:04Z

Questions

2011-02-11T20:49:42Z

Pulse Sharpener

2011-02-11T20:33:01Z

Pulse Sharpener

2011-02-11T20:32:37Z

Pulse Sharpener

2011-02-11T20:27:16Z

Pulse Sharpener

2011-02-11T20:23:03Z

Pulse Sharpener

← Older revision		Revision as of 17:38, 27 March 2011
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	:<math>\tau^{\prime} = \left ( \frac{R_1 R}{R_1+ R}\right ) C_1 = \left ( \frac{R_1 }{R_1+ R}\right )\tau</math>		:<math>\tau^{\prime} = \left ( \frac{R_1 R}{R_1+ R}\right ) C_1 = \left ( \frac{R_1 }{R_1+ R}\right )\tau</math>
		+
		+	The circuit is a filter that passes high frequency changes in V but attenating low frequency. Or in other words the power gets pushed intot he high frequncy band.

	2.) How is <math>V_{out}</math> worse than <math>V_{in}</math>( 10 pnts.)		2.) How is <math>V_{out}</math> worse than <math>V_{in}</math>( 10 pnts.)

← Older revision		Revision as of 17:35, 27 March 2011
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	2.) How is <math>V_{out}</math> worse than <math>V_{in}</math>( 10 pnts.)		2.) How is <math>V_{out}</math> worse than <math>V_{in}</math>( 10 pnts.)
		+
		+	:<math>V_{out} = V_0^{\prime} \left ( 1 - e^{-t/\tau^{\prime}}\right ) = \left ( \frac{R_1 }{R_1+ R}\right ) V_0\left ( 1 - e^{-t/\tau^{\prime}}\right )</math>


	[[Forest_Electronic_Instrumentation_and_Measurement]]		[[Forest_Electronic_Instrumentation_and_Measurement]]

← Older revision		Revision as of 17:34, 27 March 2011
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	1.) Qualitativly, why is <math>\tau^{\prime} < \tau</math>?( 10 pnts.)		1.) Qualitativly, why is <math>\tau^{\prime} < \tau</math>?( 10 pnts.)
		+
		+	:<math>\tau^{\prime} = \left ( \frac{R_1 R}{R_1+ R}\right ) C_1 = \left ( \frac{R_1 }{R_1+ R}\right )\tau</math>

	2.) How is <math>V_{out}</math> worse than <math>V_{in}</math>( 10 pnts.)		2.) How is <math>V_{out}</math> worse than <math>V_{in}</math>( 10 pnts.)

@@ Line 133: / Line 133: @@
 ==Questions==
-.) Qualitatify, why is <math>\tau^{\prime} < \tau</math>?( 10 pnts.)
+.) Qualitativly, why is <math>\tau^{\prime} < \tau</math>?( 10 pnts.)
 .) How is <math>V_{out}</math> worse than <math>V_{in}</math>( 10 pnts.)

@@ Line 59: / Line 59: @@
 .) What happens if the amplitude of <math>V_{in}</math> is doubled.
 .) What happens if R is doubled and C is halved?
 =Integrator=

← Older revision		Revision as of 20:49, 11 February 2011
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	:<math>R = \tau/C = 60 \times 10^{-6} s/ 1.210 \times 10^{-6} F =49.6 \Omega</math>		:<math>R = \tau/C = 60 \times 10^{-6} s/ 1.210 \times 10^{-6} F =49.6 \Omega</math>
		+
		+	A second measurement:
		+
		+	:<math>\tau = 70 \times 10^{-6}</math>
		+
		+	:<math>R = \tau/C = 70 \times 10^{-6} s/ 1.210 \times 10^{-6} F =57.9 \Omega</math>
		+
		+	I would say the input impedance of the BK precision function generator is about 50 <math>\Omega</math>.

	2.) The output should be given by		2.) The output should be given by

@@ Line 103: / Line 103: @@
 I measure the times constant by looking how long it take the 2 Volt pulse to rise to (1-e^{-1}) =1.26 V
-:\tau = 60 \times 10^{-6}
+:<math>\tau = 60 \times 10^{-6}</math>
 :<math>R = \tau/C = 60 \times 10^{-6} s/  1.210 \times 10^{-6} F =49.6 \Omega</math>

@@ Line 105: / Line 105: @@
 :\tau = 60 \times 10^{-6}
-:<math>R = \taus/C = 60 \times 10^{-6} s/  1.210 \times 10^{-6} F =49.6 \Omega</math>
+:<math>R = \tau/C = 60 \times 10^{-6} s/  1.210 \times 10^{-6} F =49.6 \Omega</math>
 .) The output should be given by

← Older revision		Revision as of 20:27, 11 February 2011
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	I measure the times constant by looking how long it take the 2 Volt pulse to rise to (1-e^{-1}) =1.26 V		I measure the times constant by looking how long it take the 2 Volt pulse to rise to (1-e^{-1}) =1.26 V
		+
		+	:\tau = 60 \times 10^{-6}
		+
		+	:<math>R = \taus/C = 60 \times 10^{-6} s/ 1.210 \times 10^{-6} F =49.6 \Omega</math>

	2.) The output should be given by		2.) The output should be given by

@@ Line 95: / Line 95: @@
 :<math>R_{out}</math> = impedance of the function generator at output which produces V_{in}
 : <math>C_{out}</math> = capacitor shorting the function generator output to ground (not shown in the above picture)
 .) The output should be given by