Lab 3 RS - Revision history

Shaproma: /* 5. What is the phase shift \theta for a DC input and a very-high frequency input?(5 pnts) */

2011-01-28T05:23:04Z

5. What is the phase shift \theta for a DC input and a very-high frequency input?(5 pnts)

Shaproma: /* 5. What is the phase shift \theta for a DC input and a very-high frequency input?(5 pnts) */

2011-01-28T05:20:22Z

5. What is the phase shift \theta for a DC input and a very-high frequency input?(5 pnts)

Shaproma: /* 5. What is the phase shift \theta for a DC input and a very-high frequency input?(5 pnts) */

2011-01-28T05:20:09Z

5. What is the phase shift \theta for a DC input and a very-high frequency input?(5 pnts)

Shaproma: /* 6. Calculate and expression for the phase shift \theta as a function of \nu, R, C and graph \theta -vs \nu. (20 pnts) */

2011-01-28T00:09:51Z

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

Shaproma: /* 6. Calculate and expression for the phase shift \theta as a function of \nu, R, C and graph \theta -vs \nu. (20 pnts) */

2011-01-27T23:57:12Z

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

Shaproma: /* method 1. Using fitting line */

2011-01-27T07:11:02Z

method 1. Using fitting line

Shaproma: /* method 1. Using fitting line */

2011-01-27T07:10:24Z

method 1. Using fitting line

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

2011-01-27T07:09:30Z

1-50 kHz filter (20 pnts)

Shaproma at 07:01, 27 January 2011

2011-01-27T07:01:33Z

Shaproma: /* Compare the theoretical and experimentally measured break frequencies. (5 pnts) */

2011-01-27T07:00:35Z

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

@@ Line 134: / Line 134: @@
   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
+  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)==

@@ Line 134: / Line 134: @@
   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
 ==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)==

← Older revision		Revision as of 05:20, 28 January 2011
Line 133:		Line 133:

	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

	==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)==

← Older revision		Revision as of 00:09, 28 January 2011
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−	[[File:RS phase shift m1.png \| ~~500~~ px]]	+	[[File:RS phase shift m1.png \| 600 px]]

← Older revision		Revision as of 23:57, 27 January 2011
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−	[[File:RS phase shift m1.png \| ~~600~~ px]]	+	[[File:RS phase shift m1.png \| 500 px]]

← Older revision		Revision as of 07:11, 27 January 2011
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	Theoretical break frequency: <math>12.13\ \mbox{kHz}</math>		Theoretical break frequency: <math>12.13\ \mbox{kHz}</math>
−
−	~~Experimentally measured break frequency: <math>9.59\ \mbox{kHz}</math>~~

	The fit line equation from the plot above is <math>\ y=1.071-1.005\cdot x</math>.		The fit line equation from the plot above is <math>\ y=1.071-1.005\cdot x</math>.

@@ Line 81: / Line 81: @@
 ===method 1. Using fitting line===
-:Theoretical break frequency: <math>12.13\ \mbox{kHz}</math>
+Theoretical break frequency: <math>12.13\ \mbox{kHz}</math>
-:Experimentally measured break frequency: <math>9.59\ \mbox{kHz}</math>
+Experimentally measured break frequency: <math>9.59\ \mbox{kHz}</math>
-  Q: The above was read off the graph?  Why not use fit results?
+The fit line equation from the plot above is <math>\ y=1.071-1.005\cdot x</math>.
-  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.
+From intersection point of line with x-axis we find:
 :<math>\mbox{log}(f_{exper})=\frac{1.071}{1.005} = 1.066</math>
@@ Line 97: / Line 93: @@
-:The error is:
+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>

← Older revision		Revision as of 07:09, 27 January 2011
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−	[[File:RS lab3 voltage gain m1.png \| 800 px]]	+	[[File:RS lab3 voltage gain m2.png \| 800 px]]

	=phase shift (10 pnts)=		=phase shift (10 pnts)=

@@ Line 105: / Line 105: @@
 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:
@@ Line 123: / Line 123: @@
 :<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]]
-==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>
@@ Line 136: / Line 136: @@
 [[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.
-==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

@@ Line 77: / Line 77: @@
 =Questions=
-==Compare the theoretical and experimentally measured break frequencies. (5 pnts)==
+==1. Compare the theoretical and experimentally measured break frequencies. (5 pnts)==
 ===method 1. Using fitting line===
@@ Line 104: / Line 104: @@
 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)==