Class Wiki - User contributions [en]

Key Facts from Reading Chapter 1

2010-03-01T03:24:47Z

Michaelvier:

The following facts are not profound and are possibly very obvious. Nonetheless, they might help cement certain concepts. Please add things you think would be helpful.
==Transistors==
[[Image:mosfet.jpg|thumb|right|400px|MOSFET diagram <ref> [http://www.msm.cam.ac.uk/doitpoms/tlplib/semiconductors/images/mosfet.jpg University of Cambridge] </ref>]]
*Conduction in n-type material is from free electrons.
*Conduction in p-type material is from holes (positive particles).
*The function of metal-oxide-semiconductor field-effect transistors (MOSFETs) depends on the voltage applied to the gate.
**Certain ranges of voltage allow no current to flow between the drain and the source. In this way, the MOSFET acts like an open switch.
**Another particular range of voltage allows current to easily flow from the source to the drain.
**When the voltage is in between the ranges of open and closed switch, the MOSFET can smoothly control the amount of current flowing.
*Bipolar Junction Transistors (BJTs) can act as either switches or current controls as well.
==Amplifiers==
*An inverting amplifier has a negative voltage gain, <math>\ {A_{v}}</math>.
*A noninverting amplifier has a positive voltage gain, <math>\ {A_{v}}</math>. (If you get this, you deserve a cookie)
*The power gain,<math>\ G</math>, is the ration of the output power to the input power
**<math>G=\frac{P_{o}}{P_{i}}=\frac{V_o{I_o}}{V_i{I_i}}=A_v{A_{i}}=(A_v)^2\frac{R_{i}}{R_{L}}</math>
***<math>\ R_i</math> is the amplifier's input resistance and <math>\ R_o</math> is the amplifier's output resistance.

==Decibel Conversion==
*Power gain, <math>\ G</math>, can be converted to decibels:

<math>G=(A_v)^2\frac{R_{i}}{R_{L}}</math>

<math>\ G_{db}=10 log{A_{v}}^2+10 log R_i - 10 log R_L</math>

<math>\ G_{db}=10 log |A_v|+10 log R_i - 10 log R_L</math>
*By the equation above, we can say voltage gain in decibels can be found with this equation: <math>\ A_{v dB}=20 log |A_v|</math>, where <math>\ A_v</math> is the voltage gain and <math>\ A_{v dB}</math> is the voltage gain is decibels.
*Similarly, we can say current gain in decibels can be found with this equation: <math>\ A_{i dB}=20 log |A_i|</math>, where <math>\ A_i</math> is the current gain and <math>\ A_{i dB}</math> is the current gain in decibels.

==Differential Amplifiers==
*Differential Amplifiers have two inputs. The terminal marked with a "+" is the noninverting input, and the terminal marked with a "-" is the inverting input.
*The differential input signal is given by <math>\ v_{id}=v_{i1}-v_{i2} </math>
*The output of an ideal differential amplifier is given by <math>\ v_o=A_d v_{id} </math>.
*The common-mode input signal is given by the following: <math>v_{icm}=\frac{1}{2}(v_{i1}+v_{i2})</math>.
*A real differential amplifier's output is given by <math>\ v_o=A_d v_{id}+A_{cm} v_{icm} </math>, where <math>\ A_{cm}</math> is the common-mode signal gain.

==References==
<references/>

==Contributors==
*[[Lau, Chris | Christopher Garrison Lau I]]

==Readers==
*[[Shepherd,Victor]]

==Reviewers==

*[[Ben Henry|Henry,Ben]]

*[[Michael Vier|Vier, Michael]]

*The sentence that says "voltage gain and is the voltage gain is decibels." I think you mean "in decibels." Other than that nice work!

Key Facts from Reading Chapter 1

2010-03-01T03:24:32Z

Michaelvier:

The following facts are not profound and are possibly very obvious. Nonetheless, they might help cement certain concepts. Please add things you think would be helpful.
==Transistors==
[[Image:mosfet.jpg|thumb|right|400px|MOSFET diagram <ref> [http://www.msm.cam.ac.uk/doitpoms/tlplib/semiconductors/images/mosfet.jpg University of Cambridge] </ref>]]
*Conduction in n-type material is from free electrons.
*Conduction in p-type material is from holes (positive particles).
*The function of metal-oxide-semiconductor field-effect transistors (MOSFETs) depends on the voltage applied to the gate.
**Certain ranges of voltage allow no current to flow between the drain and the source. In this way, the MOSFET acts like an open switch.
**Another particular range of voltage allows current to easily flow from the source to the drain.
**When the voltage is in between the ranges of open and closed switch, the MOSFET can smoothly control the amount of current flowing.
*Bipolar Junction Transistors (BJTs) can act as either switches or current controls as well.
==Amplifiers==
*An inverting amplifier has a negative voltage gain, <math>\ {A_{v}}</math>.
*A noninverting amplifier has a positive voltage gain, <math>\ {A_{v}}</math>. (If you get this, you deserve a cookie)
*The power gain,<math>\ G</math>, is the ration of the output power to the input power
**<math>G=\frac{P_{o}}{P_{i}}=\frac{V_o{I_o}}{V_i{I_i}}=A_v{A_{i}}=(A_v)^2\frac{R_{i}}{R_{L}}</math>
***<math>\ R_i</math> is the amplifier's input resistance and <math>\ R_o</math> is the amplifier's output resistance.

==Decibel Conversion==
*Power gain, <math>\ G</math>, can be converted to decibels:

<math>G=(A_v)^2\frac{R_{i}}{R_{L}}</math>

<math>\ G_{db}=10 log{A_{v}}^2+10 log R_i - 10 log R_L</math>

<math>\ G_{db}=10 log |A_v|+10 log R_i - 10 log R_L</math>
*By the equation above, we can say voltage gain in decibels can be found with this equation: <math>\ A_{v dB}=20 log |A_v|</math>, where <math>\ A_v</math> is the voltage gain and <math>\ A_{v dB}</math> is the voltage gain is decibels.
*Similarly, we can say current gain in decibels can be found with this equation: <math>\ A_{i dB}=20 log |A_i|</math>, where <math>\ A_i</math> is the current gain and <math>\ A_{i dB}</math> is the current gain in decibels.

==Differential Amplifiers==
*Differential Amplifiers have two inputs. The terminal marked with a "+" is the noninverting input, and the terminal marked with a "-" is the inverting input.
*The differential input signal is given by <math>\ v_{id}=v_{i1}-v_{i2} </math>
*The output of an ideal differential amplifier is given by <math>\ v_o=A_d v_{id} </math>.
*The common-mode input signal is given by the following: <math>v_{icm}=\frac{1}{2}(v_{i1}+v_{i2})</math>.
*A real differential amplifier's output is given by <math>\ v_o=A_d v_{id}+A_{cm} v_{icm} </math>, where <math>\ A_{cm}</math> is the common-mode signal gain.

==References==
<references/>

==Contributors==
[[Lau, Chris | Christopher Garrison Lau I]]

==Readers==
[[Shepherd,Victor]]

==Reviewers==

[[Ben Henry|Henry,Ben]]

*[[Michael Vier|Vier, Michael]]

*The sentence that says "voltage gain and is the voltage gain is decibels." I think you mean "in decibels." Other than that nice work!

Key Facts from Reading Chapter 1

2010-03-01T03:23:57Z

Michaelvier:

The following facts are not profound and are possibly very obvious. Nonetheless, they might help cement certain concepts. Please add things you think would be helpful.
==Transistors==
[[Image:mosfet.jpg|thumb|right|400px|MOSFET diagram <ref> [http://www.msm.cam.ac.uk/doitpoms/tlplib/semiconductors/images/mosfet.jpg University of Cambridge] </ref>]]
*Conduction in n-type material is from free electrons.
*Conduction in p-type material is from holes (positive particles).
*The function of metal-oxide-semiconductor field-effect transistors (MOSFETs) depends on the voltage applied to the gate.
**Certain ranges of voltage allow no current to flow between the drain and the source. In this way, the MOSFET acts like an open switch.
**Another particular range of voltage allows current to easily flow from the source to the drain.
**When the voltage is in between the ranges of open and closed switch, the MOSFET can smoothly control the amount of current flowing.
*Bipolar Junction Transistors (BJTs) can act as either switches or current controls as well.
==Amplifiers==
*An inverting amplifier has a negative voltage gain, <math>\ {A_{v}}</math>.
*A noninverting amplifier has a positive voltage gain, <math>\ {A_{v}}</math>. (If you get this, you deserve a cookie)
*The power gain,<math>\ G</math>, is the ration of the output power to the input power
**<math>G=\frac{P_{o}}{P_{i}}=\frac{V_o{I_o}}{V_i{I_i}}=A_v{A_{i}}=(A_v)^2\frac{R_{i}}{R_{L}}</math>
***<math>\ R_i</math> is the amplifier's input resistance and <math>\ R_o</math> is the amplifier's output resistance.

==Decibel Conversion==
*Power gain, <math>\ G</math>, can be converted to decibels:

<math>G=(A_v)^2\frac{R_{i}}{R_{L}}</math>

<math>\ G_{db}=10 log{A_{v}}^2+10 log R_i - 10 log R_L</math>

<math>\ G_{db}=10 log |A_v|+10 log R_i - 10 log R_L</math>
*By the equation above, we can say voltage gain in decibels can be found with this equation: <math>\ A_{v dB}=20 log |A_v|</math>, where <math>\ A_v</math> is the voltage gain and <math>\ A_{v dB}</math> is the voltage gain is decibels.
*Similarly, we can say current gain in decibels can be found with this equation: <math>\ A_{i dB}=20 log |A_i|</math>, where <math>\ A_i</math> is the current gain and <math>\ A_{i dB}</math> is the current gain in decibels.

==Differential Amplifiers==
*Differential Amplifiers have two inputs. The terminal marked with a "+" is the noninverting input, and the terminal marked with a "-" is the inverting input.
*The differential input signal is given by <math>\ v_{id}=v_{i1}-v_{i2} </math>
*The output of an ideal differential amplifier is given by <math>\ v_o=A_d v_{id} </math>.
*The common-mode input signal is given by the following: <math>v_{icm}=\frac{1}{2}(v_{i1}+v_{i2})</math>.
*A real differential amplifier's output is given by <math>\ v_o=A_d v_{id}+A_{cm} v_{icm} </math>, where <math>\ A_{cm}</math> is the common-mode signal gain.

==References==
<references/>

==Contributors==
[[Lau, Chris | Christopher Garrison Lau I]]

==Readers==
[[Shepherd,Victor]]

==Reviewers==

[[Ben Henry|Henry,Ben]]

[[Ben Henry|Henry,Ben]]

*The sentence that says "voltage gain and is the voltage gain is decibels." I think you mean "in decibels." Other than that nice work!

Laplace Transform of a Triangle Wave

2010-01-25T21:00:23Z

Michaelvier:

[[Image:triangle wave.jpg|500px|thumb|right|Triangle wave with period T=2 and amplitude A=2]]
'''This page is still in progress'''
==Introduction==

This article explains how to transform a periodic function (in this case a triangle wave). This is especially useful for analyzing circuits which contain triangle wave voltage sources.

==Define F(t)==

<math>m1=\frac{2+2}{.5+.5}=4</math>

<math>m2=\frac{-2-2}{1.5-.5}=-4</math>
So,

[[Image:definitionofF.jpg|140px|left]]

Using the theorem for the transform of a periodic function,

<math>L\left\{ F\left( t \right) \right\}=\frac{1}{1-e^{-2s}}\left[ \int_{-.5}^{.5}{4te^{-st}dt}+\int_{.5}^{1.5}{\left( -4t+4.5 \right)e^{-st}dt} \right]</math>

<math>L\left\{ F\left( t \right) \right\}=\frac{1}{1-e^{-2s}}\left[ \int_{-.5}^{.5}{4te^{-st}dt}+\int_{.5}^{1.5}{-4te^{-st}dt}+\int_{.5}^{1.5}{4.5e^{-st}dt} \right]</math>

<math>\int_{-.5}^{.5}{4te^{-st}}=\; \frac{4e^{.5s}-2se^{.5s}-2se^{-.5s}-4e^{-.5s}}{s^{2}}</math>

<math>\int_{.5}^{1.5}{-4te^{-st}}=\frac{6se^{-1.5s}+4e^{-1.5s}-2se^{-.5s}-4e^{-.5s}}{s^{2}}</math>

<math>\int_{.5}^{1.5}{4.5e^{-st}}=\frac{4.5se^{-.5s}-4.5se^{-1.5s}}{s^{2}}</math>

<math>F\left( s \right)=\frac{1}{1-e^{-2s}}\left( \frac{-8e^{-.5s}+4e^{.5s}+4e^{-1.5s}}{s^{2}}+\frac{.5e^{-.5s}-2e^{.5s}-1.5e^{-1.5s}}{s} \right)</math>
==Author==

*[[Vier, Michael|Michael Vier]]
==Reviewers==
==Readers==

Laplace Transform of a Triangle Wave

2010-01-25T20:59:22Z

Michaelvier:

[[Image:triangle wave.jpg|500px|thumb|right|Triangle wave with period T=2 and amplitude A=2]]
'''This page is still in progress'''
==Introduction==

This article explains how to transform a periodic function (in this case a triangle wave). This is especially useful for analyzing circuits which contain triangle wave voltage sources.

==Define F(t)==

<math>m1=\frac{2+2}{.5+.5}=4</math>

<math>m2=\frac{-2-2}{1.5-.5}=-4</math>
So,

[[Image:definitionofF.jpg|140px|left]]

Using the theorem for the transform of a periodic function,

<math>L\left\{ F\left( t \right) \right\}=\frac{1}{1-e^{-2s}}\left[ \int_{-.5}^{.5}{4te^{-st}dt}+\int_{.5}^{1.5}{\left( -4t+4.5 \right)e^{-st}dt} \right]</math>

<math>L\left\{ F\left( t \right) \right\}=\frac{1}{1-e^{-2s}}\left[ \int_{-.5}^{.5}{4te^{-st}dt}+\int_{.5}^{1.5}{-4te^{-st}dt}+\int_{.5}^{1.5}{4.5e^{-st}dt} \right]</math>

<math>\int_{-.5}^{.5}{4te^{-st}}=\; \frac{4e^{.5s}-2se^{.5s}-2se^{-.5s}-4e^{-.5s}}{s^{2}}</math>

<math>\int_{.5}^{1.5}{-4te^{-st}}=\frac{6se^{-1.5s}+4e^{-1.5s}-2se^{-.5s}-4e^{-.5s}}{s^{2}}</math>

<math>\int_{.5}^{1.5}{4.5e^{-st}}=\frac{4.5se^{-.5s}-4.5se^{-1.5s}}{s^{2}}</math>

<math>F\left( s \right)=\frac{-8e^{-.5s}+4e^{.5s}+4e^{-1.5s}}{s^{2}}+\frac{.5e^{-.5s}-2e^{.5s}-1.5e^{-1.5s}}{s}</math>
==Author==

*[[Vier, Michael|Michael Vier]]
==Reviewers==
==Readers==

Laplace Transform of a Triangle Wave

2010-01-25T20:05:02Z

Michaelvier:

[[Image:triangle wave.jpg|500px|thumb|right|Triangle wave with period T=2 and amplitude A=2]]
'''This page is still in progress'''
==Introduction==

This article explains how to transform a periodic function (in this case a triangle wave). This is especially useful for analyzing circuits which contain triangle wave voltage sources.

==Define F(t)==

<math>m1=\frac{2+2}{.5+.5}=4</math>

<math>m2=\frac{-2-2}{1.5-.5}=-4</math>
So,

[[Image:definitionofF.jpg|140px|left]]

Using the theorem for the transform of a periodic function,

<math>L\left\{ F\left( t \right) \right\}=\frac{1}{1-e^{-2s}}\left[ \int_{-.5}^{.5}{4te^{-st}dt}+\int_{.5}^{1.5}{\left( -4t+4.5 \right)e^{-st}dt} \right]</math>

<math>L\left\{ F\left( t \right) \right\}=\frac{1}{1-e^{-2s}}\left[ \int_{-.5}^{.5}{4te^{-st}dt}+\int_{.5}^{1.5}{-4te^{-st}dt}+\int_{.5}^{1.5}{4.5e^{-st}dt} \right]</math>

<math>\int_{-.5}^{.5}{4te^{-st}}=\; \frac{4e^{.5s}-2se^{.5s}-2se^{-.5s}-4e^{-.5s}}{s^{2}}</math>

<math>\int_{.5}^{1.5}{-4te^{-st}}=\frac{6se^{-1.5s}+4e^{-1.5s}-2se^{-.5s}-4e^{-.5s}}{s^{2}}</math>

<math>\int_{.5}^{1.5}{4.5e^{-st}}=\frac{4.5se^{-.5s}-4.5se^{-1.5s}}{s^{2}}</math>
==Author==

*[[Vier, Michael|Michael Vier]]
==Reviewers==
==Readers==

Laplace Transform of a Triangle Wave

2010-01-25T20:04:36Z

Michaelvier:

[[Image:triangle wave.jpg|500px|thumb|right|Triangle wave with period T=2 and amplitude A=2]]
'''This page is still in progress'''
==Introduction==

This article explains how to transform a periodic function (in this case a triangle wave). This is especially useful for analyzing circuits which contain triangle wave voltage sources.

==Define F(t)==

<math>m1=\frac{2+2}{.5+.5}=4</math>

<math>m2=\frac{-2-2}{1.5-.5}=-4</math>
So,

[[Image:definitionofF.jpg|140px|left]]

Using the theorem for the transform of a periodic function,

<math>L\left\{ F\left( t \right) \right\}=\frac{1}{1-e^{-2s}}\left[ \int_{-.5}^{.5}{4te^{-st}dt}+\int_{.5}^{1.5}{\left( -4t+4.5 \right)e^{-st}dt} \right]</math>

<math>L\left\{ F\left( t \right) \right\}=\frac{1}{1-e^{-2s}}\left[ \int_{-.5}^{.5}{4te^{-st}dt}+\int_{.5}^{1.5}{-4te^{-st}dt}+\int_{.5}^{1.5}{4.5e^{-st}dt} \right]</math>

<math>\int_{-.5}^{.5}{4te^{-st}}=\; \frac{4e^{.5s}-2se^{.5s}-2se^{-.5s}-4e^{-.5s}}{s^{2}}</math>

<math>\int_{.5}^{1.5}{-4te^{-st}}=\frac{6se^{-1.5s}+4e^{-1.5s}-2se^{-.5s}-4e^{-.5s}}{s^{2}}</math>

<math>\int_{.5}^{1.5}{4.5e^{-st}}=\frac{4.5se^{-.5s}-4.5se^{-1.5s}}{s^{2}}}</math>
==Author==

*[[Vier, Michael|Michael Vier]]
==Reviewers==
==Readers==

Laplace Transform of a Triangle Wave

2010-01-25T20:04:03Z

Michaelvier:

[[Image:triangle wave.jpg|500px|thumb|right|Triangle wave with period T=2 and amplitude A=2]]
'''This page is still in progress'''
==Introduction==

This article explains how to transform a periodic function (in this case a triangle wave). This is especially useful for analyzing circuits which contain triangle wave voltage sources.

==Define F(t)==

<math>m1=\frac{2+2}{.5+.5}=4</math>

<math>m2=\frac{-2-2}{1.5-.5}=-4</math>
So,

[[Image:definitionofF.jpg|140px|left]]

Using the theorem for the transform of a periodic function,

<math>L\left\{ F\left( t \right) \right\}=\frac{1}{1-e^{-2s}}\left[ \int_{-.5}^{.5}{4te^{-st}dt}+\int_{.5}^{1.5}{\left( -4t+4.5 \right)e^{-st}dt} \right]</math>

<math>L\left\{ F\left( t \right) \right\}=\frac{1}{1-e^{-2s}}\left[ \int_{-.5}^{.5}{4te^{-st}dt}+\int_{.5}^{1.5}{-4te^{-st}dt}+\int_{.5}^{1.5}{4.5e^{-st}dt} \right]</math>

<math>\int_{-.5}^{.5}{4te^{-st}}=\; \frac{4e^{.5s}-2se^{.5s}-2se^{-.5s}-4e^{-.5s}}{s^{2}}</math>

<math>\int_{.5}^{1.5}{-4te^{-st}}=\frac{6se^{-1.5s}+4e^{-1.5s}-2se^{-.5s}-4e^{-.5s}}{s^{2}}</math>

<math>\int_{.5}^{1.5}{4.5e^{-st}}=\frac{4.5se^{-.5s}-se^{-1.5s}}{s^{2}}</math>
==Author==

*[[Vier, Michael|Michael Vier]]
==Reviewers==
==Readers==

Laplace Transform of a Triangle Wave

2010-01-25T20:00:28Z

Michaelvier:

[[Image:triangle wave.jpg|500px|thumb|right|Triangle wave with period T=2 and amplitude A=2]]
'''This page is still in progress'''
==Introduction==

This article explains how to transform a periodic function (in this case a triangle wave). This is especially useful for analyzing circuits which contain triangle wave voltage sources.

==Define F(t)==

<math>m1=\frac{2+2}{.5+.5}=4</math>

<math>m2=\frac{-2-2}{1.5-.5}=-4</math>
So,

[[Image:definitionofF.jpg|140px|left]]

Using the theorem for the transform of a periodic function,

<math>L\left\{ F\left( t \right) \right\}=\frac{1}{1-e^{-2s}}\left[ \int_{-.5}^{.5}{4te^{-st}dt}+\int_{.5}^{1.5}{\left( -4t+4.5 \right)e^{-st}dt} \right]</math>

<math>L\left\{ F\left( t \right) \right\}=\frac{1}{1-e^{-2s}}\left[ \int_{-.5}^{.5}{4te^{-st}dt}+\int_{.5}^{1.5}{-4te^{-st}dt}+\int_{.5}^{1.5}{4.5e^{-st}dt} \right]</math>

<math>\int_{-.5}^{.5}{4te^{-st}}=\; \frac{4e^{.5s}-2se^{.5s}-2se^{-.5s}-4e^{-.5s}}{s^{2}}</math>

<math>\int_{.5}^{1.5}{-4te^{-st}}=\frac{6se^{-1.5s}+4e^{-1.5s}-2se^{-.5s}-4e^{-.5s}}{s^{2}}</math>

<math>\int_{}^{}{4.5e^{-st}}=\; L\left\{ 4.5 \right\}=\frac{4.5}{s}</math>
==Author==

*[[Vier, Michael|Michael Vier]]
==Reviewers==
==Readers==

Laplace Transform of a Triangle Wave

2010-01-25T19:47:00Z

Michaelvier:

[[Image:triangle wave.jpg|500px|thumb|right|Triangle wave with period T=2 and amplitude A=2]]
'''This page is still in progress'''
==Introduction==

This article explains how to transform a periodic function (in this case a triangle wave). This is especially useful for analyzing circuits which contain triangle wave voltage sources.

==Define F(t)==

<math>m1=\frac{2+2}{.5+.5}=4</math>

<math>m2=\frac{-2-2}{1.5-.5}=-4</math>
So,

[[Image:definitionofF.jpg|140px|left]]

Using the theorem for the transform of a periodic function,

<math>L\left\{ F\left( t \right) \right\}=\frac{1}{1-e^{-2s}}\left[ \int_{-.5}^{.5}{4te^{-st}dt}+\int_{.5}^{1.5}{\left( -4t+4.5 \right)e^{-st}dt} \right]</math>

<math>L\left\{ F\left( t \right) \right\}=\frac{1}{1-e^{-2s}}\left[ \int_{-.5}^{.5}{4te^{-st}dt}+\int_{.5}^{1.5}{-4te^{-st}dt}+\int_{.5}^{1.5}{4.5e^{-st}dt} \right]</math>

<math>\int_{-.5}^{.5}{4te^{-st}}=\; \frac{4e^{.5s}-2se^{.5s}-2se^{-.5s}-4e^{-.5s}}{s^{2}}</math>

<math>\int_{}^{}{-4te^{-st}}=\; L\left\{ -4t \right\}=-\frac{4}{s^{2}}</math>

<math>\int_{}^{}{4.5e^{-st}}=\; L\left\{ 4.5 \right\}=\frac{4.5}{s}</math>
==Author==

*[[Vier, Michael|Michael Vier]]
==Reviewers==
==Readers==

Laplace Transform of a Triangle Wave

2010-01-25T19:38:28Z

Michaelvier:

[[Image:triangle wave.jpg|500px|thumb|right|Triangle wave with period T=2 and amplitude A=2]]
'''This page is still in progress'''
==Introduction==

This article explains how to transform a periodic function (in this case a triangle wave). This is especially useful for analyzing circuits which contain triangle wave voltage sources.

==Define F(t)==

<math>m1=\frac{2+2}{.5+.5}=4</math>

<math>m2=\frac{-2-2}{1.5-.5}=-4</math>
So,

[[Image:definitionofF.jpg|140px|left]]

Using the theorem for the transform of a periodic function,

<math>L\left\{ F\left( t \right) \right\}=\frac{1}{1-e^{-2s}}\left[ \int_{-.5}^{.5}{4te^{-st}dt}+\int_{.5}^{1.5}{\left( -4t+4.5 \right)e^{-st}dt} \right]</math>

<math>L\left\{ F\left( t \right) \right\}=\frac{1}{1-e^{-2s}}\left[ \int_{-.5}^{.5}{4te^{-st}dt}+\int_{.5}^{1.5}{-4te^{-st}dt}+\int_{.5}^{1.5}{4.5e^{-st}dt} \right]</math>

<math>\int_{}^{}{4te^{-st}}=\; L\left\{ 4t \right\}=\frac{4}{s^{2}}</math>

<math>\int_{}^{}{-4te^{-st}}=\; L\left\{ -4t \right\}=-\frac{4}{s^{2}}</math>

<math>\int_{}^{}{4.5e^{-st}}=\; L\left\{ 4.5 \right\}=\frac{4.5}{s}</math>
==Author==

*[[Vier, Michael|Michael Vier]]
==Reviewers==
==Readers==

Laplace Transform of a Triangle Wave

2010-01-25T19:28:43Z

Michaelvier:

Feedback in Amplifiers

2010-01-20T18:06:02Z

Michaelvier:

== Basics of Op Amps ==
Operational amplifiers are a very common component in electrical systems. The term Op-Amp specifically refers to an amplifier with two inputs, a positive input and an inverted input, and one amplified output, produced from the difference between the inputs Vout=G*(V+-V-), where G is the gain from the amp.

== Feedback ==
[[Image:Inverting_Op_Amp.jpg|300px|thumb|right|This negative feedback Op Amp setup will amplify and invert Vin on its way to Vout. When Vin increases, the output from the Op Amp becomes negative. With the appropriate value for R1 (a very large resistance since gain will be high), VToAmp will be pulled down until equilibrium is reached and VToAmp=0.]]
Since the gain in an op amp is extremely high, usually hundreds of thousands in magnitude<ref>"Op Amps." ''www.williamson-labs.com.'' Williams Labs, 2007. Web. 10 Jan 2010.</ref>, very small differences in input voltages can cause the output to reach the amp's maximum gain. This makes op amps useful as voltage comparators when no feedback is present. But for most practical uses, feedback from the output into the inverted input is needed in order to control the input voltage difference and keep the output from reaching its maximum. When the positive input gets higher, the output goes up, causing the inverted input to rise, thus re-lowering the output. Output and input change for a while as they stabilize at their new values, at which point the positive and inverted inputs are very nearly equal. Since the gain is extremely large, this minute difference in input voltages, which we approximate to be 0, amplifies to give a voltage at Vout.

== References ==

<references/>

==Contributers==

--[[Vier, Michael|Michael Vier]] (Author)

==Reviewers==
[[Greg Fong]]
*<math>V_{out}=G(V_+-V_-)\,</math>
*< ref>[URL:URL caption] Notes </ref>
**URL:www.williamson-labs.com
**URL Caption: Op Amps
**Notes: Williams Labs, 2007. Wed. 10 Jan 2010.
*"The output is 0" - Feedback paragraph, last line. Are you sure?
*[[Ben Henry|Henry,Ben]]

==Readers==

Laplace Transform of a Triangle Wave

2010-01-20T02:03:02Z

Michaelvier:

Laplace Transform of a Triangle Wave

2010-01-20T02:02:27Z

Michaelvier:

[[Image:triangle wave.jpg|500px|thumb|right|Triangle wave with period T=2 and amplitude A=2]]
'''In progress'''
==Introduction==

This article explains how to transform a periodic function (in this case a triangle wave). This is especially useful for analyzing circuits which contain triangle wave voltage sources.

==Define F(t)==

<math>m1=\frac{2+2}{.5+.5}=4</math>

<math>m2=\frac{-2-2}{1.5-.5}=-4</math>
So,

[[Image:definitionofF.jpg|140px|left]]

==Author==

*[[Vier, Michael|Michael Vier]]
==Reviewers==
==Readers==

Laplace Transform of a Triangle Wave

2010-01-20T02:01:57Z

Michaelvier:

[[Image:triangle wave.jpg|500px|thumb|right|Triangle wave with period T=2 and amplitude A=2]]
==Introduction==

This article explains how to transform a periodic function (in this case a triangle wave). This is especially useful for analyzing circuits which contain triangle wave voltage sources.

==Define F(t)==

<math>m1=\frac{2+2}{.5+.5}=4</math>

<math>m2=\frac{-2-2}{1.5-.5}=-4</math>
So,

[[Image:definitionofF.jpg|140px|left]]

==Author==

*[[Vier, Michael|Michael Vier]]
==Reviewers==
==Readers==

File:DefinitionofF.jpg

2010-01-20T01:57:47Z

Michaelvier:

Laplace Transform of a Triangle Wave

2010-01-20T01:57:17Z

Michaelvier:

[[Image:triangle wave.jpg|500px|thumb|right|Triangle wave with period T=2 and amplitude A=2]]
==Introduction==

This article explains how to transform a periodic function (in this case a triangle wave). This is especially useful for analyzing circuits which contain triangle wave voltage sources.

==Define F(t)==

<math>m1=\frac{2+2}{.5+.5}=4</math>

<math>m2=\frac{-2-2}{1.5-.5}=-4</math>
So,

<math>F\left( t \right)=\left\{\begin{array}{cc} 4t & -.5\leq t<.5 \\ -4t+4 & .5\leq t<1.5 \end{array}\right
</math>

==Author==

*[[Vier, Michael|Michael Vier]]
==Reviewers==
==Readers==

Laplace Transform of a Triangle Wave

2010-01-20T01:37:50Z

Michaelvier:

Laplace Transform of a Triangle Wave

2010-01-20T01:32:48Z

Michaelvier: New page: Triangle wave with period T=2 and amplitude A=2

[[Image:triangle wave.jpg|500px|thumb|right|Triangle wave with period T=2 and amplitude A=2]]

File:Triangle wave.jpg

2010-01-20T01:31:26Z

Michaelvier: Triangle with period T=2 and amplitude A=2

Triangle with period T=2 and amplitude A=2

Winter 2010

2010-01-20T01:17:24Z

Michaelvier:

Put links for your reports here. Someone feel free to edit this better.

Links to the posted reports are found under the publishing author's name. (Like it says above, if you hate it...change it! I promise I won’t cry. Brandon)

1. [[Biesenthal, Dan]]

2. [[Blackley, Ben]]

3. [[Cruz, Jorge]]
<math>\Rightarrow</math> [[Fourier Example]]

4. [[Fullerton, Colby]]
<math>\Rightarrow</math> [[Laplace Transform]]
<math>\Rightarrow</math> [[Class Notes 1-5-2010]]

5. [[Grant, Joshua]]

6. [[Gratias, Ryan]]

7. [[Hawkins, John]]

<math>\Rightarrow</math> [[Exercise: Sawtooth Wave Fourier Transform]]
<math>\Rightarrow</math> [[Exercise: Sawtooth Redone With Exponential Basis Functions]]
8. [[Lau, Chris]]
<math>\Rightarrow</math> [[Fourier Series: Explained!]]

9. [[Roath, Brian]]
<math>\Rightarrow</math> [[Laplace Transform]]
<math>\Rightarrow</math> [[Class Notes 1-5-2010]]

10. [[Robbins, David]]

11. [[Roth, Andrew]]
<math>\Rightarrow</math> [[Example: LaTex format (0 points)]]
<math>\Rightarrow</math> [[Chapter 22--Fourier Series: Fundamental Period, Frequency, and Angular Frequency]]

12. [[Vazquez, Brandon]]

13. [[Vier, Michael]]
<math>\Rightarrow</math> [[Fourier Series: Explained!]]

14. [[Wooley, Thomas]]

15. [[Jaymin, Joseph]]
<math>\Rightarrow</math> [[Basic_Laplace_Transforms]]

16. [[Starr, Brielle]]

17. [[Starr, Tyler]]

==Draft Articles==
These articles are not ready for reading and error checking. They are listed so people will not simultaneously write about similar topics.
* [[Gibbs Phenomenon]]

* [[Linear Time Invariant Overview]]

* [[Solving Series RLC Circuit Using Laplace Transforms]]

* [[Laplace Transform of a Triangle Wave]]

Vier, Michael

2010-01-20T01:16:48Z

Michaelvier:

==Linear Network Analysis==

*[[Fourier Series: Explained!]]

*[[Laplace Transform of a Triangle Wave]]

== Engineering Electronics ==

*[[Feedback in Amplifiers]]

Winter 2010

2010-01-20T00:54:14Z

Michaelvier:

Winter 2010

2010-01-20T00:50:52Z

Michaelvier:

Parallel RLC circuits: Laplace Transform Method

2010-01-20T00:49:16Z

Michaelvier: New page: -Under Construction- ==Authors== *Michael Vier

-Under Construction-

==Authors==

*[[Vier, Michael|Michael Vier]]

Vier, Michael

2010-01-20T00:48:18Z

Michaelvier:

==Linear Network Analysis==

*[[Fourier Series: Explained!]]

*[[Parallel RLC circuits: Laplace Transform Method]]

== Engineering Electronics ==

*[[Feedback in Amplifiers]]

Gibbs Phenomenon

2010-01-19T04:24:12Z

Michaelvier:

==Overview==
The Gibbs phenomenon is the the tendency for Fourier sums to "jump" higher than expected at discontinuities. It is named after the American physicist J. Willard Gibbs.

==The Phenomenon==
The identifying characteristic of the Gibbs phenomenon is the spike past where the Fourier series is summing to. As my colleagues previously stated, "notice how the summation function resembles the original periodic function more as more functions are added."<ref>[http://fweb/class-wiki/index.php/Fourier_Series:_Explained! Fourier Series: Explained!]</ref>
While this is true, it can also be seen that the jump does not diminish as the frequency of additional functions is increased. In fact the spike reaches a finite limit.
[[Image:Gibbs_phenomenon_50.jpg|300px|thumb|right|Showing the spike at a discontinuity.]]

==References==
<references/>

==Contributor==
*[[Grant, Joshua | Joshua Grant]]

===Reviewed By===

*[[Vier, Michael | Michael Vier]]

===Read By===

Golden Rules

2010-01-14T18:36:23Z

Michaelvier:

When analyzing operational amplifiers, there are a few rules that need to be taken into consideration in order to solve circuits using KVL or KCL.
# Because the output voltage does not depend on the output current, the output impedance equals zero.
# The input impedance <math>Z_+=Z_-=0</math>.
# When there is a negative feedback, both inputs have the same voltage. In other words, <math>V_+</math> is equal to <math>V_-</math>.
# When solving circuits using the nodal analysis, write node equations at <math>V_+</math> and <math>V_-</math>, but not at <math>V_o</math>.

==Sources==
*[http://people.wallawalla.edu/~Rob.Frohne/ClassNotes/ENGR356/2010/Keystone/index.php Class Notes]
*[http://hyperphysics.phy-astr.gsu.edu/HBASE/electronic/opampi.html Ideal Op-amp]
*[http://mechatronics.mech.northwestern.edu/design_ref/electrical_design/opamps.html Operational Amplifiers]

==Reviewer==
[[Lau, Chris | Christopher Garrison Lau I]]
*[[Vier, Michael | Michael Vier]] -- What do you mean "the input impedance.. =0?" I think that is wrong.

Exercise: Sawtooth Wave Fourier Transform

2010-01-14T18:33:31Z

Michaelvier:

==Problem Statement==

Find the Fourier Tranform of the sawtooth wave given by the equation

 

<center><math>x(t)=t-\lfloor t \rfloor</math></center>

 

 

==Solution==

As shown in class, the general equation for the Fourier Transform for a periodic function with period <math>T</math> is given by

 

<center><math> x(t)=\frac{a_0}{2}+\sum^\infty_{n=1} \left[a_n\cos\frac{2\pi nt}{T}+b_n\sin\frac{2\pi nt}{T}\right]</math></center>

 

where

 

<center><math>\begin{cases}

a_n=\frac{2}{T}\int_c^{c+T}x(t)\cos\frac{2\pi nt}{T}dt\\

b_n=\frac{2}{T}\int_c^{c+T}x(t)\sin\frac{2\pi nt}{T}dt

\end{cases} \ \ \ \ n=0,1,2,3\dots</math></center>

 

For the sawtooth function given, we note that <math>T=1</math>, and an obvious choice for <math>c</math> is 0 since this allows us to reduce the equation to <math>x(t)=t</math>. It remains, then, only to find the expression for <math>a_n</math> and <math>b_n</math>. We proceed first to find <math>b_n</math>.

 

<center><math>b_n=\frac{2}{1}\int_0^1t\sin 2\pi nt\ dt</math></center>

 

which is solved easiest with integration by parts, letting

 

<center>

<math>u=t\qquad\Rightarrow\qquad du=dt</math>

 

<math>dv=\sin2\pi nt\ dt\qquad\Rightarrow\qquad v=-\frac{1}{2\pi n}\cos 2\pi nt</math>

</center>

 

so

 

<center>

<math>b_n=2\left[t\left(-\frac{1}{2\pi n}\right)\cos 2\pi nt\bigg|_0^1+\frac{1}{2\pi n}\int_0^1\cos 2\pi nt\ dt \right]</math>

 

<math>=2\left[\left(-\frac{1}{2\pi n}\cos 2\pi n - 0\right)+\left(\frac{1}{2\pi n}\right)^2\sin 2\pi nt \bigg|_0^1\right]</math>

 

<math>=2\left[-\frac{1}{2\pi n}(1)+0\right]</math>

 

<math>=-\frac{1}{\pi n}</math>

</center>

Now, for <math>a_n</math> we must consider the case when <math>n=0</math> separately.

 

<center>

<math>a_0=\frac{2}{1}\int_0^1t\ dt=t^2\bigg|_0^1=1</math>

</center>

 

For <math>n=1,2,3\dots</math>, we have

 

<center>

<math>a_n=\frac{2}{1}\int_0^1t\cos 2\pi nt\ dt</math>

</center>

 

which again is best solved using integration by parts, this time with

 

<center>

<math>u=t\qquad\Rightarrow\qquad du=dt</math>

 

<math>dv=\cos 2\pi nt\ dt\qquad \Rightarrow\qquad v=\frac{1}{2\pi n}\sin 2\pi nt</math>

 

</center>

so

 

<center>

<math>a_n=2\left[t\left(\frac{1}{2\pi n}\right)\sin 2\pi nt\bigg|_0^1-\int_0^1\frac{1}{2\pi n}\sin 2\pi nt\ dt\right]</math>

 

<math>=2\left[\left(\frac{1}{2\pi n}\sin 2\pi n-0\right)-\left[-\left(\frac{1}{2\pi n}\right)^2\cos 2\pi nt\right]_0^1\right]</math>

 

<math>=2\left[0+\left(\frac{1}{2\pi n}\right)^2\left(\cos 2\pi n-\cos 0\right)\right]</math>

 

<math>\ =0</math>

</center>

 

Therefore, the Fourier Transform representation of the sawtooth wave given is:

 

<center><math>x(t)=\frac{1}{2}-\sum_{n=1}^\infty \frac{1}{\pi n}\sin 2\pi nt</math></center>

 

==Solution Graphs==
The figures below graph the first few iterations of the above solution. The first graph shows the solution truncated after the first 100 terms of the infinite sum, as well as each of the contributing sine waves with offset. The second figure shows the function truncated after 1, 3, 5, 10, 50, and 100 terms. The last figure shows the Error between the Fourier Series truncated after the first 100 terms and the function itself. These figures were constructed using the following matlab code: [[SawToothFourier]].

 

[[Image:First_100_Terms.jpg|thumb|800px|center]]

 
[[Image:First_n_Terms.jpg|thumb|800px|center]]

 
[[Image:Error.jpg|thumb|800px|center]]

==Author==

John Hawkins

==Read By==

==Reviewed By==
Colby Fullerton
*[[Vier, Michael | Michael Vier]]

Vier, Michael

2010-01-14T18:31:19Z

Michaelvier:

==Linear Network Analysis==

*[[Fourier Series: Explained!]]

== Engineering Electronics ==

*[[Feedback in Amplifiers]]

Basic Op Amp circuits

2010-01-14T18:30:13Z

Michaelvier:

==Buffer Amplifier==
[[Image:BufferAmplifier.PNG|thumb|300px|Buffer Amplifier ]]
*Used to transfer voltage but not current to the following circuit. This amplifier can be used to negate the loading effects. No current flows through the amplifier, thus there is no voltage drop through the input resistor (going to the buffer amplifier).

 
 
 
 
 
 
 

==Inverting Amplifier==
[[Image:InvertingAmplifier.png|thumb|300px|Inverting Amplifier]]
*Uses negative feedback to invert and amplify voltage. Using nodal analysis at the negative terminal, the gain is found to be <math>-\frac{R_2}{R_1}</math>
*<math>R_{bias}=\frac{R_1R_2}{R_1+R_2}</math>
*To get rid of unwanted DC components, a capacitor can be added inbetween <math>R_1\,</math> and <math>V_{in}\,</math>. In this case <math>R_{bias}=R_2\,</math>
 
 
 
 

==Summing Amplifier==
[[Image:Summing_Amplifier.PNG‎|thumb|300px|Summing Amplifier]]
*<math>V_o=-R_f \left( \frac{V_3}{R_3}+\frac{V_2}{R_2}+\frac{V_1}{R_1}\right)</math>
*If all resistances are equal, then the output voltage is the (negative) sum of the input voltages
*<math>\frac{1}{R_{bias}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_f}</math>

 
 
 
 
 
 
 
 
 
==Noninverting Amplifier==
[[Image:Noninverting_Amplifier.PNG‎|thumb|300px|Noninverting Amplifier]]
*<math>V_o=V_{in} \left(1+\frac{R_2}{R_1}\right)</math>
*<math>R_{bias}=\frac{R_1R_2}{R_1+R_2}</math>
*<math>R_{bias}\,</math> goes between the positive terminal and <math>V_{in}\,</math>
*To get rid of unwanted DC components, a capacitor can be added inbetween the positive terminal and <math>V_{in}\,</math>. The bias resistor has the same value, and is placed inbetween the positive input terminal and ground.
 

==Differential Amplifier==
[[Image:Differential_Amplifier_2.PNG‎|thumb|300px|Differential Amplifier ]]
*<math>V_o=V_2\frac{(R_1+R_f)R_g}{(R_2+R_g)R_1}-V_1\frac{R_f}{R_1}</math>
*If you let <math>R_1=R_2\,</math> and <math>R_g=R_f\,</math> then the equation simplifies to <math>V_o=\frac{R_f}{R_1}(V_2-V_1)</math>

 
 
 
 
 
 
 
 

==Possible circuits to add in the future==
*Voltage-to-current converter
*Current-to-voltage converter
*Current amplifier
*[[Integrator_Amplifier | Integrator]]
*Differentiator

==Reviewers==
*Victor Shepherd
*[[Vier, Michael | Michael Vier]]

Chapter 1

2010-01-12T19:27:32Z

Michaelvier:

=Amplifier Models=
*These are purely models, and cannot be replicated in a real world environment. They are meant to explain.
*Trans stands for transfer (from voltage to current or visa versa).
*The inputs and outputs can be either current or voltage. This leads to 4 amplifier models.
*You can use any of these models, though some may be easier to work with (if you are given the Thevenin or Norton equivalent).

{| class="wikitable" border="1"
|+ Amplifier models
! Amplifier type Gain parameter Gain equation
! Voltage input
! Current input
|- align="center"
! Voltage output
| Voltage Open-circuit voltage gain <math>A_{voc}=\frac{v_{ooc}}{v_i}</math>
| Transresistance Open-circuit transresistance gain <math>R_{moc}=\frac{v_{ooc}}{i_i}</math>
|- align="center"
! Current output
| Transconductance Short-circuit transconductance gain <math>G_{msc}=\frac{i_{osc}}{v_i}</math>
| Current Short-circuit current gain <math>A_{isc}=\frac{i_{osc}}{i_i}</math>
|}

{| class="wikitable" border="1"
|+ Characteristics of ideal amplifiers
! Amplifier Type !! Input Impedance !! Output Impedance !! Gain Parameter
|-align="center"
! Voltage
| <math>\infty</math>
| 0
| <math>A_{voc}\,</math>
|-align="center"
! Current
| 0
| <math>\infty</math>
| <math>A_{isc}\,</math>
|-align="center"
! Transconductance
| <math>\infty</math>
| <math>\infty</math>
| <math>G_{msc}\,</math>
|-align="center"
! Transresistance
| 0
| 0
| <math>R_{moc}\,</math>
|}

=Differential Amplifiers=
[[Image:Differential Amplifier.PNG|thumb|300px| Differential Amplifier inputs]]
*Differential amplifiers take two (or more) input sources that produce an output voltage proportional to the difference between the input voltages.
*Instead of expressing the input voltages in terms of <math>v_{1}\,</math> and <math>v_{i}\,</math>, we can express them in terms of the differential and common-mode input.
**Differential input signal is the difference between the input voltages. <math>v_{d}=v_{1}-v_{2}\,</math>
**Common-mode input signal is the average of the input voltages. <math>v_{cm}=\frac{1}{2}(v_{1}+v_{2})</math>
**<math>v_{1}=v_{cm}+\frac{v_{d}}{2}</math>, if <math>v_{1}\,</math> is voltage at the positive terminal.
**<math>v_{2}=v_{cm}-\frac{v_{d}}{2}</math>, if <math>v_{2}\,</math> is voltage at the negative terminal.
*<math>v_o=A_d v_{d} + A_{cm} v_{cm}\,</math>, where <math>A_d\,</math> is the differential gain and <math>A_{cm}\,</math> is the common mode gain.
*The common-mode rejection ratio (CMRR) is the ratio of the magnitude of the differential gain to the magnitude of the common-mode gain.
**In decibels, <math> CMRR = 20 \log \frac{| A_d |}{| A_cm|}</math>

=Definitions=
*Input Resistance: <math>R_i</math> of an amplifier is the equivalent resistance seen when looking into the input terminals.
*Output Resistance: <math>R_o</math> is the Thevenin resistance seen when looking back into the output terminals of an amplifier.
*Open-circuit voltage gain: the ratio of output amplitude to input amplitude with the output terminals open circuited.
*Short-circuit current gain: the current gain with the output terminals of the amplifier short circuited.

=Reviewers=
*[[Lau, Chris | Christopher Garrison Lau I]]
*[[Vier, Michael | Michael Vier]] -- Make sure you subscript the 'm' in the CMRR formula.

Chapter 1

2010-01-12T19:27:05Z

Michaelvier:

=Amplifier Models=
*These are purely models, and cannot be replicated in a real world environment. They are meant to explain.
*Trans stands for transfer (from voltage to current or visa versa).
*The inputs and outputs can be either current or voltage. This leads to 4 amplifier models.
*You can use any of these models, though some may be easier to work with (if you are given the Thevenin or Norton equivalent).

{| class="wikitable" border="1"
|+ Amplifier models
! Amplifier type Gain parameter Gain equation
! Voltage input
! Current input
|- align="center"
! Voltage output
| Voltage Open-circuit voltage gain <math>A_{voc}=\frac{v_{ooc}}{v_i}</math>
| Transresistance Open-circuit transresistance gain <math>R_{moc}=\frac{v_{ooc}}{i_i}</math>
|- align="center"
! Current output
| Transconductance Short-circuit transconductance gain <math>G_{msc}=\frac{i_{osc}}{v_i}</math>
| Current Short-circuit current gain <math>A_{isc}=\frac{i_{osc}}{i_i}</math>
|}

{| class="wikitable" border="1"
|+ Characteristics of ideal amplifiers
! Amplifier Type !! Input Impedance !! Output Impedance !! Gain Parameter
|-align="center"
! Voltage
| <math>\infty</math>
| 0
| <math>A_{voc}\,</math>
|-align="center"
! Current
| 0
| <math>\infty</math>
| <math>A_{isc}\,</math>
|-align="center"
! Transconductance
| <math>\infty</math>
| <math>\infty</math>
| <math>G_{msc}\,</math>
|-align="center"
! Transresistance
| 0
| 0
| <math>R_{moc}\,</math>
|}

=Differential Amplifiers=
[[Image:Differential Amplifier.PNG|thumb|300px| Differential Amplifier inputs]]
*Differential amplifiers take two (or more) input sources that produce an output voltage proportional to the difference between the input voltages.
*Instead of expressing the input voltages in terms of <math>v_{1}\,</math> and <math>v_{i}\,</math>, we can express them in terms of the differential and common-mode input.
**Differential input signal is the difference between the input voltages. <math>v_{d}=v_{1}-v_{2}\,</math>
**Common-mode input signal is the average of the input voltages. <math>v_{cm}=\frac{1}{2}(v_{1}+v_{2})</math>
**<math>v_{1}=v_{cm}+\frac{v_{d}}{2}</math>, if <math>v_{1}\,</math> is voltage at the positive terminal.
**<math>v_{2}=v_{cm}-\frac{v_{d}}{2}</math>, if <math>v_{2}\,</math> is voltage at the negative terminal.
*<math>v_o=A_d v_{d} + A_{cm} v_{cm}\,</math>, where <math>A_d\,</math> is the differential gain and <math>A_{cm}\,</math> is the common mode gain.
*The common-mode rejection ratio (CMRR) is the ratio of the magnitude of the differential gain to the magnitude of the common-mode gain.
**In decibels, <math> CMRR = 20 \log \frac{| A_d |}{| A_cm|}</math>

=Definitions=
*Input Resistance: <math>R_i</math> of an amplifier is the equivalent resistance seen when looking into the input terminals.
*Output Resistance: <math>R_o</math> is the Thevenin resistance seen when looking back into the output terminals of an amplifier.
*Open-circuit voltage gain: the ratio of output amplitude to input amplitude with the output terminals open circuited.
*Short-circuit current gain: the current gain with the output terminals of the amplifier short circuited.

=Reviewers=
[[Lau, Chris | Christopher Garrison Lau I]]
[[Vier, Michael | Michael Vier]] -- Make sure you subscript the 'm' in the CMRR formula.

Winter 2010

2010-01-12T07:42:27Z

Michaelvier:

Winter 2010

2010-01-12T07:41:53Z

Michaelvier:

Fourier Series: Explained!

2010-01-12T07:37:23Z

Michaelvier:

==A Brief Introduction==
A Fourier series is a mathematical tool that takes a periodic function and turns it into a sum of simple oscillating functions (i.e. sines and cosines)<ref> [http://en.wikipedia.org/wiki/Fourier_series Fourier Series]</ref>. These series were discovered by Joseph Fourier to solve a heat equation in a metal plate.

==How They Work==
Fourier Series represents a periodic function through a sum of sines or cosines. Each term in the summation has a frequency n. The first term has the same frequency as the periodic function, the second term has twice the frequency of the periodic function, and so on. The more functions added, the more the summation resembles the step function. Observe the animation; notice how the summation function resembles the original periodic function more as more functions are added.
[[Image:Square Wave.jpg|500px|thumb|right|Square Wave with similar periods to the cosine function]]
[[Image:Fourier Animated.gif|500px|thumb|left|Fourier series animated to show increasing accuracy as evaluation bounds are increased.]]
==References==
<references/>
==Helpful Links==
[http://www.fourier-series.com/fourierseries2/flash_programs/fourier_series_sin_cos/index.html A very helpful game]
==Contributors==
*[[Lau, Chris | Christopher Garrison Lau I]]
*[[Vier, Michael|Michael Vier]]

Feedback in Amplifiers

2010-01-12T07:37:09Z

Michaelvier:

== Basics of Op Amps ==
Operational amplifiers are a very common component in electrical systems. The term Op-Amp specifically refers to an amplifier with two inputs, a positive input and an inverted input, and one amplified output, produced from the difference between the inputs Vout=G*(V+-V-), where G is the gain from the amp.

== Feedback ==
[[Image:Inverting_Op_Amp.jpg|300px|thumb|right|This negative feedback Op Amp setup will amplify and invert Vin on its way to Vout. When Vin increases, the output from the Op Amp becomes negative. With the appropriate value for R1 (a very large resistance since gain will be high), VToAmp will be pulled down until equilibrium is reached and VToAmp=0.]]
Since the gain in an op amp is extremely high, usually hundreds of thousands in magnitude<ref>"Op Amps." ''www.williamson-labs.com.'' Williams Labs, 2007. Web. 10 Jan 2010.</ref>, very small differences in input voltages can cause the output to reach the amp's maximum gain. This makes op amps useful as voltage comparators when no feedback is present. But for most practical uses, feedback from the output into the inverted input is needed in order to control the input voltage difference and keep the output from reaching its maximum. When the positive input gets higher, the output goes up, causing the inverted input to rise, thus re-lowering the output. Output and input change for a while as they stabilize at their new values, at which point the positive and inverted inputs are equal and the output is 0.

== References ==

<references/>

==Contributers==

--[[Vier, Michael|Michael Vier]] (Author)

==Reviewers==

==Readers==

Feedback in Amplifiers

2010-01-12T07:36:42Z

Michaelvier:

== Basics of Op Amps ==
Operational amplifiers are a very common component in electrical systems. The term Op-Amp specifically refers to an amplifier with two inputs, a positive input and an inverted input, and one amplified output, produced from the difference between the inputs Vout=G*(V+-V-), where G is the gain from the amp.

== Feedback ==
[[Image:Inverting_Op_Amp.jpg|300px|thumb|right|This negative feedback Op Amp setup will amplify and invert Vin on its way to Vout. When Vin increases, the output from the Op Amp becomes negative. With the appropriate value for R1 (a very large resistance since gain will be high), VToAmp will be pulled down until equilibrium is reached and VToAmp=0.]]
Since the gain in an op amp is extremely high, usually hundreds of thousands in magnitude<ref>"Op Amps." ''www.williamson-labs.com.'' Williams Labs, 2007. Web. 10 Jan 2010.</ref>, very small differences in input voltages can cause the output to reach the amp's maximum gain. This makes op amps useful as voltage comparators when no feedback is present. But for most practical uses, feedback from the output into the inverted input is needed in order to control the input voltage difference and keep the output from reaching its maximum. When the positive input gets higher, the output goes up, causing the inverted input to rise, thus re-lowering the output. Output and input change for a while as they stabilize at their new values, at which point the positive and inverted inputs are equal and the output is 0.

== References ==

<references/>

==Contributers==

--[[User:Vier, Michael|Michael Vier]] (Author)

==Reviewers==

==Readers==

Fourier Series: Explained!

2010-01-12T07:34:56Z

Michaelvier:

==A Brief Introduction==
A Fourier series is a mathematical tool that takes a periodic function and turns it into a sum of simple oscillating functions (i.e. sines and cosines)<ref> [http://en.wikipedia.org/wiki/Fourier_series Fourier Series] </ref>. These series were discovered by Joseph Fourier to solve a heat equation in a metal plate.
==How They Work==
Fourier Series represents a periodic function through a sum of sines or cosines. Each term in the summation has a frequency n. The first term has the same frequency as the periodic function, the second term has twice the frequency of the periodic function, and so on. The more functions added, the more the summation resembles the step function. Observe the animation; notice how the summation function resembles the original periodic function more as more functions are added.
[[Image:Square Wave.jpg|500px|thumb|right|Square Wave with similar periods to the cosine function]]
[[Image:Fourier Animated.gif|500px|thumb|left|Fourier series animated to show increasing accuracy as evaluation bounds are increased.]]
==References==
<references/>
==Helpful Links==
[http://www.fourier-series.com/fourierseries2/flash_programs/fourier_series_sin_cos/index.html A very helpful game]
==Contributors==
*[[Lau, Chris | Christopher Garrison Lau I]]
*[[User:Vier,_Michael|Michael Vier]]

File:Square Wave.jpg

2010-01-12T07:28:08Z

Michaelvier: uploaded a new version of "Image:Square Wave.jpg"

Square Wave

Fourier Series: Explained!

2010-01-12T07:09:09Z

Michaelvier:

===A Brief Introduction===
A Fourier series is a mathematical tool that takes a periodic function and turns it into a sum of simple oscillating functions (i.e. sines and cosines)<ref> [http://en.wikipedia.org/wiki/Fourier_series Fourier Series] </ref>. These series were discovered by Joseph Fourier to solve a heat equation in a metal plate.
===How They Work===
Fourier Series represents a periodic function through a sum of sines or cosines. Each term in the summation has a frequency n. The first term has the same frequency as the periodic function, the second term has twice the frequency of the periodic function, and so on. The more functions added, the more the summation resembles the step function. Observe the animation; notice how the summation function resembles the original periodic function more as more functions are added.
[[Image:Square Wave.jpg|500px|thumb|right|Square Wave with similar periods to the cosine function]]
[[Image:Fourier Animated.gif|500px|thumb|left|Fourier series animated to show increasing accuracy as evaluation bounds are increased.]]

Fourier Series: Explained!

2010-01-12T07:08:14Z

Michaelvier:

===A Brief Introduction===
A Fourier series is a mathematical tool that takes a periodic function and turns it into a sum of simple oscillating functions (i.e. sines and cosines)<ref> [http://en.wikipedia.org/wiki/Fourier_series Fourier Series] </ref>. These series were discovered by Joseph Fourier to solve a heat equation in a metal plate.
===How They Work===
Fourier Series represents a periodic function through a sum of sines or cosines. Each term in the summation has a frequency n. The first term has the same frequency as the periodic function, the second term has twice the frequency of the periodic function, and so on. The more functions added, the more the summation resembles the step function. Observe the animation; notice how the summation function resembles the original periodic function more as more functions are added.
[[Image:Square Wave.jpg|300px|thumb|right|Square Wave with similar periods to the cosine function]]
[[Image:Fourier Animated.gif|500px|thumb|left|Fourier series animated to show increasing accuracy as evaluation bounds are increased.]]

Fourier Series: Explained!

2010-01-12T06:57:44Z

Michaelvier:

Fourier Series: Explained!

2010-01-12T06:56:39Z

Michaelvier:

Fourier Series: Explained!

2010-01-12T06:56:00Z

Michaelvier:

File:Square Wave.jpg

2010-01-12T06:50:10Z

Michaelvier: Square Wave

Square Wave

File:Fourier Animated.gif

2010-01-12T06:19:53Z

Michaelvier: Fourier Series Animation

Fourier Series Animation

Electronics Score Pages

2010-01-11T22:33:01Z

Michaelvier:

===Class List===
====2010====
#[[Ben Henry|Henry, Ben]]
#[[Vier, Michael]] - 20
#[[Lau, Chris]] - 20

==Keeping Score==

=====Author:=====
*1 point/line
*1 point/(citation of other work)
*3 points/(citation of this work)
*1/20 point/line/reader
*-1/3 point/line for errata (Transfer points to the finder of the errata.)

=====Reviewer:=====
*1/6 points/line
*-1/10 points/line for errata (Transfer points to the finder of the errata.)

=====Reader:=====
*1/10 point/line
*-1/20 point/line for errata (Transfer points to the finder of the errata.)

Feedback in Amplifiers

2010-01-11T21:43:05Z

Michaelvier:

== Basics of Op Amps ==
Operational amplifiers are a very common component in electrical systems. The term Op-Amp specifically refers to an amplifier with two inputs, a positive input and an inverted input, and one amplified output, produced from the difference between the inputs Vout=G*(V+-V-), where G is the gain from the amp.

== Feedback ==
[[Image:Inverting_Op_Amp.jpg|300px|thumb|right|This negative feedback Op Amp setup will amplify and invert Vin on its way to Vout. When Vin increases, the output from the Op Amp becomes negative. With the appropriate value for R1 (a very large resistance since gain will be high), VToAmp will be pulled down until equilibrium is reached and VToAmp=0.]]
Since the gain in an op amp is extremely high, usually hundreds of thousands in magnitude<ref>"Op Amps." ''www.williamson-labs.com.'' Williams Labs, 2007. Web. 10 Jan 2010.</ref>, very small differences in input voltages can cause the output to reach the amp's maximum gain. This makes op amps useful as voltage comparators when no feedback is present. But for most practical uses, feedback from the output into the inverted input is needed in order to control the input voltage difference and keep the output from reaching its maximum. When the positive input gets higher, the output goes up, causing the inverted input to rise, thus re-lowering the output. Output and input change for a while as they stabilize at their new values, at which point the positive and inverted inputs are equal and the output is 0.

== References ==

<references/>

==Contributers==

--[[User:Vier,_Michael|Michael Vier]] (Author)

==Reviewers==

==Readers==