Lesson+2

Lesson+2 - How engineers facilitate human evolution 1...

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Lesson 2 1 How engineers facilitate human evolution.

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Nodal Analysis Review Review Ch. 4 Lesson 2 2 Lesson 2
V0 R R/2 R/4 R/4 Challenge 1 Lesson 2 3 Using voltage dividers , it follows that: V 4 = V 0 V 3 = (R/2+R/4+R/4) V 0 /(R+R/2+R/4+R/4)= R V 0 /2R = V 0 /2 V 2 =(R/4+R/4) V 0 /(R+R/2+R/4+R/4)= (R/2) V 0 /2R= V 0 /4 V 1 =(R/4) V 0 /(R+R/2+R/4+R/4) = =(R/4) V 0 /2R = V 0 /8 V 0 V 4 V 3 V 2 V 1 Circuit 1 What is it good for?

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Challenge 1 Lesson 2 4 Circuit 1 logarithmically covers the range 0 to V 0 { V 0 /2, V 0 /4, V 0 /8,…}. This is sometimes called companding . Signals of small amplitude are finely resolved (expanded) while those of large amplitude are more coarsely resolved (compressed). An application is telephony where speech data is logarithmically compressed before being presented to the li s tener. Out Input Eustachian tube Auditory bones Where do I get one?
V0 R R R R Challenge 1 Lesson 2 5 Again, using voltage dividers , it follows that: V 4 = V 0 V 3 =(3R) V 0 /4R = =3R V 0 /4R=0.75 V 0 V 2 =(2R) V 0 /(4R)= =(2R) V 0 /4R=0.5 V 0 V 1 =(R) V 0 /(4R)= =(R) V 0 /4R = 0.25 V 0 V 4 V 3 V 2 V 1 V 0 What is it good for? Circuit 2

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Challenge 1 Circuit 2 lineally quantizes a signal over the range 0 to V 0 into discrete values { V 0 , 0.75 V 0 , 0.5 V 0 , 0.25 V 0 , …}. This is characteristic of a linear analog to digital converter (ADC). Example: Maxum/Dallas MAX104 MAX105 MAX1106 MAX153 MAX196 Lesson 2 6 Flash Converter
Side Bar Introduction to Linear Algebra Lesson 2 7 Solution to 1st order equation y=Ax y =[ y 1 ], x =[ x 1 ], A =[ a 11 ], A -1 =[1/ a 11 ], and x 1 = A -1 y 1 =[1/ a 11 ] y 1 . Solution to 2nd order equation y=Ax y = , x = , A = , A -1 = , and x = A -1 y . Solution to 3rd order equation y=Ax y = , x = , A = , and x = A -1 y . (several methods can be used to invert A ; Cramer’s Method, Gaussian Elimination, etc.), or the slacker friend, MATLAB. 2 1 y y 2 1 x x 22 21 12 11 a a a a 21 12 22 11 11 21 12 22 a a a a a a a a 3 2 1 y y y 3 2 1 x x x 33 32 31 23 22 21 13 12 11 a a a a a a a a a Matrix Supplement On-Line

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Lesson 2 Lesson 2 8 Nodal Analysis Process Assume, that only current sources are present. Given an N node circuit : Arbitrarily assign the 0 node to be the reference node . Arbitrarily assign branch currents and ( N-1 ) inter-node voltages. Write ( N-1 ) KCLs at each non-reference node. Solve the ( N-1 ) linearly independent KCL in terms of ( N-1 ) unknown node voltages.
Lesson 2 Lesson 2 9 Analyze the following 3 node circuit containing only current sources using the nodal analysis procedure. 10A

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Lesson 2 Lesson 2 10 KCL @node_1 = ( I 1 - I 2 ) – I 3 I 4 =0 or ( I 1 - I 2 ) = ( I 3 + I 4 ) = ( V 10 /R 1 + V 12 /R 2 ); ( I 1 , I 2 ) known V 10 V 12 10A Ohm’s Law V 12 Sources V 20
Lesson 2 Lesson 2 11 KCL @node_2 = I 2 + I 4 I 5 =0 or I 2 = - I 4 + I 5 = ( -V 12 /R 2 + V 20 /R 3 ); I 2 known Oh oh – 2 equations in 3 unknowns { V 10 , V 12 , V 20 } V 12 V 20 10A V 10 Source

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Lesson 2 Lesson 2 12 Moral – pick 2 and only 2 node voltages and stay with them!
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