F07%20314%20HW03%20solutions%20all

F07%20314%20HW03%20solutions%20all - EECS 314 Fall 2007 HW...

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EECS 314 Fall 2007 HW 03 Solutions Problem 1
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Part 2 In this problem, the ± oltage signals Vs1 linearly depend on the temperature from sensor and equal Vs1, 0F = 4.0V and Vs1, 100F = 3.2V We need to obtain the output voltage equal to Vout, 0F = 0.0V and Vout, 100F = 10.0V with Rf = 100 ²³ ´µ¶ ·¸¹ º » ¹ ·¼ From the information above, we may set up two linear equations to solve two unknown K1 and K2: Vout, 0F = Vs1, 0F ½ ¾¿ À Vs2 ½ ¾¹ Vout, 100F = Vs1, 100F ½ ¾¿ À Vs2 ½ ¾¹ 0 = 4.0K 1 -2K 2 10 = 3.2K 1 -2K 2 K1 = - 25 / 2 and K2 = - 25 R1 = - Rf / K1 = 100k ³Á¹Â¹Ã º Ä ²³ R2 = - Rf / K2 = 100k ³Â¹Ã º Å ²³ Vout, 50F = Vs1, 50F ½ ¾¿ À Vs2 ½ ¾¹ º ƼÇÁÈ » ¹ÃÂ¹É À È»¹ÉÁÈ»¹ÃÉ º à · EECS 314 Fall 2007 HW 03 Solutions Problem 1
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Part1: First, define nodes A and B. Then, define the currents going into and out of the nodes. Using KCL and the golden rules, we know that the current going into node A equals the current leaving node A. Similarly, the current going into node B equals the current leaving node B. Represented as equations, we have: 2 1 I I 4 3 I I Notice that no current goes into the op-amp. Using ohm’s law, we get: 4 3 I I 4 3 2 R V V R V V OUT A A S Solving for Vout, we get: V S1 V OUT R 1 R 4 R 2 V S2 R 3 A B I 1 I 2 I 4 I 3 EECS 314 Fall 2007 HW 03 Solutions Problem 2
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3 4 2 4 3 ) ( R R V R R V V S A OUT Similarly: 2 1 I I 2 1 1 R V R V V B B S 2 1 2 1 R R R V V S B By the Golden Rules, we know that .
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This note was uploaded on 04/10/2008 for the course EECS 314 taught by Professor Ganago during the Fall '07 term at University of Michigan.

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F07%20314%20HW03%20solutions%20all - EECS 314 Fall 2007 HW...

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