Lesson+15

# Lesson+15 - Through this portal you will enter a...

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Through this portal you will enter a Post-Exam #1 world of hidden mysteries. Lesson 15 1

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Transfer Function Chapter 13.4-5 Challenge 15 Lesson 15 Lesson 15 “It has been estimated that two-thirds of the increase in productivity in America in recent decades is attributable to advancements in science and engineering.” National Academy of Engineering 2
Lesson 15 Lesson 15 It is further assumed that you know enough to design me a heart pacemaker (assumes I have a heart). V=6 R C=1microF SCR To date you supposedly have the ability to analyze RLC and KRLC circuits using ODEs, Laplace transforms, and impedance–based methods. 3

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Lesson 15 Your research would tell you that a silicon controlled rectifier (CSR) is a type of thyrister. When voltage is ascending, but less than 5V, the SCR is an open circuit. At 5v, the SCR becomes a 50 A current source which persists until the terminal voltage reaches 0.2V, at which point the SCR shuts off. 4 Open circuit
Lesson 15 Assume the desired heart rate is approximately 1 beat per second . The SCR provides different charging and discharge rates. V=6 R C=1microF S C R Timing Circuit 5

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Lesson 15 V=6 R C=1microF Initially assume that the SCR is an open circuit. Slow charging circuit: v C (t) = 6-6e -t/RC . Want time from 0.2V to 5V to be approximately t 2 - t 1 = 1 sec. Requires RC = 0.569 6
Lesson 15 When SCR is conducting. Fast discharge circuit. v C (t) = (6-IR) + (IR-1)e -(t-1)/RC V=6 R C=1microF I=50 micro 7

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Lesson 15 6 5 0.2 0 1 sec. 0.11 sec. t 1 t 2 8
Lesson 15 … and if you screw up the heart pacemaker, you hope that another engineer got the design of a defibrillator right, or at least has a good lawyer on speed dial. Defibrillator: 3kV 0 0 ms 10 V=6000 R C S L V 0 (t) R=patient Resistance ~50 L=50mH C=31 F But the world is more than 1 st order circuits.. 9

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Lesson 15 What is needed is a strategy for high order continuous-time linear systems that can codify the circuit information embedded in an ODEs, Laplace transforms, or impedance analysis strategies. We will show that the sought after tool is called a transfer function . 10
Transfer Function Lesson 15 A transfer function H(s) assumes a system is at-rest (ICs = 0) and that you can compute of define the circuit’s input X(s), output Y(s) and form that H(s)=Y(s)/X(s). H(s) = Y(s)/X(s) H(s) X(s) Y(s) 11

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Lesson 15 12 Done with transfer functions. Lesson 15 will develop transfer functions from a synthesis (vs. analysis ) viewpoint.
Lesson 15 Classic transfer function problem: DC motor motor: Controller: s s Motor s PID s s Motor s PID / ) ( ) ( / ) ( ) ( 1 H(s) M(s) = 13

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Lesson 15 Response: Why does the actual response differ from the desired response? 14
Lesson 15 A special case is H(s) = Y(s)/X(s); X(s) = 1 resulting h(t)=L -1 (H(s)) that is called an impulse response . - Note: If you can do step responses, you can do impulse response, simply: It will be shown that performing linear filtering is a fundamental and critically important engineering operation (e.g. bandpass filter) which is

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