Lesson+8 - 1 Lesson 8 Lesson 8 RLC Step Response RLC...

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Lesson 8 1
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RLC Step Response RLC Circuits Ch.8:3-4 Lesson 8 2 Lesson 8
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Challenge 7 Lesson 8 Given v(0) =15 and dv(t)/dt| t=0 = -15x10 4 , determine v(t). The distinct roots of the characteristic equation (Lesson 7) are: s 1 = -5,000, and s 2 = -20,000 (i.e., overdamped ) Assume the solution form: v(t) = A 1 e -5,000t + = A 2 e -20,000t 3
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Challenge 7 Lesson 8 For v(t) = A 1 e -5,000t + A 2 e -20,000t Satisfy initial conditions (by the book): v C (0): A 1 + A 2 = 15; dv C (0)/dt: -5000 A 1 -20000 A 2 = -15 x 10 4 ; » a=[1 1; -5000 -20000] 1 1 -5000 -20000 » b=[15; -150000] 15 -150000 » inv(a)*b 10 5 Solution: A 1 = 10; A 2 = 5 v(t) = 10 e -5,000t + 5 e -20,000t 4
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Challenge 7 Lesson 8 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 5 10 15 v(t) = 10 e -5000t + 5 e -20000t » i=0:10^-5:10^-3; » v=10*exp(-5000*i) + 5*exp(-20000*i); » plot(i,v) V(0)=15 dV(0)/dt = -15x10 4 5
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Lesson 8 Lesson 8 Suppose we are given an at-rest 2 nd order linear system defined below. d 2 y(t)/dt 2 + 2  0 y(t)+ 0 2 =Au(t) ( Step response ); y(0)=0, dy(0)/dt=0 What information is needed to accurately predict the system’s step response ? 6
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Lesson 8 Lesson 8 Previously, natural solutions (a.k.a., unforced or homogeneous) of a parallel RLC circuit were studied in Lesson 7 A more important and pervasive case is a forced (a.k.a., inhomogeneous) response. Step inputs are commonly found in digital and analog control systems. 7
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Lesson 8 Lesson 8 ) ( / ) ( ) ( ; / ) ( ) ( ; / ) ( : : physics Sidebar KCL dt t Cdv t i R t v t i dt t Ldi v I i i i C R L C R L Manipulate ( see text ): I dt t di LC dt t di R L i L L L 2 2 / ) ( ) ( / ) ( Simplify: LC I i LC dt t di RC dt t di L L L 1 / ) ( 1 / ) ( 2 2 8
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Time (sec.) Amplitude Step Response 0 5 10 15 -4 -2 0 2 4 6 8 10 12 From: U(1) To: Y(1) Lesson 8 Lesson 8 9 0 100 200 300 400 500 600 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Mag. Freq. Response 2 nd order filter 0 2 4 6 8 10 12 14 16 18 -0.05 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Unforced 2 nd filter response (underdamped case) y(t)=A 1 e - +j +A 2 e - -j (Lesson 7) 2 nd filter step response (underdamped case) y(t)= K 0 +B 1 e - +j +B 2 e - -j (Lesson 7) Expected step response of a 2 nd order filter.
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Lesson 8 Lesson 8 Author’s indirect approach (read it) Direct approach: Assume the solutions are of the form: i(t) = I f (t) + {assumed natural solution form} (I f = final or steady-state output) v(t) = V f (t) + {assumed natural solution form} (V f = final or steady-state output) 10
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Lesson 8 Lesson 8 Overdamped step response I = 24 mA R = 400 ICs, v C (0)=0,I L (0)=0 at-rest The initial conditions are: 0 and 0 0 0 t L L dt t di dt t Ldi v v / ) ( , / ) ( ) ( ; 000 , 80 ; 000 , 20 ; 10 5 2 1 ; 10 16 1 2 1 4 8 2 0 s s RC LC Roots to the characteristic equation are real and distinct (i.e., overdamped) 11 Resist current change
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Lesson 8 Lesson 8 Overdamped step response recipe Assume the form of the solution: t s t s f L e A e A I t i 2 1 ' 2 ' 1 ) ( Given knowledge that the input is a constant current step function, compute I f
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