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Ch06 Higher Order Networks 1s09 - Chapter 6 Second-Order...

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Higher Higher- Order Networks Order Networks Chapter 6 Department of Electrical and Electronics Engineering University of the Philippines - Diliman Artemio P. Magabo Professor of Electrical Engineering Professor of Electrical Engineering Revised by Luis G. Sison, Jan 23, 2004 Revised by Luis G. Sison, Jan 23, 2004 Revised by Jhoanna Pedrasa , July 2005 Department of Electrical and Electronics Engineering p2 Second-Order Transients The solution can be shown to be an exponential of the form st K x ε = Consider the homogeneous differential equation 0 cx dt dx b dt x d a 2 2 = + + with initial conditions x(0)=X 0 and =X’ 0 . ) 0 ( xt dx where K and s are constants. Substitution gives 0 cK bsK K as st st st 2 = ε + ε + ε Department of Electrical and Electronics Engineering p3 After canceling the exponential term, we get the characteristic equation 0 c bs as 2 = + + Using the quadratic formula, we get the two roots a 2 4ac - b b - s , s 2 2 1 ± = Assuming the roots are real and distinct, the solution will consist of two exponentials. Thus t s 2 t s 1 2 1 K K ) t ( x ε + ε = K 1 and K 2 can be evaluated using x(0) and (0). dt dx Department of Electrical and Electronics Engineering p4 Source-Free Series RLC Network Consider the circuit shown. From KVL, we get for t 0 0 idt C 1 Ri dt di L = + + Differentiating, we get 0 i C 1 dt di R dt i d L 2 2 = + + This is a homogeneous second-order differential equation. R i E + - v C + - L C t=0 Department of Electrical and Electronics Engineering p5 The characteristic equation is 0 C 1 Rs Ls 2 = + + 0 LC 1 s L R s 2 = + + or From the quadratic formula, we get the two roots LC 1 L 2 R L 2 R s , s 2 2 1 - ± - = Note: There are three types of root depending on the value of the term inside the square root sign. Department of Electrical and Electronics Engineering p6 1. Overdamped Case: The roots are real and distinct when LC 1 L 2 R 2 > t s 2 t s 1 2 1 K K ) t ( x ε + ε = The solution is the sum of two exponential terms 2. Critically Damped Case: The roots are real but repeated when LC 1 L 2 R 2 =
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Department of Electrical and Electronics Engineering p7 st 2 1 ) K t K ( ) t ( x ε + = The solution can be shown to be LC 1 L 2 R 2 < 3. Underdamped Case: The roots are complex conjugates when d 2 1 j - s , s ω ± α = ) t sin K t cos K ( ) t ( x d 2 d 1 t - ω + ω ε = α If the roots are , the solution can be shown to be Department of Electrical and Electronics Engineering p8 Comparison of Responses overdamped critically damped underdamped underdamped envelope response e - α t Department of Electrical and Electronics Engineering p9 From KVL, we get for t 0, 0 = + + c v Ri L dt di Since the circuit has reached steady-state at t=0, i(0 + )=0 and v C (0 + )=12V. Substitution gives A/s dt di 12 0 0 - = - = + + L ) ( v ) ( c Example: The circuit has reached steady-state when the switch is moved at t =0. Find ) 0 ( dt di + i(0+) and .
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