Chapter 8 - SJTU 1 Chapter 8 Second-Order Circuit SJTU 2 A...

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Unformatted text preview: SJTU 1 Chapter 8 Second-Order Circuit SJTU 2 A second-order circuit is characterized by a second- order differential equation. It consists of resistors and the equivalent of two energy storage elements. What is second-order circuit? Typical examples of second-order circuits: a) series RLC circuit, b) parallel RLC circuit, c) RL circuit, d) RC circuit SJTU 3 1. The Series RLC Circuit 2. The Parallel RLC Circuit 3. Second-Order Circuit Step Response SJTU 4 1. The Series RLC Circuit FORMULATING SERIES RLC CIRCUIT EQUATIONS Eq.(7- 33) SJTU 5 The initial conditions To solve second-order equation, there must be two initial values. SJTU 6 ZERO-INPUT RESPONSE OF THE SERIES RLC CIRCUIT With vT=0(zero-input) Eq.(7-33) becomes Eq.(3-37) try a solution of the form then Eq.(7- 39) characteristic equation SJTU 7 In general, a quadratic characteristic equation has two roots: Eq.(7-40) three distinct possibilities: Case A: If , there are two real, unequal roots Case B: If , there are two real, equal roots Case C: If , there are two complex conjugate roots SJTU 8 A source-free series RLC circuit Special case: Vc(0)=V , I L (0)=0 V(t) V I(t) t M SJTU 9 t M >t>0 t > t M What happens when R=0? SJTU 10 Second Order Circuit with no Forcing Function vc(0) = Vo , iL(0) = Io. I. OVER DAMPED: R=2 , L= 1/3 H, C=1.5F, Vo=1V, Io=2A iL(t) = -0.7 e -0.354t +2.7 e -5.646t A vc(t) = 1.318 e -0.354t -0.318 e -5.646t V SJTU 11 SJTU 12 SJTU 13 II. CRITICALLY DAMPED: R=0.943 , L= 1/3 H, C=1.5F, Vo=1V, Io=2A iL(t) = 2e -1.414t -5.83t e -1.414t A vc(t) = e -1.414t+ 2.75 t e -1.414t V SJTU 14 SJTU 15 SJTU 16 III. UNDER DAMPED: R=0.5 , L= 1/3 H, C=1.5F, Vo=1V, Io=2A iL(t) =4.25 e -0.75t cos(1.2t + 1.081) A vc(t) = 2 e -0.75t cos(1.2t - 1.047) V SJTU 17 SJTU 18 SJTU 19 IV. UNDAMPED: R=0 , L= 1/3 H, C=1.5F, Vo=1V, Io=2A iL(t) =2.915 cos(1.414t + 0.815) A vc(t) =1.374 cos(1.414t - 0.756) V SJTU 20 SJTU 21 SJTU 22 EXAMPLE 7-14 A series RLC circuit has C=0.25uF and L=1H. Find the roots of SOLUTION: For T R =8.5kohm, the characteristic equation is whose roots are These roots illustrate case A. The quantity under the radical is positive, and there are two real, unequal roots at S1=-500 and S2=-8000....
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Chapter 8 - SJTU 1 Chapter 8 Second-Order Circuit SJTU 2 A...

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