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Natural and Step Responses of RLC Circuits EE3301 Kamran Kiasaleh Learning Objectives 1. Be able to determine the natural responses of parallel and series RLC circuits 2. Be able to determine the step responses of parallel and series RLC circuits 3. Be able to determine the responses (both natural and transient) of second order circuits with op amps Parallel RLC circuit The step response of a parallel RLC circuit. A series RLC circuit. The step response of a series RLC circuit. The Natural Response of a Parallel RLC 1. Using KCL, This will disappear dv (t ) 1 t v (t ) /R + ∫ 0 v (s)ds + C + I0 = 0 L dt 2 d v (t ) 1 dv (t ) v (t ) + + =0 2 dt RC dt LC The Natural Response of a Series RLC 1. Using KVL, This will disappear di ( t ) 1 t Ri ( t ) + ∫ i ( s ) ds + L + V0 = 0 C 0 dt 2 d i ( t ) R di ( t ) i ( t ) + + = 0 2 dt L dt LC The Natural Response of a Parallel/Series RLC v ( t ) = Ae st ⇐ parallel i ( t ) = Ae st ⇐ series 1 1 s + s+ =0⇒ RC LC R 1 =0⇒ s + s+ L LC 2 2 1 1 1 s=− ± − 2 RC 2 RC LC 2 2 1 R R ± s=− − 2L 2 L LC The Natural Response of a Parallel/Series RLC There are 3 distinct cases. 1 R α= orα = Let 2 RC 2L Then, Neper frequency ω0 = 1 LC s = −α ± α 2 − ω0 2 1)α = ω 0 Critically-damped 2)α < ω 0 Under-damped 3)α > ω 0 Over-damped Resonant radiant frequency The Natural Response of a Parallel RLC There are 3 distinct cases. 1)α = ω 0 ⇒ s1 = s2 = −α s = −α + j ω 2 − α 2 1 0 2)α < ω 0 ⇒ s1 = −α − j ω 0 2 − α 2 s = −α + α 2 − ω 2 0 3)α > ω 0 ⇒ 1 s1 = −α − α 2 − ω 0 2 Critically-damped Under-damped Over-damped The Natural Response of a Parallel/Series RLC How does the response look like? v (t ) = A1e + A2e s1 t i(t ) = A1e + A2e s1 t s2 t s2 t ⇐ parallel ⇐ series Initial conditions must be used to evaluate How do we solve for the unknowns for over-damped (parallel)? v (0 )= A + A Two equations with two unknowns dv (0 ) = As +A s ⇒ + 1 2 + 1 1 dt 2 2 + dv (0 ) ic (0 ) 1 v (0 ) = = − − I0 C dt C R + + Initial voltage across the capacitor Initial current through the inductor How do we solve for the unknowns for the over-damped case(series)? i(0 + ) = A1 + A2 di(0 + )= A s + A s 1 1 2 2 Two equations with two unknowns ⇒ dt + + di(0 ) v L (0 ) 1 = = −Ri(0 + )− V0 dt L L [ Initial current through the inductor ] Initial voltage across the capacitor Example 8.9 The circuit for Example 8.2. I0 = 30mA; v (0 + ) = 12 Figure 8.7 Example 8.2. s1 = −5000;rad /sec s2 = −20,000;rad /sec The Natural Response of an underdamped Parallel/Series RLC How does the response look like? v (t ) = B1e −αt cos(ω d t )+ B2e −αt sin(ω d t ) ⇐ parallel i(t ) = B1e −αt cos(ω d t )+ B2e −αt sin(ω d t ) ⇐ series ωd = ω − α 2 0 Initial conditions must be used to evaluate 2 How do we solve for the unknowns for under-damped (parallel)? v (0 )= B dv (0 ) =ω + Two equations with two unknowns 1 + dt d B2 − αB1 ⇒ + dv (0 ) ic (0 ) 1 v (0 ) = = − − I0 C dt C R + + Initial voltage across the capacitor Initial current through the inductor How do we solve for the unknowns for under-damped (series)? i(0 + ) = B1 di(0 + dt di(0 dt + )=ω Two equations with two unknowns d B2 − αB1 ⇒ ) = v (0 ) = 1 + L L Initial current through the inductor −Ri(0 )− V ] [ L + 0 Initial voltage across the capacitor The circuit for Example 8.4 I0 = −12.25mA; v (0 + )= 0 The voltage response for Example 8.4 s1 = −200 + j979.80 s2 = −200 − j979.80 The Natural Response of a Critically Damped Parallel/Series RLC How does the response look like? v (t ) = D1te + D2e st st Initial conditions must be used to evaluate s1 = s2 = s How do we solve for the unknowns for critically-damped (parallel)? v (0 )= D dv (0 ) = D − αD + 2 + 1 dt 2 Two equations with two unknowns ⇒ + dv (0 ) ic (0 ) 1 v (0 ) = = − − I0 dt C C R + + Initial voltage across the capacitor Initial current through the inductor How do we solve for the unknowns for the critically-damped (series)? i(0 + ) = D2 di(0 + ) = D − αD 1 2 Two equations with two unknowns ⇒ dt + + di(0 ) v L (0 ) 1 = = −Ri(0 + )− V0 dt L L [ Initial current through the inductor ] Initial voltage across the capacitor Example 8.5 (critically-damped)-R has been changed to make this happen ω 0 = 10 α = ω 0 ⇒ R = 4kΩ 6 s = −1000 v (t ) = 98000te −1000t A circuit used to describe the step response of a parallel RLC circuit The Step Response of a Parallel RLC 1. Using KCL, vL ( t ) dvL ( t ) + iL ( t ) + C =I R dt 2 d iL ( t ) L diL ( t ) + iL ( t ) + LC =I 2 R dt dt 2 d iL ( t ) 1 diL ( t ) 1 I + + iL ( t ) = 2 dt RC dt LC LC The Step Response of a Parallel RLC (direct method) 1. First find the natural response 2. Add to the natural response the final value 3. Use the initial conditions to solve for coefficients d iL ,n (t ) 1 diL ,n (t ) 1 + + iL ,n (t ) = 0 2 dt RC dt LC iL (t ) = I f + iL,n (t ) 2 The Step Response of a Parallel RLC (direct method) iL (0 ), + diL (0 + ) ⇐ known dt ' s1 t ' s2 t iL (t ) = I f + A1e + A2e iL (t ) = I f + B e ' −αt 1 iL (t ) = I f + D te ' 1 cos(ω d t )+ B e −αt ' −αt 2 ' −αt 2 +De sin(ω d t ) A circuit used to illustrate the step response of a series RLC circuit. The Step Response of a Series RLC 1. Using KVL, dvc ( t ) d vc ( t ) RC + vc ( t ) + LC =V 2 dt dt diL ( t ) 1 t ' ' L + RiL ( t ) + ∫ iL (t )dt = V dt C 2 d vc ( t ) R dvc ( t ) 1 V + + vc ( t ) = 2 dt L dt LC LC 2 The Step Response of a Series RLC (direct method) v c (0 ), + dv c (0 + ) ⇐ known dt ' s1 t ' s2 t v c (t ) = V f + A1e + A2e v c (t ) = V f + B e ' −αt 1 v c (t ) = V f + D te ' 1 cos(ω d t )+ B e −αt ' −αt 2 ' −αt 2 +De sin(ω d t ) The circuit for Example 8.12 v c (0 + ) = 0, dv c (0 + ) =0 dt v c (t ) = 48 + B1'e −αt cos(ω d t )+ B2'e −αt sin(ω d t ) α = −1400;ω d = 4800 Second order circuits with op amps d 2v 0 (t ) 1 1 = v g (t ) 2 dt R1C1 R2C2 Second order circuits with op amps • • • • This is a variation of the second order system The output is the double integration of the input Depending on the initial charges on the capacitors, the response will vary For a constant input, the output will increase indefinitely d 2v 0 (t ) 1 1 = v g (t ) 2 dt R1C1 R2C2 v g (t ) = V0 V0 v 0 (t ) = t2 2R1C1R2C2 Second order circuits with op amps-imperfect integrator dv0 (t ) RbC2 v 01(t ) = −RbC2 − v (t ) dt τ2 0 dv01(t ) Ra C1 v g (t ) = −RaC1 − v 01(t ) dt τ1 τ1 = R1C1;τ2 = R2C2 v g (t ) d 2v 0 (t ) 1 1 dv0 (t ) 1 + + + v t = 0( ) dt 2 τ τ dt τ τ RaC1RbC2 1 2 1 2 −1 −1 s1 = ;s2 = τ1 τ2 v g (t ) = V0U (t ) v 0 (t ) = V f + A1e s1 t + A2e s2 t ⇔ v 0 (0+ ) = V0τ1τ 2 R1R2V0 = Vf = Ra C1RbC2 Ra Rb dv0 (0+ ) dt =0