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# 509 Class Note for ECE 20100 with Professor Tan at Purdue

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This 7 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Purdue University taught by a professor in Fall. Since its upload, it has received 18 views.

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Date Created: 02/06/15

ECE201 Linear Circuit Analysis I Prof H Z Tan Topic Second order circuits RLC constant inputs 1 Review of series or parallel source free RLC circuits R L W OR Circuit C 2 d x dx 2 c1rcu1t eguatlon 2 2039 mnx 0 dt dt characteristic eguation s2 203 a s SIXS 32 0 3172 039iquot0392 a R 1 6 and n dam in coefficient 039 series or 039 arallel u p g M ZRC p 1 undamped oscillation frequency on E 3 forms ofresponses i 6 gt on overdamped xt K163quot K2e32 81 sz real Ifc gt 0 then s1lt 0 s lt 0 ii 6 on critically damped xt K1 KJquot11 s1 sz 7 6 real Ifc gt 0 then s1 s lt 0 iii 6 lt on underdamped xt ed A cosadtB sinadt ma lm 0392 A amp B are real nal value of the response X1t For all 3 forms of Xt Xoo 0 ifs gt 0 In other words without any inputs any initial energy stored in the capacitor andor inductor will eventually be dissipated by the resistor 2 Now add a constant input to the RLC circuit R Wm OR Circuit Vs C Is a L E T C 2 d x dx c1rcu1t eguatlon 2 2039 mix F ch ch ie the only change is to add a constant F to the right side of the equation characteristic equation same as before ie the characteristic equation is determined by circuit elements R L C not inputs 6 and on same as before response xgtl Let Xnt be the response to the sourcefree circuit 2 d x20 20 dxn t dt a22x t0 dt ie Then xt x t X F where XF is a constant How to find XE i Xoo Xnoo X1 Assume 6 gt 0 then Xn00 0 regardless of 6 Therefore ie XF is the final value XF can be found by noting that for constant inputs capacitors look like open circuits amp inductors look like short circuits after a long time on OR ii Plug Xt Xnt XF into the circuit equation we get d2 d 2039 Lanxn a2nXF F dt F X F 2 a This method requires thatF be computed f1rst How to solve for Km 2 From the 6 lt Determine XF using either method i or ii on relationship determine the proper form of Xnt 0 Determine K1 K2 or AB from the initial conditions xt0 and x39t0 3 An Example V50 7 60 11H 60 ut 60 util Find vca for t 2 0 open at t25 sec lt9 39 7 1F Step 0 Relevant times Step 1 60V C U 69 Analysis at t 0 jiv 3Q l iL039 VstV i mom 120 7 60 tsec L 0 1L0 60 1 i 407 V 69 mm Vc03960V 60 i iLO39 10A limo39 Step 2 Analysis att 0 Equivalent circuit at t 0 Find ic0 because we need it later lt69 Vc0 O I39 i lt3 Q iLO By replacing C amp L with and L at a particular time we emphasize the fact that the values of vc0 amp iL0 should be treated as given during circuit analysis Step 3 Find the characteristic equation and the form of vct Deactivate the independent source to nd equivalent RLC circuit 33 Step 4 Find the constants in the response for 0 S t lt 1 Treat the problem as though t 2 0 and VS E 60V Step 5 Analysis att 1 Step 6 IZOVCD Step 7 Analysis att 1 9 Vow 3Q iL1 Find the response for 1 St lt 25 Find ic1 Step 8 Compute response fort 2 25 sec Capacitor Voltage in V Time in seconds

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