Step-by-Step Pulse Response

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Presentation transcript:

Step-by-Step Pulse Response Engineering 43 Chp 7 Step-by-Step Pulse Response Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

1st Order Ckts: Step-by-Step This Approach Relies On The Known Form Of The Solution But Finds The ODE Parameters Using Basic Circuit Analysis Tools This Method Eliminates the Need For The Determination Of The Differential Equation Model Most Useful When Variable of Interest is NOT vC or iL

Basic Concept Recall The form of the ODE Solution for a Ckt w/ One E-Storage Element and a Constant Driving Ckt Where K1  The final Condition for the Variable of Interest Can Be determined by Analyzing the Ciruit in Steady State; i.e., t→ x(0+)  The Initial Condition for the Variable Provides the Second Eqn for Calculating K2   Ckt Time Constant Determine By Finding RTH Across the Storage Element

The General Approach Obtain The Voltage Across The Capacitor or The Current Through The Inductor Thevenin With This Analysis Find Time Constant using RTH Final Condition using vTH  FC

The Steps: 1-4 STEP 1. Assume The Form Of The Solution Determine x() STEP 3: Draw The Circuit At t = 0+ The CAPACITOR Acts As a VOLTAGE SOURCE The INDUCTOR Acts As a CURRENT SOURCE Determine The VARIABLE of INTEREST At t=0+ Determine x() STEP 4: Draw The Circuit a Loooong Time After Switching to Determine The Variable In Steady State Determine x(0+) STEP 2: Draw The Circuit In Steady State just PRIOR To Switching And Determine Capacitor-Voltage Or Inductor-Current

The Steps: 5-6 STEP 5: determine the time constant With These 3-Parameters Write the Solution For the Variable of interest using The Assumed Solution RTH Determined at Cap/Ind Connection Terminals Step-By-Step DOWNside Do NOT have ODE So Can NOT easily Check Solution Can usually chk the FINAL Condition STEP 6: Determine The Constants K1 & K2,

Step-By-Step: Inductor Example For the Circuit Below Find vO for t>0 STEP-1: The Form of the Soln STEP-2: Initial inductor current (L is Short to DC) Note That vO is NOT Directly Related to The Storage Element → Use Step-by-Step

Inductor Example cont. STEP 3: Determine output at 0+ By Inductor Physics Note That at t=0+ The 6V Source is DISCONNECTED from the Ckt Elements No Connection on Supply Side Single Loop Ckt At t=0+, Replace The L with a 3A Current Src

Inductor Example cont.2 STEP 4: Find Output In Steady State After The Switching By Inductor Physics In Steady State L is SHORT to DC Recall at t=0+The 6V Source is DISconnected from the Ckt Elements The Ckt Has NO Power Source Over A long Time All the Energy Stored by The Inductor Will be Dissipated as HEAT by The Resistors, Hence

Inductor Example cont.3 STEP 5: Find Time Constant After Switch Find RTH With Respect to the L Terminals Then The Time Constant,  RTH by Series Calc

Inductor Example cont.4 STEP 6: Find The Solution Then The Solution Alternatively use x = v in:

WhiteBoard Work Let’s Work This 1st Order Cap Problem Power Source DISengaged 7e prob 6.42

Pulse Response Consider The Response Of Circuits To A Special Class Of SINGULARITY functions VOLTAGE STEP CURRENT STEP TIME SHIFTED STEP

Pulse Construction Pulse = Sum of Steps Examples

PieceWise Transient Repsonse This expression will hold on ANY interval where the sources are CONSTANT The values of the constants may be different and must be evaluated for each interval The values at the END of one interval will serve as INITIAL conditions for the NEXT interval Non-Zero Initial Condition (std ODE) The Response is Shifted From the Time Origin by an Amount t0 For CONSTANT fTH, The Time-Shifted Exponential Solution

PieceWise Example The Switch is Initially At a. At Time t=0 It Moves To b, and At t=0.5 it moves back to a. Find vO(t) for t>0 On Each Interval Where The Source is Constant The Response Will Be of the Form Piecewise constant source

PieceWise Example cont For 0t<0.5 (Switch at b) t0 = 0 Assume Solution Now Piece-2 (Switch at a) t0=0.5S Find Parameters And Piece-1 Solution

PieceWise EndPoints MUST Match

WhiteBoard Work Let’s Work This 1st Order Cap Problem R1→4 = 2 kΩ Power Source ENgaged IF we Have Time 7e prob 6.36