1 / 19

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

Engineering 43. Chp 7 Step-by-Step Pulse Response. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu. 1 st 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

adonai
Download Presentation

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Engineering 43 Chp 7 Step-by-StepPulse Response Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

  2. 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

  3. 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

  4. FC 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

  5. STEP 1. Assume The Form Of The Solution The Steps: 1-4 • 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

  6. STEP 5: determine the time constant The Steps: 5-6 • 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,

  7. Step-By-Step: Inductor Example • STEP-1: The Form of the Soln • For the Circuit Below Find vO for t>0 • 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

  8. STEP 3: Determine output at 0+ By Inductor Physics Inductor Example cont. • 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

  9. STEP 4: Find Output In Steady State After The Switching By Inductor Physics In Steady State L is SHORT to DC Inductor Example cont.2 • 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

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

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

  12. WhiteBoard Work • Let’s Work This 1st Order Cap Problem • Power Source DISengaged

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

  14. Pulse = Sum of Steps Pulse Construction • Examples

  15. Non-Zero Initial Condition (std ODE) 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 • The Response is Shifted From the Time Origin by an Amount t0 • For CONSTANT fTH, The Time-Shifted Exponential Solution

  16. Piecewise constant source 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

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

  18. PieceWise EndPoints MUST Match

  19. WhiteBoard Work • Let’s Work This 1st Order Cap Problem • R1→4 = 2 kΩ • Power Source ENgaged • IF we Have Time

More Related