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

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. 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

5. 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

6. 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,

7. 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

8. 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

9. 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

10. 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

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

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

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

14. Pulse Construction
Pulse = Sum of Steps
Examples

15. 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

16. 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

17. 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

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
7e prob 6.36