ENGR-25_Prob_2-24_Solution.ppt 1 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods Bruce Mayer, PE Registered Electrical & Mechanical Engineer Engineering 43 Chp 2 Tutorial Problem 2-24 Solution

ENGR-25_Prob_2-24_Solution.ppt 2 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods Amplifier Driving Speaker  Consider an Amplifier Circuit connected to a Speaker Driving Circuit a.k.a. the “SOURCE” Speaker a.k.a. the “LOAD”

ENGR-25_Prob_2-24_Solution.ppt 3 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods Circuit Simplification  Thévenin’s Equivalent Circuit Theorem (c.f. ENGR43) Allows Tremendous Simplification of the Amp Ckt Thevenin +  R S V S

ENGR-25_Prob_2-24_Solution.ppt 4 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods Maximum Power Transfer  The Simplest Model for a Speaker is to Consider it as a RESISTOR only  Since the “Load” Does the “Work” We Would like to Transfer the Maximum Amount of Power from the “Source” to the “Load” BASIC MODEL FOR THE ANALYSIS OF POWER TRANSFER + ─ R S V S SPEAKER MODEL R L  Anything Less Results in Lost Energy in the Driving Ckt in the form of Heat

ENGR-25_Prob_2-24_Solution.ppt 5 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods The Final Ckt Model  Driving Circuit  The Speaker

ENGR-25_Prob_2-24_Solution.ppt 6 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods Electrical Power Physics  For ANY Electrical Device with a: Potential, V, across it A current, I, thru it  V I  Then the Power Used by the Device:  Now OHM’s Law Relates the Voltage- across and Current- Thru a resistor

ENGR-25_Prob_2-24_Solution.ppt 7 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods Voltage Division  Recall the Reduced Ckt Model  This SINGLE LOOP Ckt effectively divides V S across R S and R L  Analysis of this “Voltage Divider” Ckt produces a Relationship between V S & V L

ENGR-25_Prob_2-24_Solution.ppt 8 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods Summary to This Point  What we KNOW By Thévenin Analysis of the Driving Ckt we determined V S & R S  Note that V S & R S are FIXED and beyond our Control as Speaker Designers  The Speaker Designer CAN, however control the Load Resistance, R L  Thus Our Goal Find R L such the Driving Ckt Operates at the Highest Efficiency; i.e., we seek R L that will MAXIMIZE Driver → Load Power Transfer

ENGR-25_Prob_2-24_Solution.ppt 9 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods Analytical Game Plan  Goal Find R L to Maximize P L (R L )  From the Physics we Know

ENGR-25_Prob_2-24_Solution.ppt 10 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods The MATLAB Problem  RS = 10Ω, 15Ω, 20Ω, 25Ω  RL = 10Ω, 15Ω, 20Ω, 25Ω, 30Ω  And  Define Transfer Ratio, r  Then  So to Maximize P L need to Maximize r

ENGR-25_Prob_2-24_Solution.ppt 11 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods MATLAB Game Plan  Concept  Test ALL possible Resistor Combinations then Check for Best  Because we have a small number of allowable values for RS and RL, the most direct way to choose RL is to compute the values of r for each combination of RS and RL. Since there are four possible values of RS and five values of RL, there are 4(5) = 20 combinations.

ENGR-25_Prob_2-24_Solution.ppt 12 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods MATLAB Plan (2)  We can use an array operation in MATLAB to compute r for each of these combinations by defining two 5 × 4 2D-Arrays R_L and R_S. The five rows of R_L contain the five values of RL, and its four columns are identical. The four columns of R_S contain the four values of RS, and its five rows are identical.

ENGR-25_Prob_2-24_Solution.ppt 13 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods MATLAB Plan (3)  The Arrays we Need These Arrays MUST have the same size so that we can perform element-by- element operations with them.

ENGR-25_Prob_2-24_Solution.ppt 14 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods The MATLAB Code % Bruce Mayer, PE % ENGR22 * 20Jan07 * Rev. 13Sep08 % Prob 2.24 * file Demo_Prob2_24_0809.m % % Since all COLUMNS in RL are the same, Define one Col and Replicate in Row Vector % Define RL col a = [10;15;20;25;30]; % Make Array R_L by using a in 4-element Row Vector R_L = [a,a,a,a] % % Since all ROWS in RS are the same, Define one Row and Replicate in Col Vector % Define RS row b = [10,15,20,25]; % Make Array R_S by using a in 5-element Col Vector R_S=[b;b;b;b;b] % % Use Element-by-Element Operations to Calc r % First Sum RS & RL for the 20 combos Rsum = R_S+R_L % Now sq the 20 sums RsumSq = Rsum.^2 % need "dot" as this is element-by-element % Finally Divide RL by SQd sums r = R_L./RsumSq % % Use the max(A) command to find the max value in each COL, and the ROW in in Which the max Values Occurs [max_ratio, row] = max(r)

ENGR-25_Prob_2-24_Solution.ppt 15 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods The.m-File OutPut R_L = R_S = Rsum = r = max_ratio = row = RS = 10RS = 15RS = 20RS = 25 RL = 10 RL = 15 RL = 20 RL = 25 RL = 30

ENGR-25_Prob_2-24_Solution.ppt 16 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods

ENGR-25_Prob_2-24_Solution.ppt 17 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods

ENGR-25_Prob_2-24_Solution.ppt 18 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods 3 3

ENGR-25_Prob_2-24_Solution.ppt 19 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods

ENGR-25_Prob_2-24_Solution.ppt 20 Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods