bruce mayer, pe registered electrical & mechanical engineer bmayer@chabotcollege

[email protected] • ENGR-25_Prob_2-24_Solution.ppt1

Bruce Mayer, PE ENGR/MTH/PHYS25: Computational Methods

Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 43

Chp 2 Chp 2 TutorialTutorial

Problem 2-24Problem 2-24SolutionSolution



Amplifier Driving SpeakerAmplifier Driving Speaker Consider an Amplifier Circuit

connected to a Speaker

DrivingCircuita.k.a. the

“SOURCE”

Speakera.k.a. the

“LOAD”



Circuit SimplificationCircuit Simplification Thévenin’s Equivalent Circuit Theorem (c.f. ENGR43)

Allows Tremendous Simplification of the Amp Ckt

Thevenin +

RS

VS



Maximum Power TransferMaximum 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

+

─

RS

VS SPEAKERMODEL

RL

Anything Less Results in Lost Energy in the Driving Ckt in the form of Heat



The Final Ckt ModelThe Final Ckt Model

Source Load

Driving Circuit The Speaker



Electrical Power PhysicsElectrical 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

VIP

RIV



Voltage DivisionVoltage Division Recall the Reduced

Ckt ModelSource Load

This SINGLE LOOP Ckt effectively divides VS across RS and RL

Analysis of this “Voltage Divider” Ckt produces a Relationship between VS & VL

LS

LSL RR

RVV



Summary to This PointSummary to This Point What we KNOW

• By Thévenin Analysis of the Driving Ckt we determined VS & RS

Note that VS & RS are FIXED and beyond our Control as Speaker Designers

The Speaker Designer CAN, however control the Load Resistance, RL

Thus Our Goal

Find RL such the Driving Ckt Operates

at the Highest Efficiency; i.e., we seek RL that will

MAXIMIZE Driver→Load

Power Transfer



Analytical Game PlanAnalytical Game Plan

Goal Find RL to Maximize PL(RL)

From the Physics we Know

32

1

LS

LSL

LLL

LLL

RR

RVV

IRV

IVP



The MATLAB ProblemThe MATLAB ProblemSource Load

RS = 10Ω, 15Ω, 20Ω, 25Ω

RL = 10Ω, 15Ω, 20Ω, 25Ω, 30Ω

And 2

2 S

LS

LL V

RR

RP

Define Transfer Ratio, r

2LS

L

RR

Rr

Then 2SL rVP

So to Maximize PL need to Maximize r



MATLAB Game PlanMATLAB 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.



MATLAB Plan (2)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.



MATLAB Plan (3)MATLAB Plan (3)

The Arrays we Need

30303030

25252525

20202020

15151515

10101010

LR

25201510

25201510

25201510

25201510

25201510

SR

• These Arrays MUST have the same size so that we can perform element-by- element operations with them.

2LS

L

RR

Rr



Th

e MA

TL

AB

Co

de

Th

e MA

TL

AB

Co

de

% 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 cola = [10;15;20;25;30];% Make Array R_L by using a in 4-element Row VectorR_L = [a,a,a,a]%% Since all ROWS in RS are the same, Define one Row and Replicate in Col Vector% Define RS rowb = [10,15,20,25];% Make Array R_S by using a in 5-element Col VectorR_S=[b;b;b;b;b]%% Use Element-by-Element Operations to Calc r%% First Sum RS & RL for the 20 combosRsum = R_S+R_L%% Now sq the 20 sumsRsumSq = Rsum.^2 % need "dot" as this is element-by-element%% Finally Divide RL by SQd sumsr = 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)



The .m-File OutPutThe .m-File OutPutR_L =

10 10 10 10 15 15 15 15 20 20 20 20 25 25 25 25 30 30 30 30

R_S =

10 15 20 25 10 15 20 25 10 15 20 25 10 15 20 25 10 15 20 25

Rsum =

20 25 30 35 25 30 35 40 30 35 40 45 35 40 45 50 40 45 50 55

r =

0.0250 0.0160 0.0111 0.0082 0.0240 0.0167 0.0122 0.0094 0.0222 0.0163 0.0125 0.0099 0.0204 0.0156 0.0123 0.0100 0.0187 0.0148 0.0120 0.0099

max_ratio =

0.0250 0.0167 0.0125 0.0100

row =

1 2 3 4

RS = 10 RS = 15 RS = 20 RS = 25RL =

10RL = 15RL = 20RL = 25RL = 30



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bruce mayer, pe registered electrical & mechanical engineer bmayer@chabotcollege

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