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

1. Engr/Math/Physics 25 Chp4 MATLABProgramming-4 Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. Please HELP Rm 3906A Lab • Please do NOT SAVE ANY Files to the DESKTOP on the computers in Rm3906A Lab • Saving to the machine DeskTop Leads to Clutter and Glitchy Computers • Thank You

3. Learning Goals • Write MATLAB Programs That can MAKE “Logical” Decisions that Affect Program Output • Write Programs that Employ LOOPing Processes • For→ No. Loops know a priori • while → Loop Terminates based on Logic Criteria

4. The conditional statements (if, else, elseif) we learned last time allowed us to determine at run-time whether or not to execute a block of code. What these Decision Statements Do NOT do is to allow us to execute a block more than once The TWO Things that Computers Do Better than People STORE Massive Amounts of Data REPEAT operations Loop Structures

5. A “LOOP” is a Program Structure that REPEATS Until some CONDITION is MET The NUMBER of Loops may Be Known a priori (ahead of time) No. of Loops Determined by simple COUNTING Determined Dynamically No. of Loops Determined by a DECISION statement The Loop consists of A Condition Test A Repeated Statement-Block Repetition → LOOPs

6. PreTest Loop Test vs Statement Locations • The key feature → we test to see whether or not to continue before executing the body of the loop. • i.e., The Loop May Not Execute at All • Good if Potential Zero Executions is Desired • a.k.a. “While DO”

7. PostTest Loop Test vs Statement Locations • The Key feature → Do Not Test Until the Block Executes at Least Once • Use if Design Calls for at Least-One Repetition • a.k.a. “DO While”

8. MidTest Loop Test vs Statement Locations • The generalization of both the pre-test and the post-test loops • Empty Block-1 → PreTest Loop • Empty Block-2 → PostTest Loop

9. A PreTested, COUNTED Loop for Loop Statement Start Set k = m • No. Repetitions Known • MATLAB Syntax • forCounter = Start : Increment: Endstatements • end k ≤ n? Increment kby s True Statements-1 False end Statements

10. Given for Loop Counting Variable: k=m:s:n The step value s may be negative Example: k = 10:-2:4 produces k = 10, 8, 6, 4 If s is omitted, the step value defaults to +1 If s is positive, the loop will not be executed if m is greater than n If s is negative, the loop will not be executed if m is less than n If mequalsn, the loop will be executed only once If the step value s is not an integer, round-off errors can cause the loop to execute a different number of passes than intended for Loop Rules

11. For Loop Example • Construct a 23x11 2D Array filled with RANDOM integer between −99 and +99 • Game Plan: • Use Nested for Loops along with rand, round, &fix commands • Track the No. of Construction Steps • The MATLAB Code Time For Live Demo

12. The continue statement passes control to the next iteration of the loop in which it appears, skipping any remaining statements in the body of the loop. The Following Code Uses a continue The continue Statement statement to avoid taking the log of a negative number. x = [10,1000,-10,100]; y = NaN*x; for k = 1:length(x) if x(k) < 0 continue end y(k) = log10(x(k)); end • The Result: y = 1, 3, NaN, 2

13. Let’s Fine Tune the No-Neg-Log Code by COMMENTING OUT the if-continue Commands Remove continue Statement x = [10,1000,-10,100]; y = NaN*x; for k = 1:length(x) %if x(k) < 0 %continue %end y(k) = log10(x(k)); end • The Result: y = 1.0000 3.0000 1.0000 + 1.3644i 2.0000

14. The use of loops and branching can often be avoided, thus creating simpler and faster programs by using a logical array as a mask that selects elements of another array. Any elements not selected will remain unchanged. The following session creates the logical array D from the 3x3 numeric array B Use of a Logical MASK

15. Logical Mask Session Use of a Logical MASK cont >> B = [0, -1, 4; 9, -14, 25; -34, 49, 64] B = 0 -1 4 9 -14 25 -34 49 64 >> D = (B >= 0) D = 1 0 1 1 0 1 0 1 1 Mask Array →a Logical that “masks out” Negative numbers

16. Original B = 0 -1 4 9 -14 25 -34 49 64 Logical MASK cont • Logical Mask Session cont >> B(D) = sqrt(B(D)) B = 0 -1 2 3 -14 5 -34 7 8 >> B(~D) = B(~D) + 50 B = 0 49 2 3 36 5 16 7 8 Negative Values Unchanged → Masked OUT by D(m,n) = 0 Positive Values Unchanged → Masked OUT by D(m,n) = 1

17. Logical Masking Subtlety >> x = [-7 0 8 5 -2] x = -7 0 8 5 -2 >> nn = x>=0 % the logical mask nn = 0 1 1 1 0 >> y = x(nn) y = 0 8 5 >> sqrt1 = sqrt(x(nn)) sqrt1 = 0 2.8284 2.2361 >> sqrt2 = x % make starting copy of x sqrt2 = -7 0 8 5 -2 >> sqrt2(nn) = sqrt(sqrt2(nn)) sqrt2 = -7.0000 0 2.8284 2.2361 -2.0000 ONLY the Three Sq-Roots the Three Sq-Roots AND the two NON-Roots

18. The while loop is used when the looping process terminates because a specified condition is satisfied, and thus the number of passes is not known in advance. A simple example of a while loop is while Loops x = 5; while x < 25 disp(x) x = 2*x - 1; end • Results from the disp statement are 5, 9, and 17.

19. A PreTested DYNAMIC Loop Start while Loop Statement Set Loop VarInitial value LogicalDecision • No. Repetitions UNknown • MATLAB Syntax • whileLogical Expression • statements • end True Statements(MUST IncrementLoop Variable) False end Statements

20. For the while loop to function properly two conditions must occur Start while Loop Statement Set Loop VarInitial value LogicalDecision • The loop variable must have a value BEFORE the while statement is executed (initialize) • The loop variable must be changed somehow by the statements INSIDE the Loop True Statements(MUST IncrementLoop Variable) False end Statements

21. A simple while loop k = 1 x = 9 k = 2 x = 17 k = 3 x = 33 while Loop Build Vector • The Results x = 5;k = 0; while x < 25 k = k + 1 y(k) = 3*x; x = 2*x-1 end >> y y = 15 27 51 • The loop variable x is initially assigned the value 5, and it keeps this value until the statement x = 2*x - 1 is encountered the first time. Its value then changes to 9. Before each pass through the loop, x is checked to see if its value is less than 25. If so, the pass is made. If not, the loop terminates

22. Write a .m- file to determine The min. number of terms required for the sum of the series 5k2 – 2k; k = 1, 2, 3, … to just exceed 10,000. the sum for this number of terms The .m-file and the Results Another while Loop Example tot = 0;k = 0; while tot < 10e3 k = k + 1; tot = 5*k^2 - 2*k + tot; end disp('No. terms = ') disp(k) disp('The Sum = ') disp(tot) No. Terms = 18 Sum = 10203

23. Demos: for & while • Prob 4-22 → Evaluate with for • Also list the value of the individual Terms • Use whileto find the number of terms, qmax, such that Time For Live Demo

24. The switch structure provides an alternative to using the if, elseif, and else commands. Anything programmed using switch can also be programmed using if structures. However, for some applications the switch structure producesmore readable code than when using the if structure. The switch Structure

25. switch input expression (which can be a scalar or string). case value1 statement group 1 case value2 statement group 2 . . . otherwise statement group n end MATLAB switch Syntax

26. This switch Block displays the High School Class-Name that Corresponds to a Given Grade Level switch Example grade_level = input('Hi-School Grade Level.: '); switch grade_level case 9 disp(' Freshman') case 10 disp(' Sophomore') case 11 disp(' Junior') case 12 disp(' Senior') otherwise disp(' NOT a Hi-Schl Grade Lvl') end

27. switch Example Results Hi-School Grade Level.: 9 Freshman Hi-School Grade Level.: 11 Junior Hi-School Grade Level.: 13 NOT a Hi-Schl Grade Lvl Hi-School Grade Level.: 10 Sophomore

28. Consider an Electrical Diode → I V REALBehavior IDEALModel OFFSETModel LINEARModel Example: Prob 4.27 • We can MODEL the V-I Behavior of this Device in Several ways

29. Problem-27 cont • The Diode exhibits a form of RECTIFICATION • i.e., It allows current to Flow in the FORWARD direction, But NOT in the REVERSE direction • Think of a diode as a “Check-Valve” for Electrical Current”

30. Problem-27 cont • Now Let’s Connect the Diode to • A Power Source, Vs • A Useful Load, RL +VL- • Next Assume that Vs is a Decaying Sinusoidal, Alternating Current (AC) Voltage-Source modeled mathematically as

31. Problem-27 → Plot Vs +VL- % Bruce Mayer, PE * 08Sep11 % ENGR25 * Problem 4-27 % file = Prob4_27_Vs_plot.m % INPUT SECTION tmax = input('Max time in sec = '); Vmax = input('Max Supply Potential in V = '); %CALCULATION SECTION % use linspace command to generate 500 time pts t = linspace(0,tmax,500); % Use for-Loop to generate plotting vector, vs for k = 1:500 % Calc SUPPLY V-Level vsup = Vmax*exp(-t(k)/3)*sin(pi*t(k)); vs(k) = vsup; end % PLOT SECTION plot(t,vs),ylabel('Load Voltage (V)'),xlabel('Time (sec)'),... title('Ideal-Diode Rectifier'), grid disp('Plot Complete')

32. Problem-27 → Plot Vs +VL- Diode ON

33. Recall the Ideal-Diode Model → IDEALModel Prob 27 cont • With This Diode Behavior weExpect Load a Voltage in this form +VL- • Write a MATLAB Program to Plot VL vs t for: 0 t  10s

34. Problem-27 → Plot VL Ideal % Bruce Mayer, PE * 08Sep11 % ENGR25 * Problem 4-27a % file = Prob4_27a_ideal_diode.m % INPUT SECTION tmax = input('Max time in sec = '); Vmax = input('Max Supply Potential in V = '); % CALCULATION SECTION % use linspace command to generate 500 time pts t = linspace(0,tmax,500); % Use for-Loop to generate plotting vector, vL for k = 1:500 % Calc SUPPLY V-Level at the current t(k) vs = Vmax*exp(-t(k)/3)*sin(pi*t(k)); % chkFwd or Rev condition by if-else if vs > 0 vL(k) = vs; % diode absorbs NO voltage else vL(k) = 0; % diode BLOCKS ALL Current end end plot(t,vL),ylabel('Load Voltage (V)'),xlabel('Time (sec)'),... title('Ideal-Diode Rectifier'), grid +VL- VS

35. IDEALModel Problem-27 → Plot VL Ideal

36. Recall the OffSet-Diode Model → OFFSETModel Prob 27 cont • With This Diode Behavior weExpect Load Voltage in this form +VL- • Write a MATLAB Program to Plot VL vs t for: 0 t  10s

37. Problem-27 → Plot VL Offset % Bruce Mayer, PE * 08Sep11 % ENGR25 * Problem 4-27b % file = Prob4_27b_offset_diode.m % INPUT SECTION tmax = input('Max time in sec = '); Vmax = input('Max Supply Potential in V = '); % CALCULATION SECTION % use linspace command to generate 500 time pts t = linspace(0,tmax,500); % Use for-Loop to generate plotting vector, vL for k = 1:500 % Calc SUPPLY V-Level at current t(k) vs = Vmax*exp(-t(k)/3)*sin(pi*t(k)); % chkFwd or Rev condition by if-else if vs > 0.6 vL(k) = vs-0.6; % diode absorbs 0.6V else vL(k) = 0; % diode BLOCKS All current end end plot(t,vL),ylabel('Load Voltage (V)'),xlabel('Time (sec)'),... title('Offset-Diode Rectifier'), grid +VL- VS

38. OFFSETModel Problem-27 → Plot VL Offset

39. Compare Plots Side-by-Side +VL- Prob 27 Analysis • 0.6V Offset has a large affect when the Vs amplitude is only 3V • OffSet is 20% of amplitude

40. Plots for 24V amplitude Prob 24 Analysis +VL- • Makes less difference • Note different vertical scales

41. All Done for Today SinusoidalHalfWaveRectifier

42. Engr/Math/Physics 25 Appendix Time For Live Demo Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu