1. February 21, 2005 Lecture 10 - By Paul Lin 1 CPET 190 Problem Solving with MATLAB Lecture 10 http://www.etcs.ipfw.edu/~lin

2. February 21, 2005 Lecture 10 - By Paul Lin 2 Lecture 10: MATLAB Program Design & Flow Control - III 10-1 SWITCH Construct 10-2 Problem Solving - MATH Applications 10-3 Problem Solving - Electrical Circuit Applications

3. February 21, 2005 Lecture 10 - By Paul Lin 3 10-1 SWITCH, CASE, WHATEVER SWITCH Construct Another form of flow control or branching Another form of flow control or branching General form General form switch (expression) case exp_1, statement 1-1 statement 1-1 statement 1-2 statement 1-2 **** **** case exp_2, case exp_2, statement 2-1 statement 2-1 statement 2-2 statement 2-2 **** **** case exp_3, case exp_3, statement 3-1 statement 3-1 statement 3-2 statement 3-2 **** ****end

4. February 21, 2005 Lecture 10 - By Paul Lin 4 10-1 SWITCH, CASE, WHATEVER SWITCH Construct Example Example N = input(‘Enter a number - less than 10-: ‘); switch (N) case {1, 3, 5, 7, 9}, disp(‘Odd number is matched’);, disp(‘Odd number is matched’);, case {2, 4, 6, 8, 10}, disp(‘Even number is matched’); disp(‘Even number is matched’); otherwise, otherwise, disp(‘No match is found’); disp(‘No match is found’);end

5. February 21, 2005 Lecture 10 - By Paul Lin 5 10-2 Problem Solving - MATH Applications Example 10-1: Evaluating a Function of Two Variables Example 10-1: Evaluating a Function of Two Variables Write a MATLAB program to evaluate a function f(x,y) for any two user-specified values x and y. The function is defined as follows: Write a MATLAB program to evaluate a function f(x,y) for any two user-specified values x and y. The function is defined as follows:

6. February 21, 2005 Lecture 10 - By Paul Lin 6 10-2 Problem Solving - MATH Applications Example 10-1: Evaluating a Function of Two Variables Example 10-1: Evaluating a Function of Two Variables Solution: Solution: 1.State the problem 2.Define the inputs and outputs 3.Design the algorithm 4.Translate the algorithm into MATLAB statements 5.Test the program 6.Document the process and solution.

7. February 21, 2005 Lecture 10 - By Paul Lin 7 10-2 Problem Solving - MATH Applications Example 10-1: Evaluating a Function of Two Variables (continue) Example 10-1: Evaluating a Function of Two Variables (continue) Solution: Solution: 1.State the problem – as the question 2.Define the inputs and outputs User input two values for X and Y User input two values for X and Y Output is a function value Output is a function value 3.Design the algorithm Read the input X and Y Read the input X and Y Calculate f(x,y), based on the X and Y values Calculate f(x,y), based on the X and Y values Print output Print output

8. February 21, 2005 Lecture 10 - By Paul Lin 8 10-2 Problem Solving - MATH Applications Example 10-1: Evaluating a Function of Two Variables (continue) 4.Translate the algorithm into MATLAB statements %func_evs.m% n = true; while n == true x = input('Enter the x coefficient: '); x = input('Enter the x coefficient: '); y = input('Enter the y coefficient: '); y = input('Enter the y coefficient: '); if x >= 0 && y >= 0 f_xy = x + y; f_xy = x + y; elseif x >= 0 && y = 0 && y < 0 f_xy = x + y^2; f_xy = x + y^2; elseif x == 0 && y >= 0 elseif x == 0 && y >= 0 f_xy = x^2 + y; f_xy = x^2 + y; else else f_xy = x^2 + y^2; f_xy = x^2 + y^2; end end fprintf('x = %f, y = %f, f(x,y) = %f\n', x, y, f_xy); fprintf('x = %f, y = %f, f(x,y) = %f\n', x, y, f_xy); disp('Enter Ctrl C one or two times to exit the program'); disp('Enter Ctrl C one or two times to exit the program'); disp(''); disp('');end

9. February 21, 2005 Lecture 10 - By Paul Lin 9 10-2 Problem Solving - MATH Applications Example 10-1: Evaluating a Function of Two Variables (continue) Example 10-1: Evaluating a Function of Two Variables (continue) 5.Test the program (x,y)  f(x,y)(x,y)  f(x,y) (2,3) -> f(x,y) = 5(2,3) -> f(x,y) = 5 (2, -3) -> f(x, y) = 11(2, -3) -> f(x, y) = 11 (-2, 3) -> f(x,y) = 7(-2, 3) -> f(x,y) = 7 (-2,-3) -> 13(-2,-3) -> 13 How about x, y => a very large number, and a very small number (accuracy, and computer’s limit) How about x, y => a very large number, and a very small number (accuracy, and computer’s limit)

10. February 21, 2005 Lecture 10 - By Paul Lin 10 10-3 Problem Solving - Electrical Circuit Applications Example 10-2: Cardioid Microphone Example 10-2: Cardioid Microphone Domain Knowledge on Microphone Pick-up patterns Domain Knowledge on Microphone Pick-up patterns Omni pattern – the most common patternOmni pattern – the most common pattern Cardioid pattern – A directional sound pick-up pattern for use on a stageCardioid pattern – A directional sound pick-up pattern for use on a stage Super Cardioid patternSuper Cardioid pattern Source: http://homerecording.about.com/library/weekly/aa032597.htm

11. February 21, 2005 Lecture 10 - By Paul Lin 11 10-3 Problem Solving - Electrical Circuit Applications Example 10-2: Cardioid Microphone Example 10-2: Cardioid Microphone Cardioid pattern – A directional sound pick-up pattern for use on a stageCardioid pattern – A directional sound pick-up pattern for use on a stage Gain EquationGain Equation It varies as a function of angle according to the equation It varies as a function of angle according to the equation g is the constant associated with a particular microphone; (value of g?; assume g = 0.5) g is the constant associated with a particular microphone; (value of g?; assume g = 0.5) θ is the angle from the axis of the microphone to the sound source (range of angle?) θ is the angle from the axis of the microphone to the sound source (range of angle?) Questions Questions Calculate gain versus angle andCalculate gain versus angle and prepare a polar plot (what is this?)prepare a polar plot (what is this?)

12. February 21, 2005 Lecture 10 - By Paul Lin 12 10-3 Problem Solving - Electrical Circuit Applications Example 10-2: Cardioid Microphone - continue Example 10-2: Cardioid Microphone - continue POLAR(THETA, RHO) THETA  theta in Gain equation THETA  theta in Gain equation RHO -> Gain value RHO -> Gain value POLAR Polar coordinate plot.  POLAR(THETA, RHO) makes a plot using polar coordinates of the angle THETA, in radians, versus the radius RHO.  POLAR(THETA,RHO,S) uses the linestyle specified in string S.

13. February 21, 2005 Lecture 10 - By Paul Lin 13 10-3 Problem Solving - Electrical Circuit Applications Example 10-2: Cardioid Microphone – the Program Example 10-2: Cardioid Microphone – the Program%cardioid_mic.m %Define Variables: % g -- Microphone gain constant % gain - Gain(theta) of the microphone % theta - Angle in radians from % microphone axis % Calculate gain versus angle g = 0.5; theta = 0:pi/20:2*pi; gain = 2*g*(1 +cos(theta)); % Polar plot Gain vs Angle polar(theta, gain, 'r-'); title('\bfGain vs Angle \theta'); Verifying Answrs Verifying Answrs

14. February 21, 2005 Lecture 10 - By Paul Lin 14 10-3 Problem Solving - Electrical Circuit Applications Example 10-2: Cardioid Microphone – Verifying the answer Example 10-2: Cardioid Microphone – Verifying the answer Verifying Answers: Gain = 2*g*(1+cos(theta)) Verifying Answers: Gain = 2*g*(1+cos(theta)) g = 0.5, theta = 0, gain = 2*0.5*(1+1) = 2g = 0.5, theta = 0, gain = 2*0.5*(1+1) = 2 g = 0.5, theta = 60 degree, gain = 2*0.5*(1 +0.5) = 1.5g = 0.5, theta = 60 degree, gain = 2*0.5*(1 +0.5) = 1.5 Theta = 0, gain = 2 Theta = 60 degree, gain = 2

15. February 21, 2005 Lecture 10 - By Paul Lin 15 10-3 Problem Solving - Electrical Circuit Applications Example 10-3: Plot the amplitude and frequency response of Low-Pass Filter A simple RC low- pass filter A simple RC low- pass filter Components: Vi, R and C are in series Components: Vi, R and C are in series R = 16 kΩ, C = 1 μF, R = 16 kΩ, C = 1 μF, Xc = 1/(2*pi*f*C) Xc = 1/(2*pi*f*C) Complex number calculation on Amplitude and Phase angle of the Gain = Vo/Vi Complex number calculation on Amplitude and Phase angle of the Gain = Vo/Vi Circuit Equations? Circuit Equations?

16. February 21, 2005 Lecture 10 - By Paul Lin 16 10-3 Problem Solving - Electrical Circuit Applications Example 10-3: Circuit Equations? Example 10-3: Circuit Equations? Xc = -j*1/(2*pi*f*C); Xc = -j*1/(2*pi*f*C); as f ↑, Xc↓ Vo = Vc = Vi/(R –jXc) * (- jXc) Vo = Vc = Vi/(R –jXc) * (- jXc) Gain of response is Vo/Vi = (-jXc)/(R-jXc) Gain of response is Vo/Vi = (-jXc)/(R-jXc) Vo/Vi = 1/(1 + j2πfRC) Frequency range? Frequency range? R*C = 16k * 1 uF = 16 milli-secondsR*C = 16k * 1 uF = 16 milli-seconds F = 1/RC about 60 HzF = 1/RC about 60 Hz Use 0 to 1000 Hz as rangeUse 0 to 1000 Hz as range

17. February 21, 2005 Lecture 10 - By Paul Lin 17 10-3 Problem Solving - Electrical Circuit Applications Example 3: MATLAB Program Example 3: MATLAB Program%rc_lpf.m% R = 1600; % 1600 ohms C = 1E-6; % 1 uF f = 1:2: 1000; % Frequency varies from % 1 to 1000 Hz % Response Av = 1./ (1 + j*2*pi*f*R*C); Av_amp = abs(Av); % amplitude % Phase angle phase = angle(Av) % Plots figure(1), subplot(2,1,1), loglog(f, Av_amp); title('Amplitude Response - Gain'); xlabel('Frequency - Hz'); ylabel('Av = Vo/Vi'); grid on figure(1), subplot(2,1,2), semilogx(f, phase); title('Phase Response'); xlabel('Frequency - Hz'); ylabel('Angle of Av = Vo/Vi in Radians'); grid on

18. February 21, 2005 Lecture 10 - By Paul Lin 18 10-3 Problem Solving - Electrical Circuit Applications

19. February 21, 2005 Lecture 10 - By Paul Lin 19 Summary SWITCH, CASE, WHATEVER SWITCH, CASE, WHATEVER Problem Solving - MATH Applications Problem Solving - MATH Applications Problem Solving - Electrical Circuit Applications Problem Solving - Electrical Circuit Applications