ECEn Matlab

Read and follow along with the document to get an introduction to MATLAB.

matlab_intro.docx

After completing the document, make sure that you feel comfortable with the following MATLAB topics:

Matlab Basic Operations
Matrix Operations
Array Operations
Script Files

Even if you feel comfortable, click on the links above and become familiar with the MATLAB guide
provided on the wiki page. You may need to reference it later in the course.

mt240_nr_1_6_1_power.pdf

   MT240_NR_3_2_1_T Parallel Resistors Template
   Objective: learn how to create a function file.

   Commands: function file, sum

  Exercise: Create a function file that can accept an array of resistors
            that represents resistors in parallel. The function will add them 
            together in parallel and return the value.
            Verify your function by adding the following resistors in
            parallel suing your function. [64,64,32,16,8]

mt240_nr_3_2_1_t_parallelresistorstemplate.m

   MT240_NR_3_4_1 Voltage Division
  Objective: In this activity you will learn how to create function files and use Matlab to 
             compute voltage division.

  Commands: Function File (you will learn to implement the a function file)

  Exercise: There are two images associated with this problem. Look on the
            wiki page to view them. Voltage division using resistors is a
            common technique used to obtain a certain voltage. For example.
            suppose you have a 10V voltage source, but you need to connect
            a device to 5V. You could use two resistors in series to divide
            the 10V in order to get 5V across a resistor. You can then
            attach your device in parallel to the resistor that has a 5V
            drop across it. However, The moment you attach the device in
            parallel to the resistor you notice that the voltage is no
            longer 5V. This exercise will help you understand what happens
            when you add a load resistor to a circuit that uses voltage
            division. 
     a) The first step is to reduce the circuit seen in image 1 to look like the circuit in image 2. 
        Use the parallel resistor function file created in modulo
        MT240.NR.3.2.1 to add the resistors in parallel. You may have to
        first add some resistors in series before passing the resistor
        values in the function file.
     b) Now you should have a simple circuit consisting of a 10V source,
        and two equivalent resistors. You connect a potentiomenter (RL) in
        parallel with the second equivalent resistor (see image 2), and it
        begins to draw current. Plot V2 in terms of RL as you change the
        value of RL from 0 to 100 with intervals of 1. RL = 1:100;
     c) Questions:
        1) What two values does V2 asymptotically approach
        2) What impact on V2 does RL have? Why? 

mt240_nr_3_4_1_t_voltagedivisiontemplate.m

   MT240_NR_4_3_1 Node Voltage Method

  Objective: Learn how to solve a system of equation using a matrix in
             MATLAB

  Commands: inv

  Exercise: In this exercise you will study the effects that resistor R2
            has on the voltage drop Vx. The resistor values of R2 are
            R2 = 0:5:30
            a) Using the image provided, set up the system of equation in 
               a matrix for every value of R2. See the template for help.
            b) Solve for Vx and plot Vx as a function of R2 on a graph.
            c) Question: Using the graph, estimate the value of R2 that 
               causes Vx to have 5V across it. 

mt240_nr_4_3_1_t_nodevoltagemethodtemplate.m

   MT240_NR_4_12_1_T Max Power Transfer Template

  Objective: Demonstrate the ability to solve for a T-equiv circuit and see how changes in a load
             resistor affects the power absorbed. 

  Exercise: This exercise is to help you visually understand max power
            transfer by plotting the voltage, current and power at the
            terminals ab (see image provided) as a function of the load
            resistance.
  a) Reduce the circuit using MATLAB (see image 1) to either its Norton or Thevenin
     equivalent. To do so, solve for the voltage open circuit (VOC) and current
     short circuit (ISC) to find the Thevenin resistance. 
  b) Hook the simplified circuit up to an adjustable resistor and calculate
     current, voltage, and power as a function of load resistance. The load
     resistor RL will have values of 0:1:100. 
  c) plot the power, voltage, and current across the load resistor as a
     function of RL
  d) Question: At what value of RL is power the greatest? How does this
     relate to R Thevenin?

You should be able to reduce the circuit above to look similar to one of the two circuits
shown below. RL is the load resistor that you hook up to the circuit across the terminals ab in part b.

mt240_nr_4_12_1_t_maxpowertransfertemplate.m

mt240_nr_4_12_1_s_maxpowertransfersolution.pdf

 MT240_NR_5_3_1 Inverting Op-Amp 

 Background: Op-Amps can amplify a signal up to the value of its rails. For 
             example, consider an inverting op-amp with a gain of -10 and rails
             of -10V and 10V. If the signal x = 3*cos(2*pi*t) was passed through 
             the op-amp, the rails would prevent the output from reaching a
             magnitude of 30. Instead, the rails would clip it from -10 to
             10 and the output signal would resemble a square wave. 
 
 Objective: Use MATLAB to analyse the feedback current when the op-amp goes
            from the linear region into the non-linear region. 
            i.e. when clipping occurs.



 Commands: if statement , mesh

 Exercise: For both parts of this exercise use the image provided and the
           values indicated below.
        
           Rs = 1e3;
           Rf = 1e3:1e3:6e3;
           Vpos = 20;
           Vneg = -20;
           Vin = 5*cos(200*pi*t);

 Part 1) 
      a) Write a function file that simulates and inverting op-amp and
         takes in the parameters Rs, Rf, Vpos,Vneg, and Vin. Ensure that 
         the program takes into account clipping. This means that the 
         output voltage must be within the rails, Vneg <= Vout <= Vpos.
         See the template for the inverting op-amp function file for more
         information. 
      c) Use the function from part 1a to plot Vout as a function of time
         and Rf using this given t array t = 0:.0001: 2*T-.0001;
         The Plot will require a 3D plot. Use the mesh command
      d) Question: What value of Rf causes the output signal to clip?
 
 Part 2) Change the input signal to 5V; Note that it is DC. Also, change the
         values of Rf, and calculate Vout.

         Rf = 1e3:100:5e3;
         Vin = 5;

         Use the calculate Vout to find the feedback current 'If' (see image) 
         a function of Rf. Be sure to capture the behaviour of If in the 
         linear and non-linear region. 
      a) Plot the results. 'If' vs Rf.
      b) Questions:
         1) Estimate the value of Rf when 'If' began to decrease.
         2) Why does 'If' begin to decrease?

mt240_nr_5_3_1_t_inverting_opampfunctionfiletemplate.m

mt240_nr_5_3_1_t_inverting_opamptemplate.m

 MT240_NR_5_4_1 Summing Amplifier 

 Objective: Use MATLAB to simulate a summing operational amplifier to gain
 a visual understanding of adding multiple signals together. 
 
 Commands: subplot
 
 Exercise: 
    a) create a function file that simulates a summing op-amp.
       This function will simulate a summing op-amp with an arbitrary number of
       input voltage sources, a input resistance for every voltage source, one
       feed back resistance, a positive voltage rail, and a negative voltage
       rail. The function will return the output voltage.

         The function will be passed five parameters: Vs, Rs, Rf, Vpos, Vneg.
         Vs represents the input voltages. see below for more detail
         Rs is an array of source resistance
         Rf is the feedback resistance
         Vpos is the positive voltage rail
         Vneg is the negative voltage rail
        
         Rs is an array [R1 R2 ... Rn]        
         Rs = [2e3 2e3 2e3];
         Rf = 3e3;
         Vpos = 20;
         Vneg = -20;

         Vs is a matrix with the columns representing V1, V2, ...Vn and the rows
         representing the values. If each input voltage was DC than there would
         only be one row. Rs is an array containing the input resistance of each
         input voltage. Vpos and Vneg represent the rails.

        Vs is a matrix that looks like
           V1    V2        Vn      
         | V1(1) V2(1) ... Vn(1) |
         | V1(2) V2(2) ... Vn(2) |
         |   :     :         :   |
         | V1(n) V2(n) ... Vn(n) |
        
         V1 = 4/pi*sin(2*pi*t);
         V2 = 4/(3*pi)*sin(6*pi*t);
         V3 = 4/(5*pi)*sin(10*pi*t);
         Vs = [V1' V2' V3'];

        
        See the Template for more information. 
        
    b) Using the image provided, calculate Vout as 
       a function of time. The time array (t) should be 
      (2 times the max period of the three frequencies)

         T = 1;                             this is the max period
         time_begin = 0;                     the beginning of the time array
         time_step = 0.0001;                 the time step
         time_end = 2*T-time_step;           the ending time
         t = time_begin:time_step:time_end;  array that represents time

    c) On the same figure and using the subplot command, plot all three
       input signals and Vout. 
    d) Question: 
           What type of a wave does Vout resemble? 
    e) Calculate Vout with the following input voltage sources

         V1 = 4/pi*sin(2*pi*t);
         V2 = 4/(3*pi)*sin(6*pi*t);
         V3 = 4/(5*pi)*sin(10*pi*t);
         V4 = 4/(7*pi)*sin(14*pi*t);
         V5 = 4/(9*pi)*sin(18*pi*t);
         V6 = 4/(11*pi)*sin(22*pi*t);
         V7 = 4/(13*pi)*sin(26*pi*t);
         V8 = 4/(15*pi)*sin(30*pi*t);
         V9 = 4/(17*pi)*sin(34*pi*t);
         V10 = 4/(19*pi)*sin(38*pi*t);
        
         Let the input resistance (Rs) for each input voltage source be
         2e3. Also, use the same time array you used in part b.
     f) Plot Vout obtained from part e as a function of time on a separate
        figure.
     g) Question:
           What impact does adding more input voltage sources have on Vout?

mt240_nr_5_4_1_t_sum_opampfunctionfiletemplate.m

mt240_nr_5_4_1_t_sum_opamptemplate.m

(Not required, but for fun)

 Fun application
 Any wave or signal can be made up of tons of sinusoidal signals composed of different frequencies and
 amplitudes. The program below creates a square wave from n-number of input signals. Play around with 
 the number of input signals to see the effect of adding more and more input signals.

funapplication_sumopamp.m

To run this program you will need to edit the code and use your summing op amp function.

 MT240_NR_6_3_1 Capacitor

 Objective: Understand the relationship between current, voltage, power
 and energy in a capacitor. 

 Exercise: A Capacitor of 5F with an initial voltage of 5V is attached to
           a current source. The behavior of the current source is given 
           to you below. 
           a) Calculate the voltage across the capacitor. You will need to
              use a for loop to approximate an integral.
           b) Calculate the power stored in the capacitor.
           c) Calculate the energy stored in the capacitor.
           d) Plot Voltage, Power, and Energy in the capacitor as a 
              function of time.
           e) Approximate at what point in time the energy is zero in the 
              capacitor?


 this for-loop generates an array that represents current as a function of
 time. The time array is also given.

 creates an array of length 56
	I = zeros(1,71);
	t = 0:length(I)-1;  time 
	for m = 1:length(I)
		if m < 10
			I(m) = 0;
		elseif m < 20
			I(m) = -5 +.5*m;
		elseif m < 25
			I(m) = -15+m;
		elseif m < 35
			I(m) = 10;
		elseif m < 40
			I(m) = 115-3*m;
		elseif m < 45        
			I(m) = 75-2*m;
		elseif m < 55
			I(m) = -15;
		elseif m < 70
			I(m) = -70+m;
		else
			I(m) = 0;
		end
	end

mt240_nr_6_3_1_t_capacitortemplate.m

  MT240_NR_7_2_1 Natural Response of RC Circuit

 Objective: Gain a visual understanding of how resistors affect the rate a
 capacitor dissipates energy by analysing the voltage across the
 capacitor as a function of time and tau. 

 Exercise: You have several 20mF capacitors with an initial voltage of 20V.
           You design circuits composing of one capacitor and one
           resistor. Every resistor has a different value, and the
           resistor values range from 

               R = 1e3:1e3:10e3 

           The resistor and capacitor are connected together at t = 0.
           Analyse the voltage across every capacitor as a function of
           time.

               C = 20e-3;  the value of the capacitor
               Vinit = 20;  initial voltage across the capacitor at t =
               0s



        a) Calculate the various values of tau for every RC circuit.
           tau = R*C.

           tau = INSERT CODE HERE

        b) Create the time array to use for every circuit. The duration of 
           time must be >= 5*tau of the largest tau.

             the largest tau is from the largest resistor.
             time_beg = 0;
             time_step = 10;
             time_end = 5*tau(?); index you tau array
             t = time_beg:time_step:time_end; an array of time

       c) Calculate the voltage across the capacitors as a function of
          time.

          V(t) = Vinit*exp(INSERT CODE HERE);
 

       d) Calculate the voltage across the capacitor as a function of tau;

           t_tau = 0:0.1:5;  array of tau
           V(tau) = Vinit*exp(-t_tau);

       e) Create two plots: one with the voltage as a function of time and the other 
          with voltage as a function of tau.
       f) Questions:
            1) How does resistance affect the rate at which a capacitor
               discharges?
            2) Why are the plots the same when going by tau?

mt240_nr_7_2_1_t_nat_resp_rc_circuittemplate.m

 MT240_NR_8_2_1 RLC Circuit

 Objective: Gain a visual understanding of how the resistance in an RLC
 circuit can affect the circuit's response. 

 Commands: roots, real, imag 

 Background: So far you have explored specific solutions to the three RLC
 cases: underdamped, overdamped, and critically damped. In this program
 you will explore the general solution of a parallel RLC circuit that generates
 all three cases. Your book introduces the general solution in section 8-1, 
 and we will use it to derive the necessary equation.
 The goal is to create the general solution as shown below.
 X(t) = A1*exp(s1*t) + A2*exp(s2*t)

 Below I will explain how to derive the general solution
 The second order differential equation for a parallel RLC circuit is
 d^2i/dt + (1/RC)di/dt + I/LC = 0. 
 This equation is transformed into the characteristic equation.
 s^2 + (1/RC)*S + 1/LC = 0
 Notice how the equation is a second order polynomial that can be solved.
 By solving for the roots of the characteristic equation you obtain s1, and
 s2. 
 With s1 and s2 known, you can set up a system of equations to solve for
 A1 and A2.
 X_init = A1*exp(s1*0) + A2*exp(s2*0)
 dx/dt = A1*s1*exp(s1*0) + A2*s2*exp(s2*0)
 Now that A1, A2, s1, and s2 are found, you can generate the general
 solution.
 X(t) = A1*exp(s1*t) + A2*exp(s2*t)

 Exercise: You have a parallel RLC circuit as shown in the image below.
 The current source has a value of 3A, the capacitor 100uF, the inductor
 4H, and a potentionmeter (R) assumes the values R = 70:200:1070. Assume
 that the switch has been closed for a long time before opening it at t =
 0s.
 a) Calculate the general solution to the RLC circuit for every resistor.
    There should be 6 resistor values.
 b) Plot the current through the inductor as a function of time for all
    values of R.
 c) What happens as resistance increases? And why?

mt240_nr_8_2_1_t_rlccircuittemplate.m

   MT240_NR_9_1_1 Sinusoid Source

  Objective: Gain a visual understanding of how fast a capacitor can
  discharge and charge for a given tau. Also, learn how to identify the
  voltage waveform of a capacitor.

  Background: It is possible to model a sinusoidal source using a DC source
  if the DC source is connected to a toggling switch that turns on and off.
  In this exercise you will model a sinusoidal source using a DC source. 

  Exercise: You have a circuit as described in the image below. The switch
  toggles between position a and position b. At t = 0 there is no energy
  stored in the capacitor and the switch is in position a. At t = 2*tau
  the switch moves to position b and so on as indicated in the table. 
  Note that at each switch even time starts over at t = 0.
  | time       | position | 
  | t = 0      |     a    |
  | t = 2*tau  |     b    |
  | t = 2*tau  |     a    |
  | t = 4*tau  |     b    |
  | t = tau/4  |     a    |
  | t = 4*tau  |     b    |
  | t = tau/6  |     a    |
  | t = 2*tau  |     b    |
  | t = tau/10 |     a    |
  | t = 4*tau  |    end   | End the simulation

  a) calculate the voltage as a function of time
  b) plot the voltage as a function of time with time being in ms
  c) How would you create a waveform that closely approximates a triangle?
     In other words, how fast must the switch toggle between position a and
     b?  

mt240_nr_9_1_1_t_sinusoidalsourcetemplate.m

 MT240_9_9_1 Mesh_Current Method

 Objective: Design a circuit using capacitors and resistors that cause
 Vout to be 180 degrees out of phase with Vin.

 Background: Oscillators can be constructed from op-amps and an RC network.
 The basic theory is to create a 180 degree phase shift between Vin and
 Vout. This will cuase the op-amp to continuously oscillate in attempt to
 make both inputs the same voltage level (virtual short).

 Exercise: Refer to the provided image for this problem. 
 You want to design a circuit that will oscillate at 40e3 Hz. You only
 have one resistor value (R = 1e3 ohms), but you have the various
 capacitors available (C = 1e-9:1e-10:20e-9). You decide to write a
 program to calculate the angle of Vout in reference to Vin as a function of
 Capacitance. To simplify calculations, you decide that every resistor
 must have the same value and every capacitor must have the same value.
 Also, since you are only interested in the phase shift of Vout, assume
 Vin to have a value of 1AC

                 %Variables
                 C = 1e-9:1e-10:20e-9;
                 R = 1e3;
                 w = 2*pi*40000;
                 ZC = 1./(1j*w*C);
                 Vin = 1;
                 ic = zeros(1,length(ZC));  allocate space
                
          a) Label the currents in each mesh from left to right ia,ib, and ic.
              Use mesh current method to write a system of equations.

                 System of Equations
                 Vin = ia(...) + ib(...) + ic(...)
                 0   = ia(...) + ib(...) + ic(...)
                 0   = ia(...) + ib(...) + ic(...)
                
              Put the system of equations into matrix form.

                                   Matrices
                              Impedances            Currents     A
                    | (...)    (...)    (...)    | * | ia | = | Vin |
                    | (...)    (...)    (...)    | * | ib | = |  0  |
                    | (...)    (...)    (...)    | * | ic | = |  0  |


           b) Solve for ic for every capacitor value. Remember that all
              three capacitors will have the same value. This means that
              you should have 191 different values of ic.

                  %Solve for Currents: ia, ib, ic
                 for m = 1:length(C)
                 Impedances = [INSERT CODE HERE];
                 A = [Vin;0;0];
                 Currents = INSERT CODE HERE;
                 ic(m) = Currents(3);
                 end

           c) Calculate Vout for every value of ic.
           d) Calculate the phase shift of Vout in Reference to Vin. 
           e) Plot the phase shift as a function of capacitance
           f) Question: Approximate the capacitor value that would create
              a phase shift of 180 degrees. 

mt240_9_9_1_t_meshcurrentmethodtemplate.m

   MT240_10_4_1 Complex Power
  
   Objective: Gain a visual understanding of complex power
  
   Functions to learn: refline (this could be useful) to plot avg P and
   reactive P. text (this could be useful) to label your average and
   reactive power.
  
   Exercise: You have seven black boxes(representing unknown circuits), each with terminals 
   coming out of them. You are curious to identify which circuits are more inductive, capacitive, or neither. 
   You begin by measuring the voltage and current across each terminal and measure the voltage phase shift in   
   reference to the current phase shift by setting the current phase shift to zero. From your measurements you 
   gather the following parameters: 
   V = 5*cos(w*t + thetaV), with 'w' being the frequency in rads/s (w = 100*pi),
   't' being the duration of time (t = 0:T/100:2*T - T/100), 'T' being the period 
   of the signal (T = 2*pi/w), thetaV being the voltage phase shift, 
   (thetaV = [-pi/2, -pi/4, -pi/6, 0, pi/6, pi/4, pi/2]. thetaV(1) represents the first
   black box, thetaV(2) the second and so on), and a current I = 1.25*cos(w*t). To gain a visual 
   understanding you decide to calculate and plot the average power, reactive power, and instantaneous
   power for the different voltage phase shifts using the parameters given.
   a) Calculate the voltages as a function of time for all values of thetaV.
   b) Calculate the Vrms and Irms using a for-loop to approximate an integral 
      Vrms = sqrt((1/T)*integral(V(t), from 0 to T)dt) bounds of integration are
      from 0 to T. Verify your calculation using the equation Vrms =
      Vm/sqrt(2).
   c) Calculate average, reactive, and instantaneous power for every thetaV.
   d) Plot voltage, current, average, reactive, and instantaneous power obtained from part c.
      It might be best to create a single plot for each theta since you are
      plotting a lot of information on each graph. Using a legend, titles,
      and labels will help you sort through the information.
   e) Why are all of the calculated Vrms values the same even though the
      phase shift was different?
   f) When the reactant power is zero what is the relationship between the
      phases of voltage and current? What does this tell you about the
      circuit.
   g) When the reactant power is below zero does the voltage lead or lag the current.
      Is the circuit more inductive or capacitive?
   h) When the current and voltage are in phase, why is the instantaneous
      power never below 0? When it is below zero what is happening to the
      power?

mt240_10_4_1_t_complexpowertemplate.m

   MT240_14_4_1 Cross over network

  Objective: Gain an understanding how you can use matlab to help you
  design lowpass, bandpass, and highpass filters.

  functions to learn: log10, semilogx

  Introduction: A crossover network consists of a highpass, lowpass, and
  bandpass filter. They are often used in stereo systems that separates
  a signal into three signals (bass, treble, and midrange). You will design
  a basic crossover network as depicted in the image below. 

  Exercise: Design a crossover network with the following specifications:
  |                        | Low pass | Bandpass | High pass |
  |Lower cut off frequency | N/A      |  250Hz   | 2000Hz    |
  |Upper cut off frequency | 250Hz    |  2000Hz  | N/A       | 
  a) For each filter design you will be calculating the transfer function
     of the voltage across each resistor. The equations should be simple 
     voltage division as shown in the book. See chapter 14. Design your 
     circuits choosing appropriate values for the capacitors and inductors.
  b) Find the magnitudes (|H(jw)|) for v1, v2, and v3 as a function of 'w'
    (frequency) with w being w = 0:10*2*pi:3e5*2*pi. Note that this is the 
     transfer function (H(jw) = vout/vin) thus the amplitude of the input voltage
     source isn't needed in your calculations.
    1) The midrange is a little more difficult so I provided you with the
       steps.
       a) Calculate the bandwidth. B = upper corner frequency - lower fc
       b) Solve for the inductor using the relationship B = R/L
       c) Solve for the capacitor value.
  c) Plot the magnitude in decibels vs the frequency(Hz) using a
     logarithmic scale. (use semilogx for this).
  d) How could you design a bandreject filter that rejects frequencies
     between 250Hz and 2000Hz?

mt240_14_4_1_t_crossover_networktemplate.m

   MT240_15_1_1 Active Filter

  Objective: Design a an active, low-pass filter using an op-amp

  Commands: none

  Background: While designing your crossover network you realized that
  the bass speaker's (low pass) amplifier has broken. You then decide to
  build the low pass filter and amplifier as one unit using an op-amp. 
  You decide to get fancy and use a variable capacitor so that you can
  change the corner frequency.


  Exercise: You decide to have design according to the topology shown in
  the image below.
  a) Design your circuit to have a max gain of 10.
  b) Your variable capacitor can assume values within the range: C = 1e-8:1e-7:1e-6;
  c) For every capacitor value calculate the transfer function |H(jw)| as a 
     a function of frequency using w(rads/sec) = 0:5*2*pi:3e5*2*pi;
  d) Create a bode plot of the results obtained in part c. Plot them all on
     the same graph.
  d) What value should the capacitor be if you want a corner frequency of
     about 140 Hz?
  e) What affect does the size of the capacitor have on the bandwidth?
  

mt240_15_1_1_t_activefiltertemplate.m