ECEn Matlab

 Signal Transformations

 Objective: Help student gain a visual understanding of several basic
 signal transformations.

 Commands: array operations, plot, labels, figure.

 Background Information: In MATLAB and other programming environments
 continuous signals cannot be represented by vectors or arrays, since
 these are inherently discrete. However, it is possible to approximate a
 continuous signal by taking samples of the signal, where the time
 interval between samples is small enough that the signal exhibits only
 small changes in amplitude between adjacent samples.  For example,
 consider the signal x(t) = exp(-0.5*t)*u(t).  This signal "turns on"
 at t = 0, and then exponentially decays towards 0 with an initial slope
 of -0.5 (at t = 0).  Given this initial slope, the time interval between
 samples should be significantly less than 0.5.  A nice choice might be
 100 times smaller, although this will clearly depend on the application.

 Exercise: Consider the signal x(t) = sin(2*pi/4*t).

 This is a continuous signal, "turning on" at time t = negative infinity and 
 continuing towards t = positive infinity.  Clearly we can't plot this signal 
 for all time in Matlab.  In this exercise, you should limit
 your time axis from -6 s to 6 s, and use a sampling period (interval
 between time points) of 0.01 s.

 First, set up a vector t holding your time values, and create the signal
 x(t) = sin(2*pi/4*t).  This can be done with the following code:

   t_start = -6;
   t_end = 6;
   t_interval = 0.01;

   t = t_start:t_interval:t_end;

   x = sin(2*pi/4*t);


 Then do the following:

   (1) Plot the signal x(t)

   (2) Create the following transformations of the signal x(t), and
       plot each of them in the same figure but different plots using
       the subplot command:

       (a) y1(t) = x(2*t)
       (b) y2(t) = x(0.5*t)
       (c) y3(t) = x(t-1)
       (d) y4(t) = x(-t)
       (e) y5(t) = x(4-2*t)

       Give each plot a title to distinguish the original signal from its
       transformations.

   (3) Answer the following questions:

       (a) Explain the behavior you observe in 2(a) and 2(b).
       (b) Explain in words how the signal x(t) is transformed in 2(e).
 

 Signal Power and Energy

 Objective: Help students gain a better understanding of the relation
 between frequency, total energy, and average power for sinusoidal signals
 of limited time duration.

 Commands: sum 

 Exercise: For this exercise use the function x(t) = cos(2*pi*f*t)(u(t) - u(t-6)); with t
 representing time, and f representing frequency in hz. 
 a) Create four signals with different frequencies, f = [1 10 50 1000].
     ex, y1 = cos(2*pi*f(1)*t)(u(t) - u(t-6))
 b) Use the sum command to calculate the total energy and average power of
     each signal.
 c) Create two plots that graph frequency vs total energy 
     and frequency vs average power
 d) What impact does frequency have on the total energy and average power
    of a sinusoidal signal?

A document to help gain an intuition behind convolution. convolutionguide.docx

The file below work together to create a MATLAB gui that simulates convolution

Please do not modify the following two files.
convolutiongui.m
convolutiongui.fig

You can modify the following three function files to change your signals x(t) and h(t)
x_t_gui.m This file modifies the signal x(t)
h_t_gui.m This file modifies the signal h(t)
time_parameters_gui.m This file returns the time parameters used for x(t) and h(t)

 MT380_UY2_3_1_Convolution Function File

 Objective: Create a function file in matlab that can approximate
 continuous time convolution.

 Commands: function file.

 Background Information: The convolution formula is 
 y(t) = integral (x(tau)*h(t-tau)dtau, 0, t). To implement convolution
 using MATLAB we will have to approximate it since t cannot be continuous.

 Exercise: You will create three different function files. One function
 file will contain a function for x(t), another for h(t), and the last
 function file will be your convolution function file. The parameters for
 the function files of x(t) and h(t) should only have one parameter for
 time (s).
 The headers for the two function files should look something like the ones below.
 function [x] = x_t(t)
 function [h] = h_t(t)
 They take in one value representing one instance in time, and return only
 value; the function evaluated at one value of t.

 The convolution function file header should look like the one below.
 function [y] = convolution(t_array, dtau)
 This function will take in an entire array representing time, and dtau
 which represents the time step between values of tau and call the
 functions x_t and h_t in order to perform the convolution integral. 

 Once you have created the three function files put the following function
 in both function files x_t and h_t.
 if t < 0 || t > 4
     x = 0;
 else
     x = 1;
 end
 Convolve them together using the convolution function file that you created
 by creating an appropriate time array and an appropriate value of dtau. 
 graph it and verify that the graph matches what you expect x_t*h_t to be.

 MT380.UY2.3.2 Convolution of two arrays

Objective: Understand the mechanics of convolution in order to create a
function that will approximate continuous time convolution.

Commands: for-loop, zeros, functions

 Exercise: Create a function that can approximate continuous convolution.
 The functions parameters should consist of two arrays representing signals, 
 x(t) and h(t), and a time_interval that represents the time increments in an array that
 represents time. ex: t = 0:time_interval:5. Your time interval will serve
 as dtau in y(t) = integral (x(tau)*h(t-tau)dtau, -inf, +inf). 
 The function declaration should look like
 function [y] = convolution(x, h, time_interval).
 with x, and h representing two signals.
          a) To test your function let x(t) = h(t) = u(t)-u(t-4) and convolve the
             two signals together.
          b) Plot the computed signal. 



 Some helpful information is below.

 Convolution is defined as 
 y(t) = integral (x(tau)*h(t-tau)dtau, -inf, +inf)
 Working with causal signals, convolution can be rewritten as
 y(t) = integral (x(tau)*h(t-tau)dtau, 0, t) with 0 being the lower bound
 of integration and t being the upper bound of integration. 

 In order to do Convolution in MATLAB the formula has to change since
 MATLAB can only approximate a continuous signal. Time is lost and is
 replaced by indexing.
 For example consider the code:
 t= 0:.1:1-.1;          t(s) = [0 .1 .2 .3 .4 .5 .6 .7 .8 .9];
 x = sin(pi/0.2*t);        x = [0  1  0  1  0  1  0  1  0  1]
 sin(pi/o.2*t(2)) = x(2)


 To visually demonstrate this consider the two arrays below. 
 let t = [0 1 2 3 ] then t(1) represents 0s
 let x(t) = [1 2 3 4] then x(1) = 1 at t = 0s
 let h(t) = [3 2 1 0] then h(4) = 0 at t = 3s
 let y(t) = [3 8 14 20 11 4 0] then y(1) = 3 at t = 0s

 Because indexing in MATLAB begins with 1 instead of 0, the convolution
 formula needs to be adapted for proper indexing. 
 y(index) = integral (x(tau)*h(index-tau)dtau, 1, index)
 tau cannot start at 0 since x(0) does not exist in MATLAB. This means
 that tau must start at 1. However, that creates a problem since both
 index and tau starts at 1. Thus h( (index = 1) - (tau =1 ) ) = h(0) is out of
 range. To fix this, the equation needs to be adapted even further.
 y(index) = integral (x(tau)*h(index+1 - tau)dtau, 1, index)

 Below is an example of this equation. Assume that dtau = 1.


 at index < 1    y(index <1) = integral (x(tau)*h(index+1<2-tau)dtau, 1, index)
                                 |
                                 |
 x[tau]                          |  1    2    3    4
 h[index-tau]          1   2   3 | 
 -------------------------------------------------------------
 tau          -4  -3  -2  -1   0    1    2    3    4    5    6
 During this period, the two signals do not come in contact with
 each other, thus y[index <1] must be zero. This is the property of a causal
 signal convolved with a causal system.

 at index = 1  y(1) = integral (x(tau)*h(2-tau)dtau, 1, index)
                           |
                           |
 x[tau]                    |  1    2    3    4
 h[index-tau]        1   2 |  3  
 -------------------------------------------------------------
 tau        -3  -2  -1   0    1    2    3    4    5    6
 At index = 1, and with tau ranging from 1 to index, the two signals only
 have non-zero values at tau = 1. This indicates that y(1) = 1*3.

 at index = 2   y(2) = integral (x(tau)*h(3-tau)dtau, 1, index)
                              |
                              |
 x[tau]                       |  1    2    3    4
 h[t-tau]                   1 |  2    3   
 -------------------------------------------------------------
 tau           -3  -2  -1   0    1    2    3    4    5    6
 At index = 2, and with tau ranging from 1 to index, the two signals both have non-zero
 values when tau = 1, and 2. This indicates that y(2) = 1*2 + 3*2. This is
 because y(t) depends on the integration (summation in discrete time) of
 the two signals being multiplied together. 

 at index = 3   y(3) = integral (x(tau)*h(4-tau)dtau, 1, index)
                                 |
                                 |
 x[tau]                          |  1    2    3    4
 h[t-tau]                        |  1    2    3   
 -------------------------------------------------------------
 tau          -4  -3  -2  -1   0    1    2    3    4    5    6
 At index = 3, and with tau ranging from 1 to index, the two signals both have non-zero
 values when tau = 1,2, and 3. This indicates that y(3) = 1*1 + 2*2 + 3*3.

 at index = 4   y(4) = integral (x(tau)*h(5-tau)dtau, 1, index)
                                 |
                                 |
 x[tau]                          |  1    2    3    4
 h[t-tau]                        |       1    2    3   
 -------------------------------------------------------------
 tau          -4  -3  -2  -1   0    1    2    3    4    5    6
 At index = 4, and with tau ranging from 1 to index, the two signals both have non-zero
 values when tau = 2,3 and 4. This indicates that y(4) = 2*1 + 3*2 + 4*3. 

 at index = 5   y(5) = integral (x(tau)*h(6-tau)dtau, 1, index)
                                 |
                                 |
 x[tau]                          |  1    2    3    4
 h[t-tau]                        |            1    2    3   
 -------------------------------------------------------------
 tau          -4  -3  -2  -1   0    1    2    3    4    5    6
 At index = 5, and with tau ranging from 1 to index, the two signals both have non-zero
 values when tau = 3 and 4. This indicates that y(5) = 3*1 + 2*4. 

 at index = 6   y(6) = integral (x(tau)*h(7-tau)dtau, 1, index)
                                 |
                                 |
 x[tau  ]                        |  1    2    3    4
 h[t-tau]                        |                 1    2    3
 -------------------------------------------------------------
 tau          -4  -3  -2  -1   0    1    2    3    4    5    6
 At index = 6, and with tau ranging from 1 to index, the two signals both have non-zero
 values when tau = 4. This indicates that y(6) = 4*1.

 at index > 6   y(index>6) = integral (x(tau)*h(index+1>7-tau)dtau, 1, index)
                                 |
                                 |
 x[tau]                          |  1    2    3    4
 h[t-tau]                        |                      1    2    3   
 ------------------------------------------------------------------
 tau          -4  -3  -2  -1   0    1    2    3    4    5    6    7
 At index > 6, and with tau ranging from 1 to index, the two signals only have non-zero
 values. This indicates that y(>6) = 0.

 Because of the time reversal, we need to ensure that MATLAB doesn't try
 to index a part of an array that is out of bounds. There are many ways of
 doing this. The easiest for most people will be an if-statement that
 checks to see if the index is out of bounds. Another way of doing this is
 by zero padding. Zero padding is when you concatenate an array with zeros
 to ensure that MATLAB doesn't index out of bounds.

 UY2_5_1 Convolution Properties
 
  Objective: Gain a visual understanding of the following convolution
  properties: Commutative, Associative, Distributive, Causal*Causal,
  Time-shift, Width, and Area.
 
  Commands: subplot, figure.
 
  Exercise: Use the following functions in this exercise:
  x1(t) = 2u(t) - 2u(t-2)
  x2(t) = 2r(t) -4r(t-1) + 2r(t-2)
  x3(t) = 2u(t) - 3u(t-1) + 3u(t-t) -2u(t-4)
  x4(t) = r(t)u(t)u(-(t-1))
  x5(t) = u(t) - u(t-4)
  Create two figures. In one figure plot each signal in its own subplot.
  In the other figure, plot the results to parts a to e.
  The '*' symbol denotes convolution.
  a) Compute y1(t) = x2(t)*x3(t) and y2(t) = x3(t)*x2(t)
     You should have four subplots in this first figure.
     What is the difference between y1(t) and y2(t)? If there
     is no difference, state that.
  b) Compute y3(t) = [x2(t)*x3(t)]*x4(t) and y4(t) = x2(t)*[x3(t)*x4(t)]
     What is the difference between y3(t) and y4(t)?
     If there is no difference, state that.
  c) Compute y5(t) = x3(t)*[x1(t) + x2(t)] and y6(t) = x3(t)*x1(t) + x3(t)*x2(t)
     What is the difference between y5(t) and y6(t)?
     If there is no difference, state that.
  d) Compute y7(t) = x2(t-1)*x3(t-2). Compare y7(t) with the answer in part a. 
     What is the delay in y(t), and how does it relate to the delay in the two
     signals x2(t-1) and x3(t-2)?
  e) Compute y8(t) = x1(t)*x5(t). How does the area of y8(t) relate to the
     areas of x1(t) and x5(t).
  Use the previous plots created to answer the following questions.
  What is the relation between the width of the convolved signal (y1(t)-y8(t)) and the
  width of the signals that were convolved (x1(t)-x5(t)).
  Signals x1(t)-x5(t) are all causal signals. If you convolved any two or
  more signals together, the output signal must also be (causal /
  non-causal)

Differential Equations-Euler's Method differential_equations_eulersmethod.docx

   UY2_8_1 Impulse Response of 2nd Order LCCDEs
 
  Objectives: Become familiar with Euler's method for solving 1st order
  differential equations; the effect of overdamped, critically damped, and 
  underdamped systems to a sinusoidal frequency; and approximating
  derivations in matlab.
 
  Commands: review roots.
 
  Exercise: Use the figure below for this problem. The value of the
  capacitor is 2e-3F, the value of the inductor is 1H, and the resistor
  will take on four different values R = [1,10,20,40].
  The input signal is 
    Vs = | cos(2*pi*f*t) for 0 <= t < 1(s) 
         |        0      for 1(s) <= t <= 3(s)
    and f represents the frequency in Hz, and is 22 Hz.
  This frequency is close to the resonate frequency of
  the system. 
  Vout is Vo.
 
    You will create an impulse response of the system for every value of R. 
    When you plot, create four different figures for each different
    resistor value. in each figure, create subplots: 1 for h1(t), 1 for
    h2(t), 1 for hc(t), 1 for h(t), and 1 for h(t)*Vs (Convolution)

    When you are creating the impulse responses you will have to create a
    time array for each of them. t = 0:time_interval:time_end. For the
    time_interval use 2/1000, and for time_end use 5*tau. The behavior of
    the impulse response can be captured in a time duration of 5*tau. The
    value of tau will change depending on the value of R. Remember, tau is
    the real part of the root of the characteristic equation.

    Begin by creating your 2nd order differential equation based on the image 
    provided.

    a) Split the second order differential equation into to coupled
    first-order equations. Use Euler's method to approximate h1(t) and
    h2(t). Compare them to the analytical solutions provided in the book by
    plotting them. Remember to do this for each value of R
          - analytic approach. Used for comparison
          - h1A = exp(p(1).*t);
          - h2A = exp(p(2).*t);
    b) Convolve your h1(t) and h2(t) to get hc(t). Compare it to the
    analytical solution provided in the book by plotting them.
          - analytical solution to hc
          - hcA = (1/(p(1)-p(2)))*(exp(p(1).*tc)-exp(p(2).*tc));
          - tc is the time array for hcA
    c) Approximate the derivative of hc to get h(t). Compare it to the 
    analytical solution provided in the book by plotting them.
          - analytical solution
          - hA = (1/(p(1)-p(2)))*(p(1)*exp(p(1).*tc)-p(2)*exp(p(2).*tc));
    d) Convolve your four h(t)s, one for each resistor value, with the input 
       signal and plot them. 
    e) Questions:
       1) How close of an approximation is Euler's method, and for what value
          of R is the approximation the worse and why?
       2) Using the graph that shows Vout, look at the time after the input signal
          stops (t = 1s). What is the relationship between the value of R and
          Vout after (t = 1s). Why does it behave like this?
 
  For further exploration, change the resonate frequency of the input
  signal and see what happens.

For help with this assignment see the document Differential Equations-Euler's Method

 Euler's method 2nd order


 Objective: Use Euler's method for solving 2nd order differential
 equations to approximate the displacement of a car as it drives over a
 pot hole.

 Exercise: You are driving a car at 10m/s when you run over a pot hole
 that is 0.2 meters deep and 0.5 meters wide. The car weighs 1800 kg (450
 kg per wheel), and the suspension system has a spring constant of 10^5
 N/m and a Viscous damper of 5*10^3 Ns/m.
 a) Use Euler's method to plot the path of the car starting from when it
 first hits the pothole. Hint- you will need the cart to start at level
 ground. 
 b) compare the results from part a with the analytical solution. Look at
 table 2-3 on page 71 and the example of driving over a pothole on page
 72 to compute the analytical solution. Plot it on the same graph used in
 part a.
 c) Compare and contrast the two plots

laplacetransformgui.fig laplacetransformgui.m

 S Domain Graph

 Objective: Learn how to plot the magnitude and phase of a Laplace transform
 in order to gain a visual understanding of the s-domain

 Functions: review mesh-grid and mesh

 Exercise: Apply the Laplace transform to the function
 exp(-0.9*t)*cos(2*pi*2*t). Evaluate the Laplace transform at different
 values of s. s can be split into its real part and its imaginary part; 
 s = sigma + j*omega. Let the values of sigma range from -3:.02:3; and the
 values of omega range from 2*pi*(-3:.02:3;). s represents all possible
 combinations of sigma and j*omega. In this case s should be 301x301
 matrix since both arrays of sigma and omega contain 301 elements. 
 a) Evaluate the Laplace transform at every value of s.
 b) calculate the magnitude and phase of the Laplace transform at every
    value of s.
 c) use mesh grid to plot the magnitude and phase calculated in part b.
    The x-axis should be sigma, the y-axis omega, and z-axis the magnitude or
    phase. In the graph represent the y-axis in (hz) and not omega. The
    command should look like this. mesh(sigma,omega/2/pi,Xs_magnitude);
 d) By looking at the graph, determine where the poles and zeros are. Are
    they where they should be?

 Op-amp circuit

 Objective: Graph the s-domain transfer function of an op-amp circuit to
 gain a visual understanding. Also, create a bode plot of the s-domain transfer function
 with sigma = 0 to introduce students to the Fourier domain.

 Functions to learn: none

 Exercise: Using the figure associated with the problem, solve for the
 op-amp's transfer function H(s).
 The values are shown below
 L = 10e-2; % inductor
 C = 10e-6; % capacitor 
 R1 = 10;
 R2 = 2000;
 a)Evaluate H(s) at all possible  combinations of s with
   s = sigma + j*omega, sigma = -20:1:20, and omega = 2*pi*(-1000:10:1000). 
 b) graph the magnitude of H(s) with the frequency axis in units of Hz.
 c) where are the zeros ?
 d) Evaluate H(s) with sigma = 0, and omega = 2*pi*(-1000:10:1000). Graph
   the magnitude in terms of decibels, dB = 20log(Vo/Vin); and with the frequency 
   axis in units of Hz. Also, graph the phase in terms of degrees on the
   same plot. 
 e) In part d you plotted the bode plot of the Op-Amp. This is to
    introduce you into the Fourier domain

 UY5_4_1_ComputationOfFourierSeriesCoefficients

 Objective: Use MATLAB to calculate the Fourier series coefficients of a
 continuous signal.

 Functions: review for-loops

 Exercise: Assume that the figure associated with this problem is continuous over all
 time. 
 a)Use MATLAB to find the Fourier series coefficients for 100 harmonic
   signals, or n = 100 according to equation 5.26a that's located in the book. 
 b)Use the Fourier series coefficients found in part a to calculate the 100
   harmonic signals which the original signal is composed of.
 c)Sum up all of the harmonic signals found in part b to get an approximate
   signal of the original signal. Plot the approximate and original
   signals on the same graph for one period.
 d)Plot the amplitude spectrum and phase spectrum of the signal. You do
   not need to show all 100 frequencies in this graph, but about 20. 
 e)The approximate signal does not perfectly with the original signal. Why
 is this? Does increasing or decreasing the number of harmonics fix the
 strange offset? Hint. Gibbs phenomenon. 

 UY5_12_1_Circuit Analysis Fourier Transform

 Objective: Use Fourier analysis to calculate the output to a system when
 stimulated.

 Exercise: This problem is based off of problem 5.61 that is in the book.
 You will still need to do part a by hand. The book will direct you to the
 needed figures.
 a) Derive the impulse function for the circuit. 
 b) Create a function file that models the impulse function obtained in
    part a.
 c) Create another function file that models the input signal. Remember
    that A = 5mA and T = 3.
 d) Using the functions obtained in part b and c find and plot Vout. Hint
    convolve the functions.
 e) Take a moment to appreciate how much easier that problem became by using MATLAB
    compared to if you had to convolve the two functions by hand. 

 UY6_3_1_Active Filters

 Objective: model a second order butterworth filter using an approximated
 impulse response in order to see its effectiveness in filtering signals

 functions: review roots

 Exercise: This exercise will be broken down into several parts
 part 1) in this first part you will create a low pass filter based on a
       second order butterworth filter. Create it in a function file that
       takes in two parameters (corner frequency in Hz and time interval/step).
       ex. functionName(fc , time_interval). The function file will return
       the impulse response h(t) in an array. Make the time length of the
       impulse response at least 10*tau; usually 5 tau is sufficient but
       let's be extra precise. 
       a) look at table 6-3 on page 285 in your book. This will help you
          find the coefficients for your transfer function. Look at example
          6-10 in the book if you need more help. When designing this
          transfer function, H(jw), make sure you have a dc gain of 1.
       b) look at section 2-7.3 in your book to remember how to turn a
          transfer function into a second order differential equation.
       c) review section 2-8 to transform your second order differential
          equation into an impulse response, h(t).
       If done correctly, you only need to pass the function file fc and
       the time_interval and the function file will return an array
       representing h(t).
 part 2) Now that you have constructed your filter it's time to test it. 
       a) create a low pass filter with an fc of 400 Hz using part 1 of
          the assignment. Use an appropriate time interval. time_interval
          << tau. and time_interval << 1/largest(f), look at part b for
          largest 4)
       b) create three different signals of the form x = 5*cos(2*pi*f*t). 
          the three signals will have different frequencies: f1 = 100, f2
          = 400, f3 = 4000 Hz. When creating these signals, use the same
          time_interval in part a) but have the time array last for 1 s.
       c) Convolve each signal with h(t) to simulate filtering. 
          y = x(t)*h(t), * denotes convolution. 
       d) plot h(t), and the three filtered signals.
 part 3) You will now analyse the plots to verify that your filter is
         working.
       a) you designed your filter with an fc of 400 Hz and a dc gain of
          1. Verify that the filtered signal with f1 = 100 Hz still has a max
          amplitude of 5. 
       b) Now look at the filtered signal with f2 = 400 Hz. Since the
          signal has the same frequency as the corner frequency what should
          its amplitude approximately be? 
       d) Looking at the last signal with f3 = 4000 Hz ignore the first
          part of the filtered signal that has an amplitude twice the value
          that the rest of the signal has, remember that this is just an
          approximation. f3 is ten times the frequency of f2, and you are
          using a second order filter. How much smaller should the
          amplitude of the signal with a freq(f3) be compared to the
          signal(f2)?

   UY6_12_1 Sampling Theorem With Noise

  Objective: This exercise will expose students to the process of sampling
  data, and the effects of aliasing. 

  Functions: fopen, fscan, fclose, interp, and sound. 

  Caution: The command 'sound' will clip at +/- 1V. Before playing any
  array using the sound command, make sure that every value is within the
  range. You should not have to worry about it in this lab, but it is good
  to verify so that you develop the habit. You can easily verify the max
  and min value in your workspace by specifying it to show those values, or
  you can do it by inspection of the graphs that you will plot.

  Exercise: For this exercise you will be given a file to read in. You will
            take the file and read it in. This file contains an array of
            data that represents sound. You will filter, sample, and filter
            again to gain an idea on the effects of aliasing.
            This problem is broken down into 7 parts to help simplify and
            organize it. Some of the descriptions will be long, but read
            them because they will help save you time. 
  Part 1) Get Data From the File
          You will use the command fopen to read in the file
          'PianoDataWithNoise.txt'. Everything in the file represents data used
          in generating the sound; just open up the file in notepad to see its format.
          The file is long and its length is 5512500 which
          represents 5 seconds of data. i.e. 1102500 elements represents 1
          second. When you read in the file, the format of the file is
          exponential notation. After you read the file, make sure you
          close it. 
          Reading the file will take MATLAB about 15 seconds, but it only
          needs to read it in once. After you have the file's content
          stored in a variable, comment it out so it doesn't read it every
          time you run the program. The variable will remain in the
          workspace unless you clear it. 
  Part 2) Creating the filter.
          Before you sample a signal you always want to filter it with a
          lowpass filter. In modulo MT380.UY6.8.1 you created a function
          file that generates a second order filter. Create a filter with a
          corner frequency of 500 Hz. Remember that the samples per second
          is 1102500, and the time interval/step used for creating the
          filter is 1/1102500. 
  Part 3) Filter Data
       a) You will have two arrays of data. One array will not be
          pre-filtered and the other array will be filtered. 
          Use the filter created in part 2 to filter the data obtained in
          part 1, and store the data in another array.
       b) Plot the non-pre-filtered data (NPFD) and the pre-filtered data (PFD) as a
          function of time. 
          I have given you a function file called fft_plot. If you have not
          downloaded it from the wiki page, download it now. This function
          will plot the frequency magnitude spectrum. It takes in two
          parameters: the array containing the data, and the number of
          samples that represent a second. Ex fft_plot(NPFD, 1102500). Use
          this function to plot the frequency magnitude of NPFD and PFD.
          Magnify the plots so you can see the frequencies between 0 and
          1500 Hz. Notice that each signal has a frequency at 1200 and 1500
          Hz; However, the PFD frequency at 100 and 1500 Hz is much smaller
          because of the filter. These frequencies represents possible
          noise that can get into your system. Even though the noise shows
          up in the PFD signal, it is negligibly small, and can be
          ignored. Also, other than the noise, the other frequencies are
          about 500 Hz or below. 
  Part 4) Sample Data
       a) The data that you have been given represents a continuous signal.
          In this part you will simulate sampling a continuous signal.
          Sample the NPFD signal at 1050 samples/second, and sample the PFD
          at 1050 sample/second and 500 samples/second. At this point you
          should have three different data arrays.
             1) a pre-filtered signal sampled at 1050 samples/s or Hz
             2) a pre-filtered signal sampled at 500 samples/s or Hz
             3) a non pre-filtered signal sampled at 1050 sample/s or Hz
       b) These three signals will need to be modified in order to be
          played using the sound command because the sound card can only
          play data at specific frequencies, and they are
          8000,11025, 22050, 44100, 48000, and 96000 Hz.
          To do this, you will need to interpolate the signals using the
          interp command. The signals sampled at 1050 Hz will need to be
          interpolated to have 22050 samples/s and the signal sampled at
          500 Hz will need to be interpolated to have 8000 samples/s.
          Interpolating a signal does not change the frequencies.
      c)  After you have interpolated it, plot all three signals as a
          function of time, and plot their frequency magnitude spectrum using
          the fft_plot function file provided. Compare and contrasts these
          plots to the plots created in part. 
      d)  Questions
          1) After sampling the NPFD at 1050, what frequencies did the
             noise alias to? 
          2) Why are some of the aliased frequencies attenuated?
          3) Why does the PFD signal sampled at 1050 Hz not suffer from
             aliasing?
          4) Why did the PFD signal sampled at 500 Hz become really
             distorted?
  Part 5) Create Filters for sampled data
          For the signals sampled at 1050 Hz, create a filter will a cut
          off frequency of 500 Hz. For the signal sampled at 500 Hz, create
          a filter with a cut off frequency of 250 Hz using the filter
          function file you created in modulo MT380.UY6.8.1. 
  Part 6) Filter the sampled data
       a) Filter the sampled data with their corresponding filters. 
       b) Plot each signal as a function of time, and plot their frequency magnitude 
          using the fft_plot function file provided. Remember to pass in the
          correct samples/s for each signal. 
       c) Questions
          1) Why is it important to filter after sampling data?
          2) If you have a Nyquist sampling rate of 1000 Hz, what should
             the maximum corner frequency of the low pass filter be and why?
  Part 7) Play the data
       a) You now get to play your data using the sound command. Remember
          to make sure that all values in any signal array is within 
          +/- 1.
       b) Play all signals with there corresponding frequencies. The
          signals sampled at 1050 Hz should be played at 22050 Hz, and the signal
          sampled at 500 Hz should be played at 8000 Hz
       c) Compare your results with the sound files provided on the wiki
          page.
       d) Questions
          1) You should hear a low hum in the NPFD signal. It is most
             apparent at the beginning of the sound file. Why is it
             important to pre-filter before sampling.
          2) You signal sampled at 500 Hz doesn't resemble the original
             sound. Why is it important to sample at a frequency greater
             than the Nyquist sampling rate?

fft_plot.m

This file was sampled at 1050 samples/sec, then filtered again with fc of 500 Hz.
You should be able to hear a low buz that is not in the original file

This file was first filtered with fc of 500 Hz, sampled at 1050 samples/sec, then filtered again with fc of 500 Hz.
This file should sound similar to the original just quieter due to filtering

This file was first filtered with fc of 500 Hz, sampled at 500 samples/sec, then filtered again with fc of 250 Hz.
This file shows the effects of aliasing.

   MT380.UY7.2.1 Discrete Time Signal Functions

  Objective: gain a visual understanding of the fundamental period.

  Functions: rat

  Exercise: Consider the causal signal x = cos(2*pi*f*t) with f = 5 Hz. 
           a) Calculate the period of the signal.
           b) Write a function file that can calculate the fundamental
              period of a sampled signal for any period (To) and sample rate
              (Ts). This function must be able to indicate if a fundamental
              period does not exist. For example, in my program, if the
              fundamental period does not exist i return a 0. Hint: for
              this function file, consider using the rat command.
           c) You will sample signal x at the following three sample rates
              (Ts):0.035, 0.25, 0.01, and .01/pi. Using the function file created
              in part b, determine the fundamental period of signal x
              sampled at the three different sample rates. For example, if 
              I were to sample signal x at a sample rate of 0.077 my fundamental 
              period would be 200 samples.
           d) Sample the signal x at the indicated sample rates for the duration of 1
              fundamental period. Continuing my example, my time duration would 
              be (No*Ts) or 15.4 seconds. You should have three sampled
              signals of different time duration
           e) Approximate three continuous time signals (x) by sampling at a
              very high frequency and with a time duration equal to the fundamental 
              period of all three sampling rates.Continuing my example, I
              would create a signal that approximates a continuous signal
              that begins at t = 0s and ends at t = 15.4s.
           f) Plot each sampled signal x with its corresponding continuous
              approximated signal. See solution graphs for a visual
              understanding. If the fundamental period does not exist, do
              not plot the data points, just indicate it in the plot's
              title.
           g) Questions:
              1) For each sampled signal, how many periods of the
                 continuous approximated signal fit within one fundamental
                 period of the sampled signal? 
              2) How do the numbers obtained in part 1 relate to the book 
                 equation 7.21 (No = k*(To/Ts))

   MT380_UY7_13_1_DiscreteTimeFourierSeries

  Objective: Create a Program that simulates the Discrete-Time Fourier
  Series of a sampled signal.

  Commands: none

  Exercise: The file, signal.txt, contains one period of a sampled periodic
            signal. 
            a)Open the file, import the data, and plot the original signal.
            b) Find the expansion coefficients of the given signal.
            c) Plot the magnitude line spectrum and phase line spectrum of
               the imported signal using the expansion coefficients
               obtained in part b
            d) Reconstruct the original signal using the expansion
               coefficients and plot it. 
            e) Question: What is the highest frequency in the signal?