ecen_380_matlab_assignments
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| - | ^ Chapter ^ Modulo ^ Course Topic ^ MATLAB Topic ^ created ^ | ||
| - | | Ch 1 | [[ecen_380_assignments#MT380.UY1 | MT380.UY1 ]] | Signals | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY1.1.1 | MT380.UY1.1.1 ]] | Types of Signals | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY1.2.1 | MT380.UY1.2.1 ]] | Signal Transformations | [[matlab_guide#Array Operations| Array Operations]],[[matlab_guide#Graph Function| plot]] ,[[matlab_guide#Annotation| Graph Annotation]] ,[[matlab_guide#Figures| Figures]] | yes | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY1.3.1 | MT380.UY1.3.1 ]] | Waveform Properties | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY1.4.1 | MT380.UY1.4.1 ]] | Nonperiodic Waveforms | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY1.5.1 | MT380.UY1.5.1 ]] | Signal Power and Energy | sum | yes | | ||
| - | | Ch 2 | [[ecen_380_assignments#MT380.UY2 | MT380.UY2 ]] | LTI Systems | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY2.1.1 | MT380.UY2.1.1 ]] | LTI Systems | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY2.2.1 | MT380.UY2.2.1 ]] | Impulse Response | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY2.3.0 | MT380.UY2.3.0 ]] | Convolution Aid | | yes | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY2.3.1 | MT380.UY2.3.1 ]] | Convolution with function files | [[matlab_guide#Function Files| Function Files]], [[matlab_guide#For-loop| For-loop]] | yes | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY2.3.1 | MT380.UY2.3.2 ]] | Convolution with arrays | [[matlab_guide#For-loop| For-loop]], [[matlab_guide#Ones and Zeros| Zeros]] | yes | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY2.4.1 | MT380.UY2.4.1 ]] | Graphical Convolution | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY2.5.1 | MT380.UY2.5.1 ]] | Convolution Properties | Review | yes | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY2.6.1 | MT380.UY2.6.1 ]] | Causality & BIBO Stability | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY2.7.1 | MT380.UY2.7.1 ]] | LTI Sinusoidal Response | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY2.8.0 | MT380.UY2.8.0 ]] | Imp.Resp. of 2nd Ord LCCDEs Aid | | yes | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY2.8.1 | MT380.UY2.8.1 ]] | Imp.Resp. of 2nd Ord LCCDEs | [[matlab_guide#Roots| Roots]] | yes | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY2.9.1 | MT380.UY2.9.1 ]] | Car Suspension | Review | yes | | ||
| - | | Ch 3 | [[ecen_380_assignments#MT380.UY3 | MT380.UY3 ]] | Laplace Transform | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY3.0 | MT380.UY3.0 ]] | Laplace Transform Aid | | yes | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY3.1.1 | MT380.UY3.1.1 ]] | Def. of the (Uni) L-Transform | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY3.2.1 | MT380.UY3.2.1 ]] | Poles and Zeros | [[matlab_guide#Mesh| Mesh]] | yes | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY3.3.1 | MT380.UY3.3.1 ]] | Properties of the L-Transform | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY3.4.1 | MT380.UY3.4.1 ]] | Circuit Analysis Example | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY3.5.1 | MT380.UY3.5.1 ]] | Partial Fraction Expansion | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY3.6.1 | MT380.UY3.6.1 ]] | Transfer Function | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY3.7.1 | MT380.UY3.7.1 ]] | Poles and System Stability | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY3.8.1 | MT380.UY3.8.1 ]] | Invertible Systems | | | | ||
| - | | Ch 4 | [[ecen_380_assignments#MT380.UY4 | MT380.UY4 ]] | App. of the L-Transform | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY4.1.1 | MT380.UY4.1.1 ]] | s-Domain Circuit Elem. Models | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY4.2.1 | MT380.UY4.2.1 ]] | s-Domain Circuit Analysis | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY4.5.1 | MT380.UY4.5.1 ]] | Op-Amp Circuits | Review | yes | | ||
| - | | Ch 5 | [[ecen_380_assignments#MT380.UY5 | MT380.UY5 ]] | Fourier Analysis Techniques | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY5.4.1 | MT380.UY5.4.1 ]] | Fourier Series Coefficients | [[matlab_guide#For-loop| For-loop]] | yes | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY5.7.1 | MT380.UY5.7.1 ]] | Fourier Transform | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY5.8.1 | MT380.UY5.8.1 ]] | Fourier Transform Properties | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY5.9.1 | MT380.UY5.9.1 ]] | Parseval's Thrm for Fourier T. | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY5.10.1 | MT380.UY5.10.1 ]] | Additional Attributes | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY5.11.1 | MT380.UY5.11.1 ]] | Phasor v. Laplace v. Fourier | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY5.12.1 | MT380.UY5.12.1 ]] | Circuit Analysis with Fourier | Review | yes | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY5.13.1 | MT380.UY5.13.1 ]] | The Importance of Phase Info. | | | | ||
| - | | Ch 6 | [[ecen_380_assignments#MT380.UY6 | MT380.UY6 ]] | Apps of the Fourier Trans. | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY6.1.1 | MT380.UY6.1.1 ]] | Signal Filtering | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY6.3.1 | MT380.UY6.3.1 ]] | Active Filters | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY6.8.1 | MT380.UY6.8.1 ]] | Butterworth Filters | Review | yes | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY6.12.1 | MT380.UY6.12.1 ]] | Sampling Theorem | [[matlab_guide#fopen| fopen]],[[matlab_guide#fscanf| fscanf]],[[matlab_guide#fclose| fclose]], [[matlab_guide#interp| Interpolate]],[[matlab_guide#sound| sound]] | yes | | ||
| - | | Ch 7 | [[ecen_380_assignments#MT380.UY7 | MT380.UY7 ]] | Discrete-Time Signals and Syst | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY7.1.1 | MT380.UY7.1.1 ]] | Discrete Notation and Property | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY7.2.1 | MT380.UY7.2.1 ]] | D-Time Signal Functions | rat | yes | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY7.3.1 | MT380.UY7.3.1 ]] | D-Time LTI Systems | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY7.4.1 | MT380.UY7.4.1 ]] | Properties of D-T LTI Sys. | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY7.5.1 | MT380.UY7.5.1 ]] | D-Time Convolution | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY7.6.1 | MT380.UY7.6.1 ]] | Z-Transform | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY7.7.1 | MT380.UY7.7.1 ]] | Properties of Z-Transform | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY7.8.1 | MT380.UY7.8.1 ]] | Inverse Z-Transform | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY7.9.1 | MT380.UY7.9.1 ]] | Solving Diff Eq. with Int. Cond.| | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY7.10.1 | MT380.UY7.10.1 ]] | System Transfer Function | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY7.11.1 | MT380.UY7.11.1 ]] | BIBO Stability | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY7.12.1 | MT380.UY7.12.1 ]] | System Frequency Response | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY7.13.1 | MT380.UY7.13.1 ]] | D-Time Fourier Series | Review | yes | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY7.14.1 | MT380.UY7.14.1 ]] | D-Time Fourier Transform | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY7.15.1 | MT380.UY7.15.1 ]] | Discrete Fourier Transform | | | | ||
| - | | ::: | [[ecen_380_assignments#MT380.UY7.16.1 | MT380.UY7.16.1 ]] | Fast Fourier Transform | | | | ||
| - | |||
| - | ===== Signal Transformation ===== | ||
| - | |||
| - | ==== MT380.UY1.2.1 ==== | ||
| - | |||
| - | == Discription == | ||
| - | |||
| - | <file> | ||
| - | 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). | ||
| - | |||
| - | </file> | ||
| - | |||
| - | == Solution Image == | ||
| - | |||
| - | {{:380matlab:ch1:ul_1_2_si_signaltransformationssolutionimage.jpg?400|}} | ||
| - | |||
| - | <ifauth @admin,@380ta> | ||
| - | |||
| - | == Solution == | ||
| - | |||
| - | {{:380matlab:ch1:ul_1_2_signaltransformations.m|}} | ||
| - | |||
| - | </ifauth> | ||
| - | |||
| - | |||
| - | ===== Signal Power and Energy ===== | ||
| - | |||
| - | ==== MT380.UY1.5.1 ==== | ||
| - | ==Description == | ||
| - | |||
| - | <file> | ||
| - | 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? | ||
| - | </file> | ||
| - | ==Solution Image == | ||
| - | |||
| - | {{:380matlab:ch1:uy_1_5_1_si1_poweravgsolutionimage1.jpg?400|}} | ||
| - | |||
| - | {{:380matlab:ch1:uy_1_5_1_si2_poweravgsolutionimage2.jpg?400|}} | ||
| - | |||
| - | <ifauth @admin,@380ta> | ||
| - | |||
| - | == Solution == | ||
| - | |||
| - | {{:380matlab:ch1:uy_1_5_1_poweravg.m|}} | ||
| - | |||
| - | </ifauth> | ||
| - | |||
| - | |||
| - | ===== Convolution Aid ===== | ||
| - | |||
| - | ==== MT380.UY2.3.0 ==== | ||
| - | |||
| - | A document to help gain an intuition behind convolution. | ||
| - | {{:380matlab:ch2:convolutionguide.docx|}} | ||
| - | |||
| - | The file below work together to create a MATLAB gui that simulates convolution | ||
| - | |||
| - | Please do not modify the following two files. \\ | ||
| - | {{:380matlab:ch2:convolutiongui.m|}} \\ | ||
| - | {{:380matlab:ch2:convolutiongui.fig|}} \\ | ||
| - | |||
| - | You can modify the following three function files to change your signals x(t) and h(t) \\ | ||
| - | {{:380matlab:ch2:x_t_gui.m|}} This file modifies the signal x(t) \\ | ||
| - | {{:380matlab:ch2:h_t_gui.m|}} This file modifies the signal h(t) \\ | ||
| - | {{:380matlab:ch2:time_parameters_gui.m|}} This file returns the time parameters used for x(t) and h(t) \\ | ||
| - | |||
| - | ===== Convolution ===== | ||
| - | |||
| - | ==== MT380.UY2.3.1 ==== | ||
| - | |||
| - | <file> | ||
| - | 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. | ||
| - | </file> | ||
| - | |||
| - | == Solution Image == | ||
| - | |||
| - | {{:380matlab:ch2:mt380_uy2_3_1_si_convolutionffsolutionimage.jpg?400|}} | ||
| - | |||
| - | <ifauth @admin,@380ta> | ||
| - | |||
| - | == Solution == | ||
| - | |||
| - | {{:380matlab:ch2:h_t.m|}} \\ | ||
| - | {{:380matlab:ch2:x_t.m|}} \\ | ||
| - | {{:380matlab:ch2:mt380_uy2_3_1_convolutionff.m|}} \\ | ||
| - | |||
| - | </ifauth> | ||
| - | |||
| - | ===== Convolution ===== | ||
| - | |||
| - | ==== MT380.UY2.3.2 ==== | ||
| - | |||
| - | <file> | ||
| - | 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. | ||
| - | </file> | ||
| - | |||
| - | == Solution Image == | ||
| - | |||
| - | {{:380matlab:ch2:mt380_uy2_3_1_si_convolutionffsolutionimage.jpg?400|}} | ||
| - | |||
| - | <ifauth @admin,@380ta> | ||
| - | |||
| - | == Solution == | ||
| - | |||
| - | {{:380matlab:ch2:convolution.m|}} | ||
| - | |||
| - | </ifauth> | ||
| - | |||
| - | ===== Convolution Properties ===== | ||
| - | |||
| - | ==== MT380.UY2.5.1 ==== | ||
| - | |||
| - | == Description == | ||
| - | <file> | ||
| - | 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) | ||
| - | </file> | ||
| - | |||
| - | == Solution Image == | ||
| - | |||
| - | {{:380matlab:ch2:uy2_5_1_si1convolutionpropertiessolutionimage1.jpg?400|}} | ||
| - | |||
| - | {{:380matlab:ch2:uy2_5_1_si2convolutionpropertiessolutionimage2.jpg?400|}} | ||
| - | |||
| - | <ifauth @admin,@380ta> | ||
| - | |||
| - | == Solution == | ||
| - | |||
| - | {{:380matlab:ch2:uy2_5_1_convolutionproperties.m|}} | ||
| - | |||
| - | </ifauth> | ||
| - | |||
| - | ===== Imp.Resp. of 2nd Ord LCCDEs Aid ===== | ||
| - | |||
| - | ==== MT380.UY2.8.0 ==== | ||
| - | |||
| - | Differential Equations-Euler's Method {{:380matlab:ch2:differential_equations_eulersmethod.docx|}} | ||
| - | |||
| - | ===== Imp.Resp. of 2nd Ord LCCDEs ===== | ||
| - | |||
| - | ==== MT380.UY2.8.1 ==== | ||
| - | |||
| - | == Description == | ||
| - | |||
| - | <file> | ||
| - | 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. | ||
| - | </file> | ||
| - | |||
| - | For help with this assignment see the document [[ecen_380_assignments#MT380.UY2.8.0 | Differential Equations-Euler's Method]] | ||
| - | |||
| - | |||
| - | == Figure == | ||
| - | |||
| - | {{:380matlab:ch2:uy2_8_1_impresp2ndordercircuit.png?300|}} | ||
| - | |||
| - | == Solution Image == | ||
| - | |||
| - | {{:380matlab:ch2:uy2_8_1_si1_impresp2ndordersolutionimage1.jpg?400|}} | ||
| - | |||
| - | {{:380matlab:ch2:uy2_8_1_si2_impresp2ndordersolutionimage2.jpg?400|}} | ||
| - | |||
| - | {{:380matlab:ch2:uy2_8_1_si3_impresp2ndordersolutionimage3.jpg?400|}} | ||
| - | |||
| - | {{:380matlab:ch2:uy2_8_1_si4_impresp2ndordersolutionimage4.jpg?400|}} | ||
| - | |||
| - | <ifauth @admin,@380ta> | ||
| - | |||
| - | == Solution == | ||
| - | {{:380matlab:ch2:uy2_8_1_impresp2ndorder.m|}} | ||
| - | |||
| - | </ifauth> | ||
| - | |||
| - | ===== Car Suspension System ===== | ||
| - | |||
| - | ==== MT380.UY2.9.1 ==== | ||
| - | |||
| - | <file> | ||
| - | 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 | ||
| - | |||
| - | </file> | ||
| - | |||
| - | <ifauth @admin,@380ta> | ||
| - | |||
| - | == Solution == | ||
| - | {{:380matlab:ch2:uy2_9_1_eulermethod2ndorder.m|}} | ||
| - | |||
| - | </ifauth> | ||
| - | |||
| - | ===== Laplace Transform Aid ===== | ||
| - | |||
| - | ==== MT380.UY3.0 ==== | ||
| - | |||
| - | {{:380matlab:ch3:laplacetransformgui.fig|}} | ||
| - | {{:380matlab:ch3:laplacetransformgui.m|}} | ||
| - | |||
| - | ===== Poles and Zeros ===== | ||
| - | ==== MT380.UY3.2.1 ==== | ||
| - | |||
| - | <file> | ||
| - | 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? | ||
| - | </file> | ||
| - | |||
| - | <ifauth @admin,@380ta> | ||
| - | |||
| - | == Solution == | ||
| - | |||
| - | {{:380matlab:ch3:uy3_2_1_sdomaingraph.m|}} | ||
| - | |||
| - | </ifauth> | ||
| - | |||
| - | ===== Op-Amp Circuits ===== | ||
| - | ==== MT380.UY4.5.1 ==== | ||
| - | |||
| - | <file> | ||
| - | 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 | ||
| - | </file> | ||
| - | |||
| - | == Figure == | ||
| - | |||
| - | {{:380matlab:ch3:uy4_5_1_op_amp.png?400|}} | ||
| - | |||
| - | <ifauth @admin,@380ta> | ||
| - | |||
| - | == Solution == | ||
| - | |||
| - | {{:380matlab:ch4:uy4_5_1_op_ampcircuit.m|}} | ||
| - | |||
| - | </ifauth> | ||
| - | |||
| - | ===== Fourier Series Coefficients ===== | ||
| - | ==== MT380.UY5.4.1 ==== | ||
| - | |||
| - | <file> | ||
| - | 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. | ||
| - | </file> | ||
| - | |||
| - | == Figure == | ||
| - | {{:380matlab:ch5:uy5_4_1_computationoffourierseriescoefficients_figure.jpg?400|}} | ||
| - | |||
| - | <ifauth @admin,@380ta> | ||
| - | |||
| - | == Solution == | ||
| - | {{:380matlab:ch5:uy5_4_1_fourierseriescoefficients.m|}} | ||
| - | |||
| - | </ifauth> | ||
| - | |||
| - | ===== Circuit Analysis with Fourier ===== | ||
| - | ==== MT380.UY5.12.1 ==== | ||
| - | |||
| - | <file> | ||
| - | 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. | ||
| - | </file> | ||
| - | |||
| - | <ifauth @admin,@380ta> | ||
| - | |||
| - | == Solution == | ||
| - | {{:380matlab:ch5:uy5_12_1_circuitanalysisfouriertransform.m|}} | ||
| - | {{:380matlab:ch5:h_t.m|}} | ||
| - | {{:380matlab:ch5:x_t.m|}} | ||
| - | |||
| - | </ifauth> | ||
| - | |||
| - | |||
| - | ===== Butterworth Filters ===== | ||
| - | ==== MT380.UY6.8.1 ==== | ||
| - | |||
| - | <file> | ||
| - | 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)? | ||
| - | </file> | ||
| - | |||
| - | <ifauth @admin,@380ta> | ||
| - | |||
| - | == Solution == | ||
| - | {{:380matlab:ch6:uy6_8_1_butterworthfilter.m|}} \\ | ||
| - | The function file below creates the impulse response of a second order lowpass filter. \\ | ||
| - | {{:380matlab:ch6:h_t.m|}} \\ | ||
| - | |||
| - | </ifauth> | ||
| - | |||
| - | ===== Sampling Theorem ===== | ||
| - | ==== MT380.UY6.12.1 ==== | ||
| - | |||
| - | <file> | ||
| - | 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? | ||
| - | </file> | ||
| - | |||
| - | == Sound File == | ||
| - | |||
| - | |||
| - | |||
| - | == fft_plot function file == | ||
| - | {{:380matlab:ch6:fft_plot.m|}} | ||
| - | |||
| - | == Original Music File == | ||
| - | {{:380matlab:ch6:pianofinal_5s.wav|}} | ||
| - | |||
| - | == Music Files After Sampling == | ||
| - | 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 \\ | ||
| - | {{:380matlab:ch6:pianos1nff.wav|}} | ||
| - | |||
| - | 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 \\ | ||
| - | {{:380matlab:ch6:pianos1ff.wav|}} | ||
| - | |||
| - | 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. \\ | ||
| - | {{:380matlab:ch6:pianos8ff.wav|}} | ||
| - | |||
| - | |||
| - | ===== D-Time Signal Functions ===== | ||
| - | |||
| - | ==== MT380.UY7.2.1 ==== | ||
| - | |||
| - | |||
| - | <file> | ||
| - | 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)) | ||
| - | </file> | ||
| - | |||
| - | <ifauth @admin,@380ta> | ||
| - | |||
| - | == Solution == | ||
| - | {{:380matlab:ch7:mt380_uy7_2_1_discrete_time_signal_functions.m|}} | ||
| - | {{:380matlab:ch7:fundamentalperiod.m|}} | ||
| - | |||
| - | </ifauth> | ||
| - | |||
| - | ===== D-Time Fourier Series ===== | ||
| - | ==== MT380.UY7.13.1 ==== | ||
| - | |||
| - | == Description == | ||
| - | <file> | ||
| - | 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? | ||
| - | |||
| - | </file> | ||
| - | |||
| - | == Signal == | ||
| - | |||
| - | |||
| - | |||
| - | == Solution Image == | ||
| - | |||
| - | <ifauth @admin,@380ta> | ||
| - | |||
| - | == Solution == | ||
| - | |||
| - | {{:380matlab:ch7:mt380_uy7_13_1_discretetimefourierseries.m|}} | ||
| - | |||
| - | </ifauth> | ||
ecen_380_matlab_assignments.1441379638.txt.gz ยท Last modified: 2015/09/04 09:13 by petersen
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