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Signals and Systems Theory and Lab – Week 7 Lab ET382 W7 Lab Lab 7 – The Z-Transform · Watch video entitled “Module 7– Z-Transform in MATLAB” · Work the below lab assignment below using MATLAB. · Include answers for Problems and include MATLAB coding along with any output plots that support solutions into a Word document entitled “Lab7_StudentID”. Where your student id is substituted in the file name. · Upload file “Lab7_StudentID” Activity 1: A linear time-invariant discrete-time system has transfer function · · Use Matlab to obtain the poles of the system. Is the system stable? Explain. Matlab tip: You can find the roots of a polynomial by using the roots command. For instance, if you have the polynomial x2+ 4x + 3, then you can find the roots of this polynomial as follows: >>roots([1 4 3]) where the array is the coefficients of the polynomial. · · Compute the step response. This should be done analytically, but you can use Matlab commands like conv and residue to help you in the calculations. Matlab tip: Besides using conv to look at the response of a system, it can also be used to multiply two polynomials together. For instance, if you want to know the product (x2 + 4x + 3)(x + 1), you can do the following: >>conv([1 4 3],[1 1]) where the two arrays are the coefficients of the two polynomials. The result is >> ans = 1 5 7 3 Thus, the product of the two polynomials is x3 + 5x2 + 7x + 3. Matlab tip: The command residue does the partial fraction expansion of the ratio of two polynomials. In our case, we can obtain Y(z)/z and then use the residue command to do the partial fraction expansion. Then it is relatively easy to obtain y[n] using the tables. · Plot the first seven values of the step response. Is the response increasing or decreasing with time? Is this what you would expect, and why? . If you cannot view the images above, download a PDF of the Lab. Grading Rubric 30 Establishes an understanding of discrete-time to frequency domain conversion of signals Demonstrates the ability to perform numerical 30 and symbolic computations Solves problems using current software used Excellent in the discipline (e.g. Matlab, Multisim, IDEs, etc.) Develops a solution using software with few to no errors and incorporating advanced features of the software tool Uses graphics, numerical expressions and tables to illustrate ideas Competent Needs Improvement Develops a solution using software with no major or significant errors in use of the software tool Makes significant errors in use of software tool to solve problems. The errors are related to the use and understanding of the software tool and not specifically to the problem. 20 15 8 Excellent Competent Needs Improvement Numerical expressions, images, tables, and graphs are used as needed and are in the proper format. Numerical expressions, images, tables, and graphs are generally used as needed and are in the The need for graphical or numerical information is recognized but presentation is flawed or such information is proper format not provided but with few errors. needed to illustrate ideas. 10 Provides a well written report ( Following APA formatting, including figures, equations and answer to the questions) 10 TOTAL 100 6 SOLUTION >> % solving using MATLAB >> % Part-1: >> num = [ 1 -1 -2 ] ; % Numerator: z^-z-2 >> den = [ 1 1.5 -1] ; % Denominator: z^2 + 1.5z-1 >> H=tf ( num, den, .8) % Creates the LTI object and displays the result >> % Gain factor >> K = num(1)/den(1) Stable = (abs(poles) < 1) ; If stable fprint (‘The system is stable.\n’) ; else fprint (‘WARNING: The system is NOT stable.\n’) ; end %step response H =tf ([ 1, -1,-2],[1,1.5,-1],0.8) ; stepplot (H) 4 %to find residue num = [1 -1 -2] ; den = [1.5 -2 .5 1] ; [r,p,k] = residue ( num, den) Pzmap (H) % Getting pole-zero plot % Also we can get poles using the following command… >>Poles_H=roots ([1 1.5 -1]) Poles_H= -2.0000 0.5000 >> % Now, the system has (-2) outside the unit circle. It implies that it is unstable Point to NOTE: The roots of the numerator and denominator are called poles in this case >> >> >> % Part-2: Getting step response >> syms z >> Y(z)=(z/(z-1))*((z^2-z-2)/(z^2+1.5*z-1)) Y(z) = -(z*(-z^2+z+2))/((z-1)*(z^2+(3*z)/2-1)) >>y_ n=iztrans (Y) % Getting step response using inverse z-transform Y_ n= (8*(-2)^n)/15 + ( 9*(1/2)^n)/5-4/3 >> >> % Part-3: Plotting first seven values >> >> n =0 : 1 : 6 ; >> y_ n _7 = double (sub (y_n)) Y_n_7 = 1.0000 -1.5000 1.2500 -5.3750 7.3125 -18.3438 32.8281 >> stem (n, y_n_7, ‘linewidth’, 2) % Plotting >> grid on >> ylim ([-20 40]) >> xlim ([-1 7]) >>xlabel (‘Samples (n) ‘) >>ylabel (‘Value’) >>title (‘First 7 samples of step response’)