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Course: ENEE 241, Spring 2008
School: Maryland
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241 ENEE 02* HOMEWORK ASSIGNMENT 2 Due Mon 02/04 Consider the complex numbers u = ej (/7) , v = ej (6/7) , and w = ej (2/5) (i) (2 pts.) Plot the three numbers on the complex plane. (ii) (3 pts.) Plot the straight line described by the equation |z u| = |z v | (where z is a variable point) on the complex plane. (iii) (4 pts.) Plot the circle described by the equation |z u u2 v | = |w|. Which of the three...

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241 ENEE 02* HOMEWORK ASSIGNMENT 2 Due Mon 02/04 Consider the complex numbers u = ej (/7) , v = ej (6/7) , and w = ej (2/5) (i) (2 pts.) Plot the three numbers on the complex plane. (ii) (3 pts.) Plot the straight line described by the equation |z u| = |z v | (where z is a variable point) on the complex plane. (iii) (4 pts.) Plot the circle described by the equation |z u u2 v | = |w|. Which of the three points (i.e., u, v , and w) does the circle pass through? (iv) (4 pts.) Each of the expressions v +v 13 and v v 13 can be evaluated using a single trigonometric function (either sine or cosine). Why is this so? (Hint. What other power of v does v 13 also equal? Plot v 13 on the complex plane if necessary.) (v) (4 pts.) Which polynomial of degree n = 5 has w, w2 , w3 , w4 as its roots? (vi) (3 pts.) Based on your (hopefully) correct answer to (v) above, use the geometric sum formula to show that 1 + w + w2 + w3 + w4 = 0 (Do not explicitly and calculate add all these numbers.) Solved Examples S 2.1 (P 1.10 in textbook). Let N be an arbitrary positive integer. Evaluate the product N 1 cos k=1 k N + j sin k N , expressing your answer in Cartesian form. (Use the identity 1 + 2 + + n = n(n + 1)/2 to simplify the answer rst.) S 2.2. Show that the expression f () = ej 2 ej + 1 ej + ej 2 , where ranges over [0, 2 ), is real-valued, and obtain an alternative expression for it in terms of sines and/or cosines. S 2.3. Clearly, ej 3 = ej 3 By expanding (cos + j sin )3 and separating real and imaginary parts, obtain two identities: one for cos 3 in terms of powers of cos only, and another for sin 3 in terms of powers of sin only. S 2.4 (more dicult). Consider two complex numbers w and z , where w = 0 is xed and z is variable such that |z | = 1. Show that as z traces out the unit circle, the ratio |z w | |z (1/w)| is constant in value. (Hint : If |z | = 1, then 1/z = z .)
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Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 3Due Wed 02/06Use your calculator for algebraic and trigonometric functions only, as indicated. Trial-and-error or graphical solutions to parts (i)(iii) are not acceptable. Consider the sinusoid x(t) = A cos(t + ), where
Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 4Due Fri 02/08Consider the discrete-time sinusoids x[n] = cos 3n 2 + 4 3 and y [n] = cos n 2 + 5 3(i) (4 pts.) What is the period N of x[]? Use MATLAB to generate a discrete plot of the rst four periods of x[], (i.e., f
Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 54 3 2 1 0 1 2 3 4Due Mon 02/1100.10.20.30.40.50.60.7 The rst period of the sinusoid x(t) = A cos(t + ) is plotted above, where x(0) = 2 3. (i) (5 pts.) Determine A, and . In what follows: A and are as found in
Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 6Due Wed 02/13(i) (5 pts.) Compute the sum of the two discrete-time sinusoids: x1 [n] = cos(2n/5 + /3) and x2 [n] = cos(8n/5 + /4)(Hint : Start with reducing the angular frequency of x2 [n] to a value between 0 and , th
Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 7Due Fri 02/15(i) (2 + 2 pts.) Suppose a 3 3 matrix A satises 1 4 0 3 A 0 = 2 , A 1 = 2 0 1 0 2 Find A and compute 1 A 0 1 010 B= 0 0 1 100 and 0 0 A 0 = 1 . 1 2(ii) (10 pts.) Ifand x = [ 1 2 3 ]T , determine the fol
Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 8Due Mon 02/18(i) (10 pts.) It is known that 1 4 A 1 = 5 , 0 6 Determine the matrix A. 2 1 A 1 = 4 0 0and 2 7 A 1 = 6 2 2(ii) (10 pts.) Without using your calculator, determine so that the product of the ve matrices
Maryland - ENEE - 241
ENEE 241 02* Consider the matrixHOMEWORK ASSIGNMENT 9 abcd A= e f g h wxyz Due Wed 02/20In each of the following cases, nd matrices P and Q such that PAQ equals the matrix shown. (Note that if either P or Q is superuous, it trivially equals I.) wxyz e
Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 10Due Fri 02/22Solve by hand without using calculator matrix functions. Show all intermediate steps. Let 1 0 U= 0 0 2 1 0 1 2 1 0 1 2 0 0 1(i) (4 pts.) Solve Ux = b for arbitrary x and b. Display U1 . (ii) (6 pts.) Show
Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 11Due Wed 02/27Solve by hand without using calculator matrix functions. Show all intermediate steps. Consider the 4 4 system Ax = b, 2 1 4 0 A= 1 3 2 6 9 (i) (14 pts.) Determine x. (ii) (6 pts.) Express A in the form LU,
Maryland - ENEE - 241
ENEE 241 02HOMEWORK ASSIGNMENT 12Due Fri 02/29Solve by hand without using calculator matrix functions. Show all intermediate steps. Let 1 2 A= 1 1 4 2 1 3 4 0 3 1 4 2 3 2(i) (8 pts.) Obtain the LU factorization A = LU. (ii) (12 pts.) Determine A1 by s
Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 13Due Mon 03/03Solve by hand without using calculator matrix functions. Show all intermediate steps. Consider the following 1 1 1 1 A= 1 2 1 3 three matrices: 1 1 1 1 B= 4 8 6 91 2 3 5 234 4 7 8 5 9 12 7 12 201 2 1 2
Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 14Due Wed 03/05Solve by hand without using calculator matrix functions. Show all intermediate steps. Consider the three-dimensional vectors v(1) = [ 1 2 2 4 ]T and v(2) = [ 4 2 2 1 ]T . (i) (4 pts.) Compute v(1) and v(2)
Maryland - ENEE - 241
ENEE 241 02HOMEWORK PROBLEM 15Due Fri 03/07Solve by hand without using calculator matrix functions. Show all intermediate steps. Let v(1) = [ 5 3 1 1 ]T , v(2) = [ 5 1 7 3 ]T and v(3) = [ 1 1 10 0 ]T . Also, let s = [ 3 1 4 4 ]T . (i) (5 pts.) Determin
Maryland - ENEE - 241
ENEE 241 02HOMEWORK ASSIGNMENT 16Due Mon 03/10This problem investigates the least-squares approximation of the function s(t) = 1 (t 1)2 1 + 4(t 1)2over a discrete interval of values t using a sum of three sinusoidal functions: f1 (t) = sin(t/2) , f2 (
Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 17Due Wed 03/12This problem refers to the range estimation example in the textbook (p. 113). Suppose f1 = 3 MHz, f2 = 2.5 MHz and = 1.45. The vector s in the le dataB17.txt contains 101 samples of the noisy waveform s(t)
Maryland - ENEE - 241
ENEE 241 02 Consider the 4 3 matrix V =HOMEWORK PROBLEM 18 v(1) v(2) v(3) given byDue Fri 03/14 1.3702 4.6799 0.6379 1.8877 2.5466 4.7986 V= 2.0596 5.5447 2.1429 2.5134 3.7242 0.9872 (i) (3 pts.) Compute (e.g., in MATLAB) the inner product matrix VT V,
Maryland - ENEE - 241
ENEE 241 02HOMEWORK PROBLEM 19 v(1) v(2) v(3) given byDue Mon 03/24Consider the complex-valued matrix V = 3 a + jb 6j 3 a + jb V = 6j a + jb 6j 3 (i) (4 pts.) Determine the only pair of real values a and b such that the columns of V are orthogonal. F
Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 20Due Wed 03/26Exam 2 (Monday 03/31) will cover Assignments 10-20. Let V= v(0) v(1) v(2) v(3) v(4) v(5) be the matrix of Fourier sinusoids of length N = 6. (This is the same matrix as in the last example of Reading Assig
Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 21Due Fri 03/28Solve without using MATLAB, except to verify your results. All vectors have length N = 10. (i) (4 pts.) The time-domain vector x(1) has DFT X(1) = 0 1 0 0 0 0 0 0 0 1TWhich two Fourier frequencies does x
Maryland - ENEE - 241
ENEE 241 02HOMEWORK ASSIGNMENT 22Due Wed 04/02The real-valued signal vector s has DFT S= 6 z1 3 + 2j 7 z2 5 + 3jT(i) (2 pts.) What are the values of z1 and z2 ? (ii) (3 pts.) Without using complex algebra (or MATLAB), determine the sum of the entries
Maryland - ENEE - 241
ENEE 241 02 The signal vector s is given byHOMEWORK PROBLEM 24Due Mon 04/07s=abcdefghT(i) (8 pts.) Display the following vectors: s(1) = P2 Rs s(2) = R2 Ps s(3) = P4 s s(4) = F2 s (ii) (12 pts.) Express the following vectors in terms of P, R, F and
Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 26Due Fri 04/11Single-step the MATLAB script below, keeping a gure window open. Answer questions (i)-(viii). Print and submit the last two graphs, i.e., bar(n, real(X4) and bar(n, real(X5), only. (Make sure that your MAT
Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 27 y ofTDue Mon 04/14(i) (7 pts.) Determine the circular convolution s = x x= and y= (ii) (7 pts.) If 3 1 1 5 20 1 2TUsing the FFT function in MATLAB, verify that the DFTs X, Y and S satisfy X Y = S. 1 3 2 1 4 x0 x1
Maryland - ENEE - 241
ENEE 241 02 The signalHOMEWORK ASSIGNMENT 28TDue Wed 04/16x= has DFT X given by X=000abc000X0 X1 X2 X3 X4 X5 X6 X7 X8TExpress the following DFTs in terms of the entries of X: (i) (5 pts.) The DFT X(1) of x(1) = (ii) (5 pts.) The DFT X(2) of x(2) =
Maryland - ENEE - 241
ENEE 241 02HOMEWORK PROBLEM 29Due Fri 04/18Generate the 20-point signal vector x using the MATLAB script below. n = (0:19). ; f0 = 0.15 + 0.25*rand(1) ; A = 10*rand(1) ; phi = pi*rand(1) ; x = A*cos(2*pi*f0*n + phi) + 0.3*A*randn(size(n) ; Following th
Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 30Due Mon 04/28Each of the innite sequences x = x[ ], y = y [ ] and u = u[ ] (shown below) contains seven nonzero terms.4 3 2 1 1 x 2 1 4 3 y 2 4 3 un=0n=0n=0(i) (3 pts.) Express u[ ] as a linear combination of seve
Maryland - ENEE - 241
ENEE 241HOMEWORK ASSIGNMENT 32Due Wed 04/30Consider the FIR lter given by the input-output relationship y [n] = x[n] x[n 2] + x[n 4] x[n 6] , (Note the missing, i.e., zero-valued, coecients.) (i) (4 pts.) Show that the input sequences dened for all n b
Maryland - ENEE - 241
ENEE 241 02*HOMEWORK ASSIGNMENT 33Due Fri 05/02Consider the FIR lter whose input-output relationship is given by 1 3 3 y [n] = x[n] x[n 1] + x[n 3] x[n 4] , 2 4 8 nZ(i) Determine the response y (i) [ ] of the lter to each of the input signals given by
Maryland - ENEE - 241
ENEE 241 02HOMEWORK ASSIGNMENT 34 4 1 0 1 4TDue Mon 05/05 .Consider the FIR lter with coecient vector b =The following MATLAB script computes a segment of the lter output sequence y (1) [ ] for an input sequence x(1) [ ] which has period L = 5. Speci
Maryland - ENEE - 241
ENEE 241 02HOMEWORK ASSIGNMENT 35Due Wed 05/07Use the MATLAB script below to generate the coecient vector b of a bandpass FIR lter of order M = 44. f1 = 0.19 + .03*rand(1) ; f2 = 0.34 + .03*rand(1) ; F = [0 f1 f1+0.03 f2 f2+0.03 0.5]*2 ; G = [1 1 0 0 1
Maryland - ENEE - 241
ENEE 241 02HOMEWORK ASSIGNMENT 36Due Fri 05/09(Part A) (5 pts.) Consider the FIR lter with impulse response given by h1 [n] = [n] [n 1]. Compute the outputs y1 [n] and y2 [n] of this system when the inputs are (A.1) (2 pts.) x1 [n] = cos(n + ), and (A.
Maryland - ENEE - 241
ENEE 241 02HOMEWORK ASSIGNMENT 37Due Mon 05/12A FIR lter acts on the input sequence given by x[0 : 5] = 1 3 5 2 6 2T;x[n] = 0 for n &lt; 0 and n &gt; 5to produce the output sequence given by y [0 : 11] = 2 3 0 10 23 27 61 51 54 14 2Tand y [n] = 0 for n
Maryland - ENEE - 241
HW 1 SOLUTION_(i) Plot (-1,sqrt(3) and (1,-1) on the (x,y) plane.(ii) |z1| = sqrt(-1)^2+(sqrt(3)^2) = sqrt(1+3) = 2angle(z1) = arctan(-sqrt(3) + pi = 2*pi/3 = 2.0944|z2| = sqrt(1+1) = sqrt(2) = 2.828angle(z2) = arctan(-1/1) = -pi/4(iii) z1^3 =
Maryland - ENEE - 241
HW 2 SOLUTION_(i) The polar coordinates are |u| = 1, angle(u) = pi/7 |v| = 1, angle(v) = 6*pi/7 |w| = 1, angle(u) = 2*pi/5(ii) |z-u| = |z-v| is tha perpendicular bisector of the segmentjoining u and v. This straight line is the y-axis (x = 0) (ii
Maryland - ENEE - 241
HW 3 Solution_(i) We know that the value 2.5 occurs at t=0, 0.160 and 0.180 sec andat no other intermediate times. Since a sinusoid crosses any givenlevel (except its maximum and minimum) twice in each period, it followsthat the period T equals exact
Maryland - ENEE - 241
HW 4 Solution_(i) For x[.], we have w=3*pi/4 = (3/8)*(2*pi). Thusthe fundamental period of x[.] equals N = 8.N = 8;n = 0 : 4*N-1;x = cos(3*pi*n/4 + 2*pi/3);stem(n,x)(ii) The values of w such that cos(w*n) is periodic with fundamental period N
Maryland - ENEE - 241
HW 5 Solution_(i) Clearly, A = 4. The period T equals 0.7 sec, therefore f = 10/7 Hz andW = 20*pi/7cos(phi) = 2*sqrt(3)/A = sqrt(3)/2, therefore phi = pi/6 or -pi/6. By inspection, t=0 corresponds to a point in the last quarter-cycle of the cosine wa
Maryland - ENEE - 241
Maryland - ENEE - 241
HW 7 Solution _ (i) A = [-4 3 0 2 -2 -1 1 2 2 ] A*[1 ; 0 ; 1] = [-4 ; 1 3] (ii) (a) B*x = [2; 3; 1]. B is a transformation matrix that circularly shifts the elements of the vector x. (b) B^2*x = [3; 1; 2]. Multiplying by B circularly shifts t
Maryland - ENEE - 241
HW 8 Solution_(i) The columns of A are obtained by right-multiplying A by thethree unit vectors.We have[1 ; -1 ; 0] + [2 ; 1 ; 0] = [3 ; 0 ; 0]thereforeA*[3 ; 0 ; 0] = A*[1 ; -1 ; 0] + A*[2 ; 1 ; 0]= [-4 ; -5 ; -6] + [ 1 ; -4 ; 0] = [-3 ; -9 ;
Maryland - ENEE - 241
HW 9 Solution_Note that the dimensions of P and Q can be inferred from the result,i.e., if P*A*Q is m*n, then P is m*3 and Q is 4*n.(i) P = 0 0 10 1 01 0 0Q = I(ii) P = I Q =0 0 0 10 1 0 00 0 1 01 0 0 0(iii) P = 0 0 11 0 00
Maryland - ENEE - 241
HW 10 Solution_(i) To solve U*x = b for arbitrary x = [x1 x2 x3 x4].' and b = [b1 b2 b3 b4].':x4 = b4 -2*x4 + x3 = b3 ; therefore x3 = 2*b4 + b3x2 - 2*x3 + x4 = b2 ; therefore x2 = -b4 + 2*(2*b4+b3) + b2 = b2 + 2*b3 + 3*b4x1 - 2*x2 + x3 = b1
Maryland - ENEE - 241
HW 11 Solution_Forward elimination:m x1 x2 x3 x4b_210-1-2-2401-23-1/21-3/263/235-3697-323m x1 x2 x3 x4b_210-1-20-2107-10-262363067029m x1 x2 x3 x4b_210-1-20-2107005229-20010
Maryland - ENEE - 241
HW 12 Solution_m x1 x2 x3 x4 _ 1* 4 4 4-2 2 2 0 2-1 1 1 3 3-1 1 3 1 2_ 1 4 4 4 0-6* -8-6-1/2 0-3-1-1-1/6 0-1-3-2_ 1 4 4 4 0-6 -8-6 0 0 3* 2 5/9 0 0-5/3 -1_ 1 4 4 4 0-6 -8-6 0 0 3
Maryland - ENEE - 241
HW 13 Solution_Matrix A_p m x1 x2 x3 x4_* 1* 1 1 1 -1 1 -1 1 -1 -1 1 2 4 8 -1 1 -3 6 -9_* 1 1 1 1* 0 -2* 0 -2 1/2 0 1 3 7 -2 0 -4 5 -10_* 1 1 1 1* 0 -2 0 -2* 0 0 3* 6 -5/3 0 0 5 -6_* 1 1 1 1* 0 -2 0 -2* 0 0 3 6 0 0 0
Maryland - ENEE - 241
HW 14 Solution_(i) |v1| = |v2| = sqrt(1 + 8 + 16) = 5&lt;v1,v2&gt; = 4-8+4 = 0(ii) &lt;v1,a&gt; = 3+12+12 = 27, &lt;v2,a&gt; = 12-12+3 = 3|a| = sqrt(9 + 18 + 9) = sqrt(36) = 6f1 = ( &lt;v1,a&gt; / |v1|^2 )*v1 = (27/25)*v1 Angle between a and v1 = arccos(27/(6*5) = 0.4
Maryland - ENEE - 241
HW 15 Solutions _ v1 = [ 5 3 -1 1 ].' v2 = [ 5 -1 7 3 ].' v3 = [ 1 1 -10 0 ].' s = [ 3 -1 4 4 ].' (i) We form the matrix V = [v1 v2]: V = [ 5 5 3 -1 -1 7 1 3 ] Then V'*V = [ 36 18 18 84 ] V'*s = [ 12 56 ] Solve (V'*V)*c
Maryland - ENEE - 241
HW 17 Solution_(i) Using microseconds (and noting that 1 MHz = 1/(1 microsecond) :f1 = 3 ; f2 = 2.5; gamma = 1.45t = (0 : 0.1 : 10).' ;u1 = cos(2*pi*f1*t) ;w1 = sin(2*pi*f1*t) ;u2 = cos(2*pi*f2*t) ;w2 = sin(2*pi*f2*t) ;V = [u1 w1 u2 w2] ;
Maryland - ENEE - 241
HW 18 Solutions_V = [-2.8867 4.0276 2.1607 0.0369 0.3649 -6.1858 0.0999 -6.0291 0.4916 4.0812 0.5427 2.7973 ](i) Rounded to four significant digits:V'*V = [ 16 -32 8-32 73 -7 8 -7 29 ](ii) Forward elimination on V'*V:m x1 x2 x3_ 16*-
Maryland - ENEE - 241
HW 19 Solution_i) V = [ v1 v2 v3 ]v1'*v2 = 3(a+bj) - 18j + 6j(a-jb)v1'*v2 = 0 for orthogonality, so Re( v1'*v2 ) = 3a + 6b = 0 -&gt; a = -2bIm( v1'*v2 ) = 3b - 18 + 6a = 0 -&gt; b = -2, a = 4The same terms are obtained in v2'*v3 and v3'*v1, so these i
Maryland - ENEE - 241
HW 20 Solutions_Let a = sqrt(3) throughout.v0 = [ 1 1 1 1 1 1 ].'v1 = [ 1 (1+j*a)/2 (-1+j*a)/2 -1 (-1-j*a)/2 (1-j*a)/2 ].'v2 = [ 1 (-1+j*a)/2 (-1-j*a)/2 1 (-1+j*a)/2 (-1-j*a)/2 ].'v3 = [ 1 -1 1 -1 1 -1 ].'v4 = [ 1 (-1-j*a)/2 (-1+j*a)/2 1 (-1-j*a)/2 (-1+j
Maryland - ENEE - 241
HW 21 SOLUTION_(i) X1 = [ 0 1 0 0 0 0 0 0 0 1 ].' The Fourier frequencies present in x1 are: w = (2*pi/10) = pi/5 and w = 9*(2*pi/10) = 9*pi/5 The corresponding Fourier sinusoids are v1[n] = exp(j*pi*n/5) and v9[n] = exp(j*9*pi*n/5) = exp(-
Maryland - ENEE - 241
HW 22 SOLUTION_(i) Since s is real-valued, S has circular conjugate symmetry, i.e.,S = [6 -5-3*j 3+2*j -7 3-2*j -5+3*j ].'(ii) s[0] + . + s[6] = S[0] = 6(iii) We compute the modulus and angle of each entry of S to obtain the amplutude and phase spe
Maryland - ENEE - 241
HW 24 Solution_s = [ a b c d e f g h ].'S = [ 4 -1+2j 5*j 3+j -6 3-j -5*j -1-2j ].'S1:R*s &lt;-&gt; R*S = [ 4 -1-2j -5*j 3-j -6 3+j +5*j -1+2j ].'s1 = P*R*s &lt;-&gt; F^(-1)*R*S = S1The entries of R*S are multiplied by those of the k = 8-1 = 7th sinusoi
Maryland - ENEE - 241
HW 25 Solution_s = [ a b c d e f g h ].'S = [ 4 -1+2j 5*j 3+j -6 3-j -5*j -1-2j ].'S1:R*s &lt;-&gt; R*S = [ 4 -1-2j -5*j 3-j -6 3+j +5*j -1+2j ].'s1 = P*R*s &lt;-&gt; F^(-2)*R*S = S1The entries of R*S are multiplied by those of the k = 8-1 = 6th sinusoi
Maryland - ENEE - 241
HW 29 Solution_The problem parameters (frequency f0, amplitude A and phase phi) are randomly generated in MATLAB: n = (0:19).' ; f0 = 0.15 + 0.25*rand(1) ; A = 10*rand(1) ; phi = pi*rand(1) ; x = A*cos(2*pi*f0*n + phi) + 0.3*A*randn(size(n) ;% (i
Maryland - ENEE - 241
S 30.1 (P 4.1)_(d[.] = delta[.], the unit impulse function)(i) x[n] = d[n+2] - d[n+1] + 5d[n] - d[n-1] + d[n-2]In other words,x[0] = 5x[1] = x[-1] = -1x[2] = x[-2] = 1x[n] = x[-n] = 0 for n&gt;2The DTFT of d[n-m] equals exp(-j*m*w), soX(exp(j*
Maryland - ENEE - 241
S 32.1 (P 4.5)_i) MATLAB code:b = [1 -3 1 1 -3 1].';H = fft(b,256);A = abs(H);q = angle(H);ii) H(exp(j*w) = 1 - 3*exp(-j*w) + exp(-j*2*w) + exp(-j*3*w) - 3exp(-j*4*w) + exp(-j*5*w)= exp(-j*5*w/2)*(exp(j*5*w/2) - 3*exp(j*3*w/2) + exp(j*w/2)
Maryland - ENEE - 241
S 33.1 (P 4.4)_(i) H(z) = 1 - z^(-1 - *z^(-2) + z^(-3)(ii) H(exp(j*omega) = 1 - exp(-j*omega) - exp(-j*2*omega) + exp(-j*3*omega)= exp(-j*3*omega/2). *(exp(j*3*omega/2) - exp(j*omega/2) - exp(-j*omega/2) + exp(-j*3*omega/2)= exp(-j*3*omega/2)*(
Maryland - ENEE - 241
S 34.1 (P 4.3)_Divide by 2*pi to express the three frequencies in cycles per sample:3/28, 9/35 and 17/48Since these are rational (i.e., integer fractions), thesignal is periodic. The period is the smallest integerL such that each frequency can be
Maryland - ENEE - 241
S 35.1 (P 4.10)_(i)bar(0:36, b1), title('Filter Impulse Response'), xlabel('Time n')The impulse response is symmetric about n=M/2=20.(ii)f = (0:999)/1000;H = fft(b1,1000);Ha = abs(H);plot(f,Ha), title('Amplitude Response'), xlabel('\Omega/