Chap3StudentSolutions

# 2 c 2 b 2 2 ac 0 a 2 c 2 b 2 2 ac 0 a 2 2 ac

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Unformatted text preview: B sin ( x ) = A 2 + B 2 cos x − tan −1 A and you should get 2π n g [ n ] = 2.434 cos − 0.9845 7 (g) n n sinc sinc 4 4 g[ n ] = ∗ 22 22 In the absence of the transform methods which have not been covered yet, this convolution must be done numerically. This will be relatively simple to do analytically using transform methods. 46. Find the impulse responses of the subsystems in Figure E 46 and then convolve them to find the impulse response of the cascade connection of the two subsystems. You may find this formula for the summation of a finite series useful, , α =1 N ∑0 α = 1 − α N , α ≠ 1 . n= 1− α N −1 n x[n] y1 [n] D y2 [n] D 4 5 Figure E 46 Two cascaded subsystems Solutions 3-32 M. J. Roberts - 8/16/04 47. For the system of Exercise 43, let the excitation, x[ n ] , be a unit-amplitude complex sinusoid of DT cyclic frequency, F. Plot the amplitude of the response complex sinusoid versus F over the range, −1 < F < 1. 48. In the second-order DT system below what is the relationship between a, b and c that ensures that the system is stable? 1 x[n] y[n] a b c y[ n ] = D D x[ n ] − (b y[ n − 1] + c y[ n − 2]) a Stability is determined by the eigenvalues of the homogeneous solution. a y[ n ] + b y[ n − 1] + c y[ n − 2] = 0 The eigenvalues are −b ± b 2 − 4 ac α1,2 = 2a For stability the magnitudes of all the eigenvalues must be less than one. Therefore 2 − 2 b c b + − <1 2a 2a a and − b 1 b 2 − 4 ac < 1 + 2a 2a and − − b c b − − <1 2a 2a a b 1 b 2 − 4 ac < 1 − 2a 2a −b + b 2 − 4 ac < 2 a and −b − b 2 − 4 ac < 2 a −b + j 4 ac − b 2 < 2 a and −b − j 4 ac − b 2 < 2 a If b 2 − 4 ac < 0 , In either case or b 2 + 4 ac − b 2 < 4 a 2 ac < a 2 Solutions 3-33 M. J. Roberts - 8/16/04 From the requirement, b 2 − 4 ac < 0 we know that ac must be positive. Then we can divide both sides by the positive number, ac, yielding a >1 . c If b 2 − 4 ac ≥ 0 , (−b + b 2 − 4 ac ) 2 < 4 a2 b 2 − 2b b 2 − 4 ac + b 2 − 4 ac < 4 a 2 and −2b b 2 − 4 ac < 4 a 2 − 2b 2 + 4 ac (−b − and ) 2 < 4 a2 b 2 + 2b b 2 − 4 ac + b 2 − 4 ac < 4 a 2 2b b 2 − 4 ac < 4 a 2 − 2b 2 + 4 ac and −b b 2 − 4 ac < 2 a 2 − b 2 + 2 ac b 2 − 4 ac b b 2 − 4 ac < 2 a 2 − b 2 + 2 ac and Taken together, these two requirements lead to 2 a 2 − b 2 + 2 ac > b b 2 − 4 ac ≥ 0 2 a( a + c ) ≥ b 2 and (2a 2 − b 2 + 2 ac ) > b 2 (b 2 − 4 ac ) 2 4 a 4 + b 4 + 4 a 2c 2 − 4 a 2b 2 − 4 ab 2c + 8 a 3c > b 4 − 4 ab 2c 4 a 2 ( a 2 + c 2 − b 2 + 2 ac ) > 0 a 2 + c 2 − b 2 + 2 ac > 0 a 2 − 2 ac + c 2 > b 2 − 4 ac (a − c ) 2 > b 2 − 4 ac 49. Given the excitations, x[ n ] , and the impulse responses, h[ n ], find closed-form expressions for and plot the system responses, y[ n ] . (a) n x[ n ] = u[ n ] , 7 h[ n ] = n u[ n ] 8 1 − r N , r ≠1 with respect to r.) (Hint: Differentiate ∑ r n = 1 − r n =0 N , r =1 N −1 Solutions 3-34 M. J. Roberts -...
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## This note was uploaded on 06/19/2013 for the course ENSC 380 taught by Professor Atousa during the Spring '09 term at Simon Fraser.

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