Chap3StudentSolutions

44 find the impulse responses of these systems a 3 y n

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Unformatted text preview: ∑ x[m] . m =−∞ Taking the first backward difference of both sides of the original system equation, y[ n ] − y[ n − 1] = n +1 n +1−1 m =−∞ m =−∞ ∑ x[m] − ∑ x[m] x[ n + 1] = y[ n ] − y[ n − 1] The excitation is uniquely determined by the response. Invertible. 40. A DT system is described by n y[ n ] − 8 y[ n − 1] = x[ n ] . Classify this system as to time invariance, BIBO stability and invertibility. Stability: The homogeneous equation is or n y[ n ] = 8 y[ n − 1] y[ n ] = 8 y[ n − 1] . n Thus, as n increases without bound, y[ n ] must be decreasing because it is previous value and 8 times its n 8 approaches zero. Rearranging the original equation, n y[ n ] = x[ n ] 8 + y[ n − 1] . n n x[ n ] 8 must be bounded and y[ n − 1] n n must be getting smaller because it is a decreasing fraction of its previous value. Therefore for a bounded excitation, the response is bounded. Stable. For any bounded excitation, x[ n ] , as n gets larger, Solutions 3-29 M. J. Roberts - 8/16/04 41. A DT system is described by y[ n ] = x[ n ] . Classify this system as to linearity, BIBO stability, memory and invertibility. Invertibility: Inverting the functional relationship, Invertible. x[ n ] = y 2 [ n ] . 42. Graph the magnitude and phase of the complex-sinusoidal response of the system described by 1 y[ n ] + y[ n − 1] = e − jΩn 2 as a function of Ω. This is the steady-state solution so all we need is the particular solution of the difference equation. The equation can be written as n 1 y[ n ] + y[ n − 1] = (e − jΩ ) = α n 2 where α = e − jΩ The particular solution has the form, y[ n ] = Kα n . 43. Find the impulse response, h[ n ], of the system in Figure E43. x[n] y[n] 2 0.9 D Figure E43 DT system block diagram or y[ n ] = 2 x[ n ] + 0.9 y[ n − 1] y[ n ] − 0.9 y[ n − 1] = 2 x[ n ] The homogeneous solution (for n ≥ 0) is of the form, y[ n ] = K hα n therefore the characteristic equation is K hα n − 0.9K hα n −1 = 0 . Solutions 3-30 M. J. Roberts - 8/16/04 and the eigenvalue is α = 0.9 and, therefore, y[ n ] = K h (0.9) n We can find an initial condition to evaluate the constant, K h , by directly solving the difference equation for n = 0. y[0] = 2 x[0] + 0.9 y[−1] = 2 . Therefore 2 = K h (0.9) ⇒ K h = 2 . 0 Therefore the total solution is y[ n ] = 2(0.9) n which is the impulse response. 44. Find the impulse responses of these systems. (a) 3 y[ n ] + 4 y[ n − 1] + y[ n − 2] = x[ n ] + x[ n − 1] (b) 5 y[ n ] + 6 y[ n − 1] + 10 y[ n − 2] = x[ n ] 2 45. Plot g[ n ]. Use the MATLAB conv function if needed. (a) 2πn g[ n ] = rect1[ n ] ∗ sin 9 Write out the convolution sum. Then use sin ( x + y ) = sin ( x ) cos ( y ) + cos ( x ) sin ( y ) write out the entire summation and simplify what you get. You should ultimately get 2π n g [ n ] = 2.5321sin 9 (c) 2πn g[ n ] = rect 2 [ n ] ∗ sin 9 g[n] = 0 (d) g[ n ] = rect 3 [ n ] ∗ rect 3 [ n ] ∗ comb14 [ n ] (b) First convolve the two rectangles. Then convolve the result with the comb, thereby periodically repeating it. (e) Similar to (d) but with a different result. Solutions 3-31 M. J. Roberts - 8/16/04 2π n 7 g [ n ] = 2 cos ∗ u[n] 7 8 n (f) Write the convolution sum. Express the cos in exponential form and combine with other terms. Then use ∞ 1 ∑ rn = 1 − r , r < 1 n=0 to put the result in closed form, and simplify using cos ( x + y ) = cos ( x ) cos ( y ) − sin ( x ) sin ( y ) Finally use B A cos ( x ) +...
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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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