Chap3StudentSolutions

Since the system is both homogeneous and additive it

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Unformatted text preview: K y1[ n ] − K y1[ n − 1] + 2K y1[ n − 2] = K g[ n ] Then, equating results, 4 y 2 [ n ] − y 2 [ n − 1] + 2 y 2 [ n − 2] = 4 K y1[ n ] − K y1[ n − 1] + 2K y1[ n − 2] If this equation is to be satisfied for all n, y 2 [ n ] = K y1[ n ] . Homogeneous. Additivity: Let x1[ n ] = g[ n ] . Then 4 y1[ n ] − y1[ n − 1] + 2 y1[ n − 2] = g[ n ] Let x 2 [ n ] = h[ n ] . Then 4 y 2 [ n ] − y 2 [ n − 1] + 2 y 2 [ n − 2] = h[ n ] Let x 3 [ n ] = g[ n ] + h[ n ] . Then 4 y 3 [ n ] − y 3 [ n − 1] + 2 y 3 [ n − 2] = g[ n ] + h[ n ] Add the two first two equations. 4 ( y1[ n ] + y 2 [ n ]) − ( y1[ n − 1] + y 2 [ n − 1]) + 2( y1[ n − 2] + y 2 [ n − 2]) = g[ n ] + h[ n ] Then, equating results, 4 y 3 [ n ] − y 3 [ n − 1] + 2 y 3 [ n − 2] = 4 ( y1[ n ] + y 2 [ n ]) − ( y1[ n − 1] + y 2 [ n − 1]) + 2( y1[ n − 2] + y 2 [ n − 2]) If this equation is to be satisfied for any arbitrary excitation for all n, Additive. y 3 [ n ] = y1[ n ] + y 2 [ n ]. Since the system is both homogeneous and additive, it is linear. Since the system is linear it is also incrementally linear. Since the system is linear, it is not statically non-linear. Solutions 3-9 M. J. Roberts - 8/16/04 Time Invariance: Let x1[ n ] = g[ n ] . Then 4 y1[ n ] − y1[ n − 1] + 2 y1[ n − 2] = g[ n ] Let x 2 [ n ] = g[ n − n 0 ] . Then 4 y 2 [ n ] − y 2 [ n − 1] + 2 y 2 [ n − 2] = g[ n − n 0 ] We can re-write the first equation as 4 y1[ n − n 0 ] − y1[ n − n 0 − 1] + 2 y1[ n − n 0 − 2] = g[ n − n 0 ] Then, equating results, 4 y1[ n − n 0 ] − y1[ n − n 0 − 1] + 2 y1[ n − n 0 − 2] = 4 y 2 [ n ] − y 2 [ n − 1] + 2 y 2 [ n − 2] If this equation is to be satisfied for any arbitrary excitation for all n, then y 2 [ n ] = y1[ n − n 0 ] . Time Invariant. Stability: The eigenvalues of the system homogeneous solution are found from the characteristic equation, 4α 2 − α + 2 = 0 . They are α1,2 = 0.125 ± j 0.696 or α1,2 = 0.7071e ± j1.3931 Therefore the homogeneous solution is of the form, y h [ n ] = K h1 ( 0.7071) e+ j1.3931n + K h 2 ( 0.7071) e− j1.3931n n n and, as n approaches infinity the homogeneous solution approaches zero and the total solution approaches the particular solution. The particular solution is bounded because it consists of functions of the same form as the excitation and all its unique differences and the excitation is bounded in the BIBO stability test. Therefore if x is bounded, so is y. Stable. Causality: We can rearrange the system equation into y[ n ] = 1 (x[n] + y[n − 1] + 2 y[n − 2]) 4 showing that the response at time, n, depends on the excitation at time, n, and the response at previous times. It does not depend on any future values of the excitation. Causal. Memory: The response depends on past values of the response. The system has memory. Invertibility: Solutions 3-10 M. J. Roberts - 8/16/04 The original system equation, 4 y[ n ] − y[ n − 1] + 2 y[ n − 2] = x[ n ], expresses the excitation in terms of the response. Invertible. 13. Find the impulse responses of these systems. (a) y[ n ] = x[ n ] − x[ n − 1] The impulse response is very easily found by direct iteration to be h[ n ] = δ [ n ] − δ [ n − 1] . Also, using linearity and superposition...
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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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