Chap3StudentSolutions

Then y 2 n n gm n 0 m the first equation

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Unformatted text preview: = m =n n ∑ x[m] m =−∞ Homogeneity: Let x1[ n ] = g[ n ] . Then y1[ n ] = n ∑ g[m] m =−∞ Let x 2 [ n ] = K g[ n ] . Then y 2 [ n ] = n n m =−∞ m =−∞ ∑ K g[m] = K ∑ g[m] = K y [n] . Homogeneous. Let x1[ n ] = g[ n ] . Then y1[ n ] = n ∑ g[m] m =−∞ Solutions 3-6 1 M. J. Roberts - 8/16/04 Let x 2 [ n ] = h[ n ] . Then y 2 [ n ] = Let x 3 [ n ] = g[ n ] + h[ n ] . Then y 3 [ n ] = n ∑ m =−∞ n ∑ h[m] m =−∞ (g[m] + h[m]) = n ∑ m =−∞ g[ m] + n ∑ h[m] = y [n] + y [n] . 1 m =−∞ 2 Additive. Since the system is homogeneous and additive it is also linear. The system is also incrementally linear because it is linear. The system is not statically non-linear because it is linear. Time Invariance: Let x1[ n ] = g[ n ] . Then y1[ n ] = n ∑ g[m] . m =−∞ Let x 2 [ n ] = g[ n − n 0 ] . Then y 2 [ n ] = n ∑ g[m − n ] . 0 m =−∞ The first equation can be rewritten as y1[ n − n 0 ] = Let m = q − n 0 . Then y1[ n − n 0 ] = n − n0 ∑ g[m] m =−∞ n ∑ g[q − n ] = y [n] q =−∞ 0 2 Time invariant Stability: If the excitation is a constant, the response increases without bound. n Also the solution of the homogeneous difference equation is y h [ n ] = K (1) = K . Therefore the eigenvalue is 1 whose magnitude is not less than 1 and the system must be BIBO unstable. Unstable Causality: At any discrete time, n = n 0 , the response depends only on the excitation at that discrete time and previous discrete times. Causal. Memory: At any discrete time, n = n 0 , the response depends on the excitation at that discrete time and previous discrete times. System has memory. Invertibility: Inverting the functional relationship, Solutions 3-7 M. J. Roberts - 8/16/04 y[ n ] = n ∑ x[m] . m =−∞ Invertible. Taking the first backward difference of both sides of the original system equation, y[ n ] − y[ n − 1] = n n −1 m =−∞ m =−∞ ∑ x[m] − ∑ x[m] x[ n ] = y[ n ] − y[ n − 1] The excitation is uniquely determined by the response. Invertible. 10. Show that a system with excitation, x[ n ] , and response, y[ n ] , described by y[ n ] = rect ( x[ n ]) , is non-linear, time invariant and non-invertible. 11. Show that the system of Figure E11 is non-linear, time-invariant, static and invertible. 5 x[n] y[n] 10 Figure E11 A DT system y[ n ] = 10 x[ n ] − 5 , The system is incrementally linear because the only deviation from linearity is caused by the presence of the non-zero, zero-excitation response. Invertibility: Solving the system equation for the excitation as a function of the response, x[ n ] = Invertible. y[ n ] + 5 10 12. Show that the system of Figure E12 is time-invariant, BIBO stable, and causal. Solutions 3-8 M. J. Roberts - 8/16/04 x[n] y[n] 1 4 D D y[n-2] 1 4 1 2 Figure E12 A DT system Homogeneity: Let x1[ n ] = g[ n ] . Then 4 y1[ n ] − y1[ n − 1] + 2 y1[ n − 2] = g[ n ] Let x 2 [ n ] = K g[ n ] . Then 4 y 2 [ n ] − y 2 [ n − 1] + 2 y 2 [ n − 2] = K g[ n ] Multiply the first equation by K. 4...
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