Chap3StudentSolutions

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online