Chap3StudentSolutions

# Not causal memory at time t 2 y2 x1 therefore the

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Unformatted text preview: , depends on the excitation at a later time, t = −1. Not Causal Memory: At time, t = 2, y(2) = x(1) . Therefore the response at time, t = 2, depends on the excitation at a previous time, t = 1. System has memory. Invertibility: The system excitation at any arbitrary time, t = t0 , is uniquely determined by the system response at time, t = 2 t0 . Invertible. 4. Show that a system with excitation, x( t) , and response, y( t) , described by y( t) = cos(2πt) x( t) is time variant, BIBO stable, static and non-invertible. Time Invariance: Let x1 ( t) = g( t) . Then y1 ( t) = cos(2πt) g( t) . Let x 2 ( t) = g( t − t0 ) . Then y 2 ( t) = cos(2πt) g( t − t0 ) ≠ y1 ( t − t0 ) = cos(2π ( t − t0 )) g( t − t0 ) . Time Variant Invertibility: This system is not invertible because when the cosine function is zero the unique relationship between x and y is lost; any x produces the same y, zero. Not Invertible. 5. Show that a system whose response is the magnitude of its excitation is non-linear, BIBO stable, causal and non-invertible. y( t) = x( t) Invertibility: Any response, y, can be caused by either x or –x. Not Invertible. 6. Show that the system in Figure E6 is linear, time invariant, BIBO unstable and dynamic. x(t) 0.1 ∫ ∫ ∫ 1.4 -0.7 2.5 Solutions 3-3 y(t) M. J. Roberts - 8/16/04 Figure E6 A CT system The differential equation of the system is 10 y′′′ ( t) − 14 y′′ ( t) + 7 y′ ( t) − 25 y( t) = x( t) . Homogeneity: Let x1 ( t) = g( t) . Then 10 y1 ( t) − 14 y1 ( t) + 7 y1 ( t) − 25 y1 ( t) = g( t) . ′′ ′ ′′′ Let x 2 ( t) = K g( t) . Then 10 y′′′ ( t) − 14 y′′ ( t) + 7 y′ ( t) − 25 y 2 ( t) = K g( t) . 2 2 2 If we multiply the first equation by K, we get 10K y1 ( t) − 14 K y1 ( t) + 7K y1 ( t) − 25K y1 ( t) = K g( t) ′′′ ′′ ′ Therefore 10K y1 ( t) − 14 K y1 ( t) + 7K y1 ( t) − 25K y1 ( t) = 10 y′′′ ( t) − 14 y′′ ( t) + 7 y′ ( t) − 25 y 2 ( t) ′′′ ′′ ′ 2 2 2 This can only be true for all time for an arbitrary excitation if y 2 ( t) = K y1 ( t) . Homogeneous Additivity: Let x1 ( t) = g( t) . Then 10 y1 ( t) − 14 y1 ( t) + 7 y1 ( t) − 25 y1 ( t) = g( t) . ′′ ′ ′′′ Let x 2 ( t) = h( t) . Then 10 y′′′ ( t) − 14 y′′ ( t) + 7 y′ ( t) − 25 y 2 ( t) = h( t) . 2 2 2 Let x 3 ( t) = g( t) + h( t) . Then 10 y′′′ ( t) − 14 y′′ ( t) + 7 y′ ( t) − 25 y 3 ( t) = g( t) + h( t) 3 3 3 Adding the first two equations, 10[ y1 ( t) + y′′′ ( t)] − 14[ y1 ( t) + y′′ ( t)] + 7[ y1 ( t) + y′ ( t)] − 25[ y1 ( t) | + y 2 ( t)] = g( t) + h( t) ′′ ′ ′′′ 2 2 2 Therefore 10[ y1 ( t) + y′′′ ( t)] − 14[ y1 ( t) + y′′ ( t)] + 7[ y1 ( t) + y′ ( t)] − 25[ y1 ( t) | + y 2 ( t)] ′′ ′ ′′′ 2 2 2 = 10 y′′′ ( t) − 14 y′′ ( t) + 7 y′ ( t) − 25 y 3 ( t) 3 3 3 10[ y1 ( t) | + y 2 ( t)]′′′ − 14[ y1 ( t) | + y 2 ( t)]′′ + 7[ y1 ( t) | + y 2 ( t)]′ − 25[ y1...
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