Chap3StudentSolutions

2 2 2 the first equation can be written as 10 y1 t t0

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Unformatted text preview: ( t) | + y 2 ( t)] = 10 y′′′ t) − 14 y′′( t) + 7 y′ ( t) − 25 y 3 ( t) 3( 3 3 This can only be true for all time for an arbitrary excitation if y 3 ( t) = y1 ( t) + y 2 ( t) . Additive Since it is homogeneous and additive, it is also linear. Time Invariance: Let x1 ( t) = g( t) . Then 10 y1 ( t) − 14 y1 ( t) + 7 y1 ( t) − 25 y1 ( t) = g( t) . ′′ ′ ′′′ Let x 2 ( t) = g( t − t0 ) . Then 10 y′′′ t) − 14 y′′( t) + 7 y′ ( t) − 25 y 2 ( t) = g( t − t0 ) . 2( 2 2 The first equation can be written as 10 y1 ( t − t0 ) − 14 y1 ( t − t0 ) + 7 y1 ( t − t0 ) − 25 y1 ( t − t0 ) = g( t − t0 ) ′′ ′ ′′′ Therefore Solutions 3-4 M. J. Roberts - 8/16/04 10 y1 ( t − t0 ) − 14 y1 ( t − t0 ) + 7 y1 ( t − t0 ) − 25 y1 ( t − t0 ) ′′ ′ ′′′ = 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) = y1 ( t − t0 ) . Time Invariant Stability: The characteristic equation is 10λ3 − 14 λ2 + 7λ − 25 = 0 . The eigenvalues are λ1 = 1.7895 λ 2 = -0.1948 + j1.1658 λ 3 = -0.1948 - j1.1658 So the homogeneous solution is of the form, y( t) = K1e1.7895t + K 2e( -0.1948 + j1.1658) t + K 3e( -0.1948 - j1.1658) t . If there is no excitation, but the zero-excitation response is not zero, the response will grow without bound as time increases. Unstable Causality: The system equation can be rewritten as t λ3 λ2 t λ3 λ2 ∫ ∫ ∫ x(λ1 ) dλ1dλ 2 dλ 3 + 25 ∫ ∫ ∫ y(λ1 ) dλ1dλ 2 dλ 3 1 −∞ −∞ −∞ −∞ −∞ −∞ y( t) = t λ2 t 10 − 7 ∫ ∫ y(λ1 ) dλ1dλ 2 + 14 ∫ y(λ1 ) dλ1 −∞ −∞ −∞ So the response at any time, t = t0 , depends on the excitation at times, t < t0 and not on any future values. Causal Memory: The response at any time, t = t0 , depends on the excitation at times, t < t0 . System has memory. Invertibility: The system equation, 10 y′′′ ( t ) − 14 y′′ ( t ) + 7 y′ ( t ) − 25 y ( t ) = x ( t ) expresses the excitation in terms of the response and its derivatives. Therefore the excitation is uniquely determined by the response. Invertible. Solutions 3-5 M. J. Roberts - 8/16/04 7. Show that the system of Figure E7 is non-linear, BIBO stable, static and non-invertible. (The response of an analog multiplier is the product of its two excitations.) Analog Multiplier x[n] y[n] 2 Figure E7 A DT system 8. Show that a system with excitation, x[ n ] , and response, y[ n ] , described by y[ n ] = n x[ n ] , is linear, time variant and static. 9. Show that the system of Figure E9 is linear, time-invariant, BIBO unstable and dynamic. x[n] y[n] D Figure E9 A DT system y[ n ] = x[ n ] + y[ n − 1] y[ n − 1] = x[ n − 1] + y[ n − 2] Then, by induction, y[ n ] = x[ n ] + x[ n − 1] + y[ n − 2] ∞ y[ n ] = x[ n ] + x[ n − 1] + L + x[ n − k ] + L = ∑ x[ n − k ] k =0 Let m = n − k . Then y[ n ] = −∞ ∑ x[m]...
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