Chap3StudentSolutions

# So the solution is simply the particular solution of

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Unformatted text preview: response contains an impulse and is of the form, h ( t ) = Kδ δ ( t ) + K h e R −t L u (t ) Integrating both sides of the differential equation from 0 − to 0 + , 0+ L L L + − + ∫ h ( t ) dt = δ 0 + − δ 0 − ⇒ K h + Kδ = 0 h 0 −h 0 0− R R R =0 =0 =0 = Kh () () () () = Kδ Integrating both sides of the differential equation a second time from 0 − to 0 + , 0+ 0+ L L u 0 + − u 0 − ⇒ Kδ = 1 h ( t ) dt + Kδ ∫ u ( t ) dt = R 0∫ R − 0− =1 =0 () () = Kδ =0 Then, from the first integration, K h = − R R − Rt and h ( t ) = δ ( t ) − e L u ( t ) L L The unit-step response, h −1 ( t ) is the integral of the impulse response, t R − Rλ R − Rλ δ ( λ ) − e L u ( λ ) d λ = ∫ δ ( λ ) − e L d λ ∫ L L −∞ 0− For t < 0 the integral is obviously zero. Therefore h −1 ( t ) = 0 , t < 0 h −1 ( t ) = t t t R R −t −t R − Rλ R L − Rλ L L L h −1 ( t ) = 1 − ∫ e d λ = 1 − − e = 1 + e − 1 = e L L 0− L R 0− h −1 ( t ) = e R −t L u ( t ) = e−10 t u ( t ) , Step response 8 Solutions 3-24 , t>0 M. J. Roberts - 8/16/04 vo(t) 1 -0.01 0.04 t (µs) 28. Find the impulse response of the system in Figure E28 and evaluate its BIBO stability. x(t) ∫ ∫ y(t) 1 10 1 20 Figure E28 A two-integrator system y′′ ( t) = x( t) + 1 1 y′ ( t) − y( t) 10 20 29. Find the impulse response of the system in Figure EError! Reference source not found. and evaluate its BIBO stability. x(t) ∫ ∫ y(t) 2 3 1 8 Figure EError! Reference source not found. A two-integrator system 30. Plot the amplitudes of the responses of the systems of Exercise 19 to the excitation, e jωt , as a function of radian frequency, ω . (a) y′ ( t) + 5 y( t) = x( t) First realize that the excitation, e jωt , is periodic, that is, it has always existed and will always exist repeating periodically. Therefore there is no homogeneous solution to worry about. If the system is stable it died out a long time ago and if the system is not stable, this exercise has no useful physical interpretation. So the solution is simply the particular solution of the differential equation of the form, Solutions 3-25 M. J. Roberts - 8/16/04 y p ( t) = Ke jωt . Putting that into the differential equation and solving, 1 jω + 5 Kω e jω t + 5 Ke jω t = e kω t ⇒ K = |K| 0.2 -10 π ω 10π 31. Plot the responses of the systems of Exercise 19 to a unit-step excitation. (a) h( t) = e −5 t u( t) h −1 ( t) = t ∫ h(λ ) dλ = −∞ t ∫ −∞ t e −5λ u(λ ) dλ = ∫ e −5λ dλ = − 0 1 −5λ e 5 t = 0 1 (1 − e −5t ) , t > 0 5 h −1 ( t)0 , t < 0 h −1 ( t) = 1 (1 − e −5t ) u(t) 5 h (t) -1 0.2 1 t 32. A CT system is described by the block diagram in Figure E32. x(t) 1 4 ∫ ∫ 1 4 3 4 Figure E32 A CT system Solutions 3-26 y(t) M. J. Roberts - 8/16/04 Classify the system as to homogeneity, additivity, linearity, time-invariance, stability, causality, memory, and invertibility. 33. A system has a response that is the cube of i...
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