Chap3StudentSolutions

Xt xt yt a b solutions 3 22 yt m j roberts

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Unformatted text preview: /2 -3/2 < t < -1/2 1 1 rect(τ) and tri(t- τ) -4 τ 4 -1/2 < t < 1/2 rect(τ) and tri(t- τ) rect(τ) and tri(t- τ) 1 -4 4 1/2 < t < 3/2 τ -4 4 τ 3/2 < t rect(τ) and tri(t- τ) rect(τ) and tri(t- τ) 1 1 -4 4 τ -4 4 τ 3 , g( t) = 0. 2 t +1 t +1 t +1 τ2 3 1 If − < t < − , g( t) = ∫ 1 − τ − t dτ = ∫ (1 − (τ − t)) dτ = τ − + τt { 2 2 2 >0 − 1 1 1 − − If t < − 2 2 2 1 − 2 (t + 1) 1 2 1 g( t) = t + 1 − + ( t + 1) t − − + − − t 2 2 2 2 2 t 3t 9 g( t) = + + 228 2 1 1 If − < t < , g( t) = 2 2 t 1 2 { { ∫ 1 − τ − t dτ + ∫ 1 − τ − t dτ − 1 2 <0 t >0 1 2 1 t τ2 2 τ2 g( t) = ∫ (1 − ( t − τ )) dτ + ∫ (1 − (τ − t)) dτ = τ − τt + + τ − + τt 2 − 1 2 t 1 t − t 2 2 t2 1 t 1 1 1 t t2 g( t) = t − t 2 + + − − + − + − t + − t 2 2 2 2 8 2 8 2 2 g( t) = 32 −t 4 By symmetry, g( t) = g(− t) and Solutions 3-21 M. J. Roberts - 8/16/04 3 0, t> 2 2 3t 9 1 3 t g( t) = − + , <t< 2 2 2 2 8 1 3 2 t< 4 − t , 2 g(t) 1 -2 2 (c) g( t) = e − t u( t) ∗ e − t u( t) (d) 1 1 1 t g( t) = tri 2 t + − tri 2 t − ∗ comb 2 2 2 2 (e) t 1 1 g( t) = tri 2 t + − tri 2 t − ∗ comb( t) 2 2 This is a very complicated way of saying g ( t ) = 0 . Can you determine this without going throught the whole process of convolving them? 23. A system has an impulse response, h( t) = 4 e −4 t u( t) . Find and plot the response of the 1 system to the excitation, x( t) = rect 2 t − . 4 Express the rectangle as a difference between two unit steps to simplify the problem. 24. Change the system impulse response in Exercise 23 to h( t) = δ ( t) − 4 e −4 t u( t) and find 1 and plot the response to the same excitation, x( t) = rect 2 t − . 4 25. Find the impulse responses of the two systems in Figure E25. Are these systems BIBO stable? ∫ x(t) x(t) ∫ y(t) (a) (b) Solutions 3-22 y(t) M. J. Roberts - 8/16/04 Figure E25 Two single-integrator systems (a) y′ ( t) = x( t) ⇒ h( t) = u( t) A CT system is BIBO stable if its impulse response is absolutely integrable. Impulse response is not absolutely integrable. BIBO unstable. 26. Find the impulse response of the system in Figure E26. Is this system BIBO stable? ∫ x(t) ∫ y(t) Figure E26 A double-integrator system 27. In the circuit of Figure E27 the excitation is v i ( t) and the response is v o ( t) . (a) Find the impulse response in terms of R and L. (b) If R = 10 kΩ and L = 100 µH graph the unit step response. R + vi (t) + vo(t) L - Figure E27 An RL circuit vi ( t ) = R i ( t ) + vo ( t ) i (t ) = vo ( t ) = L vi ( t ) − vo ( t ) R d L ( i (t )) = R v′i (t ) − v′o (t ) dt vi ( t ) = vi ( t ) − vo ( t ) + L v′ ( t ) − v′ ( t ) i o R L L v′ ( t ) + vo ( t ) = v′ ( t ) o i R R Solutions 3-23 M. J. Roberts - 8/16/04 L L h′ ( t ) + h ( t ) = δ ′ ( t ) R R h (t ) = 0 , t<0 For times, t > 0 , the solution is the homogeneous solution, h (t ) = K h e R −t L , t>0 Since the highest derivatives on both sides of this differential equation are the same the impulse...
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This note was uploaded on 06/19/2013 for the course ENSC 380 taught by Professor Atousa during the Spring '09 term at Simon Fraser.

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