Lecture 8

Lecture 8 - ECE 3101 Signals and Systems Reading...

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Lecture 8, 2/21/11 (Makeup Class) ECE 3101, Copyright P. B. Luh 1 Reading Assignment: Sections 2.5 and 3.6 Problem Set 3: Due tomorrow Problem Set 4: 2.5-2, 2.5-3, 3.6-2, 3.6-3, 3.6-6 due 3/1/11 The next makeup class: Wed., 3/2/11, 6:30 pm, MSB 411 Also, I will be at NSF on Tue. 3/1/11. That class and the discuss session on Mon. 3/14/11 will be swapped Last Time: Zero-State Response of LTIC Systems Sample problems for Exam. 1 Convolution of Continuous-Time Signals Direction Integration Based on Definition Semi-graphical Method ECE 3101 Signals and Systems
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Lecture 8, 2/21/11 (Makeup Class) ECE 3101, Copyright P. B. Luh 2 Laplace Transform (Later) Properties of Convolution Examples Today: Classical Solution of Differential Equations Time-Domain Analysis of D-T Systems Zero Input Response Examples Next Time: Sections 3.7- 3.9
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3 Total Response y(t) = y ZI (t) + y ZS (t) = Σ c k e λ k t + h(t) u(t) Example 2.2. y(0-) = 0, v c (0-) = 5, u(t) = 10e -3t U(t) y(t)? y(t) = y ZI (t) + y ZS (t), then ICBS from the textbook = (-5e -t + 5e -2t ) + (-5e -t + 20e -2t - 15e -3t ) = (-10e -t + 25e -2t ) + (-15e -3t ) Natural response y n (t) Forced response y φ (t) This is the “Classical” way of looking at the solution Not that good from system analysis point of view y(t) = y ZI (t) + y ZS (t) is still preferable Nevertheless, let us find y φ (t) and y(t) in this way. How? + + - - u(t) v c (t) 0.5F 1H y(t) 3 Ω
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4 Classical Solution of Differential Equations Q(D)y(t) = P(D)u(t) y(t) = y n (t) + y φ (t) Q(D) [y n (t) + y φ (t)] = P(D)u(t) Q(D)y n (t) + Q(D)y φ (t) = P(D)u(t) Q. Q(D)y n (t) = ? Q(D)y n (t) = 0 since y n (t) is the sum of natural modes Σ c k e λ k t Q(D)y φ (t) = P(D)u(t) Q. What can be said about y φ (t)? Generally not easy to find. For specific classes of u(t), we do know how to find it The Method of Undetermined Coefficients
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Lecture 8, 2/21/11 (Makeup Class) ECE 3101, Copyright P. B. Luh 5 The Method of Undetermined Coefficients ICBS that u(t) = e ζ t , ζ≠λ i , i = 1, 2, . ., N y φ (t) = β e ζ t u(t) = e ζ t , ζ = λ i y φ (t) = β te ζ t u(t) = k (a constant) y φ (t) = β (a constant) u(t) = cos( ω t+ θ ) y φ (t) = β cos( ω t+ φ ) u(t) = (t r + α r-1 t r-1 +.. + α 1 t + α 0 )e ζ t y φ (t) = ( β r t r + β r-1 t r-1 +.. + β 1 t + β 0 )e ζ t
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6 Example. (D 2 + 5D + 6)y(t) = (D + 1)u(t) with u(t) = 6t 2 , y( 0+ ) = 25/18 and dy/dt( 0+ ) = -2/3. Find y φ (t) and y(t) Characteristic polynomial: λ 2 + 5 λ + 6 = ( λ + 2)( λ + 3) Characteristic mode: e -2t and e -3t y n (t) = c 1 e -2t + c 2 e -3t u(t) = 6t 2 , y φ (t) = ( β 2 t 2 + β 1 t + β 0 ) (D 2 + 5D + 6)y φ (t) = (D + 1)u(t) (D 2 + 5D + 6) ( β 2 t 2 + β 1 t + β 0 ) = (D + 1) 6t 2 (2 β 2 ) + (10 β 2 t + 5 β 1 ) + (6 β 2 t 2 + 6 β 1 t + 6 β 0 ) = 6t 2 + 12t
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Lecture 8 - ECE 3101 Signals and Systems Reading...

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