solmidterm08_mae143b

solmidterm08_mae143b - Midterm - MAE143B, Summer II 2008...

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Unformatted text preview: Midterm - MAE143B, Summer II 2008 SOLUTIONS Prof. R.A. de Callafon August 25, 2008 2:00pm-3:20pm, CENTER 115 open-book and open-notes midterm exam use the available space to derive your results, attach extra paper if necessary use of any electronic equipment (calculator, phone, PDA) not allowed during exam Consider a linear dynamic system G characterized by the ODE √ d d d2 u(t) − 2 y(t) − 2 y(t) dt dt dt y(t) = 10u(t) − (1) 1. Write down a transfer function model for the ODE in (1) and show that the system is stable and has a right half plane zero. [10pt] Applying Laplace transform to the ODE leads to √ 10 − s √ s2 + 2s + 1 √ √ √ √ Solving s2 + 2s + 1 = (s + 1 2)2 + 1 = 0 yields the poles p1,2 = − 1 2 ± j 1 2. Since the 2 2 2 2 real part of all poles pi is smaller than 0, system is stable. Solving 10 − s = 0 yields the zero zi = 10. Since real part of zi is positive, it is a RHP (right half plane) zero. y(s) = 10u(s) − su(s) − s2 y(t) − 2sy(t) ⇒ y(s) = G(s)u(s), G(s) = 2. Compute the final value of y(t) for both an impulse and a unit step input u(t). [15pt] System is stable, we can use Final Value Theorem. For impulse input u(s) = 1 and 10 − s √ =0 lim = lim sy(s) = lim sG(s)u(s) = lim s · t→∞ s→0 s→0 s→0 s2 + 2s + 1 For step input u(s) = 1 s and 10 − s √ = 10 s→0 s2 + 2s + 1 lim = lim sy(s) = lim sG(s)u(s) = lim t→∞ s→0 s→0 3. Consider an exponential input signal u(t) = e10t . Compute the output y(t) and show that the output y(t) satisfies limt→∞ y(t) = 0, despite the increasing input signal. [15pt] With u(t) = e10t , the Laplace transform √ 1 √ 2 1 1 √ √2 √ = 2· u(s) = making y(s) = G(s)u(s) = 1 2 + 2 + ( 1 2)2 s − 10 s 2s + 1 (s + 2 2) 2 for which the inverse Laplace transform can be computed directly via entry #20 in the Laplace table on the inside cover of the textbook: y(t) = √ 1 2e− 2 √ 2t sin( 1√ 2t) 2 1 and indeed limt→∞ y(t) = 0 due to the exponential decay e− 2 1 √ 2t . Consider a dynamic system G given by the state space realization ⎧ ˙ −1 4 1 x1 (t) ⎪ x1 (t) ⎪ = + ⎪ x (t) ⎪ 0 −6 −1 ˙2 x2 (t) ⎪ ⎨ F G G: ⎪ x1 (t) ⎪ ⎪ 3 4 + 1 u(t) y(t) = ⎪ ⎪ x2 (t) ⎩ H u(t) (2) D based on the two-dimensional state vector x(t) = [x1 (t) x2 (t)]T . 4. Compute the transfer function G(s) of the system G. [20pt] Formula for computing G(s) from state space model is G(s) = H(sI − F )−1 G + D. First: (sI − F ) −1 = −1 s + 1 −4 0 s+6 = 1 (s + 1)(s + 6) s+6 4 0 s+1 Second, this makes H(sI − F )−1 : 1 (s + 1)(s + 6) 3 4 s+6 4 0 s+1 = 1 (s + 1)(s + 6) 3s + 18 4s + 16 and third H(sI − F )−1 G: 1 (s + 1)(s + 6) 3s + 18 4s + 16 1 −1 = −s + 2 (s + 1)(s + 6) finally making our transfer function G(s) = H(sI − F )−1 G + D: −s + 2 + s2 + 7s + 6 s2 + 6s + 8 (s + 2)(s + 4) −s + 2 +1= = = (s + 1)(s + 6) (s + 1)(s + 6) (s + 1)(s + 6) (s + 1)(s + 6) 5. Explain why the poles of the transfer function G(s), in this case, are equal to the diagonal elements of the state matrix F in (2) [10pt] As always, the denominator of the transfer function G(s) = H(sI − F )−1 G + D is determined by the det(sI − F ) due to the computation of the inverse of (sI − F ). Solving det(sI − F ) = 0 to compute the poles of G(s) is thus equivalent to computing the eigenvalues of the matrix F . If, in addition, the matrix F is either an upper or lower diagonal matrix, (sI − F ) will also be an upper or lower diagonal matrix. For an upper or lower diagonal matrix (sI − F ), the computation of det(sI − F ) is simply the product of the diagonal elements of the matrix (sI − F ), e.g. det(sI − F ) = (s − f11 )(s − f22 ) · · · (s − fnn ). It can be noted that the 2 × 2 matrix F in (2) is upper diagonal, making det(sI − F ) = (s − f11 )(s − f22 ) = (s + 1)(s + 6), making the poles p1 = −1 and p2 = −6 the same as f11 = −1 and f22 = −6. 2 Consider the 1st order lag compensator given by C(s) = K τ1 s + 1 , τ2 s + 1 0 < τ1 < 2 (3) 6. Compute the approximate behavior (asymptotes) of the frequency response |C(jω)| and ∠C(jω) for the low (ω < τ12 ), middle ( τ12 < ω < τ1 ) and high (ω > τ11 ) frequencies. [20pt] 1 Notice that τ2 > τ1 so indeed 1/τ2 < 1/τ1 . Since C(jω) = K 1 + jτ1 ω 1 + jτ2 ω we have |C(jω)| = |K| 1 + (τ1 ω)2 1 + (τ2 ω)2 and ∠C(jω) = ∠K + tan−1 τ1 ω − tan−1 τ2 ω For the low, middle and high frequencies we can make the following approximations: √ 1 1 : |C(jω)| ≈ |K| √ = |K| ω< τ2 1 ∠C(jω) ≈ ∠K + tan−1 0 − tan−1 0 = ∠K √ 1 1 1 1 <ω< : |C(jω)| ≈ |K| = |K| 2 τ2 τ1 (τ2 ω) |τ2 ω| ∠C(jω) ≈ ∠K + tan−1 0 − tan−1 ∞ = ∠K − π/2 (τ1 ω)2 |τ1 | 1 |τ1 ω| = |K| ω> : |C(jω)| ≈ |K| = |K| τ1 |τ2 ω| |τ2 | (τ2 ω)2 −1 −1 ∠C(jω) ≈ ∠K + tan ∞ − tan ∞ = ∠K + π/2 − π/2 = ∠K Hence, for sketching the Bode plots (log ω versus log |C(jω)| and log ω versus ∠C(jω)) in the next question, for low frequencies (ω < τ12 ) we have a horizontal line for log |C(jω)| at log |K|. For the middle frequencies ( τ12 < ω < τ11 ) we have a line with slope -1 (also -20dB/dec) for 1 log |C(jω)| as log |K| |τ2 ω| = log |K| − log |τ2 | − log ω and for high frequencies (ω > τ11 ) we have again a horizontal line for log |C(jω)| at log |K| |τ1 || . The phase ∠C(jω) simply has 2 |τ2 horizontal lines at ∠K a low at high frequencies and a horizontal line at ∠K − π/2 at the middle frequencies. Note that ∠K can only be 0 or −π as K is a real number. In case K ≥ 0, ∠K = 0 whereas K < 0 makes ∠K = −π (as is the case in the next question). 3 7. Let K = −10, τ1 = 1 and τ2 = 10 in the lag compensator C(s) given in (3). Sketch the amplitude and phase Bode plot of C(s) in the figures below. Use solid lines to sketch actual Bode plots, use dashed lines to indicate asymptotic behavior of the Bode plots. [10pt] 2 10 1 |C(jω)| 10 0 10 −1 10 −2 10 −2 10 −1 10 0 10 1 10 1 10 10 2 ω rad/s 90 ∠C(jω) [deg] 0 −90 −180 −270 −360 −2 10 −1 10 0 10 10 ω rad/s end of midterm 4 2 ...
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