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Unformatted text preview: Midterm  MAE143B, Summer II 2008 SOLUTIONS Prof. R.A. de Callafon August 25, 2008 2:00pm3:20pm, CENTER 115 openbook and opennotes 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 twodimensional 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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