18.03 Problem Set 8: Part II Solutions
Part I points: 29. 4, 30. 0, 32. 8, 33. 4.
Part II solutions:
29. (a) [8] (1) {1, i, −i}: For large t, these functions have exponential growth rate
et
. (This means that for any a < 1 < c, eat < |f(t)| < ect for large t.) They also
show a small oscillation of approximately constant amplitude and circular frequency 1.
a b
Examples: f(t) = aet + b sin(t) (a, b = 0) with F (s) = + .≥s − 1 s2 + 1
(2) {−1 + 4i, −1 − 4i}: For large t, these functions show exponential decay like e−t
,
and oscillate with circular frequency 4. Examples: f(t) = ae−t sin(4t) (a = 0) with ≥
4a
F (s) = .
(s + 1)2 + 16
(3) {−1}: For large t, these functions decay like e−t and do not oscillate. Examples:
a
f(t) = ae−t (a = 0) with ≥F (s) = s + 1 .
(4) No poles: For large t, these functions decay to zero faster than any exponential.
Examples: f(t) = aω(t − b) (a = 0≥−
, b
bs� 0) with F (s) = ae−bt, or f(t) = a(u(t) − u(t − b))
(a = 0≥, b > 0) with F (s) = a 1 −
s
e.
(b) (i) [4] Method I: For t > 0, w(t) is a solution to the homogeneous equation. The roots
must be −2
1 ± 2
3 i, so p(s) = m �
s − (−2
1 + 2
3 i)
��
s − (−2
1 − 2
3 i)
� = m(s2 +s+ 2
5 ). To find m,
remember that w˙ (0+) = 1 (for a second order system). w˙ (t) = u(t)e−t/2(3 cos(3t/2) −
m 2
(−1
2 sin(3t/2)), so w˙ (0+) = 3
2 , m = 2
3 , and p(s) = 3
2 (s2 + s + 5
2 ).
3/2 1
Method II: W (s) = L[w(t)] = (s + (1/2))2 + (9/4) = (2/3)(s2 + s + (5/2)),
so p(s) = 2
3 (s2 + s + 5
2 ).
(ii) [4] {−1
2 ± 3 i}.
2
3/2
(iii) [4] This is a throwback problem. The complex gain is W (i�) = , so
((5/2) − �2) + i�
3/2
the gain is W (i�) = � . This is maximized when the denominator, | | ((5/2) − �2)2 + �2
or its square (5
2 − �2)2 + �2, is minimized. The derivative with respect to � is
2(5
2 − �2)(−2�) + 2�, which has roots at � = 0 and � = ±→2. So �r = →2 is the worst
frequency.
(iv) [6] The yellow box lies in the plane above the imaginary axis. The base is the
amplitude response curve. The green box lies in the plane above the real axis. Its top
lies on the graph of W (s) , and its base is the real axis. The red arrows lie above the | |
poles of W (s). The yellow diamonds are located at (±i�, |W (i�)|), and represent the
chosen value of the input circular frequency and the corresponding gain.
32. (a) [8] p(s) = s2 +4s+3 = (s+2)2 −1 has roots r1 = −1 and r2 = −3. Basic solutions
are given by x1 = e−t and x2 = e−3t
. (The order is not determined, and in fact any other
pair of linearly independent solutions count as “basic.”) x˙1 = −e−t , x˙2 = −3e−3t .
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