SM 212 Test 3, 11-2-98

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1. 1001 USNA faculty, officers, and staff are battling 1000 of the best midshipmen in a paint gun fight, the battle being modeled by:

x' = -3y -1,     x(0) = 1001

y' = -3x,          y(0) = 1000

Solve this system using Laplace transforms. Who wins? Find out when they win to 2 decimal places and estimate the losses.

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Figure 1:

2. In the network diagram given (Figure 1 above), find a system of 2 differential equations in i3 and q2. Put it in standard form but do not solve the system.

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3. Compute the

inverse laplace of [ 1/(s2 + 8s + 20) * e-pi s ](t)

and the

inverse laplace of [s-1998 * e-s](t).

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4. Let

f(t) = 0,     t<0,

-1,     0<t<1,

1,    1<t<2,

0,        t>2.

a) Express f(t) as a linear combination of translations of the unit step function.

b) Find the Laplace transform of f(t).

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5. Solve

x' +x = f(t),           x(0) = 0

where f(t) is as above.

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BONUS (5 pts) Let

f(t) = 1,     0<t<1,

2,     1<t<3,

extended periodically to the real line with period 3. Find the Laplace transform of f(t).

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SOLUTIONS

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1.

---Take Laplace of given equations

s X(s) - 1001 = -3 Y(s) - 1/s

s Y(s) - 1000 = -3 X(s)

---solve for X(s) in 2nd equation, plug into 1st equation

X(s) = (-1/3) (s Y(s) - 1000)

(-1/3) s (s Y(s) - 1000) - 1001 = -3 Y(s) - 1/s

---multiply through by -3, simplify

s^2 Y(s) - 1000 s + 3003 = 9 Y(s) + 3/s

---solve for Y(s)

Y(s) = 3/(s(s2 - 9)) + 1000s/(s2 - 9) - 3003/(s2 - 9)

---take inverse Laplace to find y(t)

y(t) = (1/3) (-e3t + 3002 e-3t - 1)

---find y'(t), plug into 2nd equation and reduce to find x(t)

y'(t) = -e-3t (e6t + 3002)

x(t) = (1/3) (e3t + 3002 e-3t)

---solve for when x(t) or y(t) = 0

x(t) will never be 0.

y(t)=0 at time 1.33.

---evaluate x(1.33) to find faculty survivors

x(1.33) = 36.5

so losses = 1001 - 36 = 965

The faculty wins at t=1.33, loses 965.

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2.

---Get formulas for left loop and right loop (use Lq'' + Rq' + (1/c)q = E(t))

Left: sin(t) = -i1 + 2q2

Right: 0 = 3i3 + 4i3' - 2q2

---Pick a node and set incoming current = outgoing current (Kirchhoff's)

For bottom center node, i1 + q'2 + i3 = 0

and i1 = -q'2 - i3

---Plug into first 2 equations so all in terms of i3 and q2

Left: sin(t) = q'2 + i3 + 2q2

Right equation not modified.

sin(t) = q'2 + i3 + 2q2

0 = 3i3 + 4i3' - 2q2

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3.

---Complete the square in the denominator

1/((s+4)2 + 22) e-pi s

---Use L16 in blue book

a = (pi), F(s) = 1/((s+4)2 + 22), so f(t) = (1/2) (e-4t sin(2t))

and f(t-a)U(t-a) = ANSWER =(1/2) (e-4 (t - pi) sin(2 (t - x)) U(t - (pi)))

---Use L16 in blue book

a = 1, F(s) =s-1998, so f(t) = (1/1997!) t1997

and f(t-a)U(t-a) = ANSWER = (1/1997!) (t - 1)1997 U(t - 1)

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4.

---Use formula: (relative jump) u(t - (x value at jump) )

f(t) = -u(t) + 2u(t-1) - u(t-2)

---Use L3 in blue book for each of the three terms

using a=0, a=1, a=2

Laplace = -(1/s) + 2(e-s/s) - e-2 s/s

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5.

---Take Laplace of left side (right side was done in question #4)

x(X(s)) - x(0) + X(s) =  -(1/s) + 2(e-s/s) - e-2 s/s

---x(0)=0, solve for X(s)

X(s) = -1/(s (s + 1)) + 2e-s/(s (s + 1) - e-2 s/(s (s + 1))

---Use L16 (with L20 for each F(s)) from blue books to take inverse Laplace

x(t) = e-t - 1 - 2 (e-(t - 1) - 1) u(t - 1) + (e-(t - 2) - 1) u(t - 2)

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BONUS

---Use L14 from blue book, break into 2 intervals (0-1 and 1-3) and use P=3