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Unformatted text preview: CHAPTER 9 PROBLEM SOLUTIONS SOLUTION TO PROBLEM 9.1. If we can compute expressions for K and q that are real, then these quantities exist by construction. Consider that A, B, K and q must satisfy the following relationship: K cos( ϖ t + θ ) = K cos( θ ) ( 29 cos( ϖ t ) + K sin( θ ) ( 29 sin( ϖ t ) ≡ A cos( ϖ t ) + B sin( ϖ t ) Therefore K cos( θ ) = A and  K sin( θ ) = B . Consequently, K cos( θ ) ( 29 2 + K sin( θ ) ( 29 2 = K 2 = A 2 + B 2 in which case K = A 2 + B 2 . Further, K sin( θ ) K cos( θ ) = tan( θ ) = B A in which case θ = tan 1 B A with due regard to quadrant. SOLUTION TO PROBLEM 9.2. For the inductor, W L ( t ) = 1 2 L ⋅ i L 2 ( t ) = 1 2 L V o C L sin 1 LC t 2 = CV 2 2 sin 2 1 LC t and for the capacitor, W C ( t ) = 1 2 C ⋅ v C 2 ( t ) = 1 2 C V o cos 1 LC t 2 = CV 2 2 cos 2 1 LC t . Hence, W C + W L = 1 2 C ⋅ v C 2 ( t ) + 1 2 L ⋅ i L 2 ( t ) = CV 2 2 sin 2 1 LC t + cos 2 1 LC t = CV 2 2 SOLUTION TO PROBLEM 9.3. Since x ( t ) = ( K 1 + K 2 t ) eα t , x ' ( t ) =  α ⋅ K 1 eα t + K 2 eα t α ⋅ t ⋅ K 2 eα t and x ' ' ( t ) = α 2 ⋅ K 1 eα t α ⋅ K 2 eα t α ⋅ K 2 eα t + α 2 ⋅ t ⋅ K 2 eα t Substituting into the differential equation, we have α 2 ⋅ K 1 eα t 2 α ⋅ K 2 eα t + α 2 ⋅ t ⋅ K 2 eα t + 2 α  α ⋅ K 1 eα t + K 2 eα t α ⋅ t ⋅ K 2 eα t [ ] + α 2 K 1 eα t + K 2 teα t [ ] = This means that the solution form satisfies the differential equation. SOLUTION TO PROBLEM 9.4. (a) Suppose x ( T ) = at some T. Then K 1 e s 1 T =  K 2 e s 2 T . Since e s i T whenever s i is real and T is finite, K 1 & K 2 must have opposite signs. (b) For this we solve for T and show there can only be one solution. Since K 1 e s 1 T =  K 2 e s 2 T and e s i T , K 1 K 2 = e s 2 T e s 1 T implies l n K 1 K 2 = l n e s 2 T e s 1 T = s 2 s 1 ( 29 T Hence the unique solution is given by T = l n K 1 K 2 s 2 s 1 ( 29 provided s 2 ≠ s 1 which is the case for distinct roots. SOLUTION TO PROBLEM 9.5. Suppose x ( T ) = at some T >0. This is true if and only if K 1 e s 1 T =  K 2 Te s 1 T (*) Since e s 1 T and T > 0, (*) is true if and only if K 1 =  K 2 T which is true if and only if K 1 & K 2 have opposite signs. SOLUTION 9.6 . (a) Denote one period of oscillation by T. Then by definition 9950 T = 2 π . Hence, T = 0.63148 ms. The time constant of decay is 1 ms. Therefore, NT = N × 0 . 6 3 1 4 8 = . Hence N = 1.5836 cycles....
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 Fall '06
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