DeCarlo Ch9 Solutions

Linear Circuit Analysis: The Time Domain and Phasor Approach

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CHAPTER 9 PROBLEM SOLUTIONS S OLUTION TO P ROBLEM 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 ) = 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. S OLUTION TO P ROBLEM 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 0 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 0 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 0 2 2 sin 2 1 LC t  + cos 2 1 LC t = CV 0 2 2 S OLUTION TO P ROBLEM 9.3. Since x ( t ) = ( K 1 + K 2 t ) e t , x '( t ) =-α⋅ K 1 e t + K 2 e t -α⋅ t K 2 e t and x ''( t ) 2 K 1 e t -α⋅ K 2 e t -α⋅ K 2 e t 2 t K 2 e t Substituting into the differential equation, we have α 2 K 1 e t - 2 α⋅ K 2 e t 2 t K 2 e t + 2 α-α⋅ K 1 e t + K 2 e t -α⋅ t K 2 e t [ ] 2 K 1 e t + K 2 te t [ ] = 0
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This means that the solution form satisfies the differential equation. S OLUTION TO P ROBLEM 9.4. (a) Suppose x ( T ) = 0 at some T. Then K 1 e s 1 T =- K 2 e s 2 T . Since e s i T 0 whenever s i is real and T is finite, K 1 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 0 , 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. S OLUTION TO P ROBLEM 9.5. Suppose x ( T ) = 0 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 0 and T > 0, (*) is true if and only if K 1 = - K 2 T which is true if and only if K 1 2 have opposite signs. S OLUTION 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 × 0.63148= . Hence N = 1.5836 cycles.
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DeCarlo Ch9 Solutions - CHAPTER 9 PROBLEM SOLUTIONS...

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