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 + 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. 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 ) = α 2 K 1 e t - α ⋅ K 2 e t - α ⋅ K 2 e t + α 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 + α 2 t K 2 e t + 2 α - α ⋅ K 1 e t + K 2 e t - α ⋅ t K 2 e t [ ] + α 2 K 1 e t + K 2 te t [ ] = 0

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