Ae104a10_solutions_midterm

Ae104a10_solutions_midterm - Ae/APh 104a Mid-term...

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Unformatted text preview: Ae/APh 104a Mid-term Solutions, 2010 1/2 [30pts] (a)[3 pts] The motion of the diaphragm is opposed by the inertia of the diaphragm, acoustic damping and an acoustic stiffness, such that, per unit area, P = ( m x + x + kx ) , with Q = x. Then for a harmonic signal, using phasors P = jm + + k j x = jm + + k j Q. Converting to the electrical domain, using I a Q and V ac P a , a V ac = jm + + k j I a so the likely form of Z eD is Z eD = 1 2 a jm + + k j . Any two of the components got you full marks. (b)[5 pts] KVL on loop 1 to find an expression for Q :- V ac + a QZ eD + a Q out Z eO = 0 , so a QZ eD = V ac- a Q out Z eO . 1 KVL on loop 2:- V ac + a QZ eD + a ( Q- Q out ) Z eC =- V ac- a Q out Z eC + a Q ( Z eD + Z eC ) = 0 . Substituting for Q :- V ac- a Q out Z eC + V ac- a Q out Z eO + V ac Z eC Z eD- Q out Z eO Z eC Z eD = 0 , V ac Z eC Z eD- Q out Z eC + Z eO + Z eO Z eC Z eD = 0 . Thus, rearranging Q out V ac = Z eC a Z eD Z eO + Z eO Z eC Z eD + Z eC and Q out V ac = 1 a Z eD Z eO Z eC + Z eO Z eD + 1 . (1) (c)[5 pts] Substitute Z eD = 1 2 a R D + jM D + 1 jC D , Z eC = 1 2 a 1 jC C , and Z eO = 1 2 a ( R O + jM O ) into equation 1 to get Q out V ac = a jC D jC D ( R D + jM D )+1 jC C ( R O + jM O ) + jC D ( R O + jM O ) jC D ( R D + jM D )+1 + 1 or Q out V ac = a jC D jC C ( R O + jM O )[ jC...
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Ae104a10_solutions_midterm - Ae/APh 104a Mid-term...

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