Ae104a10_solutions_midterm

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

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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 + λ ˙ x + kx ) , with Q = ˙ x. Then for a harmonic signal, using phasors P = jωm + λ + k jω ˙ x = jωm + λ + k jω Q. Converting to the electrical domain, using I ≡ φ a Q and V ac ≡ P φ a , φ a V ac = jωm + λ + k jω I φ a so the likely form of Z eD is Z eD = 1 φ 2 a jωm + λ + k jω . Any two of the components got you full marks. (b)[5 pts] KVL on loop 1 to find an expression for Q :- V ac + φ a QZ eD + φ a Q out Z eO = 0 , so φ a QZ eD = V ac- φ a Q out Z eO . 1 KVL on loop 2:- V ac + φ a QZ eD + φ a ( Q- Q out ) Z eC =- V ac- φ a Q out Z eC + φ a Q ( Z eD + Z eC ) = 0 . Substituting for Q :- V ac- φ a Q out Z eC + V ac- φ a Q out Z eO + V ac Z eC Z eD- φQ out Z eO Z eC Z eD = 0 , V ac Z eC Z eD- φQ out Z eC + Z eO + Z eO Z eC Z eD = 0 . Thus, rearranging Q out V ac = Z eC φ a Z eD Z eO + Z eO Z eC Z eD + Z eC and Q out V ac = 1 φ a Z eD Z eO Z eC + Z eO Z eD + 1 . (1) (c)[5 pts] Substitute Z eD = 1 φ 2 a R D + jωM D + 1 jωC D , Z eC = 1 φ 2 a 1 jωC C , and Z eO = 1 φ 2 a ( R O + jωM O ) into equation 1 to get Q out V ac = φ a jωC D jωC D ( R D + jωM D )+1 jωC C ( R O + jωM O ) + jωC D ( R O + jωM O ) jωC D ( R D + jωM D )+1 + 1 or Q out V ac = φ a jωC D jωC C ( R O + jωM O )[ jωC...
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Ae104a10_solutions_midterm - Ae/APh 104a Mid-term Solutions...

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