Pset4Solution

# Pset4Solution - − − = π The two sinusoids have...

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1 . (6.1) Note the units associated with the equation (6.3c). 8 2 2 2 8 2 10 54 . 2 05 . 0 54 . 2 ) 2 / 25 . 0 ( 100 26 . 1 10 26 . 1 × × × = × = π a a h a n L 4 10 142 . 3 × = (H) i o e L R L e 2 2 ) ( ω + = Sensitivity is given by a o a o h L L e h e = where 2 / 3 2 2 2 ] ) ( [ L R R e L e i o + = ; a a a a h L h a n h L = × = 8 2 2 10 ) ( 26 . 1 . Substitute into the above rad/s 6280 1000 2 , 1000 , V 10 = × = = = R e i , 974 . 1 = L = × + = cm V h e a o ) 54 . 2 05 . 0 974 . 1 ( ] 974 . 1 1000 [ 1000 10 5 . 1 2 2 2 - 54 . 2 395 . 0 cm V =- 54 . 2 395 . 0 cm V in cm 54 . 2 =-0.395 in V =-0.395 mil mV . 2 . (6.6) C=2.25 pF when d=0.01 in C=1.5 pF when d=0.015 in 2 ) ( 1 1 | | | | RC e e i o + = For C=2.25pF, 577 . 0 ) 10 100 2 10 25 . 2 10 ( 1 1 | | | | 2 3 12 6 = × × × × × + = i o e e For C=1.5pF, 728 . 0 ) 10 100 2 10 5 . 1 10 ( 1 1 | | | | 2 3 12 6 = × × × × × + = i o e e 3 . (7.29 5th ed. 7.34 6th ed) (a). This is a high pass filter kHz RC f c 82 . 4 ) 10 330 )( 10 100 ( 2 1 2 1 12 3 = × × = = (b) 2 ) / ( 1 / | | | | c c in out f f f f e e + = , ) / ( tan 2 1 c f f = φ )} 449 . 0 000 , 10 2 sin( 35 . 1 ) 37 . 1 1000 2 cos( 508 . 0 ) 53 . 1 200 2 sin( 207 . 0 { + + + + + = t t t e out mv.

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4. (a) (b) ] ) ( 2 cos[ 2 1 ) ( 2 cos[ 2 1 ) 2 sin( ) 2 sin( 2 1 2 1 2 1 t f f t f f t f t f +
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Unformatted text preview: − − = π The two sinusoids have frequencies 1000Hz and 101,000Hz with the same amplitude of 0.5. (c) The output signal will have frequencies 1000Hz and 101,000Hz. (d) Since 5 12 10 10 * 1000 * 000 , 10 − − = = RC , the amplitude of the 1000Hz signal is 499 . ) 1000 * 2 * ( 1 5 . 2 1 = + = RC A , the amplitude of the 101,000Hz signal is 078 . ) 101000 * 2 * ( 1 5 . 2 2 = + = RC A . 0.0002 0.0004 0.0006 0.0008 0.001-1-0.5 0.5 1...
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Pset4Solution - − − = π The two sinusoids have...

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