hw7_sol_04-05-08-32

# hw7_sol_04-05-08-32 - Φ 2 Φ 1 Φ 2 Φ 1 Φ 1 Q s(n-1)T...

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Problem Set 7 Solutions a. First order active low pass filter - + C I R 2 R 1 V i V o The integrator pole is set by C i R 2 = = = M R pF kHz R C R f i dB 59 . 1 1 2 100 1 2 1 2 2 2 3 π The passband gain is set by R 2 /R 1. R 1 =R 2 = 1.59 M

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b. The SC-filter can be analyzed in a similar manner to the RC filter. The SC circuit creates an equivalent resistance as: S s eq C f R 1 = The integrator pole is at: i s s i eq dB C C f C R f π 2 2 1 3 = = R eq of the SC filter is equal to R 2 of the RC implementation. Solve for C s : S eq s f R C 1 = fF 8 . 62 = Do same for resistor parallel to Ci …
nT s (n-½)T s (n-1)T s (n-3/2)T s c. Derive H(z) Φ 1
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Unformatted text preview: Φ 2 Φ 1 Φ 2 Φ 1 Φ 1 Q s [(n-1)T s ]=C s V i [(n-1)T s ] Q I [(n-1)T s ]=Q I [(n-3/2)T s ] Φ 2 Q s [(n-1/2)T s ]=0 Q I [n-1]T s ]=Q I [(n-3/2)T s ] + Q s [(n-1)T s ] Φ 1 Q s [nT s ]=C s V i [nT s ] Q I [nT s ]=Q I [(n-1)T s ] + Q s [(n-1)T s ] V o = -Q I /C I V i = Q s /C s C I V o (nT s )=C I V o [(n-1)T s ] - C s V i [(n-1)T s ] C I V o (z)=C I V o (z)z-1- C s V i (z)z-1 1 1 1 ) ( ) ( − − − × − = = z z C C z H z V V I s i o...
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hw7_sol_04-05-08-32 - Φ 2 Φ 1 Φ 2 Φ 1 Φ 1 Q s(n-1)T...

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