Chap6StudentSolutions

5 1 15 2 1 10 0 1 10 2 10 3 10 10 b r

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Unformatted text preview: Roberts - 8/16/04 0 |H(jω)|dB -10 -20 -30 -40 -50 -1 10 0 1 10 2 10 3 10 ω 10 Phase of H(jω) 0 -0.5 -1 -1.5 -2 -1 10 0 1 10 2 10 3 10 ω 10 (b) R = 10 Ω C = 1 µF + + vi (t) L = 1 mH - vL(t) - Vo ( f ) , of these active filters and identify them as Vi ( f ) lowpass, highpass, bandpass or bandstop. 13. Find the transfer functions, H( f ) = (a) R1 C1 C2 K + vi (t) - vx(t) + vo(t) R2 - The triangle with the “ K ” inside is an ideal voltage amplifier of gain K (not an operational amplifier). It is, in circuit theory parlance, a “voltage-dependent voltage source”. “Ideal” means its input impedance is infinite so no input current flows and its output impedance is zero so the output voltage is independent of the output current (and therefore any load connected to the output). Solutions 6-8 M. J. Roberts - 8/16/04 Writing Kirchhoff’s current law at the v X ( t ) node and then writing the relationship between Vx ( f ) and Vo ( f ) using voltage division and the voltage amplifier gain, K, 1 Vx ( f ) j 2πfC1 + + G1 − Vi ( f ) j 2πfC1 − Vo ( f )G1 = 0 1 + R2 j 2πfC2 Vx ( f ) ( G1 is R2 1 + R2 j 2πfC2 K = Vo ( f ) 1 . That is, it is the conductance of the resistor, R1 .) R1 j 2πfC2 Vx ( f ) j 2πfC1 + + G1 − Vi ( f ) j 2πfC1 − Vo ( f )G1 = 0 1 + j 2πfR2C2 Vx ( f ) j 2πfR2C2 K = Vo ( f ) 1 + j 2πfR2C2 Writing the two equations as one matrix equation, j 2πfC2 + G1 j 2πfC1 + 1 + j 2πfR2C2 j 2πfKR2C2 Vx ( f ) j 2πfC1 V ( f ) = 0 Vi ( f ) −(1 + j 2πfR2C2 ) o −G1 Solving by Cramer’s rule with the excitation voltage as a forcing function, j 2πfC2 ∆ = − j 2πfC1 + + G1 (1 + j 2πfR2C2 ) + j 2πfKG1R2C2 1 + j 2πfR2C2 ∆ = (2πf ) R2C1C2 − j 2πf [C1 + C2 + G1R2C2 (1 − K )] − G1 2 j 2πfC2 1 j 2πfC1 + + G1 Vo ( f ) = 1 + j 2πfR2C2 ∆ j 2πfKR2C2 j 2πfC1 Vi ( f ) 0 V (f) (2πf ) KR2C1C2 H( f ) = o = 2 Vi ( f ) (2πf ) R2C1C2 − j 2πf [C1 + C2 + G1R2C2 (1 − K )] − G1 2 H( f ) = (2πf ) 2 KR1R2C1C2 (2πf ) 2 R1R2C1C2 − j 2πf (R1(C1 + C2 ) + R2C2 (1 − K )) − 1 Highpass So...
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