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**Unformatted text preview: **Physics 3330 Prelab 5 solutions 1 (A &amp; B). The circuit of interest is shown in Figure 1 Figure 1: Active band pass filter. The design goals for the circuit are a resonant frequency of 16 kHz, a closed loop gain of- 1, and a Q of 10. Note that the resonant frequency is f , not which is the resonant angular frequency which is measured in rad/s rather than Hz. Also, the closed loop gain is G which is the gain of the circuit as a whole, namely V out /V in . From the theory section we therefore know: f = 1 2 LC = 16 kHz (1) Q = 1 r L C = Z r = 10 (2) G ( peak ) =- Q Z R =- 1 (3) (4) We will use the 10 mH inductor so L = 10 mH and so we can solve for the other values: C = 1 L (2 f ) 2 = 1 . 01 H(2 (16 10 3 Hz)) 2 = 9 . 9 nF = 0 . 01 F (5) Z = r L C = r . 01 H 1 10- 8 F = 1000 (6) r = Z Q = 1000 10 = 100 (7) R = QZ | G ( peak ) | = 10 1000 1 = 10 k (8) (9) The final part of part B also asks for the two 3 dB frequencies. While it is possible to solve the full gain equation for these frequencies, it is simpler to use Q = f/ f which gives f = f/Q = 16 kHz / 10 = 1 . 6 kHz. Since f = f +- f- , f = f f so f- = f- f/ 2 = 16 kHz- . 8 kHz = 15 . 2 kHz and f + = f + f/ 2 = 16 kHz- . 8 kHz = 16 . 8 kHz....

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