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**Unformatted text preview: **Physics 3330 Prelab 3 solutions 1. The cutoff frequency is given in the lab as f c = 1 / (2 πRC ) which for this circuit is equal to f c = 1 / (2 πRC ) = 1 / (2 π 10 × 10 3 Ω1000 × 10- 12 F) = 15915 Hz = 15 . 9 kHz. The Bode plot below shows the transfer function T = V out /V in . As this is a low-pass filter, the low frequency signals are transfered and the high frequency signals are blocked. The vertical dotted line is at the cutoff frequency of 15.9 kHz. Note that this lines up with the horizontal line which is at V out /V in = 0 . 707. The cutoff frequency is at 3 dB which is the half-power point which corresponds to the point where the voltage is 0.707 of the input voltage. Low pass filter 10-3 10-2 10-1 1 10 10 2 10 3 10 4 10 5 10 6 10 7 Frequency (Hz) V out / V in To derive the actual formula for T we can use the complex impedences for the resistor and the capacitor: Z R = R and Z C =- j/ ( ωC ). Then T = Z C Z R + Z C =- j/ωC R- j/ωC =- j ωRC- j . To find the absolute value we multiply by the complex conjugate and take the square root: | T | 2 = 1 ω 2 R 2 C 2 + 1 | T | = 1 √ ω 2 R 2 C 2 + 1 ....

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