Question

How do you determine Vo at a frequency one octave above the critical frequency if R=1 k ohm and C=0.01 micro Farrads in series Vin=10mV

Answer 1

How do you determine Vo at a frequency one octave above the critical frequency if R=1 k ohm and C=0.01 micro Farads in series Vin=10mV

Solution

It's first order low-pass RC filter.

First find the cutoff frequency,

$$f _ {c} = \frac {1}{2 * \pi * R * C} = \frac {1}{2 * \pi * 1 0 ^ {3} \mathrm {o h m} * 0 . 0 1 * 1 0 ^ {- 6} F} = 1 5 9 2 3 \mathrm {h e r t z}$$

One octave above, $f = 31846$ hertz.

The capacitive reactance of a capacitor in an AC circuit is given as

$$X _ {C} = \frac {1}{2 * \pi * f * C} = \frac {1}{2 * \pi * 3 1 8 4 6 \mathrm {h e r t z} * 0 . 0 1 * 1 0 ^ {- 6} F} = 5 0 0 \mathrm {o h m}.$$

The output voltage $V_{out}$ :

$$V _ {o u t} = V _ {i n} * \frac {X _ {C}}{\sqrt {(R) ^ {2} + (X _ {C}) ^ {2}}} = 1 0 * 1 0 ^ {- 3} V \frac {5 0 0}{\sqrt {1 0 0 0 ^ {2} + 5 0 0 ^ {2}}} = 0. 4 4 7 * 1 0 ^ {- 2} V = 4. 4 7 m V.$$

Answer: 4.47mV.