In the low-pass filter shown in the figure , for a cut-off frequency of 5 kHz, the value of $R_{f}$ is _____ $k Ω$ .

3.18

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Solution

The correct option is A 3.18
Apply KCL at node,

$\frac{0 - V_{o u t}}{R_{f} | | \frac{1}{(10 s) n f}} + \frac{0 - V_{i n}}{1000} = 0$

Let , $Z_{2} = R_{f} | | \frac{1}{(10 s) n F}$

and $Z_{1} = 1000 Ω$

or $\frac{V_{o u t}}{V_{i n}} = - \frac{Z_{2}}{Z_{1}} = - \frac{R_{F} | | \frac{1}{10 n F_{s}}}{1000} = - \frac{R_{F}}{1000 (i (\frac{ω}{1 / R_{F} \times 10 n F}) + 1)}$

Cut-off frequency,

$ω = \frac{1}{R_{F} \times 10 n F}$

$R_{F} = \frac{1}{10 n F ω} = \frac{1}{2 π f 10 n F} = \frac{1}{2 π \times \times 5000 \times 10 \times 10^{- 9}} = 3.183 k Ω$

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