Question 18.2: Determine the low-frequency cut-off produced by the coupling......
Determine the low-frequency cut-off produced by the coupling capacitor in the circuit in Example 18.1.
Step-by-Step
The low-frequency cut-off is determined by the value of C and the input resistance. As the input resistance is approximately equal to RG it follows (from Equation 8.11) that
f_{c} = \frac{\omega_{c}}{2 \pi} = \frac{1}{2 \pi CR} Hz (8.11)
f_{c} = \frac{1}{2 \pi CR_{G}} = \frac{1}{2 × \pi × 10^{-6} × 10^{6}} = 0.16 HzRelated Answered Questions
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