Question 6.9: Filter Design Suppose that we need a filter that passes comp...

We need to pass high-frequency components and reject low-frequency components. Therefore, we need a highpass filter. The circuit diagram for a second-or-der highpass filter is shown in Figure 6.36(a), and the corresponding transfer-function magnitude plots are shown in Figure 6.36(b). Usually, we want the transfer function to be approximately constant in the passband. Thus, we choose Q_s\cong 1. We select f_0\cong 1 \text{ kHz}, so the components above 1 kHz are passed, while lower-frequency components are (at least partly) rejected. Solving Equation 6.30, for the capacitance and substituting values, we have

\begin{matrix} C&=&\frac{1}{(2 \pi)^2f_0^2L} =\frac{1}{(2 \pi)^2 \times 10^6 \times 50 \times 10^{-3}} \\ &=&0.507 \ \mu\text{F} \end{matrix}

Solving Equation 6.31 for the resistance and substituting values, we get

R=\frac{2 \pi f_0L}{Q_s}=\frac{2\pi \times 1000 \times 50 \times 10^{-3}}{1}=314.1 \ \Omega

The circuit and values are shown in Figure 6.39.