Question 6.7: 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-order highpass filter is shown in Figure 6.35(a), and the corresponding transfer-function magnitude plots are shown in Figure 6.35(b). Usually, we want the transfer function to be approximately constant in the passband. Thus, we choose Q_s ≅ 1. We select f₀ ≅ 1 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

f_0=\frac{1}{2\pi \sqrt{LC}}    (6.30)

C=\frac{1}{(2\pi )^2 f_0^2 L}=\frac{1}{(2\pi )^2 \times 10^6 \times 50 \times 10^{-3}}=0.507~\mu\mathrm{F}

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

Q_s=\frac{2\pi f_0 L}{R}         (6.31)

R=\frac{2\pi f_0 L}{Q_s}=\frac{2 \pi \times 1000 \times 50 \times 10^{-3}}{1}=341.1~\Omega

The circuit and values are shown in Figure 6.38.