Calculate the cutoff frequency of a first-order low-pass filter in the circuit of Fig. 11.57 and the cutoff frequency of the high-pass filter circuit in Fig. 11.58.

From Fig. 11.57, the resistor $R = 2.2 \text{ k}\Omega$ and the capacitor $C = 0.05 \text{ }\mu\text{F}$.
The cutoff frequency $f_c$ for a first-order low-pass filter is given by:

$$f_c = \frac{1}{2\pi RC}$$

$$R = 2.2 \text{ k}\Omega = 2.2 \times 10^3 \Omega$$ $$C = 0.05 \text{ }\mu\text{F} = 0.05 \times 10^{-6} \text{ F}$$ $$f_c = \frac{1}{2\pi (2.2 \times 10^3 \Omega)(0.05 \times 10^{-6} \text{ F})}$$ $$f_c = \frac{1}{2\pi (0.11 \times 10^{-3})}$$ $$f_c = \frac{1}{0.00069115}$$ $$f_c \approx 1446.8 \text{ Hz}$$

The cutoff frequency is $\boxed{\text{1446.8 Hz}}$.

: Calculate the cutoff frequency of the high-pass filter circuit in Fig. 11.58.

From Fig. 11.58, the resistor $R = 20 \text{ k}\Omega$ and the capacitor $C = 0.02 \text{ }\mu\text{F}$.
The cutoff frequency $f_c$ for a first-order high-pass filter is given by:

$$f_c = \frac{1}{2\pi RC}$$

$$R = 20 \text{ k}\Omega = 20 \times 10^3 \Omega$$ $$C = 0.02 \text{ }\mu\text{F} = 0.02 \times 10^{-6} \text{ F}$$ $$f_c = \frac{1}{2\pi (20 \times 10^3 \Omega)(0.02 \times 10^{-6} \text{ F})}$$ $$f_c = \frac{1}{2\pi (0.4 \times 10^{-3})}$$ $$f_c = \frac{1}{0.00251327}$$ $$f_c \approx 397.9 \text{ Hz}$$

The cutoff frequency is $\boxed{\text{397.9 Hz}}$.

: Calculate the lower and upper cutoff frequencies of the bandpass filter circuit in Fig. 11.59.

The circuit in Fig. 11.59 is a cascaded bandpass filter, consisting of a high-pass filter stage followed by a low-pass filter stage. The lower cutoff frequency ($f_{cL}$) is determined by the high-pass stage, and the upper cutoff frequency ($f_{cU}$) is determined by the low-pass stage.

For the first stage:
Resistor $R_1 = 10 \text{ k}\Omega = 10 \times 10^3 \Omega$
Capacitor $C_1 = 0.05 \text{ }\mu\text{F} = 0.05 \times 10^{-6} \text{ F}$

$$f_{cL} = \frac{1}{2\pi R_1 C_1}$$ $$f_{cL} = \frac{1}{2\pi (10 \times 10^3 \Omega)(0.05 \times 10^{-6} \text{ F})}$$ $$f_{cL} = \frac{1}{2\pi (0.5 \times 10^{-3})}$$ $$f_{cL} = \frac{1}{0.00314159}$$ $$f_{cL} \approx 318.3 \text{ Hz}$$

For the second stage:
Resistor $R_2 = 20 \text{ k}\Omega = 20 \times 10^3 \Omega$
Capacitor $C_2 = 0.02 \text{ }\mu\text{F} = 0.02 \times 10^{-6} \text{ F}$

$$f_{cU} = \frac{1}{2\pi R_2 C_2}$$ $$f_{cU} = \frac{1}{2\pi (20 \times 10^3 \Omega)(0.02 \times 10^{-6} \text{ F})}$$ $$f_{cU} = \frac{1}{2\pi (0.4 \times 10^{-3})}$$ $$f_{cU} = \frac{1}{0.00251327}$$ $$f_{cU} \approx 397.9 \text{ Hz}$$

The lower cutoff frequency is $\boxed{\text{318.3 Hz}}$ and the upper cutoff frequency is $\boxed{\text{397.9 Hz}}$.