Design an RC band-pass filter that passes a signal with frequency \(5.00 \mathrm{kHz},\) has a ratio \(V_{\text {out }} / V_{\text {in }}=0.500,\) and has an impedance of \(1.00 \mathrm{k} \Omega\) at very high frequencies. a) What components will you use? b) What is the phase of \(V_{\text {out }}\) relative to \(V_{\text {in }}\) at the frequency of \(5.00 \mathrm{kHz} ?\)

An RC band-pass filter consists of two filters: a low-pass filter followed by a high-pass filter. The transfer function depends on the values of the resistors and capacitors used in the design. The ratio \(V_{out}/V_{in}\) can be expressed as: $$\frac{V_{out}}{V_{in}} = \frac{1}{\sqrt{1 + (RC\cdot 2\pi \cdot f)^2}}$$ Where \(R\) is the resistance in ohms, \(C\) is the capacitance in farads, and \(f\) is the frequency in hertz.

We know that at 5.00 kHz, the ratio \(V_{out}/V_{in}\) is given to be 0.500. Plugging the given values into the formula above, we get: $$0.500 = \frac{1}{\sqrt{1 + (RC \cdot 2\pi \cdot 5000)^2}}$$ To find the values of the components, let's start by simplifying the equation: $$0.500^2 = 1 - (RC \cdot 2\pi \cdot 5000)^2$$ $$(RC \cdot 2\pi \cdot 5000)^2 = 1 - 0.500^2$$ $$RC = \frac{\sqrt{1 - 0.500^2}}{2\pi \cdot 5000}$$ The impedance at very high frequencies is given to be 1.00 kΩ, which means the resistance \(R\) is also 1.00 kΩ. $$1000 \cdot C = \frac{\sqrt{1 - 0.500^2}}{2\pi \cdot 5000}$$ Now, solve for \(C\): $$C = \frac{\frac{\sqrt{1 - 0.500^2}}{2\pi \cdot 5000}}{1000}$$ $$C \approx 1.05 \times 10^{-8} F$$ The required components for the filter are a 1.00 kΩ resistor and a 10.5 nF capacitor. Answer to part a): A 1.00 kΩ resistor and a 10.5 nF capacitor

The phase difference between the input and output voltages can be calculated using the formula: $$\phi = \arctan(RC \cdot 2\pi \cdot f)$$ Using the values calculated for \(RC\): $$\phi = \arctan(\frac{\sqrt{1 - 0.500^2}}{2\pi \cdot 5000} \cdot 2\pi \cdot 5000)$$ $$\phi = \arctan(\sqrt{1 - 0.500^2})$$ $$\phi \approx 60^\circ$$ The phase difference between the input and output voltages at 5.00 kHz is approximately 60°. Answer to part b): The phase difference is approximately 60° at 5.00 kHz.