The high-pass filter shown in Figure 1) Figure SnF 50 Kn ✔Correct Part F -0.5 Con(T) V has a steady-state. The main answer to this question is to determine the steady-state of the filter. We will provide an explanation on how to obtain the steady-state of the filter.Steady-state refers to the state of a circuit when it has been on for a long period of time such that the voltages and currents have stopped changing with time. In other words, it is when the output voltage of the filter is no longer changing with time.To determine the steady-state of the high-pass filter shown in Figure 1) Figure SnF 50 Kn ✔Correct Part F -0.5 Con(T) V, we need to compute the impedance of the capacitor and the resistor at a steady-state. The impedance of a capacitor is given as:Zc = 1/jωCWhere:Zc = Capacitor impedanceω = Angular frequency (ω = 2πf)C = CapacitanceThe impedance of a resistor is given as:ZR = RWhere:ZR = Resistor impedanceR = ResistanceFrom the figure provided, we can see that the resistance R = 50 kΩ and the capacitance C = 0.5 µF. We can now calculate the impedance of the resistor and capacitor at steady-state.Zc = 1/jωC = 1/j(2πf)C = -j/(2πfC) = -j/(2π×1000×0.5×10^-6) = -j318.31 kΩZR = R = 50 kΩThe total impedance of the high-pass filter is given as:Ztotal = Zc + ZRZtotal = -j318.31 kΩ + 50 kΩ = -j268.31 kΩThe steady-state voltage gain of the high-pass filter is given as:A = Vo/Vin = -Zc/ZtotalA = -(-j318.31 kΩ)/(-j268.31 kΩ)A = 1.186The steady-state voltage gain of the high-pass filter is 1.186.