Answer Happy

Considering only the high pass filter stage of the audio equaliser, excluding the amplifier and speaker (as shown in Figure 6.4), calculate voltage Vc as a phasor when the potentiometer (or variable resistor) value is 8 kN (i.e., Rp = 8 ks) and the frequency of the input signal is 1 kHz, 2.5 kHz and 10 kHz. What happens to the magnitude of Vc as frequency increases? Does this match the behaviour of a high pass filter (as per the explanation provided in the previous page)? NOTE: To avoid repeating the same calculation 3 times, you may opt to calculate Vc as a function of w or f and then substitute the values. Answer: Vc(1 kHz) = 0.366 2 68.51° V; V.(2.5 kHz) = 0.702 2 45.45° V; V(10 kHz) = 0.97 2 14.25° V - 100 nF Vin AC 68012 Rp + Vc 2 Vpp Figure 6.4. High pass filter stage of audio equaliser. Note that Rp is a 10 kN potentiometer (or variable resistor) used to control the volume.