- Considering only the low pass filter stage of the audio equaliser, excluding the amplifier and speaker (as shown in Fi

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- Considering only the low pass filter stage of the audio equaliser, excluding the amplifier and speaker (as shown in Fi

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Considering Only The Low Pass Filter Stage Of The Audio Equaliser Excluding The Amplifier And Speaker As Shown In Fi 1
Considering Only The Low Pass Filter Stage Of The Audio Equaliser Excluding The Amplifier And Speaker As Shown In Fi 1 (154.05 KiB) Viewed 15 times
- Considering only the low pass filter stage of the audio equaliser, excluding the amplifier and speaker (as shown in Figure 6.3), calculate voltage Va as a phasor when the potentiometer (or variable resistor) value is 8 k12 (i.e., Rp = 8 kN) and the frequency of the input signal is 100 Hz, 1 kHz and 10 kHz. What happens to the magnitude of Va as frequency increases? Does this match the behaviour of a low 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 Va as a function of w or f and then substitute the values. Answer: Va(100Hz) = 0.84 2 – 4.54° V; Va(1 kHz) = 0.66 2 - 38.44° V; Va(10 kHz) = 0.105 2 - 82.82º V = w 1.5 k12 Vin AC 100 nF Rp + Va 2 Vpp Figure 6.3. Low pass filter stage of audio equaliser. Note that Rp is a 10 ks potentiometer (or variable resistor) used to control the volume.
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