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a first order low pass filter is used to attenuate noise with frequencies higher than 26khz. what is the largest cutoff frequency (in khz) to achieve -20db attenuation?

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Final answer:

A first-order low-pass filter is used to attenuate noise with frequencies higher than 26 kHz. The largest cutoff frequency to achieve -20 dB attenuation is 2.6 kHz.

Step-by-step explanation:

A first-order low-pass filter is used to attenuate noise with frequencies higher than 26 kHz, and the goal is to achieve a -20 dB attenuation. The cutoff frequency is the frequency at which the filter starts attenuating the signal significantly. In this case, we need to find the largest cutoff frequency that will give us -20 dB attenuation.

The attenuation in a first-order low-pass filter can be calculated using the formula:

Attenuation (in dB) = 20 log10(f_cutoff / f_signal)

Where f_cutoff is the cutoff frequency and f_signal is the frequency of the noise we want to attenuate. Rearranging the formula, we get:

f_cutoff = f_signal * 10^(-Attenuation/20)

Substituting the given values, we have:

f_signal = 26 kHz, Attenuation = 20 dB

Calculating f_cutoff:

f_cutoff = (26 kHz) * 10^(-20/20)

= 26 kHz * 0.1

= 2.6 kHz

Therefore, the largest cutoff frequency to achieve -20 dB attenuation is 2.6 kHz.

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