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assume a user is granted with a channel that has a bandwidth of 3 khz for data 3m transmission with a signal to noise ratio is 3162. what is the theoretical highestbit rate that user can achieve?

User Whoi
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Final answer:

Using the Shannon-Hartley theorem, the theoretical highest bit rate is approximately 34.9 Kbps for a channel that has a bandwidth of 3 kHz and a signal-to-noise ratio of 3162.

Step-by-step explanation:

The correct answer is that the theoretical highest bit rate this user can achieve can be calculated using the Shannon-Hartley theorem. The formula is C = B log2(1 + S/N), where C is the channel capacity in bits per second (bps), B is the bandwidth of the channel in hertz, S/N is the signal-to-noise ratio. In this case, B is 3000 Hz (3 kHz), and the signal-to-noise ratio (S/N) is 3162, which in terms of decibels is 10 log10(3162) is approximately 35 dB. Plugging these values into the Shannon-Hartley theorem yields:

C = 3000 log2(1 + 3162) ≈ 3000 × 11.63 ≈ 34900 bps or 34.9 Kbps. Hence, the highest bit rate that can be theoretically achieved by the user is 34.9 Kbps.

To calculate the observed frequency when the source is moving towards the observer, we use the formula:

Observed frequency = (Speed of sound + Speed of the source) / Speed of sound * Frequency of the source

Given: Speed of sound = 331 m/s, Speed of the source = 1.0 × 10² m/s, and Frequency of the source = 3.0 kHz = 3.0 × 10³ Hz

Plugging in the values, we get:

Observed frequency = (331 + 1.0 × 10²) / 331 * 3.0 × 10³

User Tyczj
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