Using Shannon's Theorem, Calculate The Maximum Bit Rate

Question

Using Shannons theorem calculate the maximum bit rate that Using Shannons theorem,calculate the maximum bit rate that Using Shannon's theorem, calculate the maximum data transmission rate (bit rate) on a fiber optic link with a signal-to-noise ratio (SNR) of 40 dB. The central wavelength of the light used is 850 nm, with a bandwidth of 10 nm on either side.

Dr. Jack HW Helper · Accepted Answer

Shannon's theorem provides a fundamental limit on the maximum data rate (channel capacity) that can be transmitted over a communication channel without error, given the bandwidth and the signal-to-noise ratio (SNR). The theorem states that: $$ C = B \times \log_{2}(1 + \text{SNR}) $$ where: - $ C $ is the channel capacity in bits per second (bps), - $ B $ is the bandwidth in Hertz (Hz), - $ \text{SNR} $ is the linear signal-to-noise ratio (not in decibels). In this context, we are asked to determine the maximum bit rate achievable over a fiber optic link using Shannon's theorem, given specific parameters. Step 1: Convert the SNR from decibels to a linear ratio Given SNR in dB: 40 dB $$ \text{SNR}_{\text{linear}} = 10^{\frac{\text{SNR}_{dB}}{10}} = 10^{\frac{40}{10}} = 10^{4} = 10,000 $$ Step 2: Determine the bandwidth $ B $ The bandwidth of the optical signal is derived from the central wavelength of 850 nm and a bandwidth of 10 nm on either side, making a total spectral bandwidth of $ 20 $ nm. The spectral bandwidth in frequency terms can be computed by: $$ B = \frac{\Delta \lambda}{\lambda^{2}} \times c $$ where: - $ \Delta \lambda $ is the spectral bandwidth in meters, - $ \lambda $ is the central wavelength in meters, - $ c $ is the speed of light ($ \approx 3 \times 10^{8} $ m/s). Converting wavelengths to meters: $$ \lambda = 850\, \text{nm} = 850 \times 10^{-9}\, \text{m} $$ $$ \Delta \lambda = 20\, \text{nm} = 20 \times 10^{-9}\, \text{m} $$ Calculating the bandwidth: $$ B = \frac{20 \times 10^{-9}}{(850 \times 10^{-9})^{2}} \times 3 \times 10^{8} $$ Calculating denominator: $$ (850 \times 10^{-9})^{2} = 850^{2} \times 10^{-18} = 722,500 \times 10^{-18} = 7.225 \times 10^{-13} $$ Now, substitute into $ B $: $$ B = \frac{20 \times 10^{-9}}{7.225 \times 10^{-13}} \times 3 \times 10^{8} $$ Simplify numerator: $$ 20 \times 10^{-9} = 2 \times 10^{-8} $$ Division: \[ \frac{2 \times 10^{-8}}{7.225 \times 10^{-13}} \approx \frac{2}{7.225} \times

Using Shannons theorem,calculate the maximum bit rate that

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