Question

Consider a message signal with bandwidth B=4 KHz and a normalized power S = = 0.25. It is required to transmit this signal via a channel that attenuates the transmitted signal by 20 dB. The channel noise is AWGN with spectral density N₀/2 = 10⁻⁹ W/Hz. It is desirable to have an SNR of at least 50 dB at the receiver output. a) Find the minimum required transmit power Pₜ for a PM system with ∆∅ₘₐₓ = 5.

Answer 1

The student's question involves calculating the minimum required transmit power for a PM system, given the SNR, channel attenuation, bandwidth, and noise specifications. By converting the SNR and attenuation from dB to power ratios and applying the SNR formula, the minimum transmit power is found to ensure the desired SNR at the receiver output.

Step-by-step explanation:

To calculate the minimum required transmit power \( P_t \) for a Phase Modulation (PM) system with
`\( \Delta \phi_{\text{max}} = 5 \)`, we need to consider the signal-to-noise ratio (SNR) requirements.

The SNR at the receiver output
`(\( \text{SNR}_\text{out} \))`is given by the formula:

`\[ \text{SNR}_\text{out} = (P_t G)/(N_0 B) \]`

where:

- \( P_t \) is the transmit power,

- \( G \) is the channel gain,

- \( N_0 \) is the spectral density of the noise, and

- \( B \) is the bandwidth of the message signal.

The channel gain \( G \) can be determined from the attenuation in dB:

`\[ G = 10^{-\frac{\text{Attenuation (dB)}}{10}} \]`

In this case, the attenuation is given as 20 dB, so \( G = 10^{-\frac{20}{10}} = 0.1 \).

Given that the SNR at the receiver output is desired to be at least 50 dB, we can set up the inequality:

`\[ \text{SNR}_\text{out} = (P_t \cdot 0.1)/(10^(-9) \cdot 4 * 10^3) \geq 10^(50/10) \]`

Solving for \( P_t \), we get:

`\[ P_t \geq 10^(50/10) \cdot 10^(-9) \cdot 4 * 10^3 * 0.1 \]`

Simplifying this expression gives the minimum required transmit power \( P_t \).