At what power level must a signal be transmitted, on a 12 km cable, to ensure that a minimum Signal to Noise ratio of 10,000:1 is met?  The background noise is 700 nW and 10% of the power is lost per kilometer.

To solve this problem, we need to determine the power level at which a signal must be transmitted such that after traveling 12 km, the signal maintains a minimum Signal to Noise Ratio (SNR) of 10,000:1. Given the background noise of 700 nW and a 10% power loss per kilometer, we can follow these steps:

### Step 1: Understanding the signal power at the end of the cable
The power is reduced by 10% per kilometer, which means 90% of the power remains after each kilometer. The signal power at the end of 12 kilometers is given by:

\[
P_{\text{end}} = P_{\text{transmit}} \times (0.9)^{12}
\]

where:
- \(P_{\text{transmit}}\) is the initial transmitted power,
- 0.9 represents the 90% of power retained after each kilometer,
- 12 is the length of the cable in kilometers.

### Step 2: Signal to Noise Ratio (SNR)
The Signal to Noise Ratio is the ratio of the signal power at the end of the cable to the noise power. We're given:
\[
\text{SNR} = \frac{P_{\text{end}}}{P_{\text{noise}}}
\]
where \(P_{\text{noise}} = 700 \text{ nW}\) and the minimum SNR requirement is 10,000:1.

So, the equation for the minimum SNR is:
\[
\frac{P_{\text{end}}}{700 \times 10^{-9}} = 10,000
\]
\[
P_{\text{end}} = 700 \times 10^{-9} \times 10,000 = 7 \times 10^{-3} \text{ W} = 7 \text{ mW}
\]

Thus, the signal power at the end of the 12 km cable must be 7 mW to maintain the required SNR.

### Step 3: Transmitted power calculation
We can now calculate the transmitted power needed, given the power loss over 12 km:

\[
P_{\text{end}} = P_{\text{transmit}} \times (0.9)^{12}
\]
Substitute \(P_{\text{end}} = 7 \text{ mW}\):
\[
7 \times 10^{-3} = P_{\text{transmit}} \times (0.9)^{12}
\]

Now, calculate \((0.9)^{12}\):
\[
(0.9)^{12} \approx 0.2824
\]

So:
\[
7 \times 10^{-3} = P_{\text{transmit}} \times 0.2824
\]
\[
P_{\text{transmit}} = \frac{7 \times 10^{-3}}{0.2824} \approx 0.0248 \text{ W} = 24.8 \text{ mW}
\]

### Conclusion
The signal must be transmitted at a power level of approximately **24.8 mW** to ensure that after 12 km, the signal still has a minimum Signal to Noise Ratio of 10,000:1.

Would you like further clarification or more details on any part of the solution?

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Here are 5 related questions:
1. How does the required transmission power change if the cable length is increased?
2. What if the power loss per kilometer is 15% instead of 10%?
3. How is the SNR affected if the background noise increases?
4. What is the impact on transmission power if a lower SNR of 1,000:1 is acceptable?
5. How could using a cable with lower loss per kilometer improve the overall efficiency?

**Tip:** In signal transmission, minimizing power loss over long distances can be achieved by using higher-quality cables or signal amplifiers at intervals.

At what power level must a signal be transmitted, on a 12 km cable, to ensure that a minimum Signal to Noise ratio of 10,000:1 is met? The background noise is 700 nW and 10% of the power is lost per kilometer.

Find out how to calculate the power level needed to transmit a signal on a 12 km cable while ensuring a Signal to Noise Ratio (SNR) of 10,000:1. The problem involves background noise of 700 nW and a 10% power loss per kilometer, using exponential decay and SNR formulas. Ideal for engineering or physics students.

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