GNSS Range Accuracy Improvement Calculation

The relationship between range accuracy ($\Delta R$) and time accuracy ($\Delta t$) in a GNSS receiver is directly proportional to the speed of light ($c$). The core formula is:

$ \Delta R = c \times \Delta t $

This means that improving the clock's time accuracy directly improves the range accuracy.

Initial Range and Time Accuracy

We are given:

• Initial Range Accuracy ($\Delta R_1$): 300 m
• Speed of Light ($c$): $3 \times 10^8$ m/s

Using the formula, we can find the initial time error ($\Delta t_1$) associated with this range accuracy:

$ \Delta t_1 = \frac{\Delta R_1}{c} = \frac{300 \text{ m}}{3 \times 10^8 \text{ m/s}} = 1 \times 10^{-6} \text{ s} $

This initial time error is equivalent to 1 microsecond, matching the problem description.

Improved Range and Time Accuracy

For the improved scenario, we have:

• Improved Range Accuracy ($\Delta R_2$): 1 cm = 0.01 m
• Target Clock Accuracy ($\Delta t_{clock2}$): $3.33 \times 10^{-x}$ s
• Speed of Light ($c$): $3 \times 10^8$ m/s

Calculate the required time error ($\Delta t_2$) needed to achieve the improved range accuracy:

$ \Delta t_2 = \frac{\Delta R_2}{c} = \frac{0.01 \text{ m}}{3 \times 10^8 \text{ m/s}} $

$ \Delta t_2 = \frac{1 \times 10^{-2}}{3 \times 10^8} \text{ s} = \frac{1}{3} \times 10^{-10} \text{ s} \approx 0.333 \times 10^{-10} \text{ s} $

Converting this to the required format:

$ \Delta t_2 = 3.33 \times 10^{-11} \text{ s} $

Solving for Exponent x

The calculated time error ($\Delta t_2$) must match the given improved clock accuracy formula ($3.33 \times 10^{-x}$ s). By setting them equal:

$ 3.33 \times 10^{-x} \text{ s} = 3.33 \times 10^{-11} \text{ s} $

Comparing the powers of 10, we find the value of $x$:

$ x = 11 $

Therefore, the value of $x$ is 11.