Question

The range accuracy with a microsecond accurate clock in the GNSS receiver is about 300 m. If we improve the clock accuracy to $3.33 \times 10^{-x}$ s, the range accuracy becomes 1 cm. The value of $x$ is ............ (In integer).
Assume the speed of light to be $c = 3 \times 10^8$ m/s and that no other errors are being considered.
Hint: error-free range = speed of light $\times$ time of travel of the signal

Hint:

GNSS range error $\approx c\,\Delta t$ (one-way). Improving timing accuracy by $10^n$ improves range by the same factor.

Answer 1

For one-way ranging (GNSS code measurement), \[ \Delta R = c\,\Delta t. \] Given desired range accuracy $\,\Delta R = 1\;\text{cm} = 10^{-2}\,$m, \[ \Delta t = \frac{\Delta R}{c}=\frac{10^{-2}}{3\times 10^8} =3.33\ldots\times 10^{-11}\ \text{s}. \] Thus $x=11$. (Sanity check: for $\Delta t=1\,\mu$s, $\Delta R=c\Delta t=3\times 10^8\times 10^{-6}=300$ m, as stated.)