Answer:
The frequency of the signal is 2 GHz
Explanation:
Given;
period of the clock signal, T = 500 ps = 500 x 10⁻¹² s
the frequency of the signal is given by;
[tex]F= \frac{1}{T}\\\\F = \frac{1}{500*10^{-12}}\\\\F = 2*10^{9} \ Hz[/tex]
F = 2 GHz
Therefore, the frequency of the signal is 2 x 10⁹ Hz or 2 GHz
