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<math> R = \frac{cT_s}{4B} \left( f_{bd} + f_{bu} \right) = \frac{(3\times 10^8)(25\times 10^{-6})}{4(20\times 10^6)} (8.103 + 7.905)\times 10^6 = 1500.75\text{m} </math>
<math> v_r = \frac{\lambda_0}{4} \left( f_{bd} - f_{bu} \right) = \frac{0.3}{4}(8.103 - 7.905)\times 10^6 = 14850\text{m/s} </math>
<p> </p>
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