Number of false alarms 10152 Frequency kHz Hits d 4 157159

Number of false alarms = 10/152

Frequency (kHz)

Hits

d’

4

157/159

______

8

160/162

______

12

157/159

______

18

10/140

______

19

17/159

______

What is the minimum depth at which d’ 1 (You can do a linear interpolation between points of d’ >1 and d’ < 1 or fit a sigmoid to the curve)?

Solution

d\'

157 / 159 * 4 = 3.95

160/162 * 8 = 7.90

157 / 159 * 12 = 11.85

10 / 140 *18 = 1.286

17 / 159 * 19 = 2.03