Example 8.1 A channel sounder is in a car that moves along a street at 36km/h. Ir measures the channel impulse response at a carrier frequency of 2 GHz At what intenas does it to measure? What is the maximum excess delay the channel can have so as remain underspread? v 36 km/h 10 m/s (8.10) 3.108 0.15 m 2 109 The channel must be sampled in the time domain at a rate that is, at minimum, twice the maximum Doppler shift. Using Eq. (85) (8.11) 2 Umax 2 frep The sampling interval Tep is given as 7.5 ms (8.12) rep At a mobile speed of 36 kmh, this corresponds to a channel snapshot taken every 75 mm. To we make use of the fact that the channel must be calculate the maximum excess dela unders in order to be identifiable. Hence from Eq. (8.9), (8.13) 7.5 ms This is orders of magnitude larger than the maximum excess delays that occur in typical wireless channels (see Chapter 7). However, note that this is only the theoretical maximum delay spread that still guarantees identifiability. A correlative channel sounder would show a significant degradation in estimation quality for this high rmax. If rmx is smaller, then the repetition period 1rep of the sounding pulse should be increased; this would allow averaging of the snapshots and thus improvement of the SNR.