Section 5.5 · Worked examples · 3 of 3
Worked example 5.5.3: Two transmitters, one channel
About the worked examples of Section 5.5
Three short calculations tied to the applications. They use only , the delay and the of the retarded potentials, and no new physics. Numerical constants: , Earth radius , all media nonmagnetic. A good way to use them: read the given quantities, produce the formula from the corresponding application yourself, and only then put the numbers in.
Two transmitters, one channel
#Given. Two synchronised DVB-T transmitters and of equal power, apart, broadcast the same OFDM symbols on one frequency. The guard interval is (8k mode, ) or (). A receiver stands on the line , from .
Find. (a) the delay difference at the receiver, (b) whether it fits in each guard interval, (c) the largest anywhere on the line, and where the interval is enough, (d) the amplitude ratio of the two signals, in dB, from the of the retarded potentials.
Each transmitter's signal is retarded by its own , so the receiver sees two copies separated by , Eq. (5.5.51). The copy that arrives later only adds to the wanted signal if . Both potentials, and the fields far from the antennas, fall as , so the amplitude ratio is .
, :
: fine with the guard interval. : with the late signal spills into the next symbol.
By the triangle inequality can never exceed the spacing. It is reached anywhere on the line beyond or , so . Between the transmitters, at the distance from the midpoint, and . The interval covers , so it is enough for .
, that is .
At the receiver of (a), and . With the interval the largest allowed path difference is , more than the spacing, so the whole line is covered.
The spacing of the network is chosen with . Where the delay is too long, the law often rescues the receiver: the late signal is also the weak one, down here. Real terrain and antenna patterns change the ratio, but the delay is fixed by geometry and alone.
cf. Application 5.5.3