Question

Suppose the hop rate is increased to 2 hops/bit and the receiver uses square law combining the signal over two hops. The hopping bandwidth for this channel is

(a) 3.2767 MHz

(b) 13.1068 MHz

(c) 26.2136 MHz

(d) 1.6384 MHz

The question was asked in my homework.

The query is from Spread Spectrum topic in division Bipolar Junction Triodes (BJTs) of Electronic Devices & Circuits

Answer 1

The correct choice is (b) 13.1068 MHz

For explanation I would say: If the hopping rate is 2 hops/bit and the bit rate is 100 bits/sec, then, the hop rate is 200 hops/sec. The minimum frequency separation for orthogonality 2/T  400 Hz. Since there are N  32767 states of the shift register and for each state we select one of two frequencies separated by 400 Hz, the hopping bandwidth is 13.1068 MHz.