Solution

1 Calculation of a necessary channel pass band for transmission with the method QPSK is made under formula D F _QPSK = [ R (1 + α)]/2, where α is roll-off factor of spectrum. Being set by value α = 0,4, we receive

D F _QPSK =[ R (1 + α)]/2 = [64 (1+0,4)]/2 = 44,8 kHz.

2 According to the formula (B.1) it is defined limiting value of code rate

.

3 Under code tables we select the codes, satisfying to the requirement on a rate. Data about these codes are shown in table B.1. From the table it is visible, that for the solving given task can be used codes with the rate R _code = 1/2 which ensure enough big A-gain. in the table data the code with generator polynomials (133, 171) which at rate R _code = 0,5 ensures A-gain =6,99 dB is chosen for the project. Data of bit error probability calculation is given on figure 8.1 (signals BPSK and QPSK have the same noise immunity). It is visible, that the using of a such code ensures such performance: by the ratio signal/noise = 4,5 dB the bit error probability is less than 10^–5. Comparison with curves for uncoded QPSK shows that by p = 10^–5 this code ensures coding gain 6 dB.

table B.1 – Performances for a code choice

Table B.2 – Input data for the course work

variant number for an elaborating of course work should correspond to the number of student surname in the academic group register
variant number	S/N ratio , dB	modulation method	Rate of signal R, kbit/s	Bandwidth of channel F ch, kHz	Bit error probability p acc	Trellis complexity C perm
	4,0	QPSK			10-6
	5,0	QPSK			10-4
	6,0	BPSK			10-5
	6,5	BPSK			10-6
	4,0	QPSK			10-4
	7,0	QPSK			10-5
	5,0	BPSK			10-8
	6,0	QPSK			10-6
	5,0	BPSK			10-4
	4,5	QPSK			10-5
	5,5	BPSK			10-8
	4,0	QPSK			10-6
	4,0	QPSK			10-4
	5,0	BPSK			10-5
	4,0	QPSK			10-9
	7,0	QPSK			10-6
	5,0	BPSK			10-4
	4,5	QPSK			10-5
	5,5	QPSK			10-9
	6,0	BPSK			10-6
	7,5	QPSK			10-4
	6,5	QPSK			10-5
	6,0	QPSK			10-6
	4,5	BPSK			10-4
	5,0	BPSK			10-9
	6,0	QPSK			10-5
	4,5	QPSK			10-6
	5,0	BPSK			10-5
	5,5	BPSK			10-9
	4,5	QPSK			10-5
	5,0	QPSK			10-5