Solutions Communications Technology II WS 2011/2012

“Communications Technology II” – Solutions WS 2011/2012 12Solution to exercise 9 (vit14):a) Symbol clock model:( )-10.8 0.6b) Calculate two signal levels:w 11 = √ 1][0.8·(1+j)+0.6·(1+j)2w 42 = √ 1][0.8·(1−j)+0.6·(−1+j)2= √ 1 ] [1.4+1.4j 2= √ 1 ] [0.2−0.2j 2≈ 1+jc) ISI (here) leads to a total number of 16 different points in the signal space:QPSK symbol space with ISI≈ 0.14−0.14j1w 22w 21 w 12w 11imag →0.50-0.5-1w 23 w 24w 32 w 31w 13 w 14w 42w 43 w 44w 33 w 34w 41-1 -0.5 0 0.5 1real →d) (Solution of the additional exercise)The average signal power ¯σ d 2 has to be calculated from the average quadratic value ofthe signal space points. This calulation is trivial for the symbol alphabet at the input:The symbol’s average power is 1, since all symbols also have the absolute value of 1.Average signal power at the channel’s output:[¯σd 2 = 1∣16 ·4·0.2+0.2j ∣∣∣2∣ ∣ √ + 8·1.4+0.2j ∣∣∣22∣ √2+ 4·∣= 1 8 ·[( 0.2 2 +0.2 2) + 2·(1.4 2 +0.2 2) + ( 1.4 2 +1.4 2)]= 1 8 ·[4·0.2 2 +4·1.4 2] = 1.0∣ ]1.4+1.4j ∣∣∣2√2Thus the signal is neither extenuated nor amplified by the channel (“neutral due topower”).