{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW6S - ECE 402 Communications Engineering Homework 6...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
ECE 402: Communications Engineering Homework 6 Solutions (Fall 2009) 1. (a) For polar signaling, P b = Q ( p 2 E b /N 0 ). From the reference sheet, we see that Q (3 . 719) = 10 - 4 and Q (4 . 264) = 10 - 5 . Hence to obtain P b = 10 - 4 , we must have s 2 E b N 0 = 3 . 719 = E b N 0 dB = 10 log 10 ˆ (3 . 719) 2 2 ! = 8 . 4 dB . To obtain P b = 10 - 5 , we need s 2 E b N 0 = 4 . 264 = E b N 0 dB = 10 log 10 ˆ (4 . 264) 2 2 ! = 9 . 6 dB . (b) Unipolar signaling ( P b = Q ( p E b /N 0 )) requires twice as much (3 dB more) energy-per-bit than polar signaling for the same P b , so E b /N 0 = 11 . 4 dB for P b = 10 - 4 and E b /N 0 = 12 . 6 dB for P b = 10 - 5 . 2. The error probability of BPSK is P b = Q ( p 2 PT/N 0 ). From the reference sheet, we see that Q (4 . 753) = 10 - 6 so q 2 PT/N 0 = 4 . 753 = P = 11 . 3 N 0 T = 11 . 3 N 0 R Since N 0 = kT r , the required noise temperature is T r = P 11 . 3 Rk = 5 × 10 - 18 11 . 3 · 4740 · 1 . 38 × 10 - 23 = 6 . 76 o K . 3. The baseband version of 16QAM has data amplitudes a n = a + jb , where a and b are ± A T, ± 3 A T . The pulse waveform is g ( t ) = (1 / T )sinc( t/T ). For 16QAM, we have μ = E{ a n } = 1 16 X a,b = ± A T, ± 3 A T [ a + jb ] = 0 σ 2 = E{| a n - μ | 2 } = 1 16 X a,b = ± A T,
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}