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Unformatted text preview: 1 EE 7620 Solutions to Problem Set # 5 Nov. 10 1) Since the frequency band is (2500 , 3000) , the system must be a bandpass sytem. a) We have R b = 500 , W = 1000 . Thus R b /W = . 5 . Also E b N = PT b N = P R b N = P 500 × 2 × 10 5 = 100 P i) Any PSK system satisfies the R b /W requirement. BPSK and QPSK require the least power. E b N = 100 P = 9 . 12 . Thus P = . 0912 . ii) BFSK, 4FSK, 8FSK can be used. 8FSK uses the least power. For 8FSK, E b N = 100 P = 7 . 07 . Thus P = . 07 . iii) We can use any ASK system. BASK uses the least power and we get P = . 0912 . iv) All QASK systems are feasible. The least power is used by 4QASK which is the same as QPSK for which P = . 0912 . v) E b N = 100 P = 2 R b /W 1 R b /W = . 8284 Thus P = . 0082 . b) We have R b /W = 3 . 5 . Also E b N = PT b N = P R b N = P 500 × 2 × 10 5 = 100 P 7 i) 16PSK system satisfies the R b /W requirement. Then E b N = 100 P/ 7 = 52 . 4 . Thus P = 3 . 66 . ii) FSK can not be used. iii) 16ASK can be used. Then 100 P/ 7 = 177 . 8 . Thus P = 12 . 44 ....
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This note was uploaded on 03/05/2012 for the course EE 7615 taught by Professor Naragipour during the Fall '11 term at LSU.
 Fall '11
 Naragipour
 Frequency

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