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Unformatted text preview: modulator for the following input bit sequence, 010100111? Assume the first reference bit is 0. (10 pts) 100 H 100 pF 100 pF I = B log2 (1 + S/N) 19.4x10^6=6x10^6 log2 (1+S/N) S/N=8 or 9.24dB PLL generates one signal after the loop filter and the other signal after VCO. PLL can be used in synchronizing the local oscillator to the carrier frequency. It can also be used to detect FM modulation. m-max=4V, m-s=0.4V m-c (output) = 2.364V Total power = Pc + P-lsb + P-usb m=0.9 Ratio = Pc/Total power = 0.7117 or -1.48dB NF=NF1+(NF2-1)/A1+(NF3-1)/A1A2+ A1=A2=10 3dB is equivalent to a factor of 2. NF=2+(2-1)/10=2.1 or 3.22 dB C = (100p in series with 100p) L = 100 u f = 1 / 2 pi sqrt(LC) = 2.25 MHz 010100111 XOR 1 1 1 1 1 1 1 4 2 c usb lsb P m P P = =...
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- Spring '11