1. Assume that 16-ary QAM is used to transmit a signal. In 16-ary QAM we have the following mapping of bits to symbols: Bits Symbol Bits Symbol 0000 -3+j3 1000 +3+j3 0001 -3+jl 1001 +3+j1 0010 -3-j3 1010 +3-j3 0011 -3-j1 1011 +3-j1 0100 -1 + j3 1100 +1+j3 0101 -1+j1 1101 +1+j1 0110 -1-j3 1110 +1-j3 0111 -1-j1 1111 +1-j1 a. Draw the constellation diagram. Identify which bits will be represented by each symbol. b. Which symbols will be transmitted? What is the transmit power for each symbol? c. Suppose the following bits are to be transmitted: (1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1) Which symbols will be transmitted? What is the transmit power for each symbol? d. For the above bits, sketch the actual time domain waveform leaving the antennas for the first symbol if the carrier frequency is 100kHz? e. If each symbol is equally likely, calculate the average transmission power for this modulation scheme. f. Explain if the probability of error will be higher and lower for 4-ary QAM with the same average transmission power as this 16-ary QAM scheme. 2. Repeat the previous question for 16-ary PSK, which has the bit-symbol mapping below: Bits Symbol 0000 0001 -j/16 e 0011 0010 0110 0111 0101 0100 e-/2/16 e-13/16 e = = = 1 cos(/16)-j sin(/16) = cos(2/16)-j sin(2/16) cos(3/16)-j sin(3/16) e-14/16 = cos(4/16)-j sin(4/16) -j5/16 = cos(5/16) -j sin(5/16) e -j6/16 = cos(/16)-j sin(/16) e-/7/16 = cos(7/16) - j sin(7/16) cos(8/16)-j sin(8/16) e -j9/16 cos(9/16) -j sin(9/16) e -j8/16 j11/16 = cos(10/16)-j sin(10/16) == cos(11/16)-j sin(11/16) 1100 1101 1111 e -/10/16 1110 e 1010 1011 e 1001 -j13/16 e-/14/16 = 1000 cos(13/16) -j sin(13/16) cos(14/16)-j sin(14/16) e-/15/16 = cos(15/16)-j sin(15/16) e-/12/16 = cos(12/16) - j sin(12/16) =