Question 2.10: For a quadbit input of I = 0, I′ = 0, Q = 0, and Q′ = 0 (000......
For a quadbit input of I = 0, I′ = 0, Q = 0, and Q′ = 0 (0000), determine the output amplitude and phase for the 16-QAM modulator shown in Figure 33.
The inputs to the I channel 2-to-4-level converter are I = 0 and I′ = 0. From Figure 34, the output is -0.22 V. The inputs to the Q channel 2-to-4-level converter are Q = 0 and Q′ = 0. Again from Figure 34, the output is -0.22 V.
Thus, the two inputs to the I channel product modulator are -0.22 V and sin ω_{c}t. The output is
I = (-0.22)(sin ω_{c}t) = -0.22 sin ω_{c}t
The two inputs to the Q channel product modulator are -0.22 V and cos ω_{c}t. The output is
Q = (-0.22)(cos ω_{c}t) = -0.22 cos ω_{c}t
The outputs from the I and Q channel product modulators are combined in the linear summer and produce a modulated output of
summer output = -0.22 sin ω_{c}t – 0.22 cos ω_{c}t
= 0.311 sin(ω_{c}t – 135°)
For the remaining quadbit codes, the procedure is the same. The results are shown in Figure 35.
