## Q. 2.10

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. ## Verified Solution

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.  