(Solved): 4.4-1 In a QAM system (Fig. 4.19), the locally generated carrier has a frequency error and a p ...

4.4-1 In a QAM system (Fig. 4.19), the locally generated carrier has a frequency error $??$ and a phase error $?$; that is, the receiver carrier is $\cos \left[\left({?}_c+??\right)t+?\right]$ or $\sin \left[\left({?}_c+??\right)t+?\right]$. Show that the output of the upper receiver branch is $m_1(t)\cos [(??)t+?]?m_2(t)\sin [(??)t+?]$ instead of $m_1(t)$, and the output of the lower receiver branch is $m_1(t)\sin [(??)t+?]+m_2(t)\cos [(??)t+?]$ instead of $m_2(t)$. Figure 4.19 Quadrature amplitude multiplexing. Both modulated signals occupy the same band. Yet two baseband signals can be separated at the receiver by synchronous detection if two local carriers are used in phase quadrature, as shown in Fig. 4.19. This can be shown by considering the multiplier output $x_1(t)$ of the upper arm of the receiver (Fig. 4.19):