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The expression of QAM basis is \begin{align} \psi_1&= \sqrt{\frac{2}{E_g}}g(t)\cos(2\pi ft)\\ \psi_2&= -\sqrt{\frac{2}{E_g}}g(t)\sin(2\pi ft) \end{align}

  • I want to ask why is $\psi_2= -\sqrt{\frac{2}{E_g}}g(t)\sin(2\pi ft)$, instead of $\sqrt{\frac{2}{E_g}}g(t)\sin(2\pi ft)$? Why is there a "-" (i.e. minus) before $\sqrt{\frac{2}{E_g}}$?

  • And QAM signal is $\ s_m(t)=A_{mi}\sqrt{\frac{E_g}{2}}\psi_1 +A_{mq}\sqrt{\frac{E_g}{2}}\psi_2$, does the $\sqrt{\frac{E_g}{2}}$ here have a relation with its basis $\sqrt{\frac{2}{E_g}}$ ?

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Base choices are ambiguous. The $-$ is there because the author of the book you're reading decided it was better that way.

You'll probably find the cause for that in the quadrature mixer graphic that they use, where if the phase shift from $I$ to $Q$ is $+\frac\pi2$, since $\cos(x+\frac\pi2) = -\sin(x)$.

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For a detailed explanation, see the first few paragraphs of this answer of mine from some time ago. It is, as Marcus Muller implies, a matter of convention, but in my opinion, having the $-$ sign makes for a cleaner and more consistent description of QAM than using the $+$ sign; YMMV.

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You have asked two questions.

(1) The reason behind a negative sign is to be consistent in dealing with real signals and their complex representation. In terms of complex signals, a complex data symbol is $a=\{a_I, a_Q\}$ while the carrier signal is $e^{j\omega t}$. During the upconversion process and ignoring the pulse shaped symbol stream (i.e., focusing on one symbol only), $$a\cdot e^{j\omega t}=\{a_I+ja_Q\}\cdot\{\cos \omega t +j\sin \omega t\}\\ =a_I\cos \omega t -a_Q\sin \omega t+j(a_Q\cos\omega t +a_I\sin \omega t)$$

Obviously, it's just the real signal that is transmitted and hence for consistency $-\sin\omega t$ is used. On the Rx side, this signal is multiplied with $e^{-j\omega t}$ for downconversion which simply cancels the Tx carrier. In terms of real signals, multiplying with $\cos\omega t$ produces a DC term proportional to $a_I$ and multiplying this real signal with $-\sin \omega t$ produces a DC term proportional to $a_Q$ (and not $-a_Q$), the two minuses before $\sin$ thus cancel out. $$(a_I\cos \omega t -a_Q\sin \omega t)(-\sin\omega t) = -\frac{a_I}{2}2\cos \omega t\cdot \sin \omega t+a_Q\sin^2\omega t\\ = -\frac{a_I}{2}\sin 2\omega t+\frac{a_Q}{2}t-\frac{a_Q}{2}\cos 2 \omega t$$ Filtering out the double frequency terms gives the correct sign for $a_Q$.

(2) Yes, the scaling factors in the basis functions are for normalization purposes. The factor $\sqrt{2}$ when appearing at both Tx and Rx becomes $2$ and restores the correct amplitude for $a_I$ and $a_Q$ (see answer to (1) above where a factor $2$ is there in the denominator). The same is true for the factor $\sqrt{E_g}$ as it eventually cancels out.

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