# 0.4 Analog (de)modulation  (Page 7/9)

 Page 7 / 9

In AM transmission where the baseband signal and its modulated passband version are real valued,the spectrum of the modulated signal has twice the bandwidth of the baseband signal.As pictured in [link] , the spectrum of the baseband signal is nonzero only forfrequencies between $-B$ and $B$ . After modulation, the spectrum is nonzero in the interval $\left[-{f}_{c}-B,-{f}_{c}+B\right]$ and in the interval $\left[{f}_{c}-B,{f}_{c}+B\right]$ . Thus the total width of frequencies occupied by the modulated signal is twicethat occupied by the baseband signal. This represents a kind of inefficiency or redundancyin the transmission. Quadrature modulation provides one way of removing this redundancyby sending two messages in the frequency ranges between $\left[-{f}_{c}-B,-{f}_{c}+B\right]$ and $\left[{f}_{c}-B,{f}_{c}+B\right]$ , thus utilizing the spectrum more efficiently.

To see how this can work, suppose that there are two message streams ${m}_{1}\left(t\right)$ and ${m}_{2}\left(t\right)$ . Modulate one message with a cosine to create the in-phase signal, and the other with (the negative of) a sine to form the quadrature signal. These are summed These are also sometimes modeled as the “real” and the “imaginary”parts of a single “complex valued” signal. This complex representation is explored more fully inAppendix [link] . to form

$v\left(t\right)={m}_{1}\left(t\right)\mathrm{cos}\left(2\pi {f}_{c}t\right)-{m}_{2}\left(t\right)\mathrm{sin}\left(2\pi {f}_{c}t\right)$

which is then transmitted. A receiver structure that can recover the two messagesis shown in [link] . The signal ${s}_{1}\left(t\right)$ at the output of the receiver is intended to recover the first message ${m}_{1}\left(t\right)$ . Similarly, the signal ${s}_{2}\left(t\right)$ at the output of the receiver is intended to recover the (negative of the) second message ${m}_{2}\left(t\right)$ .

To examine the recovered signals ${s}_{1}\left(t\right)$ and ${s}_{2}\left(t\right)$ in [link] , first evaluate the signals before the lowpass filtering. Using the trigonometric identities [link] and [link] , ${x}_{1}\left(t\right)$ becomes

$\begin{array}{ccc}\hfill {x}_{1}\left(t\right)& =& v\left(t\right)\phantom{\rule{4pt}{0ex}}\mathrm{cos}\left(2\pi {f}_{c}t\right)\hfill \\ & =& {m}_{1}\left(t\right)\phantom{\rule{4pt}{0ex}}{\mathrm{cos}}^{2}\left(2\pi {f}_{c}t\right)-{m}_{2}\left(t\right)\phantom{\rule{4pt}{0ex}}\mathrm{sin}\left(2\pi {f}_{c}t\right)\phantom{\rule{4pt}{0ex}}\mathrm{cos}\left(2\pi {f}_{c}t\right)\hfill \\ & =& \frac{{m}_{1}\left(t\right)}{2}\left(1+\mathrm{cos}\left(4\pi {f}_{c}t\right)\right)-\frac{{m}_{2}\left(t\right)}{2}\left(\mathrm{sin}\left(4\pi {f}_{c}t\right)\right).\hfill \end{array}$ In a quadrature modulation system, two messages m 1 ( t ) and m 2 ( t ) are modulated by two sinusoids of the same frequency, sin ( 2 π f c t ) and cos ( 2 π f c t ) . The receiver then demodulates twiceand recovers the original messages after lowpass filtering.

Lowpass filtering ${x}_{1}\left(t\right)$ produces

${s}_{1}\left(t\right)=\frac{{m}_{1}\left(t\right)}{2}.$

Similarly, ${x}_{2}\left(t\right)$ can be rewritten using [link] and [link]

$\begin{array}{ccc}\hfill {x}_{2}\left(t\right)& =& v\left(t\right)\phantom{\rule{4pt}{0ex}}\mathrm{sin}\left(2\pi {f}_{c}t\right)\hfill \\ & =& {m}_{1}\left(t\right)\phantom{\rule{4pt}{0ex}}\mathrm{cos}\left(2\pi {f}_{c}t\right)\phantom{\rule{4pt}{0ex}}\mathrm{sin}\left(2\pi {f}_{c}t\right)-{m}_{2}\left(t\right)\phantom{\rule{4pt}{0ex}}{\mathrm{sin}}^{2}\left(2\pi {f}_{c}t\right)\hfill \\ & =& \frac{{m}_{1}\left(t\right)}{2}\phantom{\rule{4pt}{0ex}}\mathrm{sin}\left(4\pi {f}_{c}t\right)-\frac{{m}_{2}\left(t\right)}{2}\left(1-\mathrm{cos}\left(4\pi {f}_{c}t\right)\right),\hfill \end{array}$

and lowpass filtering ${x}_{2}\left(t\right)$ produces

${s}_{2}\left(t\right)=\frac{-{m}_{2}\left(t\right)}{2}.$

Thus, in the ideal situation in which the phases and frequencies of the modulation and the demodulation are identical, bothmessages can be recovered. But if the frequencies and/orphases are not exact, then problems analogous to those encountered with AM will occur in the quadrature modulation.For instance, if the phase of (say) the demodulator ${x}_{1}\left(t\right)$ is not correct, then there will be some distortion or attenuation in ${s}_{1}\left(t\right)$ . However, problems in the demodulation of ${s}_{1}\left(t\right)$ may also cause problems in the demodulation of ${s}_{2}\left(t\right)$ . This is called cross-interference between the two messages.

Use AM.m as a starting point to create a quadrature modulation system that implements the block diagram of [link] .

1. Examine the effect of a phase offset in the demodulating sinusoids of the receiver, so that ${x}_{1}\left(t\right)=v\left(t\right)cos\left(2\pi {f}_{c}t+\Phi \right)$ and ${x}_{2}\left(t\right)=v\left(t\right)sin\left(2\pi {f}_{c}t+\Phi \right)$ for a variety of $\Phi$ . Refer to Exercise  [link] .
2. Examine the effect of a frequency offset in the demodulating sinusoids of the receiver, so that ${x}_{1}\left(t\right)=v\left(t\right)cos\left(2\pi \left({f}_{c}+\gamma \right)t\right)$ and ${x}_{2}\left(t\right)=v\left(t\right)sin\left(2\pi \left({f}_{c}+\gamma \right)t\right)$ for a variety of $\gamma$ . Refer to Exercise  [link] .
3. Confirm that a $±{1}^{\circ }$ phase error in the receiver oscillator corresponds to more than 1% cross-interference.

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