# 0.4 Analog (de)modulation  (Page 8/9)

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TRUE or FALSE: Quadrature amplitude modulation can combine two real, basebandmessages of absolute bandwidth $B$ in a radio-frequency signal of absolute bandwidth $B$ .

Consider the schematic shown in [link] with the absolute bandwidth of the baseband signal ${x}_{1}$ of 6 MHz and of the baseband signal ${x}_{2}\left(t\right)$ of 4 MHz, ${f}_{1}=164$ MHz, ${f}_{2}=154$ MHz, ${f}_{3}=148$ MHz, ${f}_{4}=160$ MHz, ${f}_{5}=80$ MHz, $\Phi =\pi /2$ , and ${f}_{6}=82$ MHz.

1. What is the absolute bandwidth of ${x}_{3}\left(t\right)$ ?
2. What is the absolute bandwidth of ${x}_{5}\left(t\right)$ ?
3. What is the absolute bandwidth of ${x}_{6}\left(t\right)$ ?
4. What is the maximum frequency in ${x}_{3}\left(t\right)$ ?
5. What is the maximum frequency in ${x}_{5}\left(t\right)$ ?

Thus the inefficiency of real-valued double-sided AM transmission can be reduced using complex valued quadrature modulation, whichrecaptures the lost bandwidth by sending two messages simultaneously. For simplicity and clarity, the bulk of Software Receiver Design focuses on the real PAM case. There are also other ways of recapturingthe lost bandwidth: both single side band and vestigial sideband (discussed in the document Other Modulations on the website) send a single message, but use only half the bandwidth.

## Injection to intermediate frequency

All the modulators and demodulators discussed in the previous sectionsdownconvert to baseband in a single step, that is, the spectrumof the received signal is shifted by mixing with a cosine of frequency ${f}_{c}$ that matches the transmission frequency ${f}_{c}$ . As suggested in [link] , it is also possible to downconvert to somedesired intermediate frequency (IF) ${f}_{I}$ (as depicted in [link] ), and to then later downconvert to baseband by mixing with a cosineof the intermediate frequency ${f}_{I}$ . There are several advantagesto such a two-step procedure:

• all frequency bands can be downconverted to the same IF, which allows use of standardized amplifiers, modulators and filterson the IF signals, and
• sampling can be done at the Nyquist rate of the IF rather than the Nyquist rate of the transmission.

The downconversion to an intermediate frequency (followed by bandpass filtering to extract the passband around theIF) can be accomplished in two ways: by a local oscillator modulating from above the carrierfrequency (called high-side injection) or from below (low-side injection). To see this,consider the double sideband modulation (from "Amplitude Modulation with Suppressed Carrier" ) that creates the transmitted signal

$v\left(t\right)=2w\left(t\right)\mathrm{cos}\left(2\pi {f}_{c}t\right)$

from the message signal $w\left(t\right)$ and the downconversion to IF via

$x\left(t\right)=2\left[v\left(t\right)+n\left(t\right)\right]\mathrm{cos}\left(2\pi {f}_{I}t\right),$

where $n\left(t\right)$ represents interference such as noise and spurious signals from other users.By the frequency shifting property (A.33),

$V\left(f\right)=W\left(f+{f}_{c}\right)+W\left(f-{f}_{c}\right),$

and the spectrum of the IF signal is $X\left(f\right)$

$\begin{array}{cc}& =V\left(f+{f}_{I}\right)+V\left(f-{f}_{I}\right)+N\left(f+{f}_{I}\right)+N\left(f-{f}_{I}\right)\hfill \\ & =W\left(f+{f}_{c}-{f}_{I}\right)+W\left(f-{f}_{c}-{f}_{I}\right)+W\left(f+{f}_{c}+{f}_{I}\right)\hfill \\ & \phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}+\phantom{\rule{4pt}{0ex}}W\left(f-{f}_{c}+{f}_{I}\right)+N\left(f+{f}_{I}\right)+N\left(f-{f}_{I}\right).\hfill \end{array}$ Example of high-side and low-side injection to an IF at f I = 455 kHz. (a) transmitted spectrum, (b) low-side injected spectrum, and(c) high-side injected spectrum.

Consider a message spectrum $W\left(f\right)$ that has a bandwidth of 200 kHz, an upconversion carrierfrequency ${f}_{c}=850$ kHz, and an objective to downconvert to an intermediate frequency of ${f}_{I}=455$ kHz. For low-side injection, where the frequency of thelocal oscillator is ${f}_{\ell }<{f}_{c}$ , the goal is to center $W\left(f-{f}_{c}+{f}_{\ell }\right)$ in [link] at ${f}_{I}$ , so that ${f}_{\ell }={f}_{c}-455=395$ . For high-side injection (with ${f}_{\ell }>{f}_{c}$ ), the goal is to center $W\left(f+{f}_{c}-{f}_{\ell }\right)$ at ${f}_{I}$ , so that ${f}_{\ell }={f}_{c}+455=1305$ . For illustrative purposes, [link] supposes that the interference $N\left(f\right)$ consists of a pair of delta functions at $±$ 105 and 1780 kHz. [link] sketches $|V\left(f\right)|$ and $|X\left(f\right)|$ for both high-side and low-side injection. In this example, both methods end up with unwanted narrowband interferencesin the passband.

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biomolecules are e building blocks of every organics and inorganic materials.
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how did you get the value of 2000N.What calculations are needed to arrive at it
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