# 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

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

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.

how do I set up the problem?
what is a solution set?
Harshika
find the subring of gaussian integers?
Rofiqul
hello, I am happy to help!
Abdullahi
hi mam
Mark
find the value of 2x=32
divide by 2 on each side of the equal sign to solve for x
corri
X=16
Michael
Want to review on complex number 1.What are complex number 2.How to solve complex number problems.
Beyan
yes i wantt to review
Mark
use the y -intercept and slope to sketch the graph of the equation y=6x
how do we prove the quadratic formular
Darius
hello, if you have a question about Algebra 2. I may be able to help. I am an Algebra 2 Teacher
thank you help me with how to prove the quadratic equation
Seidu
may God blessed u for that. Please I want u to help me in sets.
Opoku
what is math number
4
Trista
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
can you teacch how to solve that🙏
Mark
Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
Brenna
(61/11,41/11,−4/11)
Brenna
x=61/11 y=41/11 z=−4/11 x=61/11 y=41/11 z=-4/11
Brenna
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years? Kala Reply lim x to infinity e^1-e^-1/log(1+x) given eccentricity and a point find the equiation Moses Reply A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place. Kimberly Reply Jeannette has$5 and \$10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
What is the expressiin for seven less than four times the number of nickels
How do i figure this problem out.
how do you translate this in Algebraic Expressions
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Got questions? Join the online conversation and get instant answers!