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3 years ago

What property of an electromagnetic wave changes when the wave is encoded with analog information?

Physics

1 answer:

Licemer1 [7]3 years ago

7 0

Answer:

Amplitude and Frequency

Explanation:

Analog signals are composed of continuous waves that can have any values for frequency and amplitude. These waves are smooth and curved.

Radio transmissions are a combination of two kinds of waves: audio frequency waves that represent the sounds being transmitted and radio frequency waves that "carry" the audio information. All waves have a wavelength, an amplitude and a frequency as shown in the figure. These properties of the wave allow it to be modified to carry sound information.

The two most common types of modulation used in radio are amplitude modulation (AM) and frequency modulation (FM). Frequency modulation minimizes noise and provides greater fidelity than amplitude modulation, which is the older method of broadcasting . Both AM and FM are analog transmission systems, that is, they process sounds into continuously varying patterns of electrical signals which resemble sound waves.

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Consider an embedded system which uses a battery with a 17.15-Amp-Hour capacity. What is the maximum average current draw (in mi

Marianna [84]

Answer:

85.12 μAmp

Explanation:

The battery power output = 17.15 Amp-hr

If the battery is to last 23 years, we have to calculate how many hours there are in 23 years

in one year there are 24 hours x 365 day = 8760 hrs

in 23 years there are 23 x 8760 = 201480 hours

maximum current to be drawn from the battery = (17.15 Amp-hr) ÷ (201480 hours) = 85.12 x 10^-6 Amp = 85.12 μA

7 0

3 years ago

A steady current I flows through a wire of radius a. The current density in the wire varies with r as J = kr, where k is a const

grin007 [14]

Answer:

Explanation:

we can consider an element of radius r < a and thickness dr. and Area of this element is

$dA=2\pi r dr$

since current density is given

$J=kr$

then , current through this element will be,

$di_{thru}=JdA=(kr)(2\pi\,r\,dr)=2\pi\,kr^2\,dr$

integrating on both sides between the appropriate limits,

$\int_0^Idi_{thru}=\int_0^a2\pi\,kr^2\,dr
\\\\
I=\frac{2\pi\,ka^3}{3} -------------------------------(1)$

Magnetic field can be found by using Ampere's law

$\oint{\vec{B}\cdot\,d\vec{l}}=\mu_0\,i_{enc}$

for points inside the wire ( r<a)

now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.

by applying the Ampere's law, we can write

$\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop.

Hence,

$B_{in}\times2\pi\,l=\mu_0\int_0^r(kr)(2\pi\,r\,dr)=
\\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3}
\\\\B_{in}=\frac{\mu_0kl^2}{3}
$

now using equation 1, putting the value of k,

$B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}
\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}
$

now, for points outside the wire ( r>a)

consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.

applying the Ampere's law

$\oint{\vec{B}_{out}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence

$B_{out}\times2\pi\,r=\mu_0\int_0^a(kr)(2\pi\,r\,dr)
\\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3}
\\\\B_{out}=\frac{\mu_0ka^3}{3r}
$

again using,equaiton 1,

$B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3}
\\\\B_{out} = \frac{ \mu_{0} I}{2 \pi r}$

8 0

3 years ago

The blackbody curved for a star named Beta is shown below. What is the surface temperature of the star rounded to the nearest wh

marshall27 [118]

Then, since the peak wavelength of the star Beta is 200nm, use Wein law and round 200 to the nearest WHOLE NUMBER. Hope that helps.

3 0

3 years ago

Read 2 more answers

Whats the answer ----------------------------

GenaCL600 [577]

Answer:

repel each other

Explanation:

The magnitude of the charge of an electron is called... ... If a positively-charged glass rod is suspended so that it turns easily and another positively-charged glass rod is brought close to it, the two rods will... Repel each other.

8 0

2 years ago

A microphone is located on the line connecting two speakers that are 0.513 m apart and oscillating in phase. The microphone is 1

pantera1 [17]

Answer:

frequency 1 = 334.30 Hz

frequency 2 = 1002.92 Hz

Explanation:

Given data

speaker distance y = 0.513 m

microphone distance D = 1.80 m

to find out

lowest two frequencies

solution

we know velocity of sound is 343 m/s

so we consider point x

so at 1st speaker distance from x = D + (y/2)

1st speaker distance from x = 1.80 + (0.513/2) = 2.0565 m .....1

and

at 2nd speaker distance from x = D - (y/2)

2nd speaker distance from x = 1.80 - (0.513/2) = 1.5435 m .........2

so destructive interference from 1 and 2 we know

1st - 2nd = ( m + 0.5 ) wavelength

2.0565 m - 1.5435 m = ( 0+ 0.5) wavelength

wavelength = 1.026 m

so here 1st min frequency will be

frequency 1 = velocity of sound / wavelength

frequency 1 = 343 / 1.026 =334.30 Hz

and

2nd min frequency will be

frequency 2 =

2.0565 m - 1.5435 m = ( 1 + 0.5) wavelength

wavelength = 0.342 m

frequency 2 = velocity of sound / wavelength

frequency 2 = 343 / 0.342 = 1002.92 Hz

7 0

3 years ago