# Tag Info

## Hot answers tagged radio

19

It's spherical because the main dish cannot be steered; steering is done by moving the receiver (the big thing hanging over the center of the reflector). A parabolic reflector would produce varying errors when aimed in different directions; a spherical reflector has the same error for all directions. Presumably the receiver is designed to compensate for ...

11

Did you read the Wikipedia article? It explains the signal rather well, I think. At any rate, it is called the Wow! signal because, as the picture shows someone wrote Wow! in the margin. As for the code and why they were excited, I quote the Wikipedia article, The circled alphanumeric code 6EQUJ5 describes the intensity variation of the signal. A ...

9

Say if I transmit: $\sin(2\pi x)$ And separately: $\sin(2\pi x\times 2)$ Does it end up as a single wave of: $\sin(2\pi x)+\sin(2\pi x\times 2)$? Yes, that's exactly how it works. This is called superposition. There are electromagnetic waves at hundreds of different frequencies all filling the air simultaneously. The way something like a ...

9

A human body may reflect and absorb radio frequencies, though not very efficiently. It may as well act as a resonance chamber for certain frequencies. For a signal of 100 MHz, the involved wavelength is 3 m, and so it is possible that parts of your body are acting slightly as a resonant chamber. (for an optimal resonance, you should have 1.5 m diameter, too ...

8

AM radio typically transmits at around 1 MHz, FM radio at about 90 MHz. Measurements of the RF spectrum of lightning strikes show a falloff with frequency of about 20 dB per decade in that frequency range, so with FM about 2 decades above AM, you'd expect AM to have about 40dB higher interference from a lightning strike. In addition to that, FM signals ...

7

Let your carrier signal be $A_0 \cdot \cos(\omega_c t)$ with amplitude $A_0$ and carrier frequency $\omega_c$. Let your signal be a simple wave, $\phi(t) = A_s \cdot \cos(\omega_s t)$. Then the modulated signal becomes $$A_0 A_s \cdot \cos(\omega_c t) \cdot \cos(\omega_s t)$$. In addition, as pointed out by George in the comments, the carrier also gets ...

7

They heat it, by different degrees depending on the polarization of molecules in the tissues and liquids. The molecules try to re-align after the radio-wave field and the movement dissipates as general heat. Think microwaves.. a consumer-grade microwave oven operates at the same radio wave spectrum as your home WiFi network (2.4 GHz) but much stronger. The ...

7

Radio waves are large wavelength waves, and non metal walls are transparent to the radiation at those wavelengths, depending on the thickness of the walls, because there are no energy level "receptors" to absorb them in bulk by excitation of electronic orbits. The wavelengths start from a centimetre up to kilometre, which is the width of the cycle of the ...

7

You might want to have a look at Does light induce an electric current in a conductor?. It's probably impossible for a radio aerial to emit visible light as the frequency of light is around the plasma frequency of the metal that the aerial is made of. We're not really supposed to address hypothetical questions, but if you could find some material with a ...

6

Here's a somewhat technical document talking about insterstellar beacons. A beacon would act as a "searchlight", sweeping across the sky, so that the time spent on each target star would be rather short. With a beam aperture able to illuminate 1% of the sky and working at 0.5 Hz they calculate that 6.9 GW power will be "visible" as far as 6000 light years. ...

6

A guitar string produces harmonics because it vibrates in a non-linear fashion. An electronic oscillator can be made to generate a much purer form of vibration (near sinusoidal) than a mechanical device such as the guitar string. Hence its harmonic level, while not zero, is much lower. For example, the harmonic distortion of a guitar string is probably on ...

5

It depends on what kind of radio telescope you're talking about. If you're talking about a single dish, with say, a horn feed at the focus, it behaves like a regular 'dish antenna' by focussing radio waves to the focus where it is amplified, processed in some manner (such as downconversion), digitised, and so on. A single dish can also have a feed array at ...

5

You're not boosting the signal; you're either acting as a reflector (capturing a bit more of it to feed to the antenna) or blocking a competing source, or perhaps a bit of both. By analogy, when you hold your hand to your ear to help you hear something, your hand is acting a reflector for sound waves to direct a little more energy into your ear. It can also ...

5

Yes, LED's, luminescence of phosphorous (CRT screens), most of the current screen technology, fluorescent lights. They aren't a thermal source. They heat up due to the electric current but that's not the working principle.

5

While I agree with Alfred Centauri's answer, I am not sure it gives a direct answer to your specific question for the following reason. If there is a receiving antenna somewhere, there is always some reflection, however minute. If the antenna is connected to a circuit, the conditions of reflection will change, and the transmitter can "notice" that. In ...

4

In the microwave band here are multi-element detectors, but at longer wavelengths the telescope is a single pixel. Yes it does take a while to build up an image, but radio pictures aren't usually very large - not the millions of pixels of an optical/IR image. One big advantage of radio telescopes is that you can combine telescopes 1000s of km apart to ...

4

There are ways that EM waves can be used to reconstruct a representation of matter and these techniques vary widely based on the method. Each method is very unique and its own set of issues. UV-D's reference Measurements Using Optic and RF waves provides a good overview. It explains some of the applications, the sources for their EM waves and ...

4

1) Normally, the antenna isn't the only component that distinguishes between the various competing signals received. The antenna does have a bandwidth and will attenuate signals outside that band. A typical antenna on a cell phone mast for example, may receive in the range 1.8GHz - 2.4GHz (just an example, you would have to look up manufacturer data ...

4

First, no, "radio propagation" is not "via atmosphere". Different wavelengths get absorbed, reflected, or simply passed by different parts of the atmosphere. There is no one general rule. Many of our radio communications within the atmosphere are pretty much like they would be in free space, for example. Second, all radio waves propagate infinitely in ...

4

As an intermediate step, consider a sinusoidal source driving an infinite transmission line with some characteristic impedance $Z_0 = 50\Omega$. The source "sees" a real impedance of $50\Omega$ and so, power is delivered to the line and, since the TL is infinitely long, the power is transported down the line, via an electromagnetic wave, without reflection. ...

3

Taking a step back, what do you want to study with your telescope? The telescope is just one component in your DIY astronomy lab. You are also going to need a plan for your study, a way to acquire signals from your telescope, and a way to process those signals. Some telescope / antenna stuff - A guy who built a telescope for examining Hydrogen emissions ...

3

The inductor and capacitor form a resonant circuit, which will pass only a specific frequency - the one you are tuning the radio to recieve. You normally tune it by making either the inductor or capacitor adjustable. edit: As described in How does radio receives signal from particular station? it's very much like a pendulum. Current flows freely in the ...

3

All electromagnetic signals that leave an antenna have an amplitude, i.e. there is power propagating as they spread. power is the rate at which energy is transferred, used, or transformed. For example, the rate at which a light bulb transforms electrical energy into heat and light is measured in watts—the more wattage, the more power, or equivalently ...

3

a path of a wave function that touches a peak or a trough exactly every 1/f increments, regardless of the "volume" First of all, that's not correct. A peak or trough in some signal $U(t)$ is defined by $\frac{\partial U}{\partial t}|_{t_{\mathrm{peak}}}=0$. If $U$ is now a product of some carrier $U_{\mathrm{c}}(t)$ and some modulator $M(t)$,  ...

3

The apparent surface you are describing is the photosphere, which is indeed dependent on the frequency you're looking at. The simplest answer is that the radius (as a function of frequency) is very nearly constant --- because the density profile of sun near the photosphere is very sharp. The photosphere is approximately the location where the optical depth ...

3

You have a few different but related questions so I will try to explain them in a simple, no-math way. If a radio tunes to a specific frequency, where does the excess energy go? Almost every object that has radio waves (electromagnetic waves) around it absorbs some of the radio energy. When the radio waves hit the electrons in the atoms and transfers ...

3

Modulation, whether AM, PM, FM or whatever (even CW), necessarily widens the spectrum from that of the pure tone of the carrier. Thus, in the design of any radio demodulator circuit, the bandwidth of the modulated carrier must be taken into account. Generally, the RF signal is down-converted, via a mixer and local oscillator, to an Intermediate Frequency ...

3

This isn't hypothetical. There is nothing that a radio does that can't be done in other parts of the spectrum. Many FM/AM radios operate in the optical range too. Your TV remote control uses IR. Lasers are used for high bandwidth point to point communications. And don't forget fiber optics, these are all radios that just use optics for the communication ...

3

You can make some assumptions like a typical FM receiver needs -110 dBm to work. Then assume you have an isotropic antenna in both cases because you didn't say anything about the antennas so we'll ignore the gain. Next take a look at the path loss based on the 910 W (+59.6 dBm) power. Your path loss can not exceed 59.6 + 110 = 169.6 dB (with loss dB and ...

3

Your question is pretty severely under-defined and you've really touched on the tip of the iceberg for a huge subject of information theory and network engineering. The bulk of this isn't physics so I will keep my answer brief. This is how I've interpreted your question: If I want to transmit data from point A to point B, is it better to have: 1) ...

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