# Tag Info

17

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 ...

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 ...

7

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

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

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 ...

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

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 ...

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 ...

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

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

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 ...

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

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 ...

2

A single dish aperture can operate a single pixel detector but this pixel projected onto the sky can be quite large compared to optical and infrared telescopes. The resolution is proportional the wavelength and therefore this can be 1000's of times the optical resolution. Therefore a radio telescope can take its image in a reasonable time simply because ...

2

Oneat, yes, and I understood what you meant "I mean this one". ;-) You meant that it's the frequency $\nu$ (nu) that appears in the formula $E=h\nu$ for the photon energy, as opposed to $\omega$ that appears in $E=\hbar\omega$ and that differs by a factor of $2\pi$. Whenever you hear "$\mbox{Hz}$" or any multiple of it, it means that the frequency is ...

2

Phone's radio stack is programmed to both conserve battery and maintain the connection with base station as long as possible. So, it will increase its sending power when it receives weak signal and reduce it when the received signal is strong to save battery power. It is just that changing EM field induces current in all conductors, so the signal from phone ...

2

It's possible to split this combined signal into the original components again. You can do that because the sine and cosine functions form a base of a Hilbert space, a space called $L^2(\mathbb{R})$. Now what does this mean? The word "space" is perhaps confusing, a mathematical space is essentially just a set of mathematical objects, in this case the space ...

2

A wave with amplitude modulation can be described with a carrier $A_0 cos(\omega_c t)$ and a modulating factor $\left [1+a_s cos(\omega_a t)\right ], |a_s| <1, \omega_s <<\omega_c$: $A(t)=A_0 cos(\omega_c t)\cdot \left [1+a_s cos(\omega_s t) \right ]=A_0 cos(\omega_c t)+A_0a_s cos(\omega_c t)cos(\omega_s t)$ The first term is a carrier frequency ...

2

If you are an electron in an atom and you have some energetic spectrum, a radio-wave may cause your transitions from one level to another. Your quantum character is manifested in your discrete energy levels. If you are an electron in a solid state, in a metal, for example, then your spectrum may be continuous and any energies are allowed. In this case you ...

2

This is a resonance in the circuit--- when you have a bunch of different frequencies driving a resonant system, the response is only strong for those frequencies which are close to the natural frequency of the resonant oscillator. You can see the same phenomenon in mechanical systems. If you have a mechanical mass on a spring, and you apply a force which ...

2

Wifi is around 2.4GHz so a wavelength of around 12.5cm Anything metal with holes much smaller than 12cm will work - even a mesh. But if you don't have a parabolic dish, only a spherical one, you might do better with a can antennea see eg http://www.turnpoint.net/wireless/cantennahowto.html for details of the calculation

2

The average displacement of an electron in the presence of an electric field is given by it's 'drift velocity' (link here). You can see right at the end of the page that the average displacement about a mean position for an electron is about $\approx 10^{-6} m$. This is clearly a tiny amount (in macroscopic terms) to be displaced. So how exactly does a 1 m ...

2

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 ...

2

The frequency of the signal is modulated in a relatively narrow band, and drives the audio circuit in proportion to the resonant power between the signal and a resonating circuit tuned to almost the range where the signal resides. The result is that the power in the audio circuit varies with the frequency of the driving signal Other signals are far from ...

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