51

Short answer: yes, possible. GPS does that (nearly) all of the time. Long answer: The SNR your receiver system needs depends on the type of signal you're considering. For example, good old analog color TV needs, depending on standard, some 40 dB SNR to be "viewable". Now, any receiver is, mathematically, an estimator. An estimator is a function that maps ...


34

Magnetrons are cheap, reliable, pretty efficient (65% or so- and they tolerate high temperatures so heat sinking is easy) and made with mature technology. They are also reasonably tolerant of VSWR issues (if the user does not put a proper load in the oven, for example). They don't really allow the frequency to change much without expensive mechanical tuning ...


19

So you want to transport that 2.5 GHz (or even 5 GHz ?) Wifi signal over TV COAX cable ? Indeed to the non-RF people you'd just think that would work. And it does BUT there will be almost no signal coming through that cable. The Wifi signal will be attenuated so much in that COAX cable that it will defeat the whole purpose of having an antenna on the roof. ...


19

Fundamentaly, we have the Shannon-Hartley formula for the communication capacity of a channel: $$C = B \log_2\left(1+{\rm SNR}\right).$$ \$C\$ is the error-free channel capacity in bits per second of information transferred. \$B\$ is the bandwidth of the channel in Hz. \$\rm SNR\$ is the signal to noise ratio of the channel. There is no stipulation that \$...


15

Parallel capacitor plates of 25 mm by 25 mm seperated by 4mm of glass with a relative permittivity of 4 would give a coupling capacitance of about 5 pF. That capacitance is in series with an antenna signal and at 2.5 GHz, would act as a blocking impedance of about 13 ohms so, it's feasible it could be used without disrupting the VSWR too much. k is the ...


12

The domestic microwave oven needs high power to cook the meal and high frequency to excite the water molecules. What is not needed is high stability because the water energy absorption spectrum is broad. (1, 2) The magnetron does this cheaply. The low price and low duty cycle of the domestic microwave means that they should last for many years despite ...


11

I never fully understood how or why it worked Well, it's not magic ;-) Actually it can either be done magnetically using coupled inductors. This is like a transformer without a magnetic core. Wireless charging as used in some mobile phones uses the same principle. Basically, one coil creates a magnetic field from an electrical signal which is then picked ...


11

In general, yes, higher frequencies attenuate more the further distance they travel. There are two effects that are responsible for this. First, higher frequency radio waves tend to be absorbed more readily by objects (ie: the penetration depth in the material is shorter). You may have noticed this effect with your home WiFi network. If you have a dual-band ...


10

I'm not sure what you're looking at, but you need to understand the exmples you link. None of them use the truth-value within the actual filtering. It's there so you have something to compare to with regard to the filter output. Here is the simple script: import random # intial parameters iteration_count = 500 actual_values = [-0.37727 + j * j * 0.00001 ...


10

Years ago I designed a video system that sent a composite video signal down a coax cable from the video controller to the display CRT. This system used monochrome video with a 20MHz dot clock to display text and graphics on the CRT. Note that at 20MHz each pixel of video was 50ns wide. At the time the typical font used on the CRT was 5 pixels wide and had ...


9

Serial buses are widely used for high speed transfer. This diagram shows the available data transfer / clocking rate of a number of serial and parallel buses. In addition, it is common to implement multiple "lanes" each using serial transfer. Serial buses are able, in practice, to implement higher transfer rates "per lane" and even parallel buses with very ...


8

No, transforms aren't "necessary", but they do make some types of calculations much simpler and more convenient. It is possible to do all computation and analisys of a signal in either the time domain or the frequency domain. However, some operations are much simpler and more intuitive in one than the other. This can be illustrated with something as ...


8

You need to account for the ear by using the Fletcher Munsen curve (spelling may not be perfect but google will show the graph). Basically all sound levels for different frequencies have been empirically captured onto the graph.


8

What I would really like to know is that is it possible to establish a communication channel (sending information) if the received power level of the signal, received by the receiver antenna is below the noise floor. DSSS (direct sequence spread spectrum) radio can have a power level below the prevailing noise level and still work: - It relies on "...


8

Hopefully, this picture should explain: - Picture taken from here. The spring represents the media in which a signal can travel. The signal can be a voltage/current or an electromagnetic field or, just a mechanical spring with someone shaking one end. An impulse is applied to the spring and, as you should be able to see, it passes through the media from ...


6

Write the equation in positive powers of z; factorise, if required, and simplify; identify the poles and zeroes by inspection. You've already gone most of the way, the final step is: \$X(z)=\dfrac{1}{(2z)^4(2z-1)}=\dfrac{1}{32z^4(z-0.5)}\$ Which gives four poles at \$z=0\$ and a pole at \$z=0.5\$. There are no finite zeroes. The way the algebra is 'done'...


6

You must use coaxial cable of the proper impedance. The most common impedance for coax cable is 50 ohms or 75 ohms. If the cable you want to use matches the impedance of the interface AND the antenna, then go for it. But if you use cable of the wrong impedance you will get significant attenuation of the signal to the point where it may not work at all. In ...


6

the power received by the antenna in dBW has to be higher than the noise floor in dBW "noise floor" as most people would understand it is not measured in dBW, or any other unit of power. Rather, the noise floor is defined by the noise spectral density, which is measured in watts per hertz, or equivalently watt-seconds. The noise floor can be measured with ...


6

A practical oscillator has to be designed taking into account component tolerances. In your example the ratio of RG to RF needs to be such that the gain round the loop is unity. Because of the tolerance in RG, RF and the capacitors there has to be some extra gain designed in so that in the worst case the gain is at least unity. In the typical case this ...


5

It's the stability that's impractical in the real world. Mathematically, numbers look very stable, and yes, of course we can tweak a coefficient here and get the answer we want. In the real world, complicated amplifier components are very temperature sensitive, and vary from batch to batch. Simple feedback components, a pair a resistors for example, are ...


5

As a practical adjunct to Marcus Müller's excellent answer... Ham radio has a number of digital modes suitable for successful signal reception below the noise floor. These numbers have a caveat, which I explain afterwards. PSK31 can be properly copied at 7 dB below the noise floor. Olivia can be properly copied at 10 dB below the noise floor. WSPR can be ...


5

At first I wondered what to do with your question, because there was a lot of misunderstanding in it, but finally, let's just answer the core question: It looks to me that the Signal-to-Noise ratio (SNR or S/N) is completely arbitrary and it's set by the user according to his preferences. No. That's wrong. The signal power reaching a receiver is a ...


5

Take two sine waves of periods \$T_1 = 2s\$ and \$T_2 = 3s\$. Suppose they both start at time = 0s. Then their "end of the cycle" points coincide only at the multiples of \$LCM(T_1,T_2) = LCM(2,3) = 6s \$ While adding both signals, we take summation of all respective points of both signals at a time instant, and we get a set of non-repeating values ...


5

You can think of any perfectly bandlimited signal as the superposition of a set of \$\frac{\sin(t)}{t} = \text{sinc}(t)\$ curves, with their peaks positioned uniformly along the time axis. Their spacing is \$\frac{2}{BW}\$. sinc(x) also happens to be the time-domain response of a perfect low-pass filter, and it explains how the continuous-time ...


4

All transforms are tools to make analysis easier. They are tools that engineers, scientists, and mathematicians have developed over the years to help make their jobs easier or to help them gain a greater understanding of the phenomenon they are looking at. The Laplace transform, for example, makes solving differential equations easier. The wavelet ...


4

The Kinect camera is NOT a TOF (time-of-flight) camera. Instead, it projects "structured light" (a pseudorandom pattern created by an IR laser and a hologram) into the scene, and analyzes the image it gets back through a second (IR) camera in order to determine depth.


4

What you can do instead is to bring the router to the antenna on the roof and use a pair of MoCA boxes to run Ethernet over your coax.


4

Any oscillator needs that little bit extra gain to start. Oscillations build cycle after cycle. Your example circuit is powered with a +5v supply. Oscillation peak-to-peak amplitude certainly cannot exceed that supply voltage, so peaks are clipped - there's your non-linearity that keeps amplitude from growing forever. It is not a particularly good oscillator,...


4

Usually clipping limits the amplitude, but if you want to build a low-distortion oscillator, then every component must stay in its linear range. Clipping is not allowed. Which means we have a problem: self-excited sustained oscillations are only possible by means of a marginally stable system The Wien Bridge oscillator is an example of a solution. It ...


4

Lets check whether option A is the correct choice or not. When signal x(t) goes through system A, the magnitudes of all frequency components get multiplied by +j or -j correspondingly. Hence, the output response alpha will look like: The magnitude of alpha response get multiplied by j in the next step. This response beta is then added with the original ...


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