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A short answer is that penetration depth of electromagnetic waves in a material is proportional to square root of resistivity, inversely proportional to square root of permeability and, most importantly for this context, inversely proportional to the square root of the frequency of the signal. (Reality is not quite this simple, but this is accurate enough ...


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In this circuit, every diode acts similarly to a switch. Diodes are used because they switch fast (ten million switches per second). In the simple model, the switch is either open or closed. The local oscillator determines switch timing: on one half cycle both diodes are open, on the other half-cycle both diodes are closed. The circuits below are a slight ...


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Frequency and the use of FM actually have little to do with it. Although at 100 km range, the propagation characteristics of different frequencies will come into play. The key parameters are SNR (signal-to-noise ratio) and bandwidth (BW). The Shannon-Hartley equation for channel capacity (C) tells us: $$C = \text{BW}(1 + \log_2 \text{SNR})$$ For maximum ...


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By reading some application notes, I discovered, indeed, that the 4-port S-parameters measurements file is meant for creating a model of the transformer, and it can be used with a simulator directly, such as Genesys or ADS. Those are expensive tools unavailable for hobbyists, but I also found the free and open source Qucs simulation supports n-port S-...


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If some people are interested, I wrote some come using Raspberry Pi zero and NodeJS https://github.com/360manu/kettlerUSB2BLE it was easier (for me) to implement this system with a complete system


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It uses 22kHz burst communication in this interface rev, not DC control. REF To permit slave devices to communicate back to the master, it is necessary to make more specific recommendations for the operating impedances of the bus than is required for a simple 22 kHz continuous tone signalling. In addition, it is advantageous to be able to accept a wide ...


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The fundamental frequency is determined by the commutator switching speed (which varies with motor rpm), but the rf produced by arcing can go up to GHz frequencies. Here's an example:- The blue line is before adding ferrite cores to the motor leads, the green line after. At lower radio frequencies (below 100MHz or so) small capacitors connected from each ...


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Note on page 5 of the datasheet for MIXIN and LNAOUT pins, the components are required for the matching network. They also serve other purposes. The series capacitors serve as DC blocks as mentioned in the datasheet. The inductor and resistors connected to LNAOUT also serve as a bias feed to the LNA.


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I suggest you to Download AppCAD DOS version from http://www.hp.woodshot.com/appcad/appcadcl.exe Run the APPCAD.BAT and choose impedance matching L type. This can easily solve this circuit by only 2 components basically.


0

Infrared could be an option? Probably also goes over 5m, but might even be easier to limit to a certain distance.


1

The way I did this was to use pre-made boards that use a circuit board j-pole antenna, lowest power setting and the highest speed. Once triggered they would ping-pong each other once every 2 seconds until the primary failed to receive 5 pings, at which point it would sound an alarm. This averaged out to be about 10 to 15m, which was plenty accurate for my ...


1

Without more information(competing signal levels, distance to target, etc) I can't really give specifics. But here goes... I would first try a good directional antenna. There are plenty of 2.4Ghz Yagi antenna available, I would try something with a 9dBi to 18dBi gain to start with. Then you'll want a GOOD cable to connect it to your board. This means ...


1

You did not specify your requirements, but if you need a low speed, you can use the following tips: Use the lowest speed possible Use the maximum power setting Use the external antenna shown in the pic Than check how many packets are arrive successfully. If this is 0, you are mostly out of luck, but if it is higher than 0, than you can use a protocol (I ...


2

How are my chances to modify this device to output drastically more power? The device is a transceiver and the transmit/receive arbitration circuit is inside the chip and therefore you cannot put an amplifier in the antenna feed (unless you are highly skilled and have access to good equipment) because it'll stop reception of the RF signal. Plus, adding ...


0

1) I have always been told that the quality factor is the reciprocal of the fractional bandwidth (calculated by using as reference an attenuation of 3dB with respect to the peak of the impedance/admittance of the circuit considered). Good for series RLC to parallel RLC, because we calculate Q @ resonance. 2) We insert a parallel capacitance and ...


1

Your reasoning is almost correct. In the step where you take the limit of the \${E_b}/{N_0}\$ equation, you're making a mistake when applying L'Hôpital's rule. It should go as follows: $$ \lim_{C/W \to 0^+} \frac{E_b}{N_0} = \lim_{C/W \to 0^+} \frac{2^{C/W} - 1}{C/W} = \lim_{C/W \to 0^+} \ln {(2)} 2^{C/W} = \ln{(2)} $$ Your problem arises when you take ...


0

Ok, so considering the PCS stuff first, they are basically saying that on axis in the horizontal plane of the antenna they the EIRP is ~83W, so +19dBW, allow for the ~9dB of antenna gain and the power they are feeding the thing is right around 10W = +10dBW. In the 5Ghz band they are running 2.5W = +4dBW EIRP, with an antenna gain of ~4.7dB, so they are ...


0

Capacitive reactance is a positive real quantity, Xc = 1/(2 * pi * f * C) The impedance of a capacitor, Zc = 1/(j * 2 * pi * f * C) = -j/(2 * pi * f C) The Capacitive reactance is the magnitude of the impedance. The -j in the impedance formula accounts for the 90 degrees phase lag of voltage with respect to current. The 90 degrees lag in the reactance ...


0

The impedance of a voltage multiplier isn't just the impedance of the capacitors. Because of the multiplication effect, the impedance of the multiplier is a multiple of the capacitor impedance. It can be calculated using the formula for the impedance of a Cockcroft-Walton multiplier with the number of stages set to 1. The impedance of a Cockcroft-Walton ...


1

Assume your IP3 is +20dBm. The 3rd order distortion drops 30dB, per 10dB input power reduction. Thus -10dBm input RF power drops the distortion by 3*(+20dBm - (-10dB)) or by 90dB below the IP3, or to +20- (90) to -70dBm; [ note my Initial Error in computing the distortion level: had gone 90dB below +10dBm and that is wrong; should be 90dB below the IP3, ...


1

in this book you can find Matlab implementations for FDTD. It specifically used for microstrip circuits but it might help you. I hope this helps.


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A CDM324 module is a very simple device. There isn't even a crystal or any other component on the board (see here) which takes care that the 24 GHz frequency is accurate. So the frequency will not be accurate. That means that the module will need to work over a wide frequency range. That also means that any other device within range operating at 24 GHz or a ...


2

For RF transformers/baluns, the inter-winding capacitance can play a big part in reducing common-mode signals between unbalanced input and balanced output. A 1 pF inter-winding capacitance may not seem that much but, at 1 GHz, it's an impedance of 159 ohms. So, if you don't ground the centre-tap your balun won't be working as effectively as it can do ...


1

A typical RF mixer does not multiply the RF signal with the LO (local oscillator) signal. Instead, the signal polarity of the output depends on whether the LO is negative or positive: when the LO is positive the output signal is the same as the input signal (up to some gain) and when the LO is negative, the output is the same as the input but with an ...


2

Stolen from: https://www.edn.com/design/test-and-measurement/4424529/So-you-think-you-understand-transmission-lines- But anyways, that's the impedance of some transmission line that's open on the other end. You can see it decreases in impedance initially, which is the hallmark of a capacitor. Then it sharply decreases and starts going back up, which is a ...


1

It's rated for 2W, it should do 2W unless they note caveats. I'm not even sure why they felt they had to include information about the resistor -- possibly because they needed it for mil-spec-ness. If you're worried, get one and put 10V DC on it. You should see a 200mA current flowing, and the thing shouldn't burn up or otherwise change characteristics ...


2

My bet is it's 2 watt. The 100 ohm resistor is listed, but there is probably more than one (or it would be a 100 ohm load) and the resistor is most likely tied to the case, when they are tied to a heatsink, you can run more power through them if the temperature is kept low. The most important thing would be to follow the temperature rating and keep it below ...


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1) What is Z0? There are no transmission lines here. The figure shows the power divider connected to three transmission lines. The parallel lines drawn extending away from the points marked \$V_1\$, \$V_2\$, and \$V_3\$ are representations of transmission lines. It's unfortunate for introductory material that they put the "Port" labelings at the far ends of ...


4

Your phone has NFC turned on so it is trying to read any NFC devices that might come into range.


3

It depends how accurate a calibration you need. The open standard in the cal kit has some parasitic capacitance. When you use it to calibrate a network analyzer (VNA) you'll specify this (and a couple other parasitics) to the VNA to get the most accurate calibration. It's usually possible to make a short standard that is closer to an ideal short than the ...


0

Using P = V^2/R, at Power = 0.001 watt, and R = 50 ohm, you can compute Vrms = 0.223 volts. Then scale up by 2 * 1.414, to compute Vpeakpeak = 0.632 volts. The +12dBm input is 4 factors of 2X power (3dB) stronger than 0dBm. The +12dBm input is 2 factors of 2X voltage (6.01dB) stronger than 0dBm. Thus we know the +12dBm is 2 * 2 * 0.223 voltRMS.


1

No, the transmitter's data pin isn't interpreted as a voltage, but as a zero or one logical signal....binary data. Similarly, the receiver's data pin is a logic signal. It will swing from ground (logic zero) to Vcc (logic one). You might think that such a transmission system is similar to a UART function. You might actually get it to work this way, but ...


2

These 433 MHz transmitters use OOK modulation (also called ASK), making the transmitted signal look like: So the transmitter is simply switched on (1) or off (0). Most 433 MHz receivers will "filter out" the 433 MHz so we're left with the blue signal. When no signal is received for some time, most receivers will try to decode a signal from the noise they ...


0

Becasue there isn't a max voltage, only a max power. You can put in whatever voltage you want, as long as the power referenced to 50Ω is kept below 12dBm. The requirement forces you to look at the frequency content of your signal and ensure that the total power of that signal does not exceed the rating. So measure or calculate the total power of whatever is ...


0

Given this is the ONLY RF circuit, but needing only 100 foot range, we can trace out the circulating-RF-resonance-path. I think its C4 -- L3 -- Cemitter_base -- C5 and C4 -- L3 -- C2 -- L2||C3 -- C6 but note the series LC to the antenna. With the car's antenna out in free-space. C3+R2 likely are the super-regenerative quenching-oscillation time ...


0

The leads don't change the characteristic impedance on the coaxial cable, they add to the impedance of your test subject. How would you calibrate away the effect from the coaxial cable with your measurement device? For example, with a VNA you typically connect a short, open, and a \$50~\Omega\$ load to the end of your measurement cable and tell your ...


2

The easiest way is to use TDR (Time Domain Reflectometry). If you believe that the coax is 50Ω and you want to verify it, you put 50Ω connectors on either end, connect one end to the TDR and terminate the other end with a 50Ω resistor. If you're right, the TDR display will be perfectly flat. If not, the steps on the display will tell you ...


2

When you connect to a 50 ohm device, you usually make the connection with a co-axial connector. Alligator clips, and the tails of wire leading up to them, will introduce a section of higher impedance. Whether this makes a negligible or a significant difference depends on your operating frequency, and the length of the tails. If you keep them well below 1/...


0

Only the part with the clips will be modified the rest will still be 50 ohms.


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