New answers tagged frequency
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Can FEMM be used to calculate real power of a PMSM at 50Hz frequency?
We're talking about real power and if this is true: -
Is the applied voltage a sinewave? – Andy aka
yes Mr.@Andyaka – Mohammed Siddique
And, apparently it is,
Then, to calculate real power you only need to do that at the fundamental frequency. Any currents that are at harmonics of the ...
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From Table 1 - Specifications, the maximum, minimum, and typical SPI frequencies are given. I say typical, as that's what they have used for characterisation in the datasheet so it will be guaranteed to work as specified under those conditions (not accounting for external factors, such as interference etc).
The ADE7758 does not generate an internal SPI ...
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On SPI, the clock is provided from the master to the slave, thus the slave does not have a "frequency" set, since the master provide the clock speed through the SCL(SCK) Line.
The slave will have a frequency range to which it will work, the master has to provide a clock within this range.
You can even change the frequency during communication.
You ...
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SPI Master will drive the SPI Slave clock line, there is no way of having those frequencies different.
But since the slave has its own limits, the master should not go faster than the slave is capable of supporting.
The maximum frequency for ADC SPI interface (SCLK) is shown in Table 2. See the allowed high and low pulse minimum widths. Both are 50 ns, that ...
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After some helpful clarification, here's a more complete answer.
First, to the direction question: how is the light shifted in a heterodyne detection system for a laser Doppler interferometer?" They typically use a Bragg cell (AOM, or acoustic optical modulator). A Bragg cell basically uses phase modulation as a way to cause a tiny shift in the ...
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Any method that provides a non-linearity or modulation function of the incoming optical signal by another (usually RF signal) can produce mixing products (heterodyne).
Mixing can occur between two optical signals at the receiver diode to produce the difference signal, however although this can be used to provide synchronous optical detection it is not of use ...
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There are several questions here that need some explainations to piece apart. I'm not actually totally sure what your are asking or trying to do (using heterodyning in optics isn't really used for communications, but it's used in optical metrology for things like autocorrelators or specific types of long wave infrared spectrometers) so here are some points:
...
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You wrote
My understanding is that heterodyning is achieved by shifting the measurement light frequency with respect to the reference light frequency by some carrier frequency.
I guess you have imagined the next frequency shifting scenario:
It's taken from systems which operate at microwaves or lower frequencies. It could in theory also work with light or ...
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PN junctions designed for higher current will have higher capacitance , which at RF frequencies makes the bipolar characteristics weaker with rising frequency. It also lowers the AC impedance at Vdc from junction capacitance and current is limited by the driver impedance.
Thus low junction capacitance is a desirable feature for RF diodes. Specs range for C ...
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When you combine two lights together, you add their intensities. Combined intensity is \$E_\text{s} + E_{\text{LO}}\$ and its power is proportional to square of it $$(E_\text{s} + E_{\text{LO}})^2 = E_\text{s}^2 + 2E_\text{S} E_{\text{LO}} + E_{\text{LO}}^2.$$
The \$2E_\text{S} E_{\text{LO}}\$ is important because multiplying sine-waves results in adding/...
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Given
Phase approaches \$\Phi\$ at high frequencies (or 0 frequency);
And a transfer function which you can rewrite in the form
$$
H(s) = \frac{N(s)}{D(s)} = K \cdot \frac{\prod_{i=1}^{n} (s-z_i)}{\prod_{j=1}^{m} (s-p_j)}
$$
Finding the relation between angle and K
At high frequencies, \$s \rightarrow \jmath\infty\$ ans so \$s-c_i \approx s\$. So,
$$
\begin{...
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It all depends on the numerator phase angle. This is either \$\small\phi =atan\:(\frac{-\omega}{\omega_{1}})=-90^o\$, or \$\small \phi =atan\:(\frac{\omega}{-\omega_{1}})=90^o\$, as \$\small\omega\rightarrow \infty \$. The denominator phase angle will always be \$\small -270^o\$, as \$\small\omega\rightarrow \infty \$.
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Negative frequencies are not intuitive. They appear in Fourier transforms because those transforms present signals as sums of complex exponentials exp(j(2Pi)ft).
Presenting signals as sums of usual sines would be perfectly possible and they would not need negative frequencies, but algebraic manipulations and formula derivations would become much more ...
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It's not a dumb question. In the case of your graph, the frequency is negative with respect to the centre frequency. So if centre frequency is 10MHz, and we talk about the 3dB points being +/-1MHz, they are 9 and 11MHz.
If you want to put a 100MHz square wave through an opamp, you will indeed need a bandwidth much more than 100MHz. Square waves have odd ...
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Could be your neighbors are sucking down available bandwidth, not leaving enough space for Bluetooth to do its hopping thing.
Use 5 GHz for your computer if you can. This will free up space on 2.4 for Bluetooth.
There’s a tool from Metageek called inSSIDer that will allow you to see what devices are sharing the Wi-Fi space with you.
Other than that, no. ...
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Is it true that frequency deviation over 2*pi is the phase deviation of an FM/PM signal?
No. The phase deviation is the integral with respect to time of the frequency deviation. Or, put another way, the frequency deviation is the derivative of the phase deviation with respect to time.
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Understood you have simple waveform that has Umax to Umin = 5ms
Period of any simple wave form is MAX to MAX or MIN TO MIN or zero crossing going up to zero crossing going up or zero crossing going down to zero crossing going down.
so it means if it is simple that Umax to Umin = 5ms also Umin to Umax should be = 5ms
If period is Umax to Umax that means T is ...
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You have been given the time difference between the maximum and minimum value of the signal. Have a look below:
Can you see where you are going wrong?
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The 'coffee machine' part of the rating is simply a resistive heater. This will draw power according to the square of the voltage. You can see this as the range of power drawn is 1090/915 = 1.2ish, and the range of input voltage is 240/220 = 1.1ish. Frequency does not affect this at all.
If you set your inverter to 200 V output, you would expect the heater ...
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Power is Volts multiplied by Amps. Since the load will most likely be resistive heaters, the voltage also will affect the power consumed. The frequency should have little effect. Also note there is some tolerance, so given a specific voltage the manufacturer is only saying the power will be with that range.
I’d suggest you get a bigger inverter.
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They are describing the equivalent series resistance of crystals, it's just a property of the material design for relatively low frequency crystals.
The fundamental frequency is commonly the series resonance (since that is the impedance minimum and it stays relatively constant as you add load capacitance), but it's probably better to explicitly specify ...
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You’ll have to derate Imax and Ipp ripple significantly @ 2MHz due to core loss.
This means L may reduce 10% for a 50’C rise.
*updated with new info on SRF
L won’t change at all unless you overheat it or get within the Q BW effects at resonance. The higher the Q factor, the closer you may operate near resonance but core loss rises with f and impedance ...
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You typically would like to be operating at <10x below SRF. When you're at <10x SRF, the inductance is pretty much the nominal inductance. As you approach the SRF, inductance increases - but you'll have to measure what it is. It might be OK, it might not.
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With the inductance specified at 100KHz, you're likely going to operating this well outside its design intent at 2.2MHZ. You generally want to be well away from the SRF.
You should pick a different inductor technology than this one if you intend to run it at 2.2MHZ.
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