22

This is like asking "how bright is a red light?". It's as bright as it is. You can make a bright red light or a dim red light. As another answer points out, the energy per photon of an electromagnetic signal depends on the frequency of the signal. But you can make a brighter or dimmer (higher or lower power) source at any frequency by emitting more or fewer ...


4

The problem is that you statement: "The output signal will be like this:" is rather wrong. What you have shown there is the output of a rectifier circuit. I don't know where that came from but it is NOT the output of the given circuit. You get that output waveform if you replace the R with a diode, but in that case you no longer have a Linear Circuit.


3

The correct estimate is a sum tolerance stackup of all variables for the crystal. (30+50+2 = 82ppm) The 1st @ room temp is the just the initial tolerance of size of the crystal to the designed centre frequency. The 2nd is the angle of the cut of the XTAL in minutes (1/60th) of a degree that only affects the 3rd order slope at room temp, so both are ...


3

Because the voltages are in rms. Remember that the rms voltage is the equivalent DC voltage source that delivers the same amount of power to a resistive load. The rms is given by $$v_\mathrm{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} v^2 \ \mathrm{d}t}.$$ In our case \$v\$ is a sum of pure sine waves. Hence it is easier to work in the frequency domain, if you ...


2

Yes, the book is playing a bit fast and loose with conventions -- but because they're conventions and not hard-and-fast rules, you need to just roll with it and try to understand the author's intent. That expression is calculated in the Laplace or Fourier domain, to find the response of the circuit to sinusoidal signals in steady-state. So by that measure, ...


2

I don't think that's possible. Do you know what the aux port is there for? I assume it's there to give feedback to blind people using the ATM. If that's the case, the aux port probably is output only and hacking the atm this way would be pretty much like hacking the atm via its display using a video projector.


2

Yes, both are designed for so-called "universal mains" Check both labels: INPUT: 100-240V~ 50-60Hz So, it is suitable for 230 V~ 50 Hz (Australia) as well as for 120 V~ 60 Hz (Canada) as both fall within above specified range.


1

Crystal tolerances are specified to be cumulative, so +/- 30ppm initial tolerance is at +25 °C only, and from that frequency it may change over the whole temperature range the max amount of +/- 50ppm, so deviation is +/- 80ppm in total. You must also notice that depending on the crystal type the maximum error may not be at the temperature extremes but ...


1

To a first order of approximation, all the deviations add. So at the end of the first year, they're saying you can expect \$\pm\$82ppm. It's a bit more complicated than that with quartz crystals. First because the specifications are only for the crystal in a super-precision oscillator circuit, and second because in general an AT-cut crystal that's ...


1

Get some log-log graph paper and draw an X axis up to 1 MHz Draw a Y axis that covers the region from unity gain to 100 or 1000 Draw a line from (1 MHz, 1) with negative slope as per purple below draw a line a horizontal line at gain 20 (blue) until it intersects the purple line Project that intersect down to the X axis Read the frequency on the X axis.


1

What would happen to the output of a very, very small bandwidth band pass filter when a pure carrier of the matching frequency is suddenly applied at its input. Would the output of the band pass filter instantly spring into action and produce a signal at the carrier frequency or, would you see a very, very sluggish carrier build up in amplitude over several ...


1

You mustn't forget two things: The convolution property of the Fourier transform (i.e. multiplication in time domain is equivalent to convolution in frequency domain) the fact that you're multiplying your carrier with your data signal So, your signal is $$s(t) = c(t) \cdot d(t)$$ where \$c\$ is the carrier (typically, a cosine or a complex sinusoid), ...


1

The rise time of Carrier Amplitude spreads BW just as the rise time of the signal determines the -3dB BW = 0.35/ t for t= 10 to 90% rise time. So the BW of the signal translates to both sides of the carrier. (unless single sideband carrier is suppressed by design to conserve BW.)


1

The RMS signal level for Gaussian white noise is measured in units per square root of bandwidth. And RMS has the same formula as when calculating the statistical standard deviation of a series of points. So, for instance; For each point, calculate the distance from the mean value (call it 0 volts), square the distance. Repeat for all other points and sum ...


1

I assume the 180 Hz frequency you got that from the nameplate and not any dominant harmonic that you measured while you were tying to find out the frequency of the machine or you were not running the generator at overspeed. Then your generator is designed to operate at a relatively lower amount of flux density. So it is possible that your machine is has a ...


1

$$ \text{Frequency} = \frac{1} { \text{Clock time} } $$ or $$ \text{Clock time} = \frac{1} { \text{Frequency} } $$ $$ \text{Clock time} = \frac{1} { \text{Frequency} } = \frac{1} { 100 \text{ MHz} } $$ $$ = \frac{1}{ (1 \cdot 10^2 \cdot 10^6) \text{ Hz} } = \frac{1}{ (1 \cdot 10^8) \text{ Hz} } = 1 \cdot 10^{-8} \text{ s} = 1 \cdot 10^1 \cdot 10^{-9} s = ...


1

The transfer function of the shown circuit is equal to: $$\mathcal{H}\left(\text{s}\right):=\frac{\text{V}_\text{out}\left(\text{s}\right)}{\text{V}_\text{in}\left(\text{s}\right)}=\frac{\frac{1}{\text{sC}}}{\frac{1}{\text{sC}}+\text{R}}=\frac{1}{1+\text{sCR}}\tag1$$ Now, when: $$\text{v}_\text{in}\left(t\right)=\hat{\text{u}}\sin\left(\omega t+\varphi\...


1

Modulation is not typically done at the carrier frequency, but rather at a lower frequency and upconverted. In the era of digital modulation, it's not even uncommon for the key processing to be done in the context if IQ signals centered on a frequency if 0 Hz. Upconversion does require a local oscillator, but this is a sine wave rather than square, and ...


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