20

They need not, narrowband FM occupies the same bandwidth as AM, for the same audio bandwidth. Transmitted AM bandwidth is fixed at twice the audio bandwidth. However, wide deviation FM provides signal to noise ratio improvements above that which is possible for AM. It also provides the so-called capture effect, whereby a weak station near to a stronger ...


15

At some point in my life, I used to run the USB business for big semi company. The best result I remember was NEC SATA controller capable of pushing 320Mbps actual data throughput for mass storage, probably current sata drives are capable of this or slightly more. This was using BOT (some mass storage protocol runs on USB). I can give a technical detailed ...


14

There's two things you might call speed: bandwidth and latency. Latency is the duration of time needed for a signal at one node of the network to reach another node of the network. Processing time for the electronics to packetize the signal and place it on the wire often dominates the latency, but the physical medium does also affect it. As far as the ...


14

The bandwidth is the frequency for which the output is down 3 dB. It it a function of the analog portion of your circuit and does not depend on the sample rate nor the Nyquist rate which are basically digital concepts. Thus, in your experiment, the bandwith will be the frequency, fin, for which the output has dropped by 3 dB.


12

A classic layout rule for high-speed op-amps is to remove the power and ground planes beneath the nets connected to the input pins. You'll find this as the first bullet point in the PCB layout section of the datasheet for your op-amp. That means, basically, remove all copper from the plane layers underneath any copper that is connected to pins 3 or 4 of ...


12

The maximum useful datarate is a fraction of the carrier frequency. However, the difference between ZigBee and WiFi has nothing to do with that. 2.4 GHz is so high, that it is not even close to the limiting factor for data rate. Generally, there are two tradeoffs with higher data rate over the same frequency carrier: RF bandwidth and power. A perfect ...


12

That 27kHz is nothing. The LPV511 has a little brother, the LPV521, which has a gain-bandwidth product of 6.2kHz. They're not making it low-bandwidth on purpose. There is no real advantage to the low bandwidth, though it improves stability. The low gain-bandwidth product is a consequence of the low power design. The LPV521 consumes only 350nA. You already ...


11

Only an ideal sine wave with constant amplitude and single frequency that runs for infinitely long in time has extremely narrow bandwidth of a single peak. Having any changes to the ideal sine wave such as changing the amplitude means that since it cannot be represented as a single sine wave any more but a sum of many sine waves of different amplitudes and ...


10

The bandwidth specifications on an oscilloscope are for sine waves only. For example, if your oscilloscope has 100MHz bandwidth that means that it won't have any problem with 100MHz sine waves. 100MHz square waves are a different matter - generally to get a good look at a square waveform you'll need 10x the bandwidth. So, for a 100MHz square wave to show up ...


10

On-Off Keying (OOK) is equivalent to AM modulation of the carrier with the data bit-stream with a 100 % modulation index. As a result the entire spectrum of the bit stream is up-converted to the carrier frequency of the OOK modulator. Additionally, due to AM modulation an image of the data-stream exists below the carrier frequency in addition to the ...


9

The difference between the two formulas arises from the fact that the Nyquist formula uses the number of encoding levels that was explicitly given (16 levels implies 4 bits/baud), while the Shannon formula is the theoretical maximum based on the SNR of the channel (40 dB implies about 6.64 bits/baud). 3000 Hz × 2 baud/cycle × 4 bits/baud = 24000 ...


8

Theoretically a square wave has infinite bandwidth but it still looks reasonably square even if the bandwidth is severely compromised. A square wave is "made from" a series of ever-increasing harmonics. See the picture below to get an understanding: - On the right is a sinewave then as you look down you'll see that it grows into a square wave. If all we had ...


8

If your scope's input amplifier has a frequency response of a first-order RC-filter, you can roughly estimate the bandwidth from the rise time: $$BW ≈ 0.35 / t_R$$ To clarify, bandwidth is defined by the frequency which is attenuated by -3dB, and the rise time corresponds to the input signal going from 10% to 90% of its amplitude. Of course, this only ...


7

I'll try a brief answer first and THEN look at the data sheet and costing :-) It is most likely very low power and/or can run from low voltage. Low speed assists low dynamic power reduction and the lack of need to support high power and speeds allows lower power to be targeted. You may also value the low EMI aspects. OK - lets look at the data sheet ... ...


7

This type of crystal lattice is not meant to be the only source of selectivity in a circuit. At very high frequencies, the parasitic capacitances of the crystal holders and electrodes simply pass everything. It would be more typical for this sort of lattice to be incorporated into an IF chain that also has ordinary LC circuits to provide the required ...


7

A digital signal is composed of a fundamental frequency and an infinite number of odd harmonics. This is how the nice clean and sharp edges of a digital signal is made. As in the image below, you have the fundamental frequency, and as you add more and more odd harmonics, the signal begins to take the shape of an ideal digital waveform. Now you mention that ...


7

Op amps which use dominant pole compensation have a constant gain-bandwidth product. You've cited a 5Hz dominant pole, and the LM301A datasheet gives a typical open loop voltage gain of \$160\text{V}/\text{mV} = 160,000\$. This gives a gain-bandwidth product of \$160,000 \times 5\text{ Hz} = 800,000\text{ Hz}\$. You have configured this op amp with a closed ...


7

There are hundreds of different antennas so for my simple attempt at an answer I'll concentrate on the "dipole" and I'm not going to go into formulas too much. What physical factor affects the bandwidth of an antenna? The impedance presented by the antenna is a major factor. If that impedance changes with frequency, then transmitted power will also ...


7

You should realize that sample rate (1 GS/s) and Bandwidth (70 MHz) are different things!!! They are related in that a certain sample rate dictates the maximum bandwidth of the signals which can be sampled accurately. This is set by the Nyquist frequency The Bandwidth of the oscilloscope is most often limited in the frontend of the oscilloscope. The ...


7

The plots below (taken from here) show why calculating ADC 'bandwidth' from peak sample values may be problematic. The plot on the right shows what happens when the sampling frequency is below the input frequency. The signal has aliased down to a lower frequency inside the Nyquist limit, and there is no way to tell what the actual input frequency was. ...


6

This is a very old topic, but it has no answer yet. This is my attempt: Why are today's implementations not capable of streaming at 53 MB/s? The calculations are nearly fine, but you are forgetting a couple of things in the available number of bytes between frame markers: Each microframe has two thresholds called EOF1 and EOF2. No bus acivity must occur ...


6

There is a simple answer: The bandwidth for the closed-loop gain is determined by the frequency where the LOOP GAIN is 0 dB. In your example circuits the loop gain is not the same - hence, the bandwidth will not be the same. The circuit with the largest loop gain (non-inverter) has the largest bandwidth. Explanation why the Loop Gain (LG) determines ...


6

No, the actual physics and formulas involved to get this absolutely right are not totally easy, but for gaussian response oscilloscopes (quick read about differences for flat response scopes: http://cp.literature.agilent.com/litweb/pdf/5988-8008EN.pdf), we relate things to each other like this: \$R_{meas} = \sqrt{R_{signal}^2 + R_{system}^2 }\$ where \$R_{...


6

As @SpehroPefhany mentioned, we assume that the op amp has a constant gain-bandwidth product, GBWP. That is, \$\text{GBWP}=G \cdot B\$ for any gain G and bandwidth B. From the given information, we can determine that the GBWP for this op amp must be \$96 \frac V V \cdot 8 \text{ kHz} = 768 \text{ kHz}\$. Now that we have the constant GBWP, we can solve for ...


6

The signals from the sensors were tiny, and needed to be amplified. Otherwise they would be lost in the noise in the rest of the system. When "interesting" things happen, neurons fire rapidly. Filter out any signal that isn't firing rapidly. We want an amplitide, but an AC signal has an average amplitude of zero. So make everything positive. Filter out ...


6

It's very hard to say whether your estimation is right without knowing more about the system and the input signal. Looking at the rise and fall times it seems reasonable by eye, but if you want a good estimation of bandwidth, it makes much more sense to use a sinusoid waveform rather than a square one. With the square input your effectively checking the slew ...


5

The 2 comes from the need to avoid aliasing. Note that \$log_2(L)\$ is the sample size, and if we divide the bit rate by sample size we get the sample rate: \$\frac{2B\log_2(L)}{\log_2(L)}= 2B\$. The formula says that bandwidth B needs a sample rate of at least 2B. You're disagreeing and saying that it just needs a B sample rate. But the bandwidth is B, ...


5

Your actual questions don't seem to have anything to do with your first paragraph. There is a theortical limit as to how much information can be carried in a channel of some bandwidth and signal to noise ratio. Look up Nyquist-Shannon for deails. After you decide what information rate you need to transmit and the minimum signal to noise ratio you expect ...


5

There is no one to one relationship between rise time and bandwidth. A slew rate limiter is a non-linear filter, so can't directly be characterized as a low pass filter with some obvious rolloff frequency. Think of it in the time domain, and you can see that a slew rate limit effects signals proportional to amplitude. A 5 Vpp signal limited to 5 V/µ...


5

I don't know what do you mean by "presence/absence" of voltage, but in general, what you describe is exactly how it works: High voltage on the data line is (usually) defined as logical 1 Low voltage is logical 0 The "frequency of the circuit" is just the maximal number of times the transition between voltage levels can happen in a second. Please note ...


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