1
\$\begingroup\$

This question already has an answer here:

I am looking at this page and coming from little EE knowledge, it left me confused. First of all, it defines bandwidth as the following

Bandwidth is defined as the measure of a circuit or transmission channel to pass a signal without significant attenuation over a range of frequencies. Bandwidth is measured between the lower and upper frequency points where the signal amplitude falls to -3 dB below the pass-band frequency. The -3 dB points are referred to as the half-power points.

Can someone explain that more simply? Also why is -3 dB considered half-power points. This definition kinda sounds like those coaxial cable splitters you can get for your home that have a frequency range printed on them. Is that what this definition is applying to?

Then I have a question about this part:

If you input a 1 V, 100 MHz sine wave into high-speed digitizer with a bandwidth of 100 MHz, the signal will be attenuated by the digitizer’s analog input path and the sampled waveform will have amplitude of approximately 0.7 V. The value of ~0.7 V can be calculated by using the following equation: -3 dB = 20 LOG (Vppout / Vppin)

Why would inputting 1 volt peak to peak 100mhz sine wave into a digitizer output 0.7 volts peak to peak. I have heard Nyquist–Shannon sampling theorem said, but don't really understand it from a working standpoint (I assume it is applying here). If a signal is coming up at 100mhz and you are taking 100mhz samples, wouldn't that be enough to duplicate the 1 volt peak to peak signal? Also, how is the -3 dB applying to that formula? I don't see how it is involved.

\$\endgroup\$

marked as duplicate by Daniel Grillo, Fizz, Dave Tweed Oct 15 '15 at 1:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

0
\$\begingroup\$

The statement is that the digitizer has 100MHz bandwidth (notice the upper case M).

This implies that at 100MHz the amplitude response is 3dB down from the low frequency response at that frequency. It should really have been more explicit but that is common usage of "Bandwidth". You can interpret the system as a low pass filter followed by a perfect sampler.

If the response is 3dB down the amplitude response is 0.7 of the low frequency value; i.e. ~0.7V

\$\endgroup\$
  • \$\begingroup\$ I see, so that is more like a cut off point that says past this point, the voltage is going to drop off essentially filtering it out. So it doesn't really have to do with analog to digital conversion? \$\endgroup\$ – Justin Oct 6 '15 at 17:52
0
\$\begingroup\$

I will answer the half power point (I need a real keyboard for the rest).

The decibel (dB) is defined as 10log(10) P1/P2.

At 50% power, that yields 10 log(10) 0.5 which is -3 (to a very close approximation).

So the half power points are at -3dB.

Why there? By convention, signals with this power level or greater are considered to be significant.

You will often see this listed as the 'Full power bandwidth' and especially for ADC and DAC devices, the 0.1dB bandwidth is listed for newer devices which helps give a realistic idea of the useful response in precision applications.

The voltage ratio of 0.7 is because voltage ratios in dB are defined as 20log(10) V1/V2 (with the often overlooked fact that this is only true if R2 = R2.

The derivation:

P1 = V1(squared)/R1. P2 = V2(squared)/R2

So P1/P2 = V1(squared)/R2 * R1/V2(squared) (I have left out a couple of operations). If, and only if, R1 = R2, then this reduces to (for dB) 10log(10) (V1/V2)squared, which ultimately yields 20log(10) V1/V2.

HTH

\$\endgroup\$

Not the answer you're looking for? Browse other questions tagged or ask your own question.