- #1
red123
- 22
- 0
How do I find Bandwidth from looking at Graph?
Click For Summary
Discussion Overview
The discussion revolves around determining bandwidth from a graph, particularly in the context of signal processing and filter characteristics. Participants explore the relationship between cutoff frequency and bandwidth, specifically for low-pass filters and non-inverting amplifiers.
Discussion Character
- Debate/contested
Main Points Raised
- One participant suggests converting the magnitude scale into dB to find the cutoff frequency at -3 dB, proposing that the bandwidth is 2 times the cutoff frequency.
- Another participant challenges this by stating that the bandwidth is equal to the cutoff frequency in this case, indicating that it spans from 0 Hz to 100 kHz.
- A different participant agrees with the challenge, noting that the circuit in question is a non-inverting amplifier and that its performance resembles a low-pass filter, where the bandwidth equals the upper cutoff frequency.
- One participant acknowledges the previous points but introduces the idea that imaginary components must also be considered, suggesting that the bandwidth could be 2 times the cutoff frequency in cases involving modulation.
- Another participant asserts that for a low-pass filter, the bandwidth is simply the cutoff frequency, contrasting this with band-pass filters where bandwidth extends to both -3 dB frequencies.
Areas of Agreement / Disagreement
Participants express differing views on the definition of bandwidth in relation to cutoff frequency, with no consensus reached on whether bandwidth should be defined as 2 times the cutoff frequency or simply equal to the cutoff frequency.
Contextual Notes
Participants reference the conventional definition of cutoff frequency at -3 dB, indicating that this is a defined standard rather than an absolute rule. There is also mention of the need to consider imaginary components in certain contexts, which remains unresolved.
- #2
Runei
- 193
- 17
You have to convert the magnitude scale into dB
dB = 20*log10(magnitude)
Look at where it is -3 dB. That is your cutoff frequency (fc).
Your bandwidth in this case will be 2*fc. Read more on http://en.wikipedia.org/wiki/Bandwidth_(signal_processing )
EDIT: Note however, that it is only a convention, that the cutoff frequency is -3 dB. It is something we have defined.
dB = 20*log10(magnitude)
Look at where it is -3 dB. That is your cutoff frequency (fc).
Your bandwidth in this case will be 2*fc. Read more on http://en.wikipedia.org/wiki/Bandwidth_(signal_processing )
EDIT: Note however, that it is only a convention, that the cutoff frequency is -3 dB. It is something we have defined.
Last edited by a moderator:
- #3
red123
- 22
- 0
Thanks
- #4
yungman
- 5,741
- 291
I thought BW is fc in this case. In this case, it looks like the -3dB is about 100KHz. The bandwidth is 100KHz as it is from 0Hz to 100KHz. I don't think it is 2fc = 200KHz.
- #5
red123
- 22
- 0
Yes, I think you are right. The circuit for this graph is a non-inverting amplifier made from a 741 op-amp. Op-amp performance is limited at high frequency; it behaves as a low-pass filter.
"In case of a low-pass filter or baseband signal, the bandwidth is equal to its upper cutoff frequency."
http://en.wikipedia.org/wiki/Bandwidth_(signal_processing)
"In case of a low-pass filter or baseband signal, the bandwidth is equal to its upper cutoff frequency."
http://en.wikipedia.org/wiki/Bandwidth_(signal_processing)
- #6
Runei
- 193
- 17
EDIT: I see, the article says it all :) My fault.
You might be right, but I seem to remember that we have to include the imaginary components also.
The frequency response is symmetrical around f = 0, for this case. And using modulation, the signal would be positioned at a frequency f = f0 (where f0 is the carrier wave), with a symmetrical shape like the picture around this frequency. The bandwidth of the signal would then be 2*fc.
You might be right, but I seem to remember that we have to include the imaginary components also.
The frequency response is symmetrical around f = 0, for this case. And using modulation, the signal would be positioned at a frequency f = f0 (where f0 is the carrier wave), with a symmetrical shape like the picture around this frequency. The bandwidth of the signal would then be 2*fc.
- #7
berkeman
Admin
- 69,473
- 24,994
In my experience, for a LPF like that, the bandwidth is just fc. If it were a BP filter, then the bandwidth extends to both -3dB frequencies.
Similar threads
- · Replies 24 ·
- · Replies 7 ·
- · Replies 23 ·
- · Replies 9 ·
- · Replies 2 ·