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First order filter help

Discussion in 'Electronics Homework Help' started by cps13, Oct 24, 2021.

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  1. cps13

    cps13

    54
    1
    Feb 25, 2013
    Hi,

    I have the following question:

    FIGURE 2 shows the response of a first-order filter to a step input. Estimate the half-power cut-off frequency of the filter and sketch its frequency and phase response.

    My attempt at the answer is:

    The half-power cut-off frequency is when the voltage is at 0.707 of maximum where the output is at -3dB. In this example, the output is a maximum of 1V and the output takes ~2.25us to reach 0.707 of this point. Therefore, the estimated cut-off frequency is:

    2.25/10^-6 = 2.25MHz

    Can somebody tell me if this is correct? Thanks.

    Figure 2 is:

    fig 2 pag 4 balance.png

    [​IMG]

    Moderators note : uploaded picture with better white balance for better readability.
     
    Last edited by a moderator: Oct 31, 2021
  2. Harald Kapp

    Harald Kapp Moderator Moderator

    12,456
    2,985
    Nov 17, 2011
    Your answer can't be correct. Check the units in your equation:
    To the left you take the 2.25 (without adding the µs, which is also an omission), so the left side of the equation is in units of µs.
    To the right you have a result in units of MHz which is 1/µs.
    But in an equation both sides need to be equal, values as well as units. The mismatch shows that your equation can't be right.
    See e.g. here for an explanation of the relation between bandwidth and rise time.
     
  3. cps13

    cps13

    54
    1
    Feb 25, 2013
    Hi - thanks for the reply but I don't follow. The reason the 2.25 is divided by 10^-6 is so that it is in uS and balanced to the MHz side.
     
  4. Harald Kapp

    Harald Kapp Moderator Moderator

    12,456
    2,985
    Nov 17, 2011
    2.25/10^-6 = 2.25 × 106
    no units at all, no µs, no MHz.
    I assumed, maybe wrongly, that the 2.25 in your equation are the 2.25 µs from the point 2.25µs when the output reaches 0.707. In that case the equation would be 2.25 µs / 10-6 = 2.25 s, still not a frequency but a time.
    MHz = 1/µs, not µs. That is where your equation fails. Read the text in the link I gave, it explains how to proceed.
     
    Martaine2005 likes this.
  5. cps13

    cps13

    54
    1
    Feb 25, 2013
    Looking at the section of your link called "What is an example of 3 dB bandwidth estimated from rise time?"

    I believe that if I calculate my rise time, which is the time to go from a magnitude of 0.1 to 0.9, then I can relate this to my corner frequency (f3db). My rise time I calculate as:

    8us - 1us = 7us

    This is because the step change starts at 1us on the graph.

    Therefore, using the formula from the material you shared:

    Tr = 0.35/3fdb
    7us = 0.35/3fdb
    therefore

    3fdb = 0.35 / 2.75us = 127,272Hz or 127kHz

    Is this the correct way to calculate?

    Thanks
     
  6. Harald Kapp

    Harald Kapp Moderator Moderator

    12,456
    2,985
    Nov 17, 2011
    Your idea is basically correct, but how do you come from
    7us = 0.35/3fdb
    to
    3fdb = 0.35 / 2.75us
    Where do you take the 2.75 µs from when in the first equation you have 7µs?
     
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