Rise-/Fall-time of a band pass filter

Discussion in 'General Electronics Chat' started by Beduin, May 21, 2009.

  1. Beduin

    Thread Starter Member

    May 20, 2009

    I have made a narrow bandpass filter (only intersted in one freq. component). I would like to read/calculate rise-/fall-time of it.

    Reading/calculating rise-/fall-time on dc signals, I can do by the help of simulations in PSpice. I read the Vdc, @ 10% and 90%and read the time axis, and the difference between these two values on the time axis is the rise-/fall-time.

    But with ac signals? I tried with Vrms, but am not sure?

    Any input will be great :)

    Am I making sense here?
    Last edited: May 21, 2009
  2. mik3

    Senior Member

    Feb 4, 2008
    The rise and fall time are the same for all signals.

    If you have a sine wave its maximum rate of change (slope when it crosses zero) has not to be greater than the slope of the rising/falling voltage.
  3. Beduin

    Thread Starter Member

    May 20, 2009

    am sorry mik3, but that was too abstract for me :confused:

    The rise/fall time is of course the same (10%-90%) for all signals. But is it the same way to calculate or read om graph?

    I have a low pass filter with an input signal coming from a half wave rectifier. I can run a time domain simulation and read the DC's 10%-90% values (x1-x2), and subtract x1 from x2. The difference between these two values is my rise/fall time.
    Now, I run time domain simulation on my banspass filter, and I get nice wave of my fc. Obviously I cant do the same as I did with the low pass filter? :confused:
    In the low pass I have one cut off freq., for me it mean one place to consider rise/fall time. In a band pass there are cut off freq. :confused:

    where is the slope of the rising/falling voltage? :confused:

    Again thanks mik3, and excuse my noobism
    Last edited: May 21, 2009
  4. studiot

    AAC Fanatic!

    Nov 9, 2007
    The slope of sinθ = cosθ so when θ = zero (at origin or start of signal) slope = cos0 = 1.

    Now of course the slope also = Δy/Δx i.e. 1:1 or 45°

    So to reproduce a sine wave the rise time slope must be greater than 45°

    Does this help?
  5. AdrianN

    Active Member

    Apr 27, 2009
    I am not sure what you mean by this. I think we do not have the same definitions here.

    Risetime is the time it takes a signal to rise from 10% to 90% at the output of a circuit (your filter for example) when there is a sudden change at the input of the circuit.

    Having said that, when there is a signal step at the filter input (example, sudden change from 0 to 5V in a very short time), the output will rise, and the measurement between 10% and 90% at the filter output will give you the risetime of the circuit. Maybe this is the meaning of

    You switched on a DC signal at the circuit input and you measured the risetime at the circuit output.

    With AC signals is the same. A square wave at the filter input will basically be a series of steps, so the risetime is measured the same as when you switch on and off DC signals. A sine wave has a softer risetime. In fact, according to Fourier, the sine wave is a square wave with all the harmonics removed, so you measure the risetime at the circuit output the same way. If the sinewave is in the circuit bandwidth, it will go through unattenuated, so the risetime of the sinewave will be the same as in the input. If the sinewave is not in bandwidth, it will be attenuated. If the sinewave goes through a slow circuit that cannot keep pace with the change rate of the sinewave (example, an opamp with a low slewrate) the sinewave will be distorted and the risetime of the sinewave will be longer. How much longer? Risetimes add with the root-square of the sum of squares, between the signal risetime and the circuit risetime.
  6. Beduin

    Thread Starter Member

    May 20, 2009
    Yes, I can digest that! :D

    I think I will look into its step response.

    Thanks, much appreciated :)

    -We should have a "karma" function, a +1 for helpful tips :)