Impulse Response, Fourier and Laplace Transforms, and Parseval's Relation

Discussion in 'Electronic Design' started by Artist, Sep 7, 2006.

  1. Artist

    Artist Guest

    I need help resolving a contradiction relating the area under an impulse
    response curve of a system and the system's frequency response.

    According to Parseval's Relation the area under the impulse response must
    equal the area under the impulse response's Fourier Transform. Also, this
    Fourier Transform is the frequency response of the system having that
    impulse response. This is a consequence of both having to have the same
    energy. I have thought about this and it makes perfect sense to me. But when
    looking at Bode plots of a single pole low pass filter I see the area under
    the frequency response changing with changes in the time constant of the
    filter. If two such filters having different bandwidths have their outputs
    normalized to the same DC gain, the area under their impulse responses must
    also be the same because when the impulse response of each is convolved with
    a step function the final values they settle on must be the same, the DC
    gain value.

    So what I do not understand is how a Bode Plot of a filter's Laplace
    Transform can show an area under the frequency reponse curve that obviously
    changes with the time constant of the exponential decay of their impulse
    response, yet the area under the Fourier Transform does not change with
    change with time constants but at the same time the Fourier Transform must
    also be the frequency response.

    Are these two different kinds of bandwidths? I got puzzled by this apparent
    contradiction when I started thinking about how the time constant of the
    exponential decay of an impulse response relates to the square root of the
    area under a frequency response curve and resulting system output noise in
    response to white input noise.
    Artist, Sep 7, 2006
    #1
  2. Artist

    Tim Wescott Guest

    Re: Impulse Response, Fourier and Laplace Transforms, and Parseval'sRelation

    Artist wrote:

    > I need help resolving a contradiction relating the area under an impulse
    > response curve of a system and the system's frequency response.
    >
    > According to Parseval's Relation the area under the impulse response must
    > equal the area under the impulse response's Fourier Transform. Also, this
    > Fourier Transform is the frequency response of the system having that
    > impulse response. This is a consequence of both having to have the same
    > energy. I have thought about this and it makes perfect sense to me. But when
    > looking at Bode plots of a single pole low pass filter I see the area under
    > the frequency response changing with changes in the time constant of the
    > filter. If two such filters having different bandwidths have their outputs
    > normalized to the same DC gain, the area under their impulse responses must
    > also be the same because when the impulse response of each is convolved with
    > a step function the final values they settle on must be the same, the DC
    > gain value.
    >
    > So what I do not understand is how a Bode Plot of a filter's Laplace
    > Transform can show an area under the frequency reponse curve that obviously
    > changes with the time constant of the exponential decay of their impulse
    > response, yet the area under the Fourier Transform does not change with
    > change with time constants but at the same time the Fourier Transform must
    > also be the frequency response.
    >
    > Are these two different kinds of bandwidths? I got puzzled by this apparent
    > contradiction when I started thinking about how the time constant of the
    > exponential decay of an impulse response relates to the square root of the
    > area under a frequency response curve and resulting system output noise in
    > response to white input noise.
    >
    >

    Parseval's relation says that the time integral of the _square_ of the
    impulse response (i.e. the total energy) is equal to the frequency
    integral of the _square_ of the frequency response* (i.e. the total energy).

    You have added to your difficulties by trying to look at the area under
    the frequency response in a Bode plot, which is a log-log plot, not a
    linear plot.

    * times a factor of 2 pi that I always have to look up.

    --

    Tim Wescott
    Wescott Design Services
    http://www.wescottdesign.com

    Posting from Google? See http://cfaj.freeshell.org/google/

    "Applied Control Theory for Embedded Systems" came out in April.
    See details at http://www.wescottdesign.com/actfes/actfes.html
    Tim Wescott, Sep 7, 2006
    #2
  3. Re: Impulse Response, Fourier and Laplace Transforms, and Parseval'sRelation

    Artist wrote:
    > I need help resolving a contradiction relating the area under an impulse
    > response curve of a system and the system's frequency response.
    >
    > According to Parseval's Relation the area under the impulse response must
    > equal the area under the impulse response's Fourier Transform. Also, this
    > Fourier Transform is the frequency response of the system having that
    > impulse response. This is a consequence of both having to have the same
    > energy. I have thought about this and it makes perfect sense to me. But when
    > looking at Bode plots of a single pole low pass filter I see the area under
    > the frequency response changing with changes in the time constant of the
    > filter. If two such filters having different bandwidths have their outputs
    > normalized to the same DC gain, the area under their impulse responses must
    > also be the same because when the impulse response of each is convolved with
    > a step function the final values they settle on must be the same, the DC
    > gain value.
    >
    > So what I do not understand is how a Bode Plot of a filter's Laplace
    > Transform can show an area under the frequency reponse curve that obviously
    > changes with the time constant of the exponential decay of their impulse
    > response, yet the area under the Fourier Transform does not change with
    > change with time constants but at the same time the Fourier Transform must
    > also be the frequency response.
    >
    > Are these two different kinds of bandwidths? I got puzzled by this apparent
    > contradiction when I started thinking about how the time constant of the
    > exponential decay of an impulse response relates to the square root of the
    > area under a frequency response curve and resulting system output noise in
    > response to white input noise.


    Beside signal parts being passed through a fourpole,
    there are also reflected signal parts.

    Rene
    --
    Ing.Buero R.Tschaggelar - http://www.ibrtses.com
    & commercial newsgroups - http://www.talkto.net
    Rene Tschaggelar, Sep 7, 2006
    #3
  4. Artist

    kkrish Guest


    > >
    > > So what I do not understand is how a Bode Plot of a filter's Laplace
    > > Transform can show an area under the frequency reponse curve that obviously
    > > changes with the time constant of the exponential decay of their impulse
    > > response, yet the area under the Fourier Transform does not change with
    > > change with time constants but at the same time the Fourier Transform must
    > > also be the frequency response.
    > >


    Laplace transform has an atteneuation
    constant but fourier shows the frequency domain represtation of a
    signal as such.Fourier transform of a signal should give the exact
    time domain representation when inverse fourier transform is taken.
    kkrish, Sep 7, 2006
    #4
  5. Artist wrote:
    > I need help resolving a contradiction relating the area under an impulse
    > response curve of a system and the system's frequency response.
    >
    > According to Parseval's Relation the area under the impulse response must
    > equal the area under the impulse response's Fourier Transform. Also, this
    > Fourier Transform is the frequency response of the system having that
    > impulse response. This is a consequence of both having to have the same
    > energy. I have thought about this and it makes perfect sense to me. But when
    > looking at Bode plots of a single pole low pass filter I see the area under
    > the frequency response changing with changes in the time constant of the
    > filter. If two such filters having different bandwidths have their outputs
    > normalized to the same DC gain, the area under their impulse responses must
    > also be the same because when the impulse response of each is convolved with
    > a step function the final values they settle on must be the same, the DC
    > gain value.


    I think the output of a filter fed by a true step function has infinite
    energy, so you can't really sum it up. You have to use an input, like
    an impulse, that has finite energy.

    {snipped}

    --
    John
    John O'Flaherty, Sep 7, 2006
    #5
  6. Artist

    Artist Guest

    "Tim Wescott" <> wrote in message
    news:...
    > Artist wrote:
    >
    >> I need help resolving a contradiction relating the area under an impulse
    >> response curve of a system and the system's frequency response.
    >>
    >> According to Parseval's Relation the area under the impulse response must
    >> equal the area under the impulse response's Fourier Transform. Also, this
    >> Fourier Transform is the frequency response of the system having that
    >> impulse response. This is a consequence of both having to have the same
    >> energy. I have thought about this and it makes perfect sense to me. But
    >> when looking at Bode plots of a single pole low pass filter I see the
    >> area under the frequency response changing with changes in the time
    >> constant of the filter. If two such filters having different bandwidths
    >> have their outputs normalized to the same DC gain, the area under their
    >> impulse responses must also be the same because when the impulse response
    >> of each is convolved with a step function the final values they settle on
    >> must be the same, the DC gain value.
    >>
    >> So what I do not understand is how a Bode Plot of a filter's Laplace
    >> Transform can show an area under the frequency reponse curve that
    >> obviously changes with the time constant of the exponential decay of
    >> their impulse response, yet the area under the Fourier Transform does not
    >> change with change with time constants but at the same time the Fourier
    >> Transform must also be the frequency response.
    >>
    >> Are these two different kinds of bandwidths? I got puzzled by this
    >> apparent contradiction when I started thinking about how the time
    >> constant of the exponential decay of an impulse response relates to the
    >> square root of the area under a frequency response curve and resulting
    >> system output noise in response to white input noise.

    > Parseval's relation says that the time integral of the _square_ of the
    > impulse response (i.e. the total energy) is equal to the frequency
    > integral of the _square_ of the frequency response* (i.e. the total
    > energy).
    >
    > You have added to your difficulties by trying to look at the area under
    > the frequency response in a Bode plot, which is a log-log plot, not a
    > linear plot.
    >
    > * times a factor of 2 pi that I always have to look up.
    >
    > --
    >
    > Tim Wescott
    > Wescott Design Services
    > http://www.wescottdesign.com
    >
    > Posting from Google? See http://cfaj.freeshell.org/google/
    >
    > "Applied Control Theory for Embedded Systems" came out in April.
    > See details at http://www.wescottdesign.com/actfes/actfes.html


    I am not trying to calculatet the area under a Bode Plot. I use it only to
    make the observation the area under it changes.

    I had not considered the squaring. I understand it now. The filter with the
    higher bandwidth will have an impulse response with more of its area higher
    off the baseline than one with a lower bandwidth. The frequency response
    curve never gets higher off baseline than its DC gain. So with squaring the
    result will be higher energy for the filter with higher bandwidth and more
    area under its frequency response curve.

    Thanks
    Artist, Sep 8, 2006
    #6

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