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Time/Frequency Domain Contraction/Expansion

  1. Nov 16, 2011 #1
    Hey guys,
    So I'm trying to intuitively understand the conclusion that a contraction in one domain leads to an expansion in the other domain (and vice-versa). Mathematically, I can see how this would happen, e.g., a bandpass filter with -pi/2 < w < pi/2 would result in a narrower time domain signal than a bandpass filter covering -pi/4 < w < pi/4 because the fourier transform will result in a sinc function with a higher frequency, thereby tapering off quicker; but I can't grasp why. One way I've tried thinking of it is:
    If I have a small time-domain signal, the frequency domain will be wide because more frequencies can 'fit' or 'match' the time-signal; whereas if the time-domain signal were long, fewer frequencies would be able to 'fit' or 'match' the signal. Does this make sense? Is it a valid way of thinking about it?

    Thanks for your help,
  2. jcsd
  3. Nov 16, 2011 #2


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    Hi and welcome.
    Any signal can be described either in the time domain or in the frequency domain. This is true with or without the Fourier Transform but the FT helps to quantify the connection between the two domains. Hopping from one to another domain allows you to pick an appropriate one for a particular process that you might want to perform on a signal (e.g. filtering or sampling).
    It's a lot easier if you are prepared to 'go along' with the Maths and not try to kid yourself that a totally arm-waving level of description is any more valid.
    Personally, I like to look at the Fourier (and inverse) transform in the following, slightly sloppy way (my excuse is that I'm basically and Engineer and not a Mathmatician). The Fourier transform shows what you get if you correlate the time varying signal v(t) with a whole range of continuous sinusoidal signals. It takes the signal and integrates the amount that the signal correlates with / matches every sinusoidal signal with frequency from zero to ∞, over all time. This corresponds to the frequency spectrum of the signal. If the signal has no rapid changes (it must extend over a long time) then there are no high frequency components; the signal will not correlate with any high frequency sinusoids.
    Your use of the terms "narrow" and "wide" might be better replaced with 'rapidly varying' and 'slowly varying'. A 'long square wave' still has high frequency components, for instance.

    otoh, if the signal is of short duration and /or has rapid changes, then it will correlate with a much wider range of frequencies and its signal spectrum will be wider.

    When I first came across the Fourier Transform I just had to sit down and fit it to what I already knew about cross correlation between various functions. In my limited view of the Maths, it makes sense if you realise that only a continuous function at the same frequency as another frequency has a non-zero correlation over all time. So the Fourier transform 'extracts' the signal frequencies that actually exist and their amplitude in a function by this smart method.
  4. Nov 16, 2011 #3
    I see, so it's correct to think that: for long (time domain) inputs (with no high frequency components), the frequency spectrum only consists of low frequency sinusoids because those are the only components that 'match' with the signal.

    And as another example, a narrow square wave (t-domain) would result in more higher frequency components than a long square wave because it is changing faster. Is this correct?

    thanks for your help sc
  5. Nov 17, 2011 #4


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    Do not confuse the length (in time) of a square wave and the rise and fall times of the edges. If you are dealing with a repeating square wave then there will be a fundamental frequency with many harmonics. The extent of the harmonics will be governed by the shape of the edges and not the width.

    If you are just talking of a single 'square(ish)' pulse then the extent of the spectrum will just depend on the shape of the edges and not the length of the pulse.
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