Instantaneous Frequency

Discussion in 'General Electronics Chat' started by lkgan, Apr 10, 2010.

  1. lkgan

    Thread Starter Member

    Dec 18, 2009
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    Hi everyone,

    I have came across this term, Instantaneous frequency which is the derivatives of instantaneous phase, i.e., ω = dδ/dt. Anyone have idea what instantaneous frequency is? Aren't frequency suppose to be measure across a cycle of the signal? What's the purpose of instataneous frequency in analysis?

    Thanks for answering my question... :)
     
  2. Ghar

    Active Member

    Mar 8, 2010
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    When frequency varies, such as in frequency modulation (FM) radio, it becomes necessary to talk about instantaneous frequency.

    What's the frequency of something like sin(t^2 + t) ?
     
  3. lkgan

    Thread Starter Member

    Dec 18, 2009
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    Well, aren't the definition of frequency suppose to be over some time? In FM radio, even frequency varies, we measure over a range of interval and might be able to calculate the average frequency right? How can we define frequency as instantaneous which happen on the dot?
     
  4. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    We define it on the dot using that definition.

    So if you have cos(wt), the phase is wt and the frequency is w.
    Notice that your equation works in this simple situation:
    \frac{d}{dt}wt = w

    You can calculate average frequency but why ignore instantaneous? You can look at both.
    You actually do need to look at both. Average frequency is a single number while your bandwidth in FM is the range of instantaneous frequencies.

    It's not necessarily over a period of time because calculus lets us get away with it.
    There's an instantaneous radius of a path even if you're not moving in a circle for example.
    http://en.wikipedia.org/wiki/Curvature#One_dimension_in_two_dimensions:_Curvature_of_plane_curves

    It's part of the concept of a limit. Generally yes, a frequency is the reciprocal of the period of repetition. Looking at very short time frame however, you can still get a sense of how fast the signal is.
    The slope of sin(wt) is w*cos(wt) for example, and the maximum slope is simply w. If you see a slope of w you know that the frequency can't be greater than w.
    Just like that curvature example, you can still fit a circle there to find a radius even without having a real circle.
     
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  5. lkgan

    Thread Starter Member

    Dec 18, 2009
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    Marvellous, thanks for giving such a good explanation. But as for the given curvature example, we can't find every instantaneous point of the path right. We must at least find the "point P" which is tangent to the curve before we can measure the radius right?

    I do understand that the maximum slope is simple w, but what's the purpose of knowing the slope of w and know that the frequency can't be greater than w?
     
  6. Ghar

    Active Member

    Mar 8, 2010
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    The maximum slope idea was just an example of getting a bigger picture from a small piece of information.
    My attempt at helping explain calculus I guess...

    Another attempt to explain it is that phase is the real thing that matters.
    You want to know how long it takes to go through 2pi radians and that determines your period and hence frequency.

    If that rate is constant, you just have the phase = wt situation.
    Frequency is then w/2pi.

    If the rate isn't constant though, a good guess is using the slope of your phase.
    At a particular moment in time the slope lets you find out when you'd get to 2pi if it were to stay constant. This gives you instantaneous frequency which may or may not be equal to the average.

    Here's just an arbitrary graph of phase vs time. Period is how long it takes to reach 2pi because that's when a sinusoid will repeat. Remember that phase is the x in sin(x) for example.

    phase.png

    Initially in segment 1 it looks like you'll hit 2pi at about 6 seconds. That means your period is 6 seconds and your frequency is 0.17 Hz.
    However, you see that the slope then changes in segment 2 (stay with the solid line). You'll actually hit 2pi at about 8.5 seconds.
    Whatever function this is actually took 8.5 seconds to repeat (go from 0 to 2pi), so the frequency of the overall waveform, if it actually repeated, would be 1/8.5 = 0.12 Hz.

    That's average frequency over your first cycle.
    Just to confirm, let's actually take the average, where average is the sum of duration*value divided by total duration.
    First slope is 1 rad/s, while the second is 0.5 rad/s.
    Taking the average... (4s * 1 rad/s + 4.5s * 0.5 rad/s) / 8.5s = 0.74 rad/s, or 0.74/2pi = 0.12 Hz.

    The instantaneous frequency in segment 2 isn't 0.12 Hz, it's slower.
    The instantaneous frequency assumes that segment goes on forever, so we need to actually look how long it would take to do a full 2pi at that given rate.
    You can see that the slope of segment 2 takes about 12 seconds to go through 2pi, if you look at time ~9 to 20. That means the instantaneous frequency of segment 2 is 1/12 = 0.8 Hz.

    Calculus makes the segments infinitely small, via the derivative, letting you have an instantaneous frequency for any instant in time.

    And yes, with the curvature you need to know your path to find the radius, just like here you need know the phase to find your frequency.
    You need to know the phase (or path) around the area where you want your instantaneous value, yes. Having a single point isn't good enough.
     
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