Time-Scaling vs Frequency Shift

  1. I'm learning about properties of fourier transforms at the moment, and I've come across 2 terms "time-scaling" and frequency shift.

    A quick google search of time-scaling suggests that it is when I make the period of my signal longer (i.e a sine wave would take longer to finish 1 cycle)

    but no results seem to come up for frequency shift
    - but the name seems to imply changing the frequency of the signal - but isn't that time-scaling?!

    thanks again!
     
  2. jcsd
  3. jbunniii

    jbunniii 3,363
    Science Advisor
    Homework Helper
    Gold Member

    A frequency shift means that the frequency components of the function have all been "shifted" (increased or decreased) by the same offset. So for example, if ##f(x) = \sin(\omega_1 x) + \sin(\omega_2 x)##, then if we add an offset ##\omega_0## to both ##\omega_1## and ##\omega_2##, we obtain a frequency-shifted version: ##g(x) = \sin((\omega_1 + \omega_0)x) + \sin((\omega_2 + \omega_0)x)##.

    In general, this is NOT equivalent to time-scaling. However, in the special case of a function with a single frequency component, the two notions are equivalent. For example, if ##f(x) = \sin(\omega x)## and we shift the frequency by an offset of ##\omega_0##, the result is ##g(x) = \sin((\omega + \omega_0) x)##. If we define
    $$c = \frac{\omega + \omega_0}{\omega} = 1 + \frac{\omega_0}{\omega}$$
    then we may equivalently write ##g(x) = \sin(\omega cx)##, which is just ##f(cx)##, i.e., a time scaled version of ##f##.

    But as you can see from the example with two frequency components in the first paragraph, the two notions are not generally equivalent. If we time scale ##f(x) = \sin(\omega_1 x) + \sin(\omega_2 x)## to obtain ##f(cx) = \sin(\omega_1 cx) + \sin(\omega_2 cx)##, then we still have two frequency components, but the frequencies are now separated by ##c(\omega_2 - \omega_1)## whereas they were originally separated by ##\omega_2 - \omega_1##. Compare this with the frequency shifted version, where the separation remains the same: ##(\omega_2 + \omega_0) - (\omega_1 + \omega_0) = \omega_2 - \omega_1##.
     
    Last edited: Apr 27, 2014
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?

0
Draft saved Draft deleted