# Time-Scaling vs Frequency Shift

1. Apr 27, 2014

### JustStudying

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 the name seems to imply changing the frequency of the signal - but isn't that time-scaling?!

thanks again!

2. Apr 27, 2014

### jbunniii

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