I've seen some information on this topic around, but I don't quite understand it.

I have a time domain signal. I understand that if I want to time shift this signal, I can do so by multiplying its Fourier transform by $\exp(-j\omega\delta t)$, where $\delta t$ is the time delay.

However, what if I want to introduce a phase shift in the time domain signal? What would I multiply its Fourier transform by? Or do I have to do this transformation in the time domain?

For context, I am trying to apply both a time delay and phase difference to a time series to correlate with another one, given that the detectors that produced the data are a distance apart and have different orientation towards the source of the signal.

  • 2
    $\begingroup$ What is the difference between your time shift and your phase shift? $\endgroup$
    – hotpaw2
    Commented Mar 5, 2020 at 17:25

2 Answers 2


Let’s be clear on what we will refer to as time delay and phase shift. Due to the common association of individual frequencies as sinusoids many confuse delay and phase shift as being equivalent. However a phase shift in time is simply multiplying a time sample by $e^{j\phi}$ while a time delay is displacing the sample in time such as done with $x(t-\tau)$.

The same thing occurs in frequency, where a phase shift of a particular carrier is done by multiplying that carrier by $e^{j\phi}$, while a frequency displacement is done by moving the carrier to a new frequency such as $X(\omega-\omega_c)$.

As you showed, a delay in time is a linear phase versus frequency in the frequency domain.

Similarly due to the bidirectional properties of the Fourier Transform, a displacement in the frequency domain would be a linear phase versus time in the time domain, specifically $e^{j\omega t}$. Multiplying a time domain signal by $e^{j\omega t}$ with $\omega t$ as a linear phase ramp, results in a frequency translation, while multiplying a time domain signal by $e^{j\phi}$ with $\phi$ as a constant, results in a static phase shift (or if it makes it clearer: a phase displacement, or a phase rotation). This phase displacement is not directly a time displacement.

Further, any multiplication in the frequency domain would be equivalent to a convolution in the time domain (and vice versa). If you want to a frequency domain operation that results in a phase change in the time domain (to be equivalent to a multiplication of the time domain samples)- this would need to be done by a convolution if in the frequency domain.

To provide further details we need to understand more precisely what is meant or intended by a “phase shift in the time domain”, specifically if this means all time samples are rotated by the same phase (which is done by convolving the frequency domain function with a complex impulse at f=0 with magnitude = 1 and associated phase such as $\delta(\omega)e^{j\phi}$ or if the time samples have a linearly increasing or decreasing phase versus time (which is done by convolving the frequency domain function with an impulse at a frequency offset $\delta(\omega-\omega_c$).


Liking the 1 answer above. Let's note that convolution by an impulse, for a discrete set of frequency samples, is the same as multiplying all and every one of the frequency samples by this impulse, which is a constant complex number of magnitude 1 with and angle equal to the phase shift intended.


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