The Relation between Time Shift and Phase

We now study the time shift of a waveform and how it’s related to the phase.

A sinusoidal waveform is given by the expression below:

$$x(t) = A \cos (\omega t + \theta)$$

This signal has these three parameters:

• Amplitude $A$
• Frequency $\omega$
• Phase $\theta$

In general, people find it quite straightforward to understand how the amplitude $A$ represents the signal level while the frequency $\omega$ is related to the number of rotations in a second.

$$\omega = 2\pi f = \frac{2\pi}{T}$$

where $T$ is wave time period. Phase, however, is the tricky part.

Phase as a shadow of frequency

We know that a time shift to the right by $\tau$ units of a sinusoidal wave is written as:

$$\begin{align*} x(t)&=A\cos \left\{\omega (t-\tau)\right\}\\ &=A \cos (\omega t - \omega \tau)\\ &=A\cos (\omega t + \theta) \end{align*}$$

In words, the phase in terms of a time shift is given by:

$$\theta = -\omega \tau$$

This is why we say that the phase is a shadow of frequency.

Note: For the same time shift, the phase is different for each frequency!

And the time shift for a given wave phase can be written as:

$$\tau = -\frac{\theta}{\omega}$$

Let’s verify this with the help of an example.

Example

Consider a sinusoidal wave, like so:

$$x(t) = 3\cos(0.4\pi t+0.4\pi)=3\cos\left\{0.4\pi [t-(-1)]\right\}$$

• Amplitude: Clearly, the amplitude $A$ is equal to $3$.
• Frequency: The frequency is:

$$2\pi f = 0.4\pi \quad \rightarrow \quad f = 0.2$$

The time period is then:

$$T = \frac{1}{f}=5$$

• Time shift: Since $\theta=0.4\pi$, we have:

$$\tau = -\frac{0.4\pi}{0.4\pi} = -1$$

Another way to look at this is that cosine has a peak value when its argument is zero. Putting $0.4\pi t+0.4\pi=0$, we get time shift as $-1$.

We can now run through different time shifts and phases in a coding environment.