Phase, Frequency, Amplitude, and all that..

A second order equation and its solutions

UBC Calculus Online Course Notes

Phase, Frequency, Amplitude, and all that..

In our previous work, we saw that the differential equation

$\frac{d^2y}{dt} = -\omega ^2 y$

Has solutions of the form

$f(t) = A \sin(\omega t) + B \cos(\omega t)$

On this page we will spend some time understanding what these functions look like, and how they behave. In the demonstration below, you will see a trigonometric function in which several parameters can be varied.

The effect of frequency and phase

We would like to understand the behaviour of the following function:

$f(t) = A \sin(\omega t+ \phi)$

In this function, $ t $ is a variable. The other quantities are in general fixed, and each of them influences the shape of the graph of this function. Let us explore how the shape of the graph of changes as we change its three parameters called the Amplitude, $ A $ , the frequency, $\omega$ and the phase shift, $\phi$ . You can drag the nodes to see what happens as each of these three quantities are varied.

For your consideration:

(1) Describes what happens as each of the above parameters is varied ?
(2) How might understanding this function help us to understand the function which we are trying to understand at the top of this page?

You have probably noticed that the amplitude governs the heights of the peaks, the frequency governs their spacing, i.e. how many cycles the function goes through in a given time interval, and the phase shift determines where the curve crosses the axis.

Let us first consider the shape of the function

$A \sin(\omega t)$

Since our original function, $\sin(t)$ is a periodic function that goes through one complete cycle when $t=2 \pi$ , the function $A \sin(\omega t)$ will go through a complete cycle when $\omega t= 2 \pi$ , i.e. it will have completed a cycle when $t= 2 \pi/ \omega$ . We say that it has a period (which we will denote by $ T $ ) given by

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

The height of the peaks and valleys in this function will be given by its amplitude, $ A $ .

We are now ready to consider the effect of the phase-shift, $\phi$ . In fact, we can make note of the fact that the graph of the function $A \sin(\omega t + \phi)$ will cross the t axis when

$\sin(\omega t + \phi)=0$

The first time that this happens is when

$(\omega t + \phi)=0$

which corresponds to a value of t given by

$t = -\frac{\phi}{\omega}$

Thus, the graph will be shifted so that it crosses the t axis at this value. The shape of the curve does not change, only its position on the t axis.

Superimposing sines and cosines

Let us take a second look at the function we investigated above, and notice that when we apply the trigonometric identity

$\sin(\theta + \phi) = \sin(\theta) \cos(\phi)+ C \sin(\phi)\cos(\theta)$

we obtain

$C \sin(\omega t + \phi) =C \sin(\omega t) \cos(\phi)+ C \sin(\phi)\cos(\omega t)$

Remembering that $\phi$ is a constant, and therefore so is $\cos(\phi), \sin(\phi)$ , and assigning the names

$A =C \cos(\phi) \] \[ B= C \sin(\phi)$

we have found that

$A \sin(\omega t) + B \cos(\omega t)=C \sin(\omega t + \phi)$

Thus, by using a trigonometric identity for the sums of angles, we have reduced a problem we needed to understand (the question we started with, at the top of this page) with a problem that we already know how to solve. We have found that the sum of a sine and a cosine curve is actually equivalent to a sine with a phase shift.

A bit of care is required, however, since in order for this conversion to work, it must be true that

$A^2 + B^2 = C^2$

For your consideration:

(1) Explain why this relationship must hold between the constants (Hint: note the definitions of these constants in terms of the trigonometric functions applied to $\phi$ , and remember the inmportant trigonometric identity that must therefore be satisfied.
(2) If $\phi = \pi/4$ and , find the values of the constants .
(3) If $A=0.5, B = \sqrt{3}/2$ , find the value of C and $\phi$

Example:

Describe the behaviour of the function

$f(t) = \sqrt{3} \sin(5 t) + \cos(5 t)$

Solution:

We observe that the constants in front of the trigonometric functions have the values

$A =C \cos(\phi)=\sqrt{3} \] \[ B= C \sin(\phi)=1$

We would like to find the angle $\phi$ and the amplitude $ C $ that fit with this pattern. The ratio of the constants

$\frac{B}{A} =\frac{C \sin(\phi)} {C \cos(\phi)}= \tan{\phi}=\frac{1}{\sqrt{3}}$

Thus, looking up the angle that has a value of $\tan{\phi}= 1/\sqrt{3}$ we find that

$\phi = \frac{\pi}{6}$

Thus the phase shift is $\phi = \pi/6$ . We further calculate that

$C \cos(\frac{\pi}{6})= C \frac{\sqrt{3}}{2} = \sqrt{3} \] \[ C \sin(\frac{\pi}{6})= C \frac{1}{2} = 1$

which tells us that $ C = 2 $ . Thus, we conclude that

$f(t) = \sqrt{3} \sin(5 t) + \cos(5 t) = 2 \sin(5 t + \frac{\pi}{6})$

Thus the above superposition of sines and cosines is equivalent to a sine with amplitude 2, with frequency 5, and which crosses the t axis at $- \pi/30$ .