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Physics 5B Winter 2009

How to add sine functions of different amplitude and phase

In these notes, I will show you how to add two sinusoidal waves, each of different amplitude andphase, to get a third sinusoidal wave. That is, we wish to show that given

E1 = E10 sin t , (1)

E2 = E20 sin(t + ) , (2)

the sum E E1 + E2 can be written in the form:

E = E10 sin t + E20 sin(t + ) = E0 sin(t + ) (3)

where the amplitude E0 and phase are determined in terms of E10, E20 and . In these notes,we shall derive that the amplitude E0 is given by

E0 =

E210 + E220 + 2E10E20 cos

and the phase is determined modulo 2 by1

sin =E20 sin

E0, cos =

E10 + E20 cos

E0.

By definition, the amplitudes E10 and E20 are positive numbers. If we divide the last two equations,then can be determined modulo from:

tan = E20 sin E10 + E20 cos

.

To determine modulo 2, we need to supplement the result for tan with

sign(sin ) = sign(sin ) , sin = 0 ,

where sign(sin ) literally means the sign (i.e. either +1 or 1) of the quantity sin . Finally, ifsin = 0 then cos = 1, and can be fixed modulo 2 by

cos =

1 , cos = 1 ,

1 , cos = 1 , and E10 > E20 ,

1 , cos = 1 , and E10 < E20 .

Note that in the case of cos = 1 and E10 = E20, we have E = 0 in which case E0 = 0 and isno longer meaningful. In fact, this is the only circumstance in which E0 can vanish.

I shall provide two different derivations of the above formulae. Finally, in an appendix, I willprovide a mathematically more advanced derivation that makes use of complex numbers. If youare unfamiliar with complex numbers, you can skip the appendix for now and return to this lastderivation later after you take Physics 116A or an equivalent course. This last method also providesthe real motivation for the method of phasors introduced in Section 2 below.

1The phrase, is determined modulo 2, means that is determined up to an additive integer multiple of 2.This is all that is needed, since adding a multiple of 2 to the phase angle does not change the value of sin(t +).

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1. Algebraic method

First, set t = 0 in eq. (3) to obtain E20 sin = E0 sin . Solving for sin yields:

sin =

E20 sin

E0 .

Next, set t = /2 in eq.(3). Noting that sin(+ /2) = cos , it follows that E10 + E20 cos =E0 cos . Solving for cos yields:

cos =E10 + E20 cos

E0.

Finally, using cos2 +sin2 = 1, and inserting the expressions for cos and sin just obtained, onefinds:

E20 = (E10 + E20 cos )2 + E220 sin

2

= E210 + 2E10E20 cos + E220(cos

2 + sin2 )

= E210 + E

220 + 2E10E20 cos .

By definition, E0 is a non-negative number.2 Thus, we take the positive square root to obtain

E0 =

E210 + E220 + 2E10E20 cos .

This completes our derivation.

2. Geometric methodthe method of phasors

Consider a fictitious vector in a two-dimensional space, whose length is E0

, which makes an angle with respect to the x-axis as shown below:

x

y

E0 sin E0

If we project this vector onto the y-axis, then its projected length is E0 sin as shown above. Thisvector is called a phasor and represents a quantity with an amplitude E0 and an angle .

The utility of such a representation is that we can perform the sum of eq. (3) by consideringthe phasors corresponding to each sine term in the sum, and then adding the phasors vectorially!The projection of the vector sum of the two phasors onto the y-axis is just the sum of the two sinefunctions that we wish to compute. This vector sum can be carried out geometrically, and providesa second method for evaluating E0 and .

2Warning! This is a matter of convention, which Giancoli chooses not to follow (without warning you).

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To see how this works, consider the computation of eq. (3) by the method of phasors. Werepresent E1 and E2 [cf. eqs. (1) and (2)] as shown in the figure below.

Then, the phasor representation of E is just the vector sum shown above. We identify E10, E20and E0, as the lengths of the phasors representing E1, E2 and E, respectively. To evaluate E0and , we focus on the triangle in the figure above. First, using the law of cosines,

E20 = E210 + E

220 2E10E20 cos( ) ,

since is the angle between the phasors representing E1 and E2. Using cos( ) = cos , weend up with

E0 =

E210 + E220 + 2E10E20 cos ,

as before. Next, using the law of sines,

sin E20= sin( )E0

.

Using sin( ) = sin , we can solve for sin . We find that

sin =E20 sin

E0,

which again agrees with our previous result. This equation only fixes modulo . In order to fix modulo 2, we employ the law of sines again (noting that the angle in the triangle between thephasors representing E2 and E is given by ):

sin

E20=

sin( )

E10.

This equation can be rearranged in the following form:

E10E20

=sin( )

E10

=sin cos cos sin

sin

=sin

tan cos .

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One can solve this equation easily for tan to obtain

tan =E20 sin

E10 + E20 cos ,

in agreement with our previous result. Finally, we can use our results above for sin and tan to

compute cos as follows

cos =sin

tan =

E20 sin

E0

E10 + E20 cos

E20 sin

=

E10 + E20 cos

E0,

which completes the derivation.

3. The limit of equal amplitudes

As a check, consider the case of equal amplitudes, E10 = E20 E0. Then, using the aboveresults,

E0 =

2E0(1 + cos ) .Recalling the trigonometric identity, cos2(/2) = 12(1 + cos ), we end up with:

E0 = 2E0| cos(/2)| .

Note the absolute value sign, since by definition the amplitude E0 is defined to be non-negative,which means we must take the positive square root:

cos2(/2) = | cos(/2)|. If cos = 1, then

E0 = 0 and the angle is undefined. Otherwise, we may use the results derived above for sin andcos to obtain.

sin =sin

2| cos(/2)|=

2 sin(/2)cos(/2)

2| cos(/2)|=

sin(/2) , if cos(/2) > 0 ,

sin(/2) , if cos(/2) < 0 ,

cos = 1 + cos 2| cos(/2)|

= cos2(/2)

| cos(/2)|= | cos(/2)| ,

after using the trigonometric identity sin = 2 sin(/2) cos(/2). Note that

cos(/2) = 12 (1 + cos ) = 0 , for cos = 1 ,

which must hold ifE0 = 0. The results above for sin and cos imply that:

=

/2 , if cos(/2) > 0 ,

+ /2 , if cos(/2) < 0 .

In Section 34-4 of Giancoli, a convention is chosen in which E0 can be of either sign, so inorder to compare Giancolis results with ours, one must be careful if cos(/2) < 0. Nevertheless,

independently of this convention, we can write:E = E0 sin(t + ) = 2E0| cos(/2)| sin(t + )

= 2E0| cos(/2)|(sin t cos + cos t sin )

= 2E0| cos(/2)|

sin t

cos2(/2)

| cos(/2)|+ cos t

sin(/2) cos(/2)

| cos(/2)|

= 2E0 cos(/2)(sin t cos(/2) + cos t sin(/2))

= 2E0 cos(/2)sin(t + /2) ,

which agrees with Eq. 34-5c of Giancoli (on page 907).

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Appendix: Adding two sine functions of different amplitude and phase using complex

numbers

To perform the sum:

E = E10 sin t + E20 sin(t + ) = E0 sin(t + ) , (4)

we note the famous Euler formula:ei = cos + i sin .

In particular, sin is the imaginary part ofei. Thus, if we consider the equation:

E10eit + E20e

i(t+) = E0ei(t+) , (5)

then the imaginary part of this equation coincides with eq. (4). By the way, I can view a complexnumber x + iy as a vector in a two-dimensional space (called the complex plane) that points fromthe origin to the point (x, y). This vector is precisely the phasor that we employed in Section 2of these notes. In particular, in this language, eq. (5) describes the sum of two complex numbers,which is depicted by the sum of the phasors in the figure shown in Section 2. The projected lengthsof the phasors on the y-axis simply correspond to the imaginary parts of the corresponding complexnumbers.

Thus, to solve for E0 and , we can simply start from eq. (5). If I multiply this equation byeit, I obtain:

E10 + E20ei = E0e

i (6)

where E10, E20 and E0 are non-negative (and real) by definition. To compute E0, I simply takethe complex absolute value of both sides of this equation. For any complex number z = x + iy, thecomplex absolute value is given by |z| =

x2 + y2. Hence,

E20 = |E10 + E20ei|2 = |E10 + E20 cos + iE20 sin |

2

= (E10 + E20 cos )2 + E220 sin

2

= E210 + 2E10E20 cos + E

220 ,

after using sin2 + cos2 = 1. Taking the positive square roots yields

E0 =

E210 + E220 + 2E10E20 cos .

To determine , I simply take the real and imaginary parts of eq. (6):

E0 cos = E10 + E20 cos , E0 sin = E20 sin ,

Solving for cos and sin , respectively, we immediately find:

cos =E10 + E20 cos

E0

,

sin =E20 sin

E0.

Thus, we have successfully reproduced the main results obtained previously in these notes. I thinkyou must agree that this last approach is by far the simplest from a computational point of view.Once you learn how use and manipulate complex numbers, many tasks in mathematical physicsbecome much simpler!

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