Adding Phasors

Section 1.4 Adding Phasors

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Figure 1.14. Phasor diagram showing separate phasors for oscillations $y_{1}$ and $y_{2}$ rotating together with the same angular frequency.

The power and utility of the phasor representation really show up when combining oscillations. Consider two oscillations, both with the same angular frequency

ω,

but with different amplitudes and phases:

\begin{matrix} (1.12) & y_{1} (t) = A_{1} \cos (ω t + ϕ_{1}) and y_{2} (t) = A_{2} \cos (ω t + ϕ_{2}) . \end{matrix}

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The superposition of these two oscillations,

y_{t o t} = y_{1} + y_{2},

will be sinusoidal with the same angular frequency

ω,

but it is very messy to calculate the new amplitude and phase shift algebraically; it is actually much simpler using the complex phasor method. Because the oscillation's have the same frequency, the phasors rotate together as time passes, as illustrated in Figure 1.14. That is, the angle between the phasors will always be

Δ ϕ = ϕ_{2} - ϕ_{1} .

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The resultant phasor, with its own amplitude and phase, is illustrated in Figure 1.15. We can determine the amplitude and phase of the resultant phasor by adding the two input phasors the same way we add vectors. The amplitude of the resultant will be less than the sum of the two original phasor amplitudes (unless

Δ ϕ = 0

) and the phase of the resultant will be something between the phases of the two original phasors.

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Figure 1.15. The combined oscillation is described by a phasor that is the vector sum of the two separate phasors.

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Example 1.16. Rock Your Boat.

You're sitting in a boat in the middle of a calm lake. Suddenly a motor boat drives by, producing waves that would oscillate your boat up and down as follows:

\begin{matrix} (1.13) & y_{1} (t) = A_{1} \cos (ω t + ϕ_{1}), \end{matrix}

where $A_{1} = 25 cm,$ $ω = \frac{2 π}{3},$ and $ϕ_{1} = \frac{π}{6} .$ At the same time, another motor boat drives by, producing waves that would oscillate your boat up and down as follows:

\begin{matrix} (1.14) & y_{2} (t) = A_{2} \cos (ω t + ϕ_{2}), \end{matrix}

where $A_{2} = 15 cm,$ $ω = \frac{2 π}{3},$ and $ϕ_{1} = \frac{π}{3} .$ However, since both waves act on the boat simultaneously, the actual oscillation of your boat is the superposition of these two waves. How much does your boat move up and down as a result of the combination of these waves?

Solution.

The phasor diagram for these two separate oscillations is shown in Figure 1.17. The resultant phasor can be determined from the vector addition of the phasors shown in Figure 1.18.


Phasor	real part (horizontal)	imaginary part (vertical)

1	$25 \cos (\frac{π}{6}) = 21.6$	$25 \sin (\frac{π}{6}) = 12.5$

2	$15 \cos (\frac{π}{3}) = 7.5$	$15 \sin (\frac{π}{3}) = 13.0$

Total	$29.1$	$25.5$

Figure 1.17. Phasor diagram for two water waves in Example 1.16.

So, the amplitude of the resultant phasor is

\begin{matrix} (1.15) & A_{t o t} = \sqrt{{29.1}^{2} + {25.5}^{2}} = 38.7 cm, \end{matrix}

and its initial phase is

\begin{matrix} (1.16) & ϕ_{t o t} = \tan^{- 1} (\frac{25.5}{29.1}) = 0.72 rad . \end{matrix}

Figure 1.18. Phasor addition for two water waves in Example 1.16.

We can write the superposition as

\begin{aligned} y_{t o t} (t) = & A_{t o t} \cos (ω t + ϕ_{t o t}) \\ = & 38.7 \cos (ω t + 0.72), \end{aligned}

with $y$ in cm and $t$ in sec.

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Example 1.19. Two Towers.

Figure 1.20. Illustration of radio towers in Example 1.19.

Two radio towers, A and B, separated by $20 m,$ broadcast the same radio signal of wavelength $λ = 12 m .$ You're standing at the point $P$ indicated in the figure with your radio wave amplitude meter. With only tower A broadcasting, you measure a wave amplitude of 7 (in some unspecified unit). With only tower B broadcasting, you measure a wave amplitude of 5. What amplitude do you measure when both towers are broadcasting?

Solution.

Here we're interested in the superposition of the waves from the two towers. We aren't explicitly given the phase difference between the wave signals that arrive at $P$ from the two towers, but we can figure it out because we know the distances from point $P$ to each tower. If the waves leave the two towers in phase, they won't necessarily be in phase when they reach point $P$ because the waves from Tower B have to travel farther. How much farther? Well, the distance between Tower B and point $P$ is $\sqrt{50^{2} + 20^{2}} = 53.85 m,$ so the waves from Tower B have to travel $3.85 m$ farther. That means the waves from Tower A will be $3.85 m$ ahead of the waves from Tower B.

We can turn that distance difference into a phase difference for the waves. The path length difference $Δ r = 3.85 m$ produces $\frac{3.85}{12} = 0.32$ wavelengths of difference between the two waves as they arrive at point $P .$ If the waves from Tower A arrived exactly one wavelength ahead of the waves from Tower B, the signals would add constructively. The phase difference would be $2 π,$ and peaks and troughs of the two waves would be aligned. In this case, however, the waves from Tower A arrive only $0.32 wavelengths$ ahead of the waves from Tower B. Consequently, the phase difference $Δ ϕ$ is

\begin{matrix} (1.17) & Δ ϕ = 2 π \frac{Δ r}{λ} = 2 π \frac{3.85}{12} = 2.0 rad . \end{matrix}

We can then write expressions for the two oscillations at point $P$ as follows:

\begin{matrix} (1.18) & y_{A} (t) = 7 \cos (ω t + 2.0) y_{B} (t) = 5 \cos (ω t) . \end{matrix}

Why did I pick zero initial phase for the signal from Tower B? Because I could. In this case (and in most cases we'll deal with), we're only interested in the phase difference between signals. Therefore, I can always choose to describe one wave with zero initial phase, and then put the phase difference into the expression for the other wave. With one phasor on the horizontal axis, the phasor addition is just easier.

Now we can construct a phasor diagram for the two oscillations, shown in Figure 1.21.

Figure 1.21. Phasor diagram for radio waves in Example 1.19.

Once again, the resultant phasor can be determined from the vector addition of the phasors.


Phasor	real part (horizontal)	imaginary part (vertical)

1	$5 \cos (0) = 5$	$5 \sin (0) = 0$

2	$7 \cos (2.0) = - 2.9$	$7 \sin (2.0) = 6.4$

Total	$2.1$	$6.4$

So, the amplitude of the resultant phasor is

\begin{matrix} (1.19) & A_{t o t} = \sqrt{{2.1}^{2} + {6.4}^{2}} = 6.7, \end{matrix}

and its initial phase is

\begin{matrix} (1.20) & ϕ_{t o t} = \tan^{- 1} (\frac{6.4}{2.1}) = 1.25 rad, \end{matrix}

and we can thus write the superposed oscillation as

\begin{matrix} (1.21) & y_{t o t} (t) = 6.7 \cos (ω t + 1.25) . \end{matrix}