Adding sinusoids with the same frequency · TANS 1.13

Adding two sinusoidal functions

The goal: add two sinusoidal functions with the same frequency (the same $b$) and see that the result is another sinusoid with that frequency, with a new amplitude and phase.

The idea: sine is an imaginary part

For a complex number in exponential form, $r\,e^{i\theta} = r\cos\theta + i\,r\sin\theta$, so

\operatorname{Im}\left(r\,e^{i\theta}\right) = r\sin\theta \qquad\qquad \operatorname{Re}\left(r\,e^{i\theta}\right) = r\cos\theta

With $\theta = bx + \varphi$, any sinusoid can be written as the imaginary part of a product:

A\sin(bx + \varphi) = \operatorname{Im}\left(A\,e^{i(bx + \varphi)}\right) = \operatorname{Im}\left(e^{ibx}\cdot A\,e^{i\varphi}\right)

If $f_1(x) = A_1\sin(bx + \varphi_1)$ and $f_2(x) = A_2\sin(bx + \varphi_2)$, the imaginary part of a sum is the sum of the imaginary parts, and $e^{ibx}$ is a common factor because the frequency is the same:

f_1(x) + f_2(x) = \operatorname{Im}\left(e^{ibx}\left(A_1e^{i\varphi_1} + A_2e^{i\varphi_2}\right)\right)

The result

The bracket is a fixed complex number that does not depend on $x$. Writing it as $R\,e^{i\alpha}$:

f_1(x) + f_2(x) = \operatorname{Im}\left(R\,e^{i(bx + \alpha)}\right) = R\sin(bx + \alpha)

R = \left|A_1e^{i\varphi_1} + A_2e^{i\varphi_2}\right| \qquad \alpha = \arg\left(A_1e^{i\varphi_1} + A_2e^{i\varphi_2}\right)

In practice, the bracket is added on the calculator and the modulus and argument are read off.

Watch the calculator mode: if the phases are in degrees, degrees; if they are in radians, radians. And to check the modulus by hand, use the cosine rule: $R^{2} = A_1^{2} + A_2^{2} + 2A_1A_2\cos(\varphi_1 - \varphi_2)$.

Simulator · Add two sinusoids

Move the amplitudes and phases (both waves have the same $b$). On the right, the two phasors placed tip to tail: the green arrow is their sum, and its length and angle are the amplitude and phase of the sum.

Amplitude $A_1$ Phase $\varphi_1$ Amplitude $A_2$ Phase $\varphi_2$ Frequency $b$ (shared)

$f_1$$f_2$$f_1 + f_2$

Example 1 · $\sin x + 2\sin\left(x + \frac{\pi}{2}\right)$

Let $f(x) = \sin x$ and $g(x) = 2\sin\left(x + \frac{\pi}{2}\right)$, and $h(x) = f(x) + g(x)$.

Each one as an imaginary part

f(x) = \operatorname{Im}\left(e^{ix}\right) \qquad g(x) = \operatorname{Im}\left(2\,e^{i\left(x + \frac{\pi}{2}\right)}\right) = \operatorname{Im}\left(2\,e^{ix}\,e^{\frac{\pi}{2}i}\right)

h(x) = \operatorname{Im}\left(e^{ix} + 2\,e^{ix}\,e^{\frac{\pi}{2}i}\right) = \operatorname{Im}\left(e^{ix}\left(1 + 2\,e^{\frac{\pi}{2}i}\right)\right)

The bracket in exponential form

1 + 2\,e^{\frac{\pi}{2}i} = 1 + 2i, \qquad |1 + 2i| = \sqrt{5} \approx 2.24, \qquad \arg(1 + 2i) = \arctan 2 \approx 1.107

h(x) = \operatorname{Im}\left(\sqrt{5}\,e^{i(x + 1.107)}\right) = \sqrt{5}\,\sin(x + 1.107)

$h(x) \approx 2.24\,\sin(x + 1.11)$

Plotting $\sin x + 2\sin\left(x + \frac{\pi}{2}\right)$ and $2.24\sin(x + 1.11)$ on the calculator, the two graphs coincide. The amplitude of the sum is $\sqrt{5} \approx 2.24$, not $1 + 2 = 3$: the two waves are not in phase, and the amplitudes would only add up if they had the same phase.

Example 2 · Find $r$ and $\alpha$

If $f(x) = 3\sin\left(3x + \frac{\pi}{12}\right)$ and $g(x) = 4\sin\left(3x + \frac{3\pi}{4}\right)$, find $r$ and $\alpha$ such that $f(x) + g(x) = r\sin(3x + \alpha)$.

Solution

f(x) + g(x) = \operatorname{Im}\left(e^{3xi}\left(3\,e^{\frac{\pi}{12}i} + 4\,e^{\frac{3\pi}{4}i}\right)\right)

3\,e^{\frac{\pi}{12}i} + 4\,e^{\frac{3\pi}{4}i} \approx (2.898 + 0.776\,i) + (-2.828 + 2.828\,i) = 0.069 + 3.605\,i

r = \sqrt{13} \approx 3.61 \qquad \alpha \approx 1.55

$f(x) + g(x) = \sqrt{13}\,\sin(3x + 1.55)$

Check with the cosine rule: the phases differ by $\frac{3\pi}{4} - \frac{\pi}{12} = \frac{2\pi}{3}$, and $r^{2} = 9 + 16 + 2\cdot 3\cdot 4\cos\frac{2\pi}{3} = 25 - 12 = 13$.

Homework · Two alternating current sources

Question

Two alternating current sources have voltages of $100$ V and $180$ V. The first has a phase shift of $45^\circ$ and the second of $135^\circ$. The two sources are combined.

a) Find the voltage of the combined source. b) Find the phase shift of the combined source.

See the solution

Both sources as imaginary parts

V_1 = 100\sin(bt + 45^\circ) = \operatorname{Im}\left(100\,e^{ibt}\,e^{45^\circ i}\right) \qquad V_2 = 180\sin(bt + 135^\circ) = \operatorname{Im}\left(180\,e^{ibt}\,e^{135^\circ i}\right)

V_T = V_1 + V_2 = \operatorname{Im}\left(e^{ibt}\left(100\,e^{45^\circ i} + 180\,e^{135^\circ i}\right)\right)

The bracket, with the calculator in degrees

100\,e^{45^\circ i} + 180\,e^{135^\circ i} \approx (70.71 + 70.71\,i) + (-127.28 + 127.28\,i) = -56.57 + 197.99\,i

It is in the second quadrant. Modulus and argument:

R = \sqrt{56.57^{2} + 197.99^{2}} \approx 205.9 \qquad \alpha = 180^\circ - \arctan\dfrac{197.99}{56.57} \approx 180^\circ - 74.1^\circ = 105.9^\circ

$V_T \approx 206\,\sin(bt + 105.9^\circ)$: a) $206$ V b) $105.9^\circ$

A quick check: the phases differ by $135^\circ - 45^\circ = 90^\circ$, so the two vectors are perpendicular and the amplitude follows from Pythagoras, $\sqrt{100^{2} + 180^{2}} = \sqrt{42\,400} \approx 205.9$.

Correction to the board

In class we got $V_T = 162.08\sin(bt + 141.45^\circ)$. That is what the calculator gives in radian mode: it reads $45$ and $135$ as radians, not degrees. With phase shifts in degrees, the calculator must be in degrees. The right answer is $206$ V and $105.9^\circ$.