Encyclopedia of Electrical Engineering

As noted earlier in this chapter, the addition of sinusoidal voltages
and currents will frequently be required in the analysis of ac circuits.
One lengthy but valid method of performing this operation is to place
both sinusoidal waveforms on the same set of axes and add algebraically the magnitudes of each at every point along the abscissa, as shown for $c = a + b$ in Fig. 1.
**Fig. 1: **Adding two sinusoidal waveforms on a point-by-point basis.
This, however, can be a long and
tedious process with limited accuracy. A shorter method uses the
rotating radius vector shown in Fig. 2,
**Fig. 2: **Generating a sinusoidal waveform through the vertical projection of a
rotating vector
During its
rotational development of the sine wave, the phasor will, at the
instant t = 0, have the positions shown in Fig. 3(a) for each
waveform in Fig. 3(b).
**Fig. 3: **(a) The phasor representation of the sinusoidal waveforms of Fig. 3(b);

(b) finding the sum of two sinusoidal waveforms of v1 and v2.
Note in Fig. 3(b) that $v_2$ passes through the horizontal axis at
$t = 0 s$, requiring that the radius vector in Fig. 3(a) be on the horizontal axis to ensure a vertical projection of zero volts at $t = 0 s$.
Its length in Fig. 3(a) is equal to the peak value of the sinusoid as
required by the radius vector in Fig. 2. The other sinusoid has
passed through $90^\circ$ of its rotation by the time $t = 0s$ is reached and therefore has its maximum vertical projection as shown in Fig. 3(a).

Since the vertical projection is a maximum, the peak value of the sinusoid that it will generate is also attained at $t = 0 s$, as shown in Fig. 3(b). Note also that $v_T =v_1$ at $t =0 s$ since $v_2 = 0 V $ at this instant. It can be shown [see Fig. 3(a)] using the vector algebra $$1 V \angle{0^\circ} + 2 V \angle{90^\circ} = 2.236 V \angle{63.43^\circ}$$ In other words, if we convert $v_1$ and $v_2$ to the phasor form using $$v = V_m \sin(wt + \theta) = V_m \angle{\pm \theta}$$ and add them using complex number algebra, we can find the phasor form for $v_T$ with very little difficulty. It can then be converted to the time domain and plotted on the same set of axes, as shown in Fig. 3(b). Figure 3(a), showing the magnitudes and relative positions of the various phasors, is called a**phasor diagram**. It is actually
a "snapshot" of the rotating radius vectors at $t = 0 s$.
In the future, therefore, if the addition of two sinusoids is required,
they should first be converted to the phasor domain and the sum found
using complex algebra. The result can then be converted to the time
domain.
**Fig. 4: **Adding two sinusoidal currents with phase angles other than 90 degrees.
The case of two sinusoidal functions having phase angles different
from 0 and 90 appears in Fig. 4. Note again that the vertical
height of the functions in Fig. 4(b) at $t = 0 s$ is determined by the
rotational positions of the radius vectors in Fig. 4(a).
Since the rms, rather than the peak, values are used almost exclusively in the analysis of ac circuits, the phasor will now be redefined for
the purposes of practicality and uniformity as having a magnitude equal
to the rms value of the sine wave it represents. The angle associated
with the phasor will remain as previously described as the phase angle.

In general, for all of the analyses to follow, the phasor form of a sinusoidal voltage or current will be $$\text{V} =V \angle {\theta} \,\text{and}\, \text{I} =I \angle {\theta}$$ where V and I are rms values and v is the phase angle. It should be pointed out that in phasor notation, the sine wave is always the reference, and the frequency is not represented.
**Example 1: **Find the input voltage of the circuit of Fig. 5 if
$$v_a = 50 \sin(377t + 30^\circ) \, \text{and}\\ v_b = 30 \sin(377t + 60^\circ)$$
**Fig. 5: **For Example 1.
**Solution: **

Applying Kirchhoff's voltage law, we have $$e_{in} = v_a + v_b$$ Converting from the time to the phasor domain yields $$\begin{split} v_a &= 50 \sin(377t + 30^\circ) = {50 \over \sqrt{2}} \angle{30^\circ}\\ &= 35.35V \angle{30^\circ}\\ \end{split}$$ and $$e_{in} = v_a + v_b$$ Converting from the time to the phasor domain yields $$\begin{split} v_b &= 30 \sin(377t + 60^\circ) = {30 \over \sqrt{2}} \angle{60^\circ}\\ &= 21.21V \angle{60^\circ}\\ \end{split}$$ Converting from polar to rectangular form for addition yields $$V_a = 35.35V \angle{30^\circ}= 30.61 V + j 17.68 V$$ $$V_b = 21.21 V \angle{60^\circ} = 10.61 V + j 18.37 V$$ Then $$\begin{split} E_{in} &= V_a + V_b \\ &= (30.61 V + j 17.68 V) + (10.61 V + j 18.37 V)\\ &= 41.22 V + j 36.05 V\\ \end{split}$$ Converting from rectangular to polar form, we have $$E_{in} = 41.22 V + j 36.05 V = 54.76 V \angle{41.17^\circ}$$ Converting from the phasor to the time domain, we obtain $$E_{in} = 54.76 V \angle{41.17^\circ} \Rightarrow\\ e_{in} = \sqrt{2}(54.76) \sin(377t + 41.17^\circ)\\ $$ and $$e_{in} = 77.43 \sin(377t + 41.17^\circ)$$ A plot of the three waveforms is shown in Fig. 6. Note that at each instant of time, the sum of the two waveforms does in fact add up to $e_{in}$. At $t = 0 (wt = 0), e_{in}$ is the sum of the two positive values, while at a value of $wt$, almost midway between $\pi/2$ and $\pi$, the sum of the positive value of $v_a$ and the negative value of $v_b$ results in $e_{in} = 0$. **Fig. 6: **Solution for Example 1.

(b) finding the sum of two sinusoidal waveforms of v1 and v2.

Since the vertical projection is a maximum, the peak value of the sinusoid that it will generate is also attained at $t = 0 s$, as shown in Fig. 3(b). Note also that $v_T =v_1$ at $t =0 s$ since $v_2 = 0 V $ at this instant. It can be shown [see Fig. 3(a)] using the vector algebra $$1 V \angle{0^\circ} + 2 V \angle{90^\circ} = 2.236 V \angle{63.43^\circ}$$ In other words, if we convert $v_1$ and $v_2$ to the phasor form using $$v = V_m \sin(wt + \theta) = V_m \angle{\pm \theta}$$ and add them using complex number algebra, we can find the phasor form for $v_T$ with very little difficulty. It can then be converted to the time domain and plotted on the same set of axes, as shown in Fig. 3(b). Figure 3(a), showing the magnitudes and relative positions of the various phasors, is called a

In general, for all of the analyses to follow, the phasor form of a sinusoidal voltage or current will be $$\text{V} =V \angle {\theta} \,\text{and}\, \text{I} =I \angle {\theta}$$ where V and I are rms values and v is the phase angle. It should be pointed out that in phasor notation, the sine wave is always the reference, and the frequency is not represented.

Applying Kirchhoff's voltage law, we have $$e_{in} = v_a + v_b$$ Converting from the time to the phasor domain yields $$\begin{split} v_a &= 50 \sin(377t + 30^\circ) = {50 \over \sqrt{2}} \angle{30^\circ}\\ &= 35.35V \angle{30^\circ}\\ \end{split}$$ and $$e_{in} = v_a + v_b$$ Converting from the time to the phasor domain yields $$\begin{split} v_b &= 30 \sin(377t + 60^\circ) = {30 \over \sqrt{2}} \angle{60^\circ}\\ &= 21.21V \angle{60^\circ}\\ \end{split}$$ Converting from polar to rectangular form for addition yields $$V_a = 35.35V \angle{30^\circ}= 30.61 V + j 17.68 V$$ $$V_b = 21.21 V \angle{60^\circ} = 10.61 V + j 18.37 V$$ Then $$\begin{split} E_{in} &= V_a + V_b \\ &= (30.61 V + j 17.68 V) + (10.61 V + j 18.37 V)\\ &= 41.22 V + j 36.05 V\\ \end{split}$$ Converting from rectangular to polar form, we have $$E_{in} = 41.22 V + j 36.05 V = 54.76 V \angle{41.17^\circ}$$ Converting from the phasor to the time domain, we obtain $$E_{in} = 54.76 V \angle{41.17^\circ} \Rightarrow\\ e_{in} = \sqrt{2}(54.76) \sin(377t + 41.17^\circ)\\ $$ and $$e_{in} = 77.43 \sin(377t + 41.17^\circ)$$ A plot of the three waveforms is shown in Fig. 6. Note that at each instant of time, the sum of the two waveforms does in fact add up to $e_{in}$. At $t = 0 (wt = 0), e_{in}$ is the sum of the two positive values, while at a value of $wt$, almost midway between $\pi/2$ and $\pi$, the sum of the positive value of $v_a$ and the negative value of $v_b$ results in $e_{in} = 0$.

Complex Numbers
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Mathematical Operations with Complex Numbers