PRINCIPLES OF ALTERNATING CURRENT

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COMPLEX NUMBERS

PASSIVE COMPONENTS IN AC CIRCUITS

A sinusoidal voltage can be described by the equation:

v(t) = V_M sin (ωt + Φ) or v(t) = V_M cos (ωt + Φ)

Here are a few examples to illustrate the terms above.

The properties of the 220 V AC voltage in household electrical outlets in Europe:

Effective value: V_Eff = 220 V
Peak value: V_M= √2*220 V = 311 V

Frequency: f = 50 1/s = 50 Hz
Angular frequency: ω = 2*π*f = 314 1/s = 314 rad/s
Period: T = 1/f = 20 ms
Time function: v(t)=311 sin (314 t)

Let’s see the time function using TINA’s Analysis/AC Analysis/Time Function command.

The properties of the 120 V AC voltage in the household electrical outlet in the US:

Effective value: V_Eff = 120 V
Peak value: V_M=√2 120 V = 169.68 V ≈ 170 V
Frequency: f = 60 1/s = 60 Hz
Angular frequency: ω = 2*π*f= 376.8 rad/s ≈ 377 rad/s
Period: T = 1/f = 16.7 ms
Time function: v(t)=170 sin(377 t)

Note that in this case the time function could be given either as v(t)=311 sin (314 t+Φ) or v(t)=311 cos (314 t+Φ), since in the case of the outlet voltage we do not know the initial phase.

The initial phase plays an important role when several voltages are present simultaneously. A good practical example is the three-phase system, where three voltages of the same peak value, shape and frequency are present, each of which has a 120° phase shift relative to the others. In a 60 Hz network, the time functions are:

v_A(t)=170 sin (377 t)

v_B(t)=170 sin (377 t – 120°)

v_C(t)=170 sin (377 t + 120°)

The following figure made with TINA shows the circuit with these time functions as TINA’s voltage generators.

The voltage difference v_AB= v_A(t) – v_B(t) is shown as solved by TINA’s Analysis/AC Analysis/Time Function command.

Note that the peak of v_AB (t) is approximately 294 V, larger than the 170 V peaksof the v_A(t) or v_B(t) voltages, but also not simply the sum of their peak voltages. This is due to the phase difference. We will discuss how to calculate the resulting voltage (which is Ö3 * 170 @ 294 in this case) later in this chapter and also in the separate Three-phase Systems chapter.

Characteristic values of sinusoidal signals

Though an AC signal continuously varies during its period, it is easy to define a few characteristic values for comparing one wave with another: These are the peak, average and root-mean-square (rms) values.

We have already met the peak value V_M , which is simply the maximum value of the time function, the amplitude of the sinusoidal wave.

Sometimes the peak-to-peak (p-p) value is used. For sinusoidal voltages and currents, the peak-to-peak value is double the peak value.

The average value of the sine wave is the arithmetic average of the values for the positive half cycle. It is also called absolute average since it is the same as the average of the absolute value of the waveform. In practice, we encounter this waveform by rectifying the sine wave with a circuit called a full wave rectifier.

It can be shown that the absolute average of a sinusoidal wave is:

V_AV= 2 / π V_M ≅ 0.637 V_M

Note that the average of a whole cycle is zero.
The rms or effective value of a sinusoidal voltage or current corresponds to the equivalent DC value producing the same heating power. For example, a voltage with an effective value of 120 V produces the same heating and lighting power in a light bulb as does 120 V from a DC voltage source. It can be shown that the rms or effective value of a sinusoidal wave is:

V_rms = V_M / √2 ≅ 0.707 V_M

These values can be calculated the same way for both voltages and currents.

The rms value is very important in practice. Unless indicated otherwise, power line AC voltages (e.g. 110V or 220V) are given in rms values. Most AC meters are calibrated in rms and indicate the rms level.

Example 1 Find the peak value of the sinusoidal voltage in an electrical network with 220 V rms value.

V_M = 220/0.707 = 311.17 V

Example 2 Find the peak value of the sinusoidal voltage in an electrical network with 110 V rms value.

V_M = 110/0.707 = 155.58 V

Example 3 Find the (absolute) average of the sinusoidal voltage if its rms value is 220 V.

V_a = 0.637 * V_M = 0.637*311.17 = 198.26 V

Example 4 Find the absolute average of the sinusoidal voltage if its rms value is 110 V.

The peak of the voltage from Example 2 is155.58 V and hence:

V_a = 0.637 * V_M = 0.637 * 155.58 = 99.13 V

Example 5 Find the ratio between the absolute average (V_a) and rms (V) values for the sinusoidal waveform.

V/V_a = 0.707/0.637 = 1.11

Note that you cannot add average values in an AC circuit because it leads to improper results.

PHASORS

As we have already seen in the previous section, it is often necessary in AC circuits to add sinusoidal voltages and currents of the same frequency. Though it is possible to add the signals numerically using TINA, or by employing trigonometric relations, it is more convenient to use the so-called phasor method. A phasor is a complex number representing the amplitude and phase of a sinusoidal signal. It is important to note that the phasor does not represent the frequency, which must be the same for all phasors.

A phasor can be handled as a complex number or represented graphically as a planar arrow in the complex plane. The graphic representation is called a phasor diagram. Using phasor diagrams, you can add or subtract phasors in a complex plane by the triangle or parallelogram rule.

There are two forms of complex numbers: rectangular and polar.

The rectangular representation is in the forma + jb, where j = Ö-1 is the imaginary unit.

The polar representation is in the form Ae^{j j} , where A is the absolute value (amplitude) and f is the angle of the phasor from the positive real axis, in the counterclockwise direction.

We will use bold letters for complex quantities.

Now let’s see how to derive the corresponding phasor from a time function.

First, assume that all the voltages in the circuit are expressed in the form of cosine functions. (All voltages can be converted to that form.) Then the phasor corresponding to the voltage of v(t) = V_M cos( w t+f) is: V_M = V_Me ^jf , which is also called the complex peak value.

For example, consider the voltage: v(t) = 10 cos( w t+30°)

The corresponding phasor is: V

We can calculate the time function from a phasor in the same way. First we write the phasor in polar form e.g. V_M = V_Me ^jr and then the corresponding time function is

v(t)=V_M (cos(wt+r).

For example, consider the phasor V_M =10 – j20 V

Bringing it to polar form:

And hence the time function is: v(t) = 22.36 cos(wt – 63.5^°) V

Phasors are often used to define the complex effective or rms value of the voltages and currents in AC circuits. Given v(t) = V_Mcos(wt+r)= 10cos(wt+30°)

Numerically:

v(t) = 10*cos (wt-30°)

The complex effective (rms) value: V = 0.707*10* e^{– j30°} = 7.07 e^{– j30°} = 6.13 – j 3.535

Vice versa: if the complex effective value of a voltage is:

V = – 10 + j 20 = 22.36 e ^{j 116.5°}

then the complex peak value:

and the time function: v(t) = 31.63 cos ( wt + 116.5° ) V

A short justification of the above techniques is as follows. Given a time function
V_M (cos( w t+r), let’s define the complex time function as:

v (t) =V_M e ^jr e ^jwt = V_Me ^jwt = V_M (cos(r) + j sin(r))e ^jwt

where V_M =V_M e ^{j r t} = V_M (cos(r) + j sin(r)) is just the phasor introduced above.

For example, the complex time function of v(t)=10 cos(wt+30°)

v (t) = V_Me ^jwt = 10 e ^j30 e ^jwt = 10e ^jwt (cos(30) + j sin(30))= e ^jwt (8.66+j5)

By introducing the complex time function, we have a representation with both a real part and an imaginary part. We can always recover the original real function of time by taking the real part of our result: v(t) = Re {v(t)}

However the complex time function has the great advantage that, since all the complex time functions in the AC circuits under consideration have the same e^jwt multiplier, we can factor this out and just work with the phasors. Moreover, in practice we do not use the e^jwt part at all–just the transformations from the time functions to the phasors and back.

To demonstrate the advantage of using phasors, let’s see the following example.

Example 6 Find the sum and the difference of the voltages:

v₁ = 100 cos (314*t) and v₂ = 50 cos (314*t-45°)

First write the phasors of both voltages:

V_1M = 100 V_2M= 50 e ^{– j 45°} = 35.53 – j 35.35

Hence:

V_add = V_1M + V_2M = 135.35 – j 35.35 = 139.89 e^{– j 14.63°}

V_sub = V_1M – V_2M = 64.65 + j35.35 = 73.68 e ^{j 28.67°}

and then the time functions:

v_add(t) = 139.89 * cos(wt – 14.63°)

v_sub(t) = 73.68 * cos(wt + 28.67°)

As this simple example shows, the method of phasors.is an extremely powerful tool for solving AC problems.

Let’s solve the problem using the tools in TINA’s interpreter.

The amplitude and phase results confirm the hand calculations.

Now lets check the result using TINA’s AC analysis.

Before performing the analysis, let’s make sure that the Base function for AC ia set to cosine in the Editor Options dialog box from the View/Option menu. We will explain the role of this parameter at Example 8.

The circuits and the results:

Example 7 Find the sum and the difference of the voltages:

v₁ = 100 sin (314*t) and v₂ = 50 cos (314*t-45°)

This example brings up a new question. So far we have required that all time functions be given as cosine functions. What shall we do with a time function given as a sine? The solution is to transform the sine function to a cosine function. Using the trigonometric relation sin(x)=cos(x-p/2)=cos(x-90°), our example can be rephrased as follows:

v₁ = 100 cos(314t – 90°) and v₂ = 50 cos (314*t – 45°)

Now the phasors of the voltages are:

V_1M = 100 e ^{– j 90°} = -100 j V_2M= 50 e ^{– j 45°} = 35.53 – j 35.35

Hence:

V _add = V_1M + V_2M = 35.53 – j 135.35

V _sub = V_1M – V_2M = – 35.53 – j 64.47

and then the time functions:

v_add(t) = 139.8966 cos(wt-75.36°)

v_sub(t) = 73.68 cos(wt-118.68°)

Let’s solve the problem using the tools in TINA’s interpreter.

Let’s check the result with TINA’s AC Analysis

Example 8

v₁ = 100 sin (314*t ) and v₂ = 50 sin (314*t-45°)

This example brings up one more issue. What if all voltages are given as sine waves and we also wish to see the result as a sine wave?. We could of course convert both voltages to cosine functions, compute the answer, and than convert the result back to a sine function–but this isn’t necessary. We can create phasors from the sine waves in the same way that we did from cosine waves and then simply use their amplitude and phases as amplitude and phase of sine waves in the result.

This will obviously give the same result as transforming the sine waves to cosine waves. As we could see in the previous example, this is equivalent to multiplying by –j and then using the cos(x) = sin (x-90°) relation to transform it back to a sine wave. This is equivalent to multiplying by j. In other words, since –j × j = 1, we could use the phasors derived directly from the amplitudes and phases of sine waves to represent the function and then return to them directly. Also, reasoning in the same manner about the complex time functions, we could consider sine waves as the imaginary parts of the complex time functions and supplement them with the cosine function to create the full complex time function.

Let’s see the solution to this example using the sine functions as the base of the phasors (transforming sin( w t) to the real unit phasor (1) ).

V_1M = 100 V_2M= 50 e ^{– j 45°} = 35.53 – j 35.35

Hence:

V _add = V_1M + V_2M = 135.53 – j 35.35

V _sub = V_1M – V_2M = 64.47+ j 35.35

Note that the phasors are exactly the same as in Example 6 but not the time functions:

v₃(t) = 139.9sin(wt – 14.64°)

v₄(t) = 73.68sin(wt + 28.68°)

As you can see, it is very easy to obtain the result using sine functions, especially when our initial data are sine waves. Many textbooks prefer to use the sine wave as the base function of phasors. In practice, you can use either method, but don’t confuse them.

When you create the phasors, it is very important that all time functions are first converted either to sine or cosine. If you started from sine functions, your solutions should be represented with sine functions when returning from phasors to time functions. The same is true if you start with cosine functions.

Let’s solve the same problem using TINA’s interactive mode. Since we want to use sine functions as the base for creating the phasors, make sure that the Base function for AC is set to sine in the Editor Options dialog box from theView/Option menu.

and the time functions:

 

COMPLEX NUMBERS

PASSIVE COMPONENTS IN AC CIRCUITS

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