The sinusoidal waveform is one of the most important waveforms in Electrical Engineering. After many explanations about the dc circuit, let us move on to the next level material.

*Make sure to read what is the electric circuit first in order to understand the next explanation.*

Because up until now we limit the electric circuit basic theory only using dc circuit.

## What is Sinusoidal Waveform

Before we start, why we need to learn ac circuits? Although it is harder to learn, it comes with many benefits. For a history lesson, dc sources were the primary electric sources until the late 1800s.

At the beginning of the next century, the alternating current was introduced. Those two had a quite battle among the engineers.

Because ac has more efficiency and better economic transmission, thus it ended up as a winner.

The sinusoidal source is useful for ac circuit as the same with the dc source important role for dc circuit.

Make sure to read:

After we studied about the constant value of electricity over time, we will study about time-varying voltage and current. First thing, we will give our focus on sinusoidal electricity waveform or simply *sinusoid*.

A

sinusoidis a signal waveform which has the form of cosine or sine function.

For most of the time, alternating current (ac) refers to a sinusoidal waveform. Sinusoid means the value has positive and negative values in specific interval.

An ac circuit is a circuit which is driven by ac current or voltage sources.

For a starting, why do we need to learn ac circuit? What is the “better” side of ac besides it is more complex than dc? The ac sources has some “better” reasons to be learned.

**First** is its characteristic as a sinusoidal waveform. We often meet it in the motion of a pendulum, vibrate a string, or some ripples on the water surface.

**Second**, like we mentioned before, the sinusoidal signal is easy to generate and transmit. The electricity supplied to the world, homes, factories, industries, laboratories, etc has a sinusoidal waveform.

**Third**, we can implement Fourier Analysis for analysis of periodic signals which has the characteristic of a sinusoidal waveform.

**The last**, it is quite complicated to understand, but sinusoid is easy to analyze because its integral and derivative functions are also sinusoids.

After so many benefits from learning sinusoidal, do you know what mathematical theory we will use for ac circuit?

The most basic of the dc circuit analysis theory such as :

- Kirchhoff’s law AC
- Ohm’s law

will be applied again for the ac circuit. You need to understand some parameters found in the ac circuit such as :

**We will use the dc circuit analysis techniques for ac circuit such as :**

- Node analysis
- Mesh analysis

**Along with dc circuit theorems for ac circuit :**

- Superposition theorem
- Source transformation
- Thevenin and Norton

## Sinusoidal Waveform Function

Please look at the sinusoidal function of ac voltage :

(1) |

where

*V _{m}* =

*amplitude*of the sinusoid

*ω* = *angular frequency* in radians/s

*ωt* = the *argument* of the sinusoid.

The sinusoidal waveform in Figure.(1a) shows a function of its argument and Figure.(1b) as a function of time.

Figure 1. The figure of V sin _{m}ωt : (a) in function ωt, (b) in function t. |

From the figure, you will be able to grab the image of the “repeat cycle” of sinusoid every *T* seconds, hence, *T* is called the period of the sinusoid. From two figures in Figure.(1), we conclude *ωT* = 2π,

(2) |

Notice *v(t)* has repetitive value every *T* seconds is shown by replacing *t* by *t + T* in Equation.(1). We obtain

(3) |

Thus,

(4) |

that *v* has the same value at *t + T* the same as when it is at *t* and *v(t)* can be called *periodic*. In general,

A

periodic functionis one that fulfillsf(t) = f(t + nT), for alland for all integerst.n

*T* is different with *t*, where *T* is the time required for one cycle to complete or the number of seconds per cycle.

*T* is also called by period. The opposite or reciprocal value of *period (T)* is the *frequency (f)* of the sinusoid, means the number of cycles per second. Hence,

Let us see a **sinusoidal wave equation**,

(7) |

where (*ωt* + ø) is the argument and ø is the *phase*. Both argument and phase can be in radians or degrees.

There are two sinusoids,

(8) |

shown in Figure.(2). The starting point of *v _{2}* in Figure.(2) occurs first in time. From Figure.(2) we can say that

*v*leads

_{2}*v*by ø or that

_{1}*v*lags

_{1}*v*by ø.

_{2}If ø ≠ 0, we also say that *v _{1}* and

*v*are out of phase.

_{2}If ø = 0, then *v _{1}* and

*v*are called to be in phase; they reach their minima and maxima at the same exact time.

_{2}We are able to compare *v _{1}* and

*v*in this matter because they operate with the same frequency even the amplitude is different. Below we will see

_{2}**how to graph sinusoidal functions**.

Figure 2. Two sinusoids with different phases |

Every sinusoid waveform can be expressed in either cosine or sine form. When comparing two sinusoids, it is wise to express in the same waveform as either cosine or sine with positive amplitudes.

This can be achieved by using the following trigonometric function:

(9) |

Using these identities, it is easy to notice that

(10) |

With these relationships, we can transform a sinusoid from cosine form to sine or vice versa.

A graphical drawing can be used to compare or relate as an alternative approach to using the trigonometric function in Equations.(9) and (10).

Figure 3. Graphical (a) cos(ωt – 90^{o}) = sinωt, (b) sin(ωt + 180^{o}) = -sinωt |

Notice the set of axes in Figure.(3a).

The horizontal axis refers to the magnitude of cosine and the vertical axis (points down) refers to the magnitude of sine.

The angle is measured by positive value in a counterclockwise direction from the horizontal axis. This calculation method can be used to compare two sinusoids.

For example, we consider in Figure.(3a) from subtracting 90^{o} from cos*ωt* results sin*ωt*, or cos(*ωt* – 90^{o}) = sin*ωt*.

Similarly, adding 180^{o} to sin*ωt* results -sin*ωt*, or sin(*ωt* + 180^{o}) = -sin*ωt*, as can be seen in Figure.(3b).

This graphical method can add two sinusoids with the same frequency when one has sine waveform and the other has cosine waveform.

In order to add A cos*ωt* and B sin*ωt* like shown in Figure.(4a), A is the magnitude of cos*ωt* and B is the magnitude of sin*ωt*.

Figure 4. (a) Adding A cosωt and B sinωt, (b) Adding 3 cosωt and -4 sinωt. |

Hence,

(11) |

where

(12) |

For simple example, we add 3 cos*ωt* and -4 sin*ωt* like shown in Figure.(4b) and results

(13) |

From these posts, we will learn about AC Power Analysis:

We will also looking for some applications such as:

After finishing the single-phase AC circuit, we will move on to analyzing the three-phase AC circuit and read a brief explanation of how does a wattmeter work.

## Example of Sinusoidal Function

For better understanding, let us review examples below :

**1. Find the amplitude, phase, period, and frequency from**

*v(t)* = 12 cos(50*t* + 10^{o})

__Solution :__

Amplitude : *V _{m}* = 12 V

Phase : ø = 10^{o}

Angular frequency : *ω* = 50 rad/s

Period : *T* = 2π/*ω* = 2π/50 = 0.1257 s

Frequency : *f* = 1/*T* = 7.958 Hz.**2. Determine the phase angle between v_{1} = -10 cos(ωt + 50^{o}) and v_{2} = 12 sin(ωt – 10^{o}). State which one is leading.**

__Solution: __There are three ways to solve the problem. We can use trigonometric identities for the first two and use the graphical approach for the third.

** Method 1**:

*v*and

_{1}*v*are in different forms, so we need to make them have the same form, say cosine form.

_{2}(1.1) |

and

(1.2) |

It can be concluded from (1.1) and (1.2) that the phase difference between these two is 30^{o}. We can rewrite *v _{2}* as

(1.3) |

Comparing (1.1) and (1.3) shows *v _{2}* leads

*v*by 30

_{1}^{o}.

** Method 2** : Let us express

*v*in sine form

_{1}But *v _{2}* = 12 sin(

*ωt*– 10

^{o}). Comparing these two shows

*v*lags

_{1}*v*by

_{2}*ωt*, same with

*v*leads

_{2}*v*by 30

_{1}^{o}.

** Method 3** : We may state

*v*as simply -10 cos

_{1}*ωt*with +50

^{o}phase shift. We can draw

*v*as in Fig.5.

_{1}Figure 5 |

Similarly, *v _{2}* is 12 sin

*ωt*with -10

^{o}phase shift as drawn in Figure.(5). We can conclude

*v*leads

_{2}*v*by 90

_{1}^{o}– 50

^{o}– 10

^{o}= 30

^{o}.

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