# Sine Functions

### Why do we need Sine functions?

Much of what you will work with in electronics requires alternating current, meaning you will encounter sine waves, square waves, and triangle waves.  It might have been a few years since trigonometry, but you will quickly realize that mastering the basics of trigonometry is essential for your success in electrical engineering.  Sine functions are one thing you will be seeing a lot of in electrical engineering.  There is no avoiding them.  Later in the curriculum, you will even learn how to break down square waves and triangle waves into series of sine waves.

The general equation for sinusoidal voltage signal is given by

$v(t) = A \sin(\omega t + \Phi) + V_{DC} = A \sin( 2 \pi f t + \Phi) + V_{DC}$

Where A is the amplitude, $\omega$ is the angular frequency, t is time, $\Phi$ is phase shift, and $V_{DC}$ is a constant voltage.

Likewise, current follows the same general equation.

$i(t) = A \sin(\omega t + \Phi) + I_{DC} = A \sin( 2 \pi f t + \Phi) + I_{DC}$

The equation for the sine wave in Figure 1 is $v(t) = 4\cdot\sin(2 \pi \cdot 1000t) + 2V$.  The following sections will explain how to derive the equation for this sine wave as well as any other sine function that you may encounter.

### Amplitude

The amplitude (A) is found by taking the difference between the highest point and the lowest point of a waveform and dividing by two.  In Figure 2, the highest point (VH) is 6 V while the lowest point (VL) is -2 V.  The amplitude, therefore, is 4 V.

$A = \frac{V_{H} - V_{L}}{2} = \frac{6-(-2)}{2} = 4V$

### Period, Frequency, and Angular Frequency

The period (T) of a waveform is defined as the time it takes for that waveform to complete one cycle.  Figure 2 shows two periods, or cycles, of a sine wave.  For this sine wave, the period is equal to 1 ms.  Identifying the period length allows you to derive the frequency (f) and angular frequency (ω) of the signal.

$\omega = 2 \pi \cdot f$       $\omega = \frac{2\pi}{T} = \frac{2\pi}{.001} = 6283 rad/s$     $f = \frac{1}{T} = \frac{1}{.001} = 1000Hz = 1kHz$

Frequency is defined as the number of cycles a waveform completes in one second.  One cycle per second is known as a Hertz (Hz).  The waveform above has a frequency of 1000 cycles per second, or 1 kHz.

The units for angular frequency are radians per second (rad/s).  Angular frequency is similar to the frequency measured in Hertz, except now we are measuring radians rather than cycles.  Because one cycle is equal toradians (360°), the angular frequency is derived by multiplying the frequency by.  A common error in homework problems occurs when students neglect to convert one measure of frequency to another.  Pay attention to which form you should be using and always remember the equation $\omega = 2\pi \cdot f$.

### Offset Voltage

When a DC voltage is applied to an AC voltage signal, the AC signal is said to have a DC offset voltage.  A signal’s offset voltage can also be seen as the signal’s average voltage.  You can use integration to find the average value of a time-varying function, but for sine functions, it is easier to use observation.

Consider the sine wave in Figure 3.  The equation for this sine wave is $v(t) = 4 \sin(6283t)$. The waveform is oscillating around zero and has no offset.  In contrast, the sine wave in Figure 4 has an offset of 2 V.  This time, the waveform is oscillating around 2 and its equation is $v(t) = 4 \sin(6283t) + 2V$.

When determining the offset of a sine function, remember that an offset is a constant value.  Adding a constant to a sine function only changes the level about which the sine wave oscillates.

### Phase Shift

A sine function has a phase shift when it does not begin its cycle at t = 0. A phase shift is illustrated in the figure to the right. To calculate phase shift, use the equations,

$\Phi_{degrees} = \frac{t_{1} - t_{2}}{T} \cdot 360 ^ {\circ}$ $\Phi_{radians} = \frac{t_{1} - t_{2}}{T} 2\pi rad$

where t1 is the waveform’s original starting point, t2 is where the wave’s starting point has been shifted to, and T is the period.

Example: Find the phase shift of the sine function in Figure 5.

Solution: The sine wave has been shifted from its original starting position (t1) of 0 ms to its new starting point (t2) at -0.4 ms.  Therefore,
t1 = 0 ms, t2 = -0.4 ms, and T = 1 ms.  Solving for degrees, the phase shift is equal to 144°.

$\phi = \frac{0-(-0.4 ms)}{1 ms} \cdot 360^ {\circ} = 144^ {\circ}$

Defining phase shift with respect to t = 0 works fine in mathematics, but real life won’t tell you when time is equal to zero.  In labs this semester you will be observing signals on an oscilloscope.  For a single signal, you can assume that there is no phase shift.  Phase shift becomes important when more than one signal is involved.  There are times when knowing the phase shift between two signals is crucial (the design of an audio system is one instance).

In instances with multiple signals, the phase shift of one signal is measured relative to another signal. One wave is your reference point from which you measure all other waves.  Figure 6 shows V1 as the reference and V2 as the wave that has been shifted.  Figure 7 shows the opposite.  Once you decide on a reference, the same formula given above applies.

$\phi = \frac{t_{1} - t_{2}}{T} \cdot 360^{\circ}$

The sign of the phase shift will vary depending on which signal you use as a reference.  The phase shift from V1 to V2 in Figure 6 is positive because V2 is shifted to an earlier time.  This is a confusing concept for many people.  Though it appears that V1 is ahead of V2, this is not the case.  V1 starts its cycle a full 3 ms before V2 does.  V1, therefore, is leading V2.

Another way of explaining this is by using the equation.  In Figure 6, t1 – t2 must be positive because t1 is larger.  In Figure 7, t1 – t2 must be negative.

$\phi_{figure6} = \frac{3 ms}{15 ms} \cdot 360^ {\circ} = 72^ {\circ}$

$\phi_{figure7} = \frac{-3 ms}{15 ms} \cdot 360^ {\circ} = -72^ {\circ}$

Written by Ryan Eatinger (reatinge@ksu.edu).  Thank you!