## Why Sine Waves?

One of the most commonly used test functions for a circuit or system is the sine wave. This is not because sine waves are a particularly common signal. They are in fact quite rare - the transmission of electricity (a 60 Hz sine wave in the U.S., 50 Hz in much of the rest of the world) is one example. The reason sine waves are important is complex and involve a branch of Mathematics called Fourier Theory. Briefly put: any signal going into a circuit can be represented by a sum of sine waves of varying frequency and amplitude (often an infinite sum). As a simple example, consider the graphs shown below. The top graph shows a triangle wave. The middle graph shows a number of sine waves of varying frequency and amplitude. The bottom graph shows the sum of the sine waves (red) and the original triangle wave (dotted black).

Clearly even just a few sine waves are sufficient in this case to closely approximate the original function. Fourier states that any function (with some very minor restrictions that won't concern us), can be represented in this way.

This is why sine waves are important. Not because they are common, but because we can represent arbitrarily complex functions using only these very simple function.

## Determining system output given input and transfer function

Given that sine waves are important, how can we analyze the response of a circuit or system to sinusoidal inputs? There are many ways to do this, depending on your mathematical sophistication. Let's use a fairly basic explanation that uses phasors. If you are unfamiliar with phasors, you can find a description in almost any circuits or systems textbook.

Using complex impedances it is possible to find the transfer function of a circuit. For example, the circuit below is described by the transfer function, H(s), where s=

*jω*.Circuit | Transfer Function |
---|---|

Consider the case where R=1 and C=0.1. In that case:

Generally we know the input V

_{i}and want to find the output Vo. We can do this by simple multiplication
If we have a phasor representation for the input and the transfer function, the multiplication is simple (multiply magnitudes and add phases). Finding the output becomes easy. Let's look at some examples:

##### Example 1

and the transfer function evaluates to

The output is just the product of the input and the transfer function (evaluated as phasors)

Note that V

_{o}has an amplitude of 0.95 and lags V_{i}by 72°. It is a phase "lag" because the output lags, or follows, the input (the input goes up before the output so the output is following the input).##### Example 2

Change input phase

Both input and output have shifted 40° from those in Example 1.

##### Example 3

Change input frequency

Frequency has changed, magnitude of output has increased, but phase lag has decreased (to 45°).

##### Example 4

Cosine Input

Note: all angles are given in degrees. They should be changed to radians before evaluation by calculator or computer.

nice post

ReplyDelete