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).
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.
Given that sine waves are important, how can we analyze the response of a circuit or system to sinusoidal inputs (after all transients have died out  the socalled sinusoidal steady state)? 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. A technique using Laplace Transforms is given here.
For a system of the type we are studying (linear constant coefficient) if the input to a system is a sinusoid at a particular frequency, then the output of the system is also a sinusoid at the same frequency, but typically with a different amplitude or phase. Put another way, if the input to a system (described by the transfer function H(s)) is A·sin(ω·t+φ) then the output is M·A·sin(ω·t+φ+θ). This is likewise true for cosine  only the magnitude and phase of the output relative to the input changes. This is shown below.
In this diagram the magnitude of the sinusoid has changed by a factor of M (which we will take to be a positive real number) and the phase has changed by a factor of θ (a real number, not necessarily positive). It is our task to find the value of M and θ for a particular system, H(s), at a particular frequency, ω. We call M the magnitude of the system (or transfer function) at ω, and we call θ the phase of the system at that frequency.
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:
and the transfer function evaluates to
The output is just the product of the input and the transfer function (evaluated as phasors). Thus the magnitude will change by a factor of 0.316 (this is the gain of the system) and the phase will change by 71.6° (this is the phase of the system).
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).
Change input phase
Both input and output have shifted 40° from those in Example 1.
Change input frequency
Frequency has changed, magnitude of output has increased, but phase lag has decreased (to 45°).
Cosine Input
Note: all angles are given in degrees. They should be changed to radians before evaluation by calculator or computer.
Sinusoidal functions are important because functions of time can be broken down into a sum of sinusoids. Given a system given with a sinusoidal input, we can determine the output in a straightforward manner from the transfer function. These two facts, together, make the determination of a transfer functions to sinusoidal inputs a useful endeavor (and, ultimately, quite powerful).
The difficulty in representing the transfer function comes about because we need to plot a complex number, H(s) or H(jω), as a function of frequency. Consider the transfer function
To show how the magnitude and phase vary we need two plots one for magnitude (magenta, in the plot below) and one for phase (cyan).
Note: Standard Bode plots are logarithmic on the frequency axis, and plot the magnitude in dB's (deciBels). We'll explore that in the next installment.
We can use these plots to determine the solutions to the examples 1 to 4 without resorting to algebra (though this presupposes that we have the graphs shown  techniques for drawing these graphs come in subsequent sections).
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°.
Change input phase
Both input and output have shifted 40° from those in Example 1.
Change input frequency
Frequency has changed, magnitude of output has increased, but phase lag has decreased (to 45°).
Cosine input (i.e, a change of phase).
The magnitude and phase plots determine the phasor representation of the transfer function at any frequency. On the graphs below we can see that at 10 rad/sec the phasor representation of the transfer function has a magnitude of 0.707 and a phase of 45°. This means that at 10 rad/sec the magnitude of the output will be 0.707 times the magnitude of the input and the output will lag the input by 45°.
TThe frequency response of a system is presented as two graphs: one showing magnitude and one showing phase. The phasor representation of the transfer function can then be easily determined at any frequency. The magnitude of the output is the magnitude of the phasor representation of the transfer function (at a given frequency) multiplied by the magnitude of the input. The phase of the output is the phase of the transfer function added to the phase of the input.
To get a more intuitive idea of what the frequency response represents, consider the system below. (Hit start button to show animation)
Click here for an animation of an analogous electrical system.
Animation by Ames Bielenberg
The transfer function of the system is given by (with m=1, b=0.5, k=1.6, u=input to system, y=output (the position of the mass):
The magnitude and phase plots are shown below.
You can see by the animation that at low frequencies the input and output are equal in magnitude, and in phase (after the initial startup transient dies out). This is shown by a magnitude of one and a phase of zero. At intermediate frequencies the system is somewhat resonant, and the output actually gets larger than the input (but there is a growing phase lag, i.e., negative phase). As frequency increases further, the output decreases; again, you can see this both in the animation and in the magnitue plot. The outline of the peaks of the output plot is similar to the magnitude plot above. The phase is not as obvious, but it obviously starts at 0° and then decreases to 180° (you may need to zoom in to see the phase shift). At 180° phase shift the two waveforms are completely out of phase, when one is at a maximum, the other is at a minimum.
© Copyright 2005 to 2015 Erik Cheever This page may be freely used for educational purposes.
Erik Cheever Department of Engineering Swarthmore College