presented by sri vastav with Matlab illusion.
The previous document made the case that the study of sinusoidal inputs is important, and showed how phasor representations of the input and the system transfer function can be used to easily determine system output. This document will explain why the standard way of representing the transfer function is with two plots: magnitude vs. frequency and phase vs. frequency. At the bottom of this page there is an animation to help develop an intuitive understanding of the concepts describe in this page.
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 graph this, the most straightforward way (with a computer) might be to plot the value of H(s) as the frequency changes. This yields the blue line in the three-dimensional plot shown below.
It would obviously be hard to get accurate information about the real and imaginary parts of H(s) from such a plot. It is easier if we plot the real and imaginary parts as a function of frequency (the red and green projections of the blue line). Clearly, in this case, two 2-dimensional graphs (one for real and one for imaginary) are superior to a single 3-dimensional graph.
However, in the last document we showed that to easily determine the output given the input, we would like to have the transfer function in phasor notation. This means that we should make a plot of magnitude and phase. Again, we could make a single 3-dimensional plot (the blue line), but it would be easier to interpret the results if we make two 2-dimensional plots (the magenta and cyan lines).
To clarify further, lets make separate plots of the magnitude and phase.
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.
Consider the examples from the previous document.
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°.
To get a more intuitive idea of what the frequency response represents, consider the system below. (Hit start button to show animation)
For an animation of an analogous electrical system, go here.
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):
You can see by the animation that at low frequencies the input and output are equal, and in phase. At intermediate frequencies the system is somewhat resonant, and the output actually gets larger than the input (but there is a growing phase lag). As frequency increases further, the output decreases. The outline of the peaks of the output plot is similar to the magnitude plots above (the phase plot is not obvious, but it obviously starts at 0° and then decreases - if you type "ctrl +" you can zoom in to see the phase shift). In this case, the magnitude plot would start at one (output=input) at low frequencies, it would then increase, followed by a decrease.
Bode plots (introduced next) formalize a particular method for drawing magnitude and phase plots (as a function of frequency) associated with a given transfer function