Engineering problems can rarely be completely described by a single real-valued
function of one real-valued variable. (This is both good news and bad news.
It is what makes engineering interesting, and it is what makes engineering
hard!) This means that most of the skill in visualization (at least at the
level of two-dimensional plotting) is the ability to figure out what to graph,
and what not to graph, in order to convey insight into the real problem.
In general, this is accomplished by graphing a simplified version of the more
general function that describes the solution to a problem. In other words,
having solved a problem, an engineer must then find a function whose graph
conveys the essentials of the solution without overwhelming the reader with
extraneous detail.
The rest of this lesson illustrates this idea in action. We will study the
problem of forced oscillation, which shows up in all areas of
engineering. That is, we will try to understand how a physical system (such as
a car) vibrates when subjected to a repetitive external force (such as bumps in
the road). We will try to impress upon you the critical role that
two-dimensional plotting has in communicating the solutions to these kinds of
problems.
Now that you know about gain, you can easily calculate the oscillation of a car
as it bounces down the road given the gain of its suspension and the function
describing the rate at which we hit the bumps in the road. Suppose that the
road is described by the expression
A cos(2 Pi f t)
Remember that A characterizes the height of the bumps and f characterizes
their spacing. Since we are measuring f in units of Hz, which are
oscillations per second, the value of the variable t should be specified in
units of seconds.
If the transfer function of the car is given by the function G(f I), what
expression will describe the car's bouncing?
Click here for the answer
Use what you've learned so far to plot, on the same graph, the oscillation of a
road and the oscillation of a car. Let A = 1 and produce several plots for
different values of f and for different kinds of cars. You can pick some
interesting values for f by looking at the plot of the overall gain for the
car. On the resulting plots, the horizontal axis will be time (in seconds) and
the vertical axis will be amplitude.
What's unrealistic about the graphs that you have just produced? Hint: no
physical system can react instantly to an input!
Click here for the answer
In fact, the amount by which the oscillation of a particular car lags the
oscillation of a road depends upon the characteristics of the car's suspension
and upon the frequency of the bumps. (In other words, the time lag depends
upon the same two factors as the gain.) This time lag is known as the
phase delay.
The phase delay is just as easy to compute as the gain if you know the transfer
function. The phase delay at a given frequency f is the angle of
G(f I). Mathematica uses the function Arg[z] to give the angle of the complex number
z.
Graph the phase delay of each car's suspension over frequencies ranging from 0
to 20 Hz, much as you graphed the gain. As with the gain, you will want to
graph them all on the same plot.
In these plots, the horizontal axis will be in units of Hz, but what are the
units of phase delay on the vertical axis? A negative phase delay means that
the response (the bounce) will follow the signal (the bump). It gives the
number of radians by which the response follows the signal. Since there are
2&pgr; radians in a complete oscillation, a phase delay of --&pgr;, for
example, would mean that the response follows the signal by half an
oscillation.
Now try this. If you know that the phase delay of a system at a frequency of
f Hz is -p radians, by how many seconds will the response trail the
signal?
Click here for the answer
A phase delay of -p radians means that the response will trail the signal by
the following fractional part of a complete oscillation: p/(2 &pgr;). At a
frequency of f Hz, each oscillation lasts 1/f seconds. So the delay in
seconds will be p/(2 f &pgr;).
For example, a phase delay of --&pgr; radians at 10 Hz translates into a delay of
Pi/(2 10 &pgr;) = 1/20 of a second.
You already know that the formula describing oscillation is of the form
A cos(2 &pgr; f t)
You have been graphing a lot of such curves during this lesson. In all of
these graphs, what is true when t = 0?
Click here for the answer
In all of them, the curve is at its high point (i.e., it has the value A)
when t=0. This is a consequence of the fact that cos(0) = 1.
The formula describing a phase-delayed oscillation is
A cos(2 &pgr; f t -- p)
where p is the phase delay. Experiment with plotting this formula, letting
A and f be one, for various values of p between 0 and --2&pgr; radians.
It'll help to plot more than one curve on the same graph. What happens when
p is --2&pgr;?
Now you know all that you need to know to graph the response of a particular
car to a particular bumpy road. If the road is described by the formula
A cos(2 &pgr; f t)
and you know the transfer function G of the car, what formula will
describe the phase-delayed oscillation of the car?
Click here for the answer
We already know how to how to take the gain into account:
A Abs[ G(f I)] Cos[2 Pi f t]
To account for the phase delay, we simply compute and subtract the phase delay:
A Abs[G(f I)] Cos[2 Pi f t -- Arg[G(f I)]]
Now you can pick a frequency and a car and plot its response to the bumps in
a road.