Hi Neil,
Start with a small input voltage level and sweep over the frequency of interest, find the frequency which causes the maximum output. If it clips, lower the input. If it doesnt clip, raise the input until it does clip. Now you know the input that causes clipping, and the input frequency that causes a max output.
You want to keep the input below that which causes clipping in the actual application.
The graph is made by running the frequency along the x axis (horizontal axis) and the amplitude along the y axis (vertical). That will give you an output vs frequency plot.
If you want to use semi log paper you can run frequency along the log axis, and have a linear amplitude plot.
If you want to use log log paper, you'll have both running along log axises.
In practice it is typical to see log frequency and linear amplitude, or both log. It's also typical to plot the amplitude in either regular units (like volts) or in db. To get volts db from the output voltage use the formula:
V(db)=20*log10(V(volts))
where log10 is the log base 10 log.
So you apply one frequency input at a given voltage determined above, then measure the output, then plot it on the paper or else convert to db and then plot that.
Note that you need to make sure the input voltage stays at the same level for the whole plot, even though you change the frequency. This means monitoring the input level as well as the output.
Now monitoring the input and output isnt quite as simple as all that. You need to use a volt meter that is capable of measuring frequencies in the range of interest. Most low cost multimeters on the AC scale are made to measure power line frequencies, like 50Hz maybe up to 400Hz. That's not adequate for measuring frequencies in the full audio range. You'll need to find a meter that can measure AC voltages at frequencies up to maybe 15kHz or higher, preferably 20kHz or 25kHz for audio work. If you cant find one or dont want to buy one, then you might be better off using a passive half wave rectifier circuit using a Schottky diode (not a regular rectifier diode). Unfortunately, this isnt perfectly simple either because you really have to calibrate the rectifier. Perhaps an op amp based rectifier circuit, but it has to use an op amp that can work up to maybe 10MHz or something like that, although this is really a different subject area that would be best with its own topic thread.
Small note:
Log paper is sometimes used because it allows plotting the frequency along the horizontal, and often in filters we dont see much change over a wide range of frequencies so we like to squeeze the horizontal axis. Log paper allows this and gives a little nicer picture of the response. If you dont have paper with the log scale running on the horizontal though, you can make your own roughly. If you mark off several points along the horizontal axis with equal spacing (as you would with any type of plot anyway) and then instead of marking them 1, 2, 3, etc., you mark those points 1, 10, 100, 1000, 10000, 100000. You then measure the response and plot using that kind of scale instead of a linear one. If that's not enough resolution, another trick is to use 1, 2, 4, 8, 16, 32, 64, 128, etc., where you just double the frequency to get the next mark point. Of course you can probably get a pic of log paper on line and use that as a guide too.
To get closer to actual real log paper, you can use the formula:
x=log10(y)
where
log10 is the base 10 log function,
y is the desired log mark point,
x is the actual position along the axis in linear units.
For example, to plot the log scale using this method you would first mark off equal distant points along the x axis as before, but then to mark sub divisions you would use that formula:
to mark the position for 2 you would use:
x=log10(2)=0.301 approximately,
so you would mark a point that is 0.3 times the total distance between the 1 and the 10 on the scale you already have marked. Just to give you an idea where this is, it's about 1/3 of the way from the 1 to the 10 (a little less of course).
Similarly, to mark the position for 3 you would use:
x=log10(3)=0.477 approximately,
so you would make a point that is 0.477 of the distance from the 1 to the 10 (this is close to half way between the 1 and the 10).
Not too difficult i suppose
Just a few ideas.