March 1933 Radio News
[Table
of Contents]
These articles are scanned and OCRed from old editions of the Radio & Television News magazine. Here is a list of the Radio & Television
News articles I have already posted. All copyrights are hereby acknowledged.

It has been a while since I saw the quotient rule for derivatives
applied. Probably the last time was in a college text book, because
I'm pretty sure I haven't had the occasion to use it since then
 except maybe back in the days when I was writing my
RF Workbench software and needed to derive closed form solutions
for group delay in filters. This 1933 article from Radio News
magazine presented the quotient rule as part of a discussion for
finding the impedance of a load for maximum power transfer. Pure
resistances were used in the example, but the method applies as
well to complex impedances.
See all available
vintage
Radio News articles.
Mathematics in Radio
Calculus and Its Application in Radio
By J. E. Smith* Part Twenty
Figure 1.
Figure 2.
Figure 3.

In order to prove a very important and fundamental relation in
the application of calculus to radio circuits, let us investigate
the simple series circuit of Figure 1. The current through this
circuit is:
(1)
The power delivered to the output resistance r is:
(2) p_{2} = i^{2} R
Now, from (1):
(3)
and substituting this value of i^{2} in (2) we have :
(4)
In the above equation, let e = 1 and r_{0} = 10 ohms.
Tabulating the results of p_{2} for various values of r,
we have:
When r = 0, the corresponding value of p_{2} = 0
When r = 1, the corresponding value of p_{2} =
When r = 5, the corresponding value of p_{2} =
When r = 10, the corresponding value of p_{2} =
When r = 12, the corresponding value of p_{2} =
When r = 15, the corresponding value of p_{2} =
When r = 20, the corresponding value of p_{2} =
Plotting these values graphically as shown in Figure 2, we find
that the maximum power occurs when the output resistance r is equal
to 10 ohms. But this was also the value for the resistance r_{0},
so it can be concluded that the maximum power in an output resistance
in a circuit similar to Figure 1 is obtained when the output resistance
is equal to the internal resistances of the other parts of the circuit
r_{0}.
In this simple case it was rather easy to plot the values of
the variables and note the condition of maximum power. But there
are many equations which are rather involved and where this method
would be rather inconvenient. Let us investigate the possible use
of calculus for solving this type of problem. It will be necessary
at this time to point out a few additional and simple theorems of
the fundamental calculus. Let us refer to the curve of Figure 3,
which is similar to the one shown in the previous figure. The direction
of such a curve at any point can be defined to be the same as the
direction of the line tangent to the curve at that point. Therefore,
a line PT tangent to the curve at point A will have an angle expressed
as follows:
We have previously learned that this is represented as the slope
and is the value of the derivative at that point. It will be noticed
that as the function is increasing, this tangent line will in general
be making an acute angle with the axis of x. It will be equal to
a positive number, since the sin θ = 0, and cos θ =
a are in the first quadrant and have positive values:
thus:
When the function is decreasing, a tangent line drawn at any
point such as C in Figure 3 will be making an obtuse angle "θ"
with the axis of x. It will therefore be equal to a negative number;
thus:
From the above analysis, since the derivative is a positive number
up to the maximum point of the curve and a negative number beyond
this point, it follows that at the maximum point it must be equal
to zero.
This very important relation gives us a method of solving the
following equation for the appropriate value of r.
(a)
In order that the power p_{2} become a maximum, it follows:
(b)
Solving:
(c)
The solution of equation c appears to be a little complicated, but
it is simply an application of the elementary forms of differentiation.
Showing the intermediate steps of the solution, we have, first,
the derivative of a quotient which is of the form:
Solving:(d)
Second, it will be recalled:
Then (d) becomes:
(e)
This becomes:
(f)
* President, National Radio Institute.
Posted March 25, 2014
