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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.
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