September 1932 Radio News
[Table
of Contents]
Wax nostalgic about and learn from the history of early electronics.
See articles from Radio &
Television News, published 1919  1959. All copyrights hereby acknowledged.

This entry level introduction of differential calculus as it applies
to electronic circuit analysis appeared way back in a 1932 edition
of Radio News. It was written by none other than Sir Isaac
Newton himself (just kidding, of course). Author J.E. Smith
created an extensive series of lessons that began with simple component
and voltage supply descriptions and worked up through algebraic
manipulations and on finally to calculus. I remember not being the
best math student in high school (OK, one of the worst), but once
I got an appreciation for the power of mathematics for analyzing
electronics, mechanics, physics, and even economics, my motivation
level soared to where I craved more of it and ended up receiving
"As" in all my college math courses. That is truly an indication
that while not everyone can excel at math, the proper environment
can make a world of difference.
Mathematics in Radio  Differential Calculus
Calculus and Its Application in Radio
Part Eighteen
By J. E. Smith*
We are learning that the study of Calculus is not as difficult
as it first appeared to be and that it is essential to the understanding
of advanced electrical theory and practice.
The practical engineer generally asks himself to what extent
the use of Calculus will enable him to advance more rapidly in his
work. This is more effectively answered by the fact that a knowledge
of this subject, when once appreciated, induces a confidence in
one's self that is invaluable.
In order to appreciate more fully the practical applications
of Calculus to Radio, let us develop in more detail some of the
relations in Radio theory which are already known. The study of
such a subject only requires a correlation of known facts.
We have stated previously that the differential calculus deals
with the rate of change of one variable with respect to another.
Let us investigate the simple sine wave of Fig. 1, in order to bring
out a little more thoroughly the relationship involving the "rate
of change." We have here a sine wave function of "e" or "i" which
varies with the value of θ. But going a step further, the
rate of change of the sine wave at any instant can be expressed
in another way. Thus, with reference to Fig. 2 (a), a line PT has
been drawn tangent to the sine wave at point A. The slope of such
a line is defined as the ratio of 0 to a. This ratio is approximately
equal to 1. Consider Fig. 2 (b), and at point B, it is noticed that
the slope of the line is determined from the ratio of approximately
11 divisions to 37 divisions which is about 1/3. Now, if the line
PT is drawn tangent to the curve where it goes through the zero
axis at point a as shown in Fig. 3, it is apparent that its slope
is greater than for points A and B. It is seen to be in the ratio
of about 25 to 17, or approximately 1.5.
The slope of the line drawn tangent to the curve at any point
determines the rate of change of the wave. From the above analyses,
the rate of change at 0 is greater than at A. In like manner, the
rate of change at A is greater than at B, and for a point on the
curve which is its maximum, such as E max in Fig. 1, the slope must
be zero. Thus, the rate of change is maximum at zero (o) and minimum
at the top of the wave (E max).
The use of calculus will show this result immediately. Let the
wave of Fig. 1 be expressed as follows:
(I) e = E max sin θ
But it was shown in the last article that the value of the derivative
at any point on the curve is equal to the slope of the line drawn
tangent to the curve at that point.
Let us take the derivative of (I) with respect to θ:
(II)
This is of the form
(cv), the solution of which has been
shown to be equal to
.
Here, E max equals C and v
equals sin θ.
Equation (II) becomes:
(III)
= E max cos θ
Plotting III in Fig. 4, it is noticed that calculus has determined
immediately the rate of change at any instant of the sine wave.
It is also in agreement with the above analyses, as it is noticed
that the rate of change is maximum when the sine wave passes through
zero and is at minimum when the crest of the wave has been reached.
This analysis is a very important one, and further use will be made
of it in later discussions.
Reactive Electromotive Force Drop with Inductance
Let us consider how the use of calculus shows the relation of
the current with respect to the voltage in an inductive circuit.
We think of the inductance of a circuit as being associated with
the surrounding magnetic flux and also with the current flowing
through the circuit. In considering the inductive circuit of Fig.
5, we recall that there is a magnetic flux around the coil due to
the current i. The selfinductance of this circuit is simply equal
to the ratio of the total magnetic flux Φ to the current i.
This is expressed as:
(I) L =
(II) Therefore, Φ = L i
If the current in Fig. 5 is changing at every instant as would
be the case with an alternating current, the magnitude of the magnetic
flux around the coil would likewise change. This changing magnetic
flux induces an e.m.f. in the coil, and the magnitude of this induced
e.m.f, "e" at any instant is equal to the rate of change of the
magnetic flux with respect to time (t). This is expressed as:
(III) e =
The above relation can be used to advantage in connection with
equation (II) in order to show the relation of the current in an
inductive circuit with respect to the impressed voltage. Equation
(III) tells us that we can take the derivative of equation (II)
with respect to (t). Thus:
(IV)
=
Here, L is a constant, and (IV) is of the form
,
the solution of which is c
.
Then (III) becomes:
(V) e = L =
The instantaneous value of the current is expressed by the following
equation:
(VI) i = Imax sin θ = Imax sin ωt
Now, equation (V) tells us that we can take the derivative of equation
(VI) with respect to "t". Thus:
(VII)
=
(Imax sin ωt)
Here, I max is a constant, and (VII) is of the form
(cv). The various steps of differentiating are as follows:
(Imax sin ωt) = Imax
(sin ωt) = Imax cos ωt
(ωt) = ω Imax cos ωt
The solution of (V) thus becomes:
(VIII) e = ω L Imax cos ωt
This last formula, which has been easily obtained by the use
of differential calculus gives us all the information that is necessary
in order to show the relation of the current in the circuit of Fig.
5, with respect to the impressed voltage. From equation VIII, we
recognize the reactance formula for inductance, that is,
xL = 2
π
f L = ω L.
Plotting the instantaneous value of the current and the value
of the voltage e, as obtained in equation VIII, the graph, Fig.
6, immediately informs us that the current in an inductive circuit
lags behind the impressed voltage by 90°.
* President, National Radio Institute.
Posted May 4, 2014
