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February 1968 Radio-Electronics
 [Table of Contents]
Wax nostalgic about and learn from the history of early electronics.
See articles from Radio-Electronics,
published 1930-1988. All copyrights hereby acknowledged.
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Mathophobes beware; I cannot be held accountable for instances of hyperventilation,
fainting, intense sweating or nervousness, post traumatic stress, etc. When in high
school and junior high, I suffered from the common malady. However, after setting
my sights on earning an electrical engineering degree, I embraced mathematics as
a necessary skill to be successful. In fact, I did very well in all my math classes
even up through calculus, differential equations, and even higher levels. It soon
became evident that my fellow students who struggled the most - even failed - at
engineering courses were those who never had a handle on math. Since becoming comfortable
with higher mathematics, my appreciation for its abilities to describe the physical
world has constantly grown. It is also nice when reading technical articles to at
least somewhat understand the math behind the concepts. Author Norman Crowhurst
attempts in this 1968 Radio-Electronics magazine article to sooth the pain
of complex numbers a bit. Here are
Part 1,
Part 2, and
Part 3.
Imaginary Numbers Are a Cinch - Part 3
Fig. 1 - Diagram of high pass filter, and what it does in simple
math terms.
Want a low- or high-pass filter? Find out how to design your own
By Norman H. Crowhurst
When I next dropped by George's lab, he'd been doing his homework on imaginary
numbers. He'd made a neat tabulation of all the results he could find about the
simple two-element low-pass filter we had been working with and started extending
it to a three-element type. Now he wanted to know more.
"I think I could go ahead with more complicated low-pass filters, with a little
trial and error," he said. "I suppose you can apply the same method to high-pass
filters, but I couldn't see it."
"Let's just reason it out as we did for the low-pass." I sketched a high-pass
circuit and started translating its performance into an equation (Fig. 1). "The
output is an inductor in shunt with the terminating resistor, and its normalized
value is unity. Ee let the inductive reactance be represented as b units at cutoff
frequency, as we did with the shunt capacitor of the low-pass filter. Should these
units represent reactance or susceptance really, at cutoff frequency?"
"That's what had me stymied," admitted George.
"You can use either, with the proper care," I went on. "I find it simpler to
use susceptance for all shunt elements and reactance for all series elements. Then
we show its variation with frequency and phase of the elements by where we write
x and j."
"So the susceptance of the inductor will be jb/x?" queried George, who
had obviously been thinking about this.
"Last week we set a standard about phase for this kind of calculation," I reminded
him. "It was that a + j means the quantity leads the reference quantity.
Here, reference quantity is voltage across both the output resistance and the shunt
inductor, so the j applies to the inductor current."
"And current lags voltage in an inductance," went on George, "so it should be
-j, right?"
"Right. But an easier way is to just remember that the j and x always
go together; this automatically takes care of the sign." I showed him that b/jx
is the same as -j(b/x). He was already figuring out the expression for this
high-pass filter.
George was taking to the use of j like a duck takes to water. In short
order, he had the phase and amplitude expressions figured out, as well as values
for a and b for the constant-resistance type. He set these down alongside those
he had tabulated for the low-pass unit.
"Hey," he said, "the values and, when you make ab = 1, the phase angles are the
same for both, except for the sign. What does that mean?"
"Simply that the transfer phase of these two filters is 180° apart at all
frequencies, when they use the same cutoff frequency as in a crossover."
"Can you also show from this that a crossover like these makes constant resistance
and delivers constant total power?"
"For power, take the reciprocals of the amplitude-squared expressions, and add
them," I suggested, which he did (Fig. 2) almost as quickly as I said it, and found
the answer reduced to 1/1.
"Does that prove the constant-resistance property as well?"
"Not directly, you need to figure out the input admittance of each filter and
then ado the two together." I showed him step by step how to do this for the low-pass
(Fig. 3). He did the same for the high-pass, converted to x instead of 1/x as the
variable in this case, added them, and found the sum was again equivalent to 1/1.
"Show me a three-element low-pass design," George asked. I drew the design (Fig.
4-a). I put in a, b, and c to represent the normalized reactance and susceptance
of the elements at cutoff, and we started figuring.
"The voltage across the output inductor is jcx," said George, "so the
total voltage at the mid-point capacitor is (1 + jcx) , isn't it?" I nodded,
so he went on, "Then the current through the capacitor is ..." and he wrote it out:
jbx (1 + jcx). "Now what?"
"What's the current through the input inductor?" I suggested.
Fig. 2 - Power condition of the network.
Fig. 3 - Computing Input admittance of the low-pass filter.
Fig. 4 - Design (a) of low-pass filter and the graph (b) of its performance.
Fig. 5 - How to make X2 and X4 disappear.
Fig. 6 - Changing slope curve by calculus.
Fig. 7 - Transform math into a filter.
Fig. 8 - Movement of point on a curve.
Fig. 9 - 0verall amplitude-squared math expression.
"It's the current through the capacitor, plus that going to the output - is output
current 1?" he asked. I nodded, so he wrote, for current in the input inductor:
1 + jbx(1 + jcx). "Then the voltage across the input inductor is ...
and he wrote out:
jax {1 + jbx (1 + jcx) }.
"Now you just add the voltages together to get the input voltage," I suggested,
sketching in the voltages on the schematic. By this time George was working it,
to come up with the answer:
1 + jcx + jax {1 + jbx (1 + jcx) }.
"Yuk," he commented, so I took over and removed the parentheses, till we had:
1 - abx2 + j(a + c)x - jabcx3
"Now what?" asked George. "Just solve for a, b, and c, I suppose you'll say."
"That isn't as difficult as you might think, for the constant-resistance or maximal-flatness
case. First, let's figure out the amplitude-squared expression."
George squared the real and imaginary parts and added the terms in x2
and x4 to come up with
1 + [ (a + c)2 - 2ab ] x2 +
[ a2b2 - 2abc(a + c) ] x4 + a2b2c2x
which evoked another "Yuk !" when he'd finished.
"Well, we can simplify that somewhat," I said. "First we want the rolloff asymptotic
to cutoff, so the coefficient of x6 must be one."
"Come again," said George, confused by the abstract math terms.
So I sketched out the response on log/linear paper (Fig. 4-b). I explained that
the 1 of the amplitude-squared expression represents the horizontal part of the
response in the passband, while the x6 of the last term represents an
18-dB-per-octave slope, crossing the zero line at cutoff frequency.
"So that means abc must equal 1." George promptly made that substitution by crossing
out abc and a2b2c2.
"Next, for maximal flatness, the terms in x2 and x4 have
to disappear," I told him.
"So the whole thing boils down to just (1 + x6)?" asked George, and
added, "Oh, I see now what you meant by its being asymptotic."
I nodded. "'The curve approaches a straight line, but never quite gets there."
"That's negotiable," he said, writing out (a + c)2 - 2ab = 0 for x2
and a2b2 - 2(a + c) = 0 for x4 (Fig. 5). Then he
looked stumped.
"Because abc = 1, how about writing 1/c for ab?" I suggested, which he immediately
did.
"From the x4 equation, 2(a + c) = 1/c2, so (a + c) = 1/2c2,
right?" he asked. I nodded. "So I can substitute this in the x2 equation
to find c?"
He was doing it and came up with the conclusion c = 1/2, followed by ab = 2,
then a = 1.5, and finally b = 4/3.
"Now we know the coefficients, let's look at phase," I suggested. George had
the expressions for y, y2 and Φt with values of a, b and c substituted
to give a simple numerical result (Fig. 5).
"One element shifts 45° at cutoff, two elements make it 90°, does the
third element shift it to 135°?" He crossed out the x's in the arctan expression
and found it simplified down to -1, which is the tangent of 135°. Then he spotted
that 90° would be where the denominator is zero, or x2 = 1/2 and
x = 0.707, while 180° would be where the numerator is zero, or x2
= 2 and x = 1.414.
"The whole thing looks easier by the minute," he commented. "Don't bother with
the high-pass section, I'll work that some other time, for my own amusement. What
I want to get to is that filter you designed that started us off on all this. Can
you show me that?"
"Well, that's a little different," I admitted. "In the first place, all the filters
we've discussed so far are 'ideal' designs: they are exactly what we want them to
be - constant resistance, uniform total power. But there is no such thing as a perfect
linear-phase filter, so we have to find what is the best we can do with a given
number of components. Sometimes calculus will help, sometimes it's a matter for
trial and error, with diminishing deviations. So I think we should look at that
kind of problem first."
"Oh yes, you were going to show me how calculus helps," George commented. "Can
you give me a quick example?"
"Calculus deals with slopes," I explained. "It's not too easy to work out for
phase response, but it's easy with amplitude response and we use the same principles
for phase. Suppose you have a response with rolloff something like this," and I
sketched it out (Fig. 6). "Detailed analysis shows its amplitude expression is of
the form (1 + x2 + x4) which yields a 4.77-dB loss at the
90° point, where x = 1. Suppose we want to add a circuit that will reduce the
loss to zero at this frequency, with minimum deviation up to this frequency."
"Sounds difficult. How do you go about it?"
"First step is to cook up the required overall amplitude-squared response, for
which two requirements will set maximum flatness: boost equaling attenuation at
x = 1 (4.77 dB for both); and, slope at the same point equal and opposite to that
of the existing circuit, both measured on a dB/log frequency scale." I sketched
in how this worked.
"As we're interested in frequencies up to the normalizing point of x = 1," I
went on, "it's easy and useful to work in that reference, and find out how good
the result is when we get through. The attenuation of 4.77 dB (log 3 = 0.477) merely
means that 1 + x2 + x4 is 3 when x = 1 (which is fairly obvious).
We use a two-reactance filter that provides a peak before rolloff, which means its
amplitude response will be of the form y2 = 1 - ax2 + bx4.
We must solve for the two conditions: boost equal to attenuation, and equal, but
opposite slopes; both at x = 1, to find a and b. Then we can convert this information
into a designable filter circuit, by use of the j operator."
"I get the idea," said George, "but it's not all clear yet. For boost to equal
attenuation, the amplitude-squared expression should be the reciprocal at x = 1,
shouldn't it?" I nodded. "So (1 - a + b) must have a value of) 1/3," he went on.
"But how do you figure out the equal-slope part?"
"That's where calculus comes in," I replied. "The slope we are interested in
is plotted on log frequency and log amplitude (dB) scales, so the slopes we want
are expressed by
(d log y)/(d log x), writing y for the amplitude. Because we use expressions
for amplitude squared, it's easier to work in y2 and x2 as
variables, so we'll use that form." Showing him the standard differentials in a
handbook he had, I wrote the conversion
For the first expression, y2 = 1 + x2 + x4.
I figured this out to
Making x = 1 reduces this to a simple 3/3 = + 1. For the expression y = 1 + ax2'
+ bx4, the slope figures out to
Making x = 1 in this expression reduces it to (-a + 2b)/(1-a + b), which needs
to be -1.
So the equation for equal slopes may be written: a - 2b = 1 - a + b, which rearranges
to 2a - 3b = 1. And the amplitude requirement equation reduces to a - b = 2/3.
George soon had these solved, to a = 1, b = 1/3. "But how do you make that into
a filter?" he wanted to know.
"Assume it's of this configuration," and I drew it out (Fig. 7), using c and
d as constants for the inductor and capacitor. "Then you solve for values of c and
d that make the amplitude-squared equation come out to
y2 = 1 - x2 + 1/3x4."
George was glad of the exercise in using j and in short order had equations
in c and d that made a = 1 and b = 1/3. Together we solved them, using a slide-rule
to get approximate values of c = 0.395 and d = 1.462. Using the reciprocal of d,
which is 0.685, we could use a reactance chart to design such a filter for any terminating
resistance and cutoff frequency.
"How good is the correction this filter makes?" George wanted to know next.
"We can use calculus for that too. The maximum error will be a stationary point
on the graph of the amplitude-squared equation, which means the slope of the graph
is zero." I sketched this.
"Just a minute there," interjected George, "you've lost me, momentarily."
I showed him (Fig. 8) that a point moving along a curve can go up or down; or
it can momentarily do neither, in which case it is stationary at the top of a peak
or at the bottom of a dip. I pointed out that an amplitude-squared response with
terms in x2 and x4 can have one stationary point (a peak)
if the coefficient of x2 is negative.
Thus, the combined response after "correction" can have three stationary points,
one at zero frequency, followed by a dip and a peak before final roll off. "To find
these points, we differentiate the amplitude-squared expression, for convenience
using x2 as the variable rather than x, and equate to zero (for zero
slope).
"The overall amplitude-squared equation is ... " George began, when
he'd grasped this, and he wrote it out to y2 = 1 + 1/3x4 -
2/3x6 + 13xs.
I differentiated it (not bothering with log terms, because a stationary point
has zero slope no matter which scale is used) and got (see Fig. 9)
Equating this to zero, x2 must be zero, or 1/2, or 1. The point at
which x2 = 1 is the point for which we designed, s the point where x2
= 1/2, at 0.707 below cutoff, represents a dip. Substituting x2 = 1/2
into the amplitude-squared expression gives the fraction 49/48. Taking 10 times
the logarithm of 49/48 gives the deviation in dB as 0.089.
George was impressed with how small a deviation this correction could achieve
- less than 0.1 dB. Before I left, he wanted to know what else operator j
could be used for.
I suggested it could be used to analyze anything that uses vectors. We had discussed
vector analysis before.
George was convinced his new knowledge of how to use operator j was
going to prove a very useful tool. R-E
Posted April 21, 2023
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