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From: Phil Hobbs <pcdhSpamMeSenseless@electrooptical.net>
Newsgroups: sci.electronics.design
Subject: Re: Omega
Date: Sun, 30 Jun 2024 09:31:29 -0400
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On 2024-06-30 03:44, Cursitor Doom wrote:
> Gentlemen,
> 
> For more decades than I care to remember, I've been using formulae
> such as Xc= 1/2pifL, Xl=2pifC, Fo=1/2pisqrtLC and such like without
> even giving a thought as to how omega gets involved in so many aspects
> of RF.  BTW, that's a lower-case, small omega meaning
> 2*pi*the-frequency-of-interest rather than the large Omega which is
> already reserved for Ohms. How does it keep cropping up? What's so
> special about the constant 6.283 and from what is it derived?
> Just curious...
> 

As an old colleague of mine from grad school would say, "It just comes 
out in the math." ;)

The 2*pi factor comes from the time domain / frequency domain 
conversion, and the basic behavior of linear differential equations with 
constant coefficients. (That's magic.(*))  For now we'll just talk about 
LR circuits and pulses.

A 1-second pulse (time domain) has an equivalent width of 1 Hz 
(frequency domain, including negative frequencies).  That's pretty 
intuitive, and shows that seconds and cycles per second are in some 
sense the same 'size'.  The two scale inversely, e.g. a 1-ms pulse has 
an equivalent width of 1 kHz, also pretty intuitive. (Equivalent width 
is the mathematical quantity for which this 1-Hz/1-s inverse relation 
holds exactly, independent of the shape of the waveform.)

Moving gently towards the frequency domain, we have the ideas of 
resistance and reactance.  Resistance is defined by

V = IR, (1)

independent of both time and frequency.  Actual resistors generally 
behave very much that way, over some reasonable range of frequencies and 
power levels.  Either V or I can be taken as the independent variable, 
i.e. the one corresponding to the dial setting on the power supply, and 
the equation gives you the other (dependent) variable.

A 1-Hz sine wave of unit amplitude at frequency f is given by

I = sin(2 pi f t),  (2)

and the reactance of an inductance L is

X = 2 pi f L.    (3)

The reactance is analogous to resistance, except that since inductance 
couples to dI/dt rather than I.  From the definition of inductance,

V = L dI/dt.   (4)

Plugging (2) into (4), you get

V = L dI/dt = L * (2 pi f) cos(2 pi f t) = X_L cos(2 pi f t),    (5)

where X_L is the inductive reactance.

We see that the voltage dropped by the inductance is phase shifted by 
1/4 cycle. Since the cosine reaches its peak at 0, where the current 
(the independent variable) is just going positive, we can say that the 
voltage waveform is _advanced_ by a quarter cycle, i.e. that the voltage 
is doing what the imposed current was doing a quarter cycle previously. 
(This seems like a fine point, but it's crucial to keeping the sign of 
the phase shift right, especially when you're a physics/engineering 
amphibian like me--the two disciplines use opposite sign conventions.)

Besides the phase shift, the voltage across the inductance has an extra 
factor of 2 pi f.  This is often written as a Greek lowercase omega, 
which for all you slipshod HTML-mode types is ω = 2πf.

Writing the sine wave as

I = sin(ωt) (6)

is faster, but the factor of 2 pi in amplitude keeps coming up, which it 
inescapably must, and it doesn't even really simplify the math much.

For instance, if we apply a 1-V step function across a series RL with a 
time constant

tau = L/R = 1 second,    (7)

the voltage on the resistor is

V = 1-exp(-t).    (8)

In the frequency domain, the phase shift makes things a bit more 
complicated.  If we use our nice real-valued sinusoidal current waveform 
(6) that we can see on a scope, then (after a small flurry of math), the 
voltage on the resistor comes out as

V = sin(t - arctan(omega L/R)) / sqrt(1 + (omega L / R)**2). (9)

This is because sines and cosines actually are sums of components of 
both positive and negative frequency, and which don't behave the same 
way when you differentiate them:

sin(omega t) = 1/2 * (exp(j omega t) - exp(-j omega t)) (10)

and

cos(omega t) = 1/2 * (exp(j omega t) + exp(-j omega t)). (11)

By switching to complex notation, and making a gentlemen's agreement to 
take the real part of everything before we start predicting actual 
measurable quantities, the math gets much simpler.  Our sinusoidal input 
voltage becomes

Vin = exp(j omega t) (12)

and the voltage across the resistor is just the voltage divider thing:

V/Vin = R / (R + j omega L).  (13)

At low frequencies, the resistance dominates and the inductance doesn't 
do anything much, just a small phase shift

theta ~= - j omega L/R.

At high frequencies, the inductance dominates.  In the middle, the two 
effects become comparable at a frequency

omega0 = R/L.

At that frequency, the phase shift is -45 degrees and the amplitude is 
down by 1/sqrt(2) (-3 dB) and the power dissipated in the resistor falls 
to half of its DC value.

If we're using the series LR as a lowpass filter, that's the frequency 
that divides the passband, where the signal mostly gets through, from 
the stopband, where it mostly doesn't.

So when we think in the time domain, a 1-ohm/1-henry LR circuit responds 
in about a second, whereas in the frequency domain, its bandwidth rolls 
off at omega = 1, i.e. at 1/(2 pi) Hz.

With sinusoidal waveforms, we can think of 1 second corresponding to 1 
radian per second, whereas with pulses, a 1 second pulse has a 1-Hz-wide 
spectrum (counting negative frequencies).

Thing is, a sine wave varies smoothly and goes through a much more 
complicated evolution (positive to negative and back) within a cycle, so 
it just takes longer, by a factor that turns out to be 2*pi.

Cheers

Phil Hobbs


(*) Kipling, "How the Rhinoceros got his skin"

-- 
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC / Hobbs ElectroOptics
Optics, Electro-optics, Photonics, Analog Electronics
Briarcliff Manor NY 10510

http://electrooptical.net
http://hobbs-eo.com