Hein ten Horn wrote:

Hmm, let's examine this.
From the two composing oscillations you get the overall
displacement:
y(t) = sin(2 pi 220 t) + sin(2 pi 224 t)
From the points of intersection of y(t) at the time-axes you
can find the period of the function, so examine when y(t) = 0.
sin(2 pi 220 t) + sin(2 pi 224 t) = 0
(..)
(Assuming you can do the math.)
(..)
The solutions a
t = k/(220+224) with k = 0, 1, 2, 3, etc.
so the time between two successive intersections is
Dt = 1/(220+224) s.
With two intersections per period, the period is
twice as large, thus
T = 2/(220+224) s,
hence the frequency is
f = (220+224)/2 = 222 Hz,
which is the arithmetic average of both composing
frequencies.

As I said before, it might be correct to say that the average, or
effective frequency is 222 Hz. But the actual period varies from
cycle to cycle over a period of 1/(224-220).

the amplitude of which is easily plotted versus time using Mathematica,
Mathcad, Sigma Plot, and even Excel. I think you should still give that a
try.


No peculiarities found.

Perhaps you would agree that a change in period of less than 2% might
be difficult to observe - especially when you're not expecting to see
it. To more easily find the 'peculiarities' I suggest that you try
using more widely spaced frequencies.

gr, Hein

gr right back at ya,

jk

Jim Kelley wrote:
Hein ten Horn wrote:

Hmm, let's examine this.
From the two composing oscillations you get the overall
displacement:
y(t) = sin(2 pi 220 t) + sin(2 pi 224 t)
From the points of intersection of y(t) at the time-axes you
can find the period of the function, so examine when y(t) = 0.
sin(2 pi 220 t) + sin(2 pi 224 t) = 0
(..)
(Assuming you can do the math.)
(..)
The solutions a
t = k/(220+224) with k = 0, 1, 2, 3, etc.
so the time between two successive intersections is
Dt = 1/(220+224) s.
With two intersections per period, the period is
twice as large, thus
T = 2/(220+224) s,
hence the frequency is
f = (220+224)/2 = 222 Hz,
which is the arithmetic average of both composing
frequencies.

As I said before, it might be correct to say that the average, or effective
frequency is 222 Hz. But the actual period varies from cycle to cycle over
a period of 1/(224-220).

the amplitude of which is easily plotted versus time using Mathematica,
Mathcad, Sigma Plot, and even Excel. I think you should still give that a
try.


No peculiarities found.

Perhaps you would agree that a change in period of less than 2% might be
difficult to observe - especially when you're not expecting to see it. To
more easily find the 'peculiarities' I suggest that you try using more
widely spaced frequencies.

Before we go any further I'd like to exclude
that we are talking at cross-purposes.
Are you pointing at the irregularities which
can occur when the envelope passes zero?
(That phenomenon has already been
mentioned in this thread.)


gr right back at ya,

:-)
"gr" is not customary, but, when writing it satisfies
in several languages: German (gruß, grüße),
Dutch (groet, groeten) and English.

Adieu, Hein