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Unformatted text preview: EEL 4930 Audio Engineering Lecture g; P W Trigonometric Identities In Lecture 1 we needed the following trig identity:
cos(x + y) = cos(x) cos(y) — sin(x) sin(y). We noted that it’s easy to derive using the properties of complex numbers. Let
six = cos(x) +j sin(x) and ejy = cos(y) + j sin(y) ; 610‘”) = [cos(x) + i sin(x)]*[ cos(y) +1“ sin(y)] = [cos(x) cos(y) — sin(x) sin y] + j [sin(x) cos(y) + cos(x)sin(y)] Re[ej(x+y)] =Eos(x+y) = cos(x) cos(y) — sin(x) sin(y)! (1) <2>
Also, _ (3)
Adding (2) and (3) yields (4)
Subtracting (3) from (2) yields (5) For x = A+B and y = A—B, x+y = 2A and x—y = 2B, so ' sin(A) + sin(B) = 2 sin[(A+B)/2] cos[(A—B)l2] (6) EV
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: l H ezl’r I EEL 4930 Audio Engineering u, a Lecture 2. #1, P: 4/7 Forced Vibration Previously we considered free vibration, where the spring—massdamper system was disturbed and
then left to os'cillate on its own until the vibration died out due to the damping. Now we consider the case where the system is acted on by a force varying sinusoidally with time
. t . . .
given by F = Re[Fo elm ], where F0 1s a complex constant. We can write this as m dzu/dtz + b du/dt + k u = F0 ej‘”t or
m [dzu/dtz + 2y du/dt + 03,2 u] = F0 ejmt where 2y = b/m and 0002 = k/m, or dzu/dt2+ (a) /Q) du/dt + u) 2 u = F /mejwt where Q: (1) /2y = on m/b.
0 0 0 0 0 With some effort we could show that complete solutions to such system are sums of two parts. One
is the transient response, which is of the form that we found in the last lecture: x1(t) = C e'thos (colt + 4)) where y = (Do/2Q and col = V0002 — y2 = cooi/ 1 — 1/(4Q2). For ﬁnite Q values, this part will eventually
die out because of the e'yt factor. Eventually a second term will dominate; this second term is called
the forced response, and is a vibration at the same frequency as. the forcing function, with constant amplitude and phase: x = Re[A ejwt]. Lettin u(t) = A eJmt we get
\ 2 g
dzu/dtz + (mo/Q) du/dt + 0002 u = Fo/m ejmt ‘(ot . 'wt ‘oot 'cot
—Ao)2 eJ + _] (mo/Q) Aw eJ + 0002 A e] = FO/m eJ —Aoo2 + j (COO/Q) Aw + 0002 A = FO/m A((D) = (FO/m) / [(0)02  (1)2) +j(1/Q) (1)060]
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W {6’5 é4f7 9f$t44 7: aéfcrb «if/7 ‘ ‘ EEL 4930 Audio Engineering 2. o /0 Lecture 2' F» The Helmholtz Resonator is closely analogous to the mass—spring system. It consists of a
container of air with an opening. The volume of air in the opening vibrates because of the
springiness of the air in the container. You’ve probably played with one when you blew over an
empty bottle to make a low whistling tone. Let the neck of the container have length L and cross
sectional area S. Then the mass of the air in the neck is
m = pSL where p is the density of air. If the air mass moves a distance x into the container, it
compresses the air in the container so that the air volume
decreases from V to V—Sx. Thus the pressure in the
container rises from PA to PA+p. You might expect that the pressure increase would be
proportional to the volume decrease, but this is a so
called adiabatic process, so the compression affects the
temperature, and the actual pressure rise is p/PA = —y AV/V = —y Sx/V, where y s 1.4 is a constant
pertaining to speciﬁc heats and PA is the ambient
pressure. Now, the mass m feels the force of the difference in
pressures inside and outside the container, so the net
force is F = pS,
so F = ma, or dZX/dt2 = F/m. Substituting for F and m, we get
dzx/dt2 = pS/pSL = —[y S PA/VpL] x We’ve seen equations like this before. This is the differential equation for simple harmonic motion,
with frequency of oscillation 030 = VY S PA/VpL
The speed of sound is c = VyPA/p , so too = c S/(VL) or ftJ = c/21r VS/(V L) As an example, consider a 1 liter bottle with S = 3 cm2, L = 5 cm. At typical temperatures and
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 Spring '08
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 Audio Engineering, Trigonometry, Trigraph

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