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Lecture 2

# Lecture 2 - EEL 4930 Audio Engineering Lecture g P W...

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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 x1 @uagﬁrw‘01wma’64 3w: *YZW - g 3.152: F hTJl/L (Q, 75 FWan 97.?“ 05L? VM7 3/009/ 5’0 Sea/“ﬁlm’s are vu7 AW? SIMk/fﬁééa/é FM {0(4) 61 I ea «730 we I 4* ._ _ #1 _,:5: 32/ 7 ________._,_,___.4.__.___.‘..___.....__.‘.-.____.._....__._.._.__,‘.______,,_. r.______..-V. _ . A...__._. £7 ﬁlﬂﬂ/B inc Q=/ooo7 l . r‘o’f I 5‘0 M‘ my 2: e a a 72—0 / N Vac/Z400 4’6 i It; 3 m (/90) / (mo) _ i . —— ___..r. n 07‘ a 4.6 ~= : _ /o:c;((azm’ was “4.562; mob Lycés 2‘; way ‘7 4046. 54"“) Q»’/ tit fmzﬂnézﬂ /°JZL in A {m VIA/4 for? £75733“ is 2779 ! l , 3 1(5): C e—Y—T—‘(ol (Jf +42) 7‘ t’ (/J—T i l . \$1,421 flax (QT/awe Cwéufl+ﬂ)=l/ z ‘ 2 film” 15 ékx, } 3 ‘ ’ 2 i 52} Erley/ 4/!”— 62 tr = éKC/e kt] / J , z i \$n/é4nej7 Wf=if+7 73 J—ZPCC/e'}(t+ﬂ] i i x l _\,t/ -2Y(t’+'r> ‘ ch’z e Z A i «die E 774.). W 4/? /;A:. z Z ,5 L K I; ~7'Yt’ : l H e-zl’r I EEL 4930 Audio Engineering u, a Lecture 2. #1, P: 4/7 Forced Vibration Previously we considered free vibration, where the spring—mass-damper 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] This has magnitude lAI = (Fe/m) / \/{[(0002 - 002)]2 + [(l/Q) 030(1)]2} and phase angle ./_A = (b = tan‘1{[(1/Q) wochooz — (02)} IA(00)| is relatively small for (0 << 000, reaches a peak near (1) = (no, and then decreases towards zero for 00 >> (no . For 00 << too, the response and the forcing function are nearly in phase with each other; when w = (no the response lags 90° behind the forcing function; for 00 >> (no the response is 180° out of phase with the forcing function F. Lg/AMPAU 4M§ a {mug/‘Moéva. F = Eei‘kmd his“? f- El €54 qqsd Zola ' #2 M :5 x Ww7 fa HEW IVS/W % a 575E4- ZTI— (Ln/g JIM 74a m/a/fﬁou? ﬁzz/@4444 ¥ K’Q M % a POM/7f“, . . WMFWGf/ Jim/f new ,we aw; SM/Mrs/ZW \$57451,“ W flu/t: ‘I’J 73/6) - A (a) e 0 / A m Mead (4,74 4M;Zz9nr¢2) ML aim“ W? 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 pressures, the speed of sound is 344 m/s, which gives a resonant frequency about 130 Hz. 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Lecture 2 - EEL 4930 Audio Engineering Lecture g P W...

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