14.2. The Mean Value and the RootMeanSquare Value. Introduction. Prerequisites. Learning Outcomes


 Amos Baldwin Hicks
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1 he Men Vlue nd the RootMenSqure Vlue 4. Introduction Currents nd voltges often vry with time nd engineers my wish to know the men vlue of such current or voltge over some prticulr time intervl. he men vlue of timevrying function is defined in terms of n integrl. An ssocited quntity is the rootmensqure (r.m.s). For exmple, the r.m.s. vlue of current is used in the clcultion of the power dissipted by resistor. Prerequisites Before strting this Section you should... Lerning Outcomes On completion you should be ble to... be ble to clculte definite integrls be fmilir with tble of trigonometric identities clculte the men vlue of function clculte the rootmensqure vlue of function HELM (8): Workbook 4: Applictions of Integrtion
2 . Averge vlue of function Suppose timevrying function f(t) is defined on the intervl t b. he re, A, under the grph of f(t) is given by the integrl A = f(t) f(t) dt. his is illustrted in Figure 5. f(t) m b t b t () the re under the curve from t = to t = b Figure 5 (b) the re under the curve nd the re of the rectngle re equl On Figure 3 we hve lso drwn rectngle with bse spnning the intervl t b nd which hs the sme re s tht under the curve. Suppose the height of the rectngle is m. hen re of rectngle = re under curve m(b ) = f(t) dt m = b he vlue of m is the men vlue of the function cross the intervl t b. f(t) dt Key Point he men vlue of function f(t) in the intervl t b is b f(t) dt he men vlue depends upon the intervl chosen. If the vlues of or b re chnged, then the men vlue of the function cross the intervl from to b will in generl chnge s well. Exmple Find the men vlue of f(t) = t over the intervl t 3. Solution Using Key Point with = nd b = 3 nd f(t) = t men vlue = b f(t) dt = 3 3 t dt = [ t 3 3 ] 3 = 3 3 HELM (8): Section 4.: he Men Vlue nd the RootMenSqure Vlue
3 sk Find the men vlue of f(t) = t over the intervl t 5. Use Key Point with = nd b = 5 to write down the required integrl: Your solution men vlue = Answer 5 5 t dt Now evlute the integrl: Your solution men vlue = Answer 5 5 t dt = 3 [ t 3 3 ] 5 = 3 [ ] = = 3 Engineering Exmple Sonic boom Introduction Impulsive signls re described by their pek mplitudes nd their durtion. Another quntity of interest is the totl energy of the impulse. he effect of blst wve from n explosion on structures, for exmple, is relted to its totl energy. his Exmple looks t the clcultion of the energy on sonic boom. Sonic booms re cused when n ircrft trvels fster thn the speed of sound in ir. An idelized sonicboom pressure wveform is shown in Figure 6 where the instntneous sound pressure p(t) is plotted versus time t. his wve type is often clled n Nwve becuse it resembles the shpe of the letter N. he energy in sound wve is proportionl to the squre of the sound pressure. P p(t) t P Figure 6: An idelized sonicboom pressure wveform HELM (8): Workbook 4: Applictions of Integrtion
4 Problem in words Clculte the energy in n idel Nwve sonic boom in terms of its pek pressure, its durtion nd the density nd sound speed in ir. Mthemticl sttement of problem Represent the positive pek pressure by P nd the durtion by. he totl coustic energy E crried cross unit re norml to the sonicboom wve front during time is defined by E = <p(t) > /ρc where ρ is the ir density, c the speed of sound nd the time verge of [p(t)] is <p(t) > = p(t) dt () Find n pproprite expression for p(t). () () (b) Hence show tht E cn be expressed in terms of P,, ρ nd c s E = P 3ρc. Mthemticl nlysis () he intervl of integrtion needed to compute () is [, ]. herefore it is necessry to find n expression for p(t) only in this intervl. Figure 6 shows tht, in this intervl, the dependence of the sound pressure p on the vrible t is liner, i.e. p(t) = t + b. From Figure 6 lso p() = P nd p( ) = P. he constnts nd b re determined from these conditions. At t =, + b = P implies tht b = P. At t =, + b = P implies tht = P /. Consequently, the sound pressure in the intervl [, ] my be written p(t) = P (b) his expression for p(t) my be used to compute the integrl () p(t) dt = = = P ( P t + P ) dt = [ 4P 3 t3 P ( ] t + P t ) = P /3. t + P. ( 4P t 4P ) t + P dt Hence, from Eqution (), the totl coustic energy E crried cross unit re norml to the sonicboom wve front during time is E = P 3ρc. Interprettion he energy in n Nwve is given by third of the sound intensity corresponding to the pek pressure multiplied by the durtion. HELM (8): Section 4.: he Men Vlue nd the RootMenSqure Vlue 3
5 Exercises. Clculte the men vlue of the given functions cross the specified intervl. () f(t) = + t cross [, ] (b) f(x) = x cross [, ] (c) f(t) = t cross [, ] (d) f(t) = t cross [, ] (e) f(z) = z + z cross [, 3]. Clculte the men vlue of the given functions over the specified intervl. () f(x) = x 3 cross [, 3] (b) f(x) = x cross [, ] (c) f(t) = t cross [, ] (d) f(z) = z 3 cross [, ] 3. Clculte the men vlue of the following: () f(t) = sin t cross [ ], π (b) f(t) = sin t cross [, π] (c) f(t) = sin ωt cross [, π] (d) f(t) = cos t cross [ ], π (e) f(t) = cos t cross [, π] (f) f(t) = cos ωt cross [, π] (g) f(t) = sin ωt + cos ωt cross [, ] 4. Clculte the men vlue of the following functions: () f(t) = t + cross [, 3] (b) f(t) = e t cross [, ] (c) f(t) = + e t cross [, ] Answers. () (b) (c) (d) 4 (e) () (b).693 (c).948 (d) 3. () (b) (c) [ cos(πω)] π π πω (d) + sin ω cos ω (g) ω 4. () 4 (b).75 (c).75 9 π (e) (f) sin(πω) πω 4 HELM (8): Workbook 4: Applictions of Integrtion
6 . Rootmensqure vlue of function If f(t) is defined on the intervl t b, the mensqure vlue is given by the expression: b [f(t)] dt his is simply the men vlue of [f(t)] over the given intervl. he relted quntity: the rootmensqure (r.m.s.) vlue is given by the following formul. Key Point 3 RootMenSqure Vlue b r.m.s vlue = [f(t)] b dt he r.m.s. vlue depends upon the intervl chosen. If the vlues of or b re chnged, then the r.m.s. vlue of the function cross the intervl from to b will in generl chnge s well. Note tht when finding n r.m.s. vlue the function must be squred before it is integrted. Exmple 3 Find the r.m.s. vlue of f(t) = t cross the intervl from t = to t = 3. Solution r.m.s = b [f(t)] dt = 3 3 [t ] dt = 3 t 4 dt = [ t 5 5 ] HELM (8): Section 4.: he Men Vlue nd the RootMenSqure Vlue 5
7 Exmple 4 Clculte the r.m.s vlue of f(t) = sin t cross the intervl t π. Solution Here = nd b = π so r.m.s = π π sin t dt. he integrl of sin t is performed by using trigonometricl identities to rewrite it in the lterntive form ( cos t). his technique ws described in 3.7. π [ ] π ( cos t) sin t r.m.s. vlue = dt = t = π 4π 4π (π) = =.77 hus the r.m.s vlue is.77 to 3 d.p. In the previous Exmple the mplitude of the sine wve ws, nd the r.m.s. vlue ws.77. In generl, if the mplitude of sine wve is A, its r.m.s vlue is.77a. Key Point 4 he r.m.s vlue of ny sinusoidl wveform tken cross n intervl of width equl to one period is.77 mplitude of the wveform. Engineering Exmple 3 Electrodynmic meters Introduction A dynmometer or electrodynmic meter is n nlogue instrument tht cn mesure d.c. current or.c. current up to frequency of khz. A typicl dynmometer is shown in Figure 7. It consists of circulr dynmic coil positioned in mgnetic field produced by two wound circulr sttor coils connected in series with ech other. he torque on the moving coil depends upon the mutul inductnce between the coils given by: = I I dm 6 HELM (8): Workbook 4: Applictions of Integrtion
8 where I is the current in the fixed coil, I the current in the moving coil nd θ is the ngle between the coils. he torque is therefore proportionl to the squre of the current. If the current is lternting the moving coil is unble to follow the current nd the pointer position is relted to the men vlue of the squre of the current. he scle cn be suitbly grduted so tht the pointer position shows the squre root of this vlue, i.e. the r.m.s. current. Pointer Scle Moving coil Spring Fixed sttor coils Figure 7: An electrodynmic meter Problem in words A dynmometer is in circuit in series with 4 Ω resistor, rectifying device nd 4 V r.m.s lternting sinusoidl power supply. he rectifier resists current with resistnce of Ω in one direction nd resistnce of kω in the opposite direction. Clculte the reding indicted on the meter. Mthemticl Sttement of the problem We know from Key Point 4 in the text tht the r.m.s. vlue of ny sinusoidl wveform tken cross n intervl equl to one period is.77 mplitude of the wveform. Where.77 is n pproximtion of. his llows us to stte tht the mplitude of the sinusoidl power supply will be: V pek = V rms = V rms In this cse the r.m.s power supply is 4 V so we hve V pek = 4 = V During the prt of the cycle where the voltge of the power supply is positive the rectifier behves s resistor with resistnce of Ω nd this is combined with the 4 Ω resistnce to give resistnce of 6 Ω in totl. Using Ohm s lw V = IR I = V R As V = V pek sin(θ) where θ = ωt where ω is the ngulr frequency nd t is time we find tht during the positive prt of the cycle Irms = π ( ) sin(θ) π 6 HELM (8): Section 4.: he Men Vlue nd the RootMenSqure Vlue 7
9 During the prt of the cycle where the voltge of the power supply is negtive the rectifier behves s resistor with resistnce of kω nd this is combined with the 4 Ω resistnce to give 4 Ω in totl. So we find tht during the negtive prt of the cycle Irms = π π π ( ) sin(θ) 4 herefore over n entire cycle Irms = π π ( sin(θ) 6 ) + π π π We cn clculte this vlue to find I rms nd therefore I rms. Mthemticl nlysis Irms = π π ( sin(θ) 6 ( π Irms = sin (θ) π 36 ) + π π π + π π ( ) sin(θ) 4 ( ) sin(θ) 4 sin (θ) 96 ) Substituting the trigonometric identity sin (θ) cos(θ) we get I rms = = = ( π cos(θ) 4π π ( [ θ 36 sin(θ) 7 ] π + π ( π 4π 36 + π ) = A 96 + π cos(θ) 96 [ θ 96 sin(θ) 39 ] π π ) ) I rms =.3 A to d.p. Interprettion he reding on the meter would be.3 A. 8 HELM (8): Workbook 4: Applictions of Integrtion
10 Exercises. Clculte the r.m.s vlues of the given functions cross the specified intervl. () f(t) = + t cross [, ] (b) f(x) = x cross [, ] (c) f(t) = t cross [, ] (d) f(t) = t cross [, ] (e) f(z) = z + z cross [, 3]. Clculte the r.m.s vlues of the given functions over the specified intervl. () f(x) = x 3 cross [, 3] (b) f(x) = x cross [, ] (c) f(t) = t cross [, ] (d) f(z) = z 3 cross [, ] 3. Clculte the r.m.s vlues of the following: [ () f(t) = sin t cross, π ] (b) f(t) = sin t cross [, π] (c) f(t) = sin ωt cross [, π] (d) f(t) = cos t cross [ ], π (e) f(t) = cos t cross [, π] (f) f(t) = cos ωt cross [, π] (g) f(t) = sin ωt + cos ωt cross [, ] 4. Clculte the r.m.s vlues of the following functions: Answers () f(t) = t + cross [, 3] (b) f(t) = e t cross [, ] (c) f(t) = + e t cross [, ]. ().87 (b).575 (c).447 (d).7889 (e) ().4957 (b).77 (c) (d) ().77 (b).77 (c) (d).77 (e).77 (f) 4. ().58 (b).3466 (c).74 sin πω cos πω πω sin πω cos πω + πω (g) + sin ω ω HELM (8): Section 4.: he Men Vlue nd the RootMenSqure Vlue 9
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