15.6. The mean value and the rootmeansquare value of a function. Introduction. Prerequisites. Learning Outcomes. Learning Style


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1 The men vlue nd the rootmensqure vlue of function 5.6 Introduction Currents nd voltges often vry with time nd engineers my wish to know the verge vlue of such current or voltge over some prticulr time intervl. The verge vlue of timevrying function is defined in terms of n integrl. An ssocited quntity is the rootmensqure (r.m.s) vlue of current which is used,for exmple,in the clcultion of the power dissipted by resistor. Prerequisites Before strting this Block you should... Lerning Outcomes After completing this Block you should be ble to... clculte the men vlue of function clculte the rootmensqure vlue of function be ble to clculte definite integrls 2 be fmilir with tble of trigonometric identities Lerning Style To chieve wht is expected of you... llocte sufficient study time briefly revise the prerequisite mteril ttempt every guided exercise nd most of the other exercises
2 . Averge vlue of function Suppose timevrying function f(t) is defined on the intervl t b. The re, A,under the grph of f(t) is given by the integrl This is illustrted in Figure. f(t) A = f(t)dt h b t Figure. the re under the curve from t = to t = b nd the re of the rectngle re equl On Figure 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 h. Then re of rectngle = re under curve h(b ) = h = b f(t)dt f(t)dt The vlue of h is the verge or men vlue of the function cross the intervl t b. Key Point The verge vlue of function f(t) in the intervl t b is b f(t)dt The verge vlue depends upon the intervl chosen. If the vlues of or b re chnged,then the verge vlue of the function cross the intervl from to b will chnge s well. Engineering Mthemtics: Open Lerning Unit Level 5.6: Applictions of Integrtion 2
3 Exmple Find the verge vlue of f(t) =t 2 over the intervl t 3. Solution Here = nd b =3. verge vlue = = b 3 3 f(t)dt t 2 dt = 2 [ t 3 3 ] 3 = 3 3 Try ech prt of this exercise Prt () Find the verge vlue of f(t) =t 2 over the intervl 2 t 5. Here = 2 nd b =5. Answer Prt (b) Now evlute the integrl. Answer 3 Engineering Mthemtics: Open Lerning Unit Level 5.6: Applictions of Integrtion
4 More exercises for you to try. Clculte the verge vlue of the given functions cross the specified intervl. () f(t) =+t cross [0, 2] (b) f(x) =2x cross [, ] (c) f(t) =t 2 cross [0, ] (d) f(t) =t 2 cross [0, 2] (e) f(z) =z 2 + z cross [, 3] 2. Clculte the verge vlue of the given functions over the specified intervl. () f(x) =x 3 cross [, 3] (b) f(x) = cross [, 2] x (c) f(t) = t cross [0, 2] (d) f(z) =z 3 cross [, ] (e) f(t) = cross [ 3, 2] t 2 3. Clculte the verge vlue of the following: () f(t) = sin t cross [ ] 0, π 2 (b) f(t) = sin t cross [0,π] (c) f(t) = sin ωt cross [0,π] (d) f(t) = cos t cross [ ] 0, π 2 (e) f(t) = cos t cross [0,π] (f) f(t) = cos ωt cross [0,π] (g) f(t) = sin ωt + cos ωt cross [0, ] 4. Clculte the verge vlue of the following functions: () f(t) = t + cross [0, 3] (b) f(t) =e t cross [, ] (c) f(t) =+e t cross [, ] Answer Engineering Mthemtics: Open Lerning Unit Level 5.6: Applictions of Integrtion 4
5 2. 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)] 2 dt This is simply the verge vlue of [f(t)] 2 over the given intervl. The relted quntity: the rootmensqure (r.m.s.) vlue is given by the following formul. Key Point r.m.s vlue = b [f(t)] 2 dt The 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 chnge s well. Note tht when finding n r.m.s. vlue the function must be squred before it is integrted. Exmple Find the r.m.s. vlue of f(t) =t 2 cross the intervl from t =tot =3. Solution r.m.s = = = b [f(t)] b 2 dt 3 [t 3 2 ] 2 dt 3 [ t 5 t 2 4 dt = 2 5 ] 3 =4.92 Exmple Clculte the r.m.s vlue of f(t) = sin t cross the intervl 0 t 2π. 5 Engineering Mthemtics: Open Lerning Unit Level 5.6: Applictions of Integrtion
6 Solution Here = 0 nd b =2π. r.m.s = 2π 2π 0 sin 2 tdt The integrl of sin 2 t is performed by using trigonometricl identities to rewrite it in the lterntive form ( cos 2t). This technique ws described in Chpter 3 Block π ( cos 2t) r.m.s. vlue = dt 2π 0 2 2π = ( cos 2t)dt 4π 0 [ ] 2π sin 2t = t Thus the r.m.s vlue is = 4π 4π (2π) = =0.707 In the previous exmple the mplitude of the sine wve ws,nd the r.m.s. vlue ws In generl,if the mplitude of sine wve is A,its r.m.s vlue is 0.707A. Key Point The r.m.s vlue of ny sinusoidl wveform tken cross n intervl equl to one period is mplitude of the wveform. Engineering Mthemtics: Open Lerning Unit Level 5.6: Applictions of Integrtion 6
7 More exercises for you to try. Clculte the r.m.s. vlues of the functions: () f(t) =+t cross [0, 2] (b) f(x) =2x cross [, ] (c) f(t) =t 2 cross [0, ] (d) f(t) =t 2 cross [0, 2] (e) f(z) =z 2 + z cross [, 3] 2. Clculte the r.m.s. vlues of the functions: () f(x) =x 3 cross [, 3] (b) f(x) = cross [, 2] x (c) f(t) = t cross [0, 2] (d) f(z) =z 3 cross [, ] (e) f(t) = cross [ 3, 2] t 2 3. Clculte the r.m.s. vlues of the functions: () f(t) = sin t cross [ ] 0, π 2 (b) f(t) = sin t cross [0,π] (c) f(t) = sin ωt cross [0,π] (d) f(t) = cos t cross [ ] 0, π 2 (e) f(t) = cos t cross [0,π] (f) f(t) = cos ωt cross [0,π] (g) f(t) = sin ωt + cos ωt cross [0, ] 4. Clculte the r.m.s. vlues of the functions: () f(t) = t + cross [0, 3] (b) f(t) =e t cross [, ] (c) f(t) =+e t cross [, ] Answer 7 Engineering Mthemtics: Open Lerning Unit Level 5.6: Applictions of Integrtion
8 End of Block 5.6 Engineering Mthemtics: Open Lerning Unit Level 5.6: Applictions of Integrtion 8
9 t2 dt Bck to the theory 9 Engineering Mthemtics: Open Lerning Unit Level 5.6: Applictions of Integrtion
10 3 Bck to the theory Engineering Mthemtics: Open Lerning Unit Level 5.6: Applictions of Integrtion 0
11 . () 2 (b) (c) 3 (d) 4 3 (e) () 0 (b) (c) (d) (e) 6 3. () 2 (b) 2 (c) [ cos(πω)] (d) 2 (e) 0 (f) sin(πω) π π πω π πω 4. () 4 (b).752 (c) Bck to the theory (g) +sin ω cos ω ω Engineering Mthemtics: Open Lerning Unit Level 5.6: Applictions of Integrtion
12 . () (b).5275 (c) (d).7889 (e) () (b) (c) (d).0690 (e) () (b) (c) (g) + sin2 ω ω 4. ().58 (b).3466 (c) Bck to the theory sin πω cos πω (d) (e) (f) 2 2πω sin πω cos πω + 2 2πω Engineering Mthemtics: Open Lerning Unit Level 5.6: Applictions of Integrtion 2
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