Question Number 173350 by Muktarr last updated on 10/Jul/22
Answered by aleks041103 last updated on 10/Jul/22
![r.m.s. = root mean square RMS(f(x))=(√(⟨f^( 2) (x)⟩))=(√((1/T)∫_x_0 ^(x_0 +T) f^( 2) (x) dx)) where f(x)=f(x+kT), k∈Z in this case, T=1 and f(x)=2+4frac(x), where frac(x):=x−⌊x⌋ ⇒f_(rms) =(√((1/1)∫_0 ^( 1) (2+4frac(x))^2 dx)) ∫_0 ^( 1) (2+4frac(x))^2 dx=4∫_0 ^1 (1+2x)^2 dx= =2∫_0 ^1 (1+2x)^2 d(1+2x)= =(2/3)[(1+2x)^3 ]_0 ^1 =(2/3)((1+2.1)^3 −(1+2.0)^3 )= =(2/3)(27−1)=((52)/3) ⇒f_(rms) =(√((52)/3))](Q173360.png)
$${r}.{m}.{s}.\:=\:{root}\:{mean}\:{square} \\ $$$${RMS}\left({f}\left({x}\right)\right)=\sqrt{\langle{f}^{\:\mathrm{2}} \left({x}\right)\rangle}=\sqrt{\frac{\mathrm{1}}{{T}}\underset{{x}_{\mathrm{0}} } {\overset{{x}_{\mathrm{0}} +{T}} {\int}}{f}^{\:\mathrm{2}} \left({x}\right)\:{dx}} \\ $$$${where}\:{f}\left({x}\right)={f}\left({x}+{kT}\right),\:{k}\in\mathbb{Z} \\ $$$${in}\:{this}\:{case},\:{T}=\mathrm{1} \\ $$$${and}\:{f}\left({x}\right)=\mathrm{2}+\mathrm{4}{frac}\left({x}\right),\:{where}\:{frac}\left({x}\right):={x}−\lfloor{x}\rfloor \\ $$$$\Rightarrow{f}_{{rms}} =\sqrt{\frac{\mathrm{1}}{\mathrm{1}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\mathrm{2}+\mathrm{4}{frac}\left({x}\right)\right)^{\mathrm{2}} {dx}} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\mathrm{2}+\mathrm{4}{frac}\left({x}\right)\right)^{\mathrm{2}} {dx}=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} {dx}= \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} {d}\left(\mathrm{1}+\mathrm{2}{x}\right)= \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left[\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{3}} \right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{3}}\left(\left(\mathrm{1}+\mathrm{2}.\mathrm{1}\right)^{\mathrm{3}} −\left(\mathrm{1}+\mathrm{2}.\mathrm{0}\right)^{\mathrm{3}} \right)= \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{27}−\mathrm{1}\right)=\frac{\mathrm{52}}{\mathrm{3}} \\ $$$$\Rightarrow{f}_{{rms}} =\sqrt{\frac{\mathrm{52}}{\mathrm{3}}} \\ $$
Commented by Tawa11 last updated on 11/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$