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Question Number 173350 by Muktarr last updated on 10/Jul/22

Answered by aleks041103 last updated on 10/Jul/22

$${r}.{m}.{s}.\:=\:{root}\:{mean}\:{square} \\$$$${RMS}\left({f}\left({x}\right)\right)=\sqrt{\langle{f}^{\:\mathrm{2}} \left({x}\right)\rangle}=\sqrt{\frac{\mathrm{1}}{{T}}\underset{{x}_{\mathrm{0}} } {\overset{{x}_{\mathrm{0}} +{T}} {\int}}{f}^{\:\mathrm{2}} \left({x}\right)\:{dx}} \\$$$${where}\:{f}\left({x}\right)={f}\left({x}+{kT}\right),\:{k}\in\mathbb{Z} \\$$$${in}\:{this}\:{case},\:{T}=\mathrm{1} \\$$$${and}\:{f}\left({x}\right)=\mathrm{2}+\mathrm{4}{frac}\left({x}\right),\:{where}\:{frac}\left({x}\right):={x}−\lfloor{x}\rfloor \\$$$$\Rightarrow{f}_{{rms}} =\sqrt{\frac{\mathrm{1}}{\mathrm{1}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\mathrm{2}+\mathrm{4}{frac}\left({x}\right)\right)^{\mathrm{2}} {dx}} \\$$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\mathrm{2}+\mathrm{4}{frac}\left({x}\right)\right)^{\mathrm{2}} {dx}=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} {dx}= \\$$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} {d}\left(\mathrm{1}+\mathrm{2}{x}\right)= \\$$$$=\frac{\mathrm{2}}{\mathrm{3}}\left[\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{3}} \right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{3}}\left(\left(\mathrm{1}+\mathrm{2}.\mathrm{1}\right)^{\mathrm{3}} −\left(\mathrm{1}+\mathrm{2}.\mathrm{0}\right)^{\mathrm{3}} \right)= \\$$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{27}−\mathrm{1}\right)=\frac{\mathrm{52}}{\mathrm{3}} \\$$$$\Rightarrow{f}_{{rms}} =\sqrt{\frac{\mathrm{52}}{\mathrm{3}}} \\$$

Commented by Tawa11 last updated on 11/Jul/22

$$\mathrm{Great}\:\mathrm{sir} \\$$