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Mathematics 19 Online
OpenStudy (anonymous):

Determine average and RMS value of the waveform ?(see attachment)

OpenStudy (anonymous):

OpenStudy (anonymous):

its pretty easy

OpenStudy (anonymous):

just need to find the period, then describe the function in each of the different invervals

OpenStudy (anonymous):

so its period ( time to repeat is 5seconds ) , you see that from the graph

OpenStudy (anonymous):

now describe the function in the seperate intervals for 0<=t<=2 , f(t) = 2 for 2<t<=3 , f(t) = 0 for 3<t<=4 , f(t) = -3 for 4<t<=5, f(t) =0

OpenStudy (anonymous):

now the average value is defined as the integral of the function, over one period, divided by that period

OpenStudy (anonymous):

Average value =\[\frac{1}{5} ( \int\limits_{0}^{2} 2dt + \int\limits_{2}^{3}0dt + \int\limits_{3}^{4} -3 dt + \int\limits_{4}^{5} 0 dt )\]

OpenStudy (anonymous):

where the (1/5) comes from dividing by the period

OpenStudy (anonymous):

rms value is similar , excepts it the ROOT MEAN SQAURE value which means you find the mean ( ie average ) of the square of the function , then you take the sqrt of it

OpenStudy (anonymous):

you still use the same interval definitions for the function

OpenStudy (anonymous):

RMS value = \[\sqrt{ \frac{1}{5} (\int\limits_{0}^{2} (2)^2 dt + \int\limits_{2}^{3} 0^2 dt + \int\limits_{3}^{4} (-3)^2 dt + \int\limits_{4}^{5} 0^2 dt )}\]

OpenStudy (anonymous):

you can do the easy calculations

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