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Electrical & Electronic Control Question & Answers

   

Added on  2020-02-19

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Electrical and electronic control 1ELECTRICAL AND ELCTRONIC CONTROLBy NameCourseInstructorInstitutionLocationDateQuestion one : a). i), An even function is that which is symmetric about the y-axis, it is like if a reflection is taken of the graph of the function in the y-axis.Formally we say that the function is even if for all x and –x in the domain of function we have f(x) =f(-x) . and this can be given graphically as below (Brown, J, 2013 ).
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Electrical and electronic control 2Likewise we can define odd function ,as the function which satisfies the property :f(-x) = -f(x) , as and example f(x) = x3
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Electrical and electronic control 3So in context for Fourier series we have some interesting results that depend on this property of the function :Since while calculating Fourier coeff. We often come across the type of integral:Which can be simplified into nicer forms or even get vanished for odd functions.So while expressing the Fourier series representation of the function in terms of sine and cosine, we can say:Since sine is an odd function and cos is the even function, so for odd function we generally have sines and for evenThe function we have cosine in their Fourier expansion (Brown, J, 2013 ).ii) Now about half wave symmetry:In general half wave symmetry for function f(t) happens when :f(t) =- f(t-L) for some L ,
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Electrical and electronic control 4so in above figure if you do f(t-L) --> -f(t-L) you get back f(t).A half-wave symmetric function can be even, odd or neither.b)By inspection We can say :Figure b,d,e,f are even symmetric . Hence their Fourier expression will have only cosine terms.Excluding f figure b,d,e ave same period, so the fundamental is going to be same for all three. Figure a and c are neither odd nor even, they are half wave symmetric. This will result in zero amplitude for even coefficients for both sine and cos terms.Question two:a) He firing angle ά = π3Power dissipated into bulb =100watts and the voltage source (Vrms) = 230VFrequency =50HzPeak voltage = Vrms0.707Peak voltage = 2300.707 Voltage peak=325.3 V
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