2.8-2 If a periodic signal satisfies certain symmetry conditions, the evaluation of the Fourier series components is somewhat simplified. Show that: (a) If g(t) = g(−t) (even symmetry), then all the sine terms in the Fourier series vanish (bn = 0). (b) If g(t) = −g(-t) (odd symmetry), then the dc and all the cosine terms in the Fourier series vanish (a = an = 0). Further, show that in each case the Fourier coefficients can be evaluated by integrating the periodic signal over the half-cycle only. This is because the entire information of one cycle is implicit in a half-cycle due to symmetry. Hint: If ge(t) and go (t) are even and odd functions, respectively, of t, then (assuming no impulse or its derivative at the origin) 2a ge(t) dt = ge(t) dt and go(t) dt = 0 Also, the product of an even and an odd function is an odd function, the product of two odd functions is an even function, and the product of two even functions is an even function.

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2.8-2 If a periodic signal satisfies certain symmetry conditions, the evaluation of the Fourier series
components is somewhat simplified. Show that:
(a) If g(t) = g(−t) (even symmetry), then all the sine terms in the Fourier series vanish (bn = 0).
(b) If g(t) = −g(-t) (odd symmetry), then the dc and all the cosine terms in the Fourier series
vanish (a = an = 0).
Further, show that in each case the Fourier coefficients can be evaluated by integrating
the periodic signal over the half-cycle only. This is because the entire information of one
cycle is implicit in a half-cycle due to symmetry.
Hint: If ge(t) and go (t) are even and odd functions, respectively, of t, then (assuming no
impulse or its derivative at the origin)
2a
ge(t) dt =
ge(t) dt
and
go(t) dt = 0
Also, the product of an even and an odd function is an odd function, the product of two odd
functions is an even function, and the product of two even functions is an even function.
Transcribed Image Text:2.8-2 If a periodic signal satisfies certain symmetry conditions, the evaluation of the Fourier series components is somewhat simplified. Show that: (a) If g(t) = g(−t) (even symmetry), then all the sine terms in the Fourier series vanish (bn = 0). (b) If g(t) = −g(-t) (odd symmetry), then the dc and all the cosine terms in the Fourier series vanish (a = an = 0). Further, show that in each case the Fourier coefficients can be evaluated by integrating the periodic signal over the half-cycle only. This is because the entire information of one cycle is implicit in a half-cycle due to symmetry. Hint: If ge(t) and go (t) are even and odd functions, respectively, of t, then (assuming no impulse or its derivative at the origin) 2a ge(t) dt = ge(t) dt and go(t) dt = 0 Also, the product of an even and an odd function is an odd function, the product of two odd functions is an even function, and the product of two even functions is an even function.
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