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Question

A periodic signal with period T0 is real. The signal x(t) has even and odd part shown as xe(t) and x0(t) , also Ck is the kth exponential Fourier series coefficient of x(t) . If Ck=⎧⎪ ⎪⎨⎪ ⎪⎩(103|k|+3kj) k≠0 and |K|<310, K=0 Then which of the following is correct?

A
The power of odd part of x(t) is 25.50 W
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B
The power of even part of x(t) is 120.46 W
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C
The power of even part of x(t) is 127.76 W
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D
None of the above
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E
The power of odd part of x(t) is 28.50 W
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Solution

The correct option is C The power of even part of x(t) is 127.76 W If x(t) is real then Fourier coefficient of even part of x(t) i.e., xe(t) is Re {CK} so, Bk=⎧⎪⎨⎪⎩103|k| |k|<310, k=0 so, Bk=⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩10, k=0103, k=1,−1106, k=2,−2 so power of xe(t)=∞∑k=−∞|Bk|2 =100+2(1009+10036) =127.76 W If x(t) is real, the Fourier coefficient of odd part of x(t) i.e., x0(t) is img {cK } So, DK=⎧⎨⎩3K, for K≠0and|K|<30, for K=0 So , Power of x0(t) = ∞∑k=−∞|DK|2 =[(3−2)2+(3−1)2+(3)3+(32)2] =94+9+9+94=22.5 W

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