Given that, $\int_{T}^{2 T} x (t) . d t = 2$ $x (t), y (t) = \frac{d}{d x} x (t)$ signals are periodic signals of time period T = 8, with Fourier series coefficient $a_{k} a n d b_{k}$ respectively.
Which one of the following expressions are correct ?

a_{0} = 0.25, a_{4} = \frac{- j b_{4}}{π}, a_{8} = \frac{- j b_{8}}{2 π}

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a_{0} = 0.25, a_{4} = j π b_{4}, a_{8} = j 2 π b_{8}

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a_{0} = 4, a_{4} = \frac{- j b_{4}}{π}, a_{8} = \frac{- j b_{8}}{2 π}

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a_{0} = 4, a_{4} = j π b_{4}, a_{8} = j 2 π b_{8}

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Solution

The correct option is A $a_{0} = 0.25, a_{4} = \frac{- j b_{4}}{π}, a_{8} = \frac{- j b_{8}}{2 π}$
Given that, $x (t) F . S ⟷ a_{k}$

$g (t) = (\frac{d x (t)}{d t} x (t)) F . S ⟷ b_{k} = (j k \frac{2 π}{T} a_{k})$

$a_{k} = \frac{b_{k}}{j (\frac{2 π}{T}) k}, k \neq 0$

When k = 0,

$a_{k} = \frac{1}{T} \int_{0}^{T} x (t) . d t = \frac{2}{T}$

$a_{k} = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{2}{T}, & k = 0 \frac{b_{k}}{j (\frac{2 π}{T}) k}, & k \neq 0 \end{matrix}$

$∴ a_{0} = \frac{2}{8} = 0.25$

$a_{4} = \frac{b_{4}}{j (\frac{2 π}{8}) \times 4} = \frac{- j b_{4}}{π}$

$a_{8} = \frac{b_{8}}{j (\frac{2 π}{8}) \times 8} = \frac{- j b_{8}}{2 π}$

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Given that, ∫2TTx(t).dt=2 x(t),y(t)=ddxx(t) signals are periodic signals of time period T = 8, with Fourier series coefficient ak and bk respectively. Which one of the following expressions are correct ?

The correct option is A a0=0.25,a4=−jb4π,a8=−jb82πGiven that, x(t)F.S⟷ak g(t)=(dx(t)dtx(t))F.S⟷bk=(jk2πTak) ak=bkj(2πT)k,k≠0 When k = 0, ak=1T∫T0x(t).dt=2T ak=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩2T,k=0bkj(2πT)k,k≠0 ∴a0=28=0.25 a4=b4j(2π8)×4=−jb4π a8=b8j(2π8)×8=−jb82π

Given that, $\int_{T}^{2 T} x (t) . d t = 2$ $x (t), y (t) = \frac{d}{d x} x (t)$ signals are periodic signals of time period T = 8, with Fourier series coefficient $a_{k} a n d b_{k}$ respectively.
Which one of the following expressions are correct ?