Correct Answer - Option 4 : sine harmonics
Fourier series:
The Fourier series for the function f(x) in the interval α < x < α + 2π is given by
\(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\cos nx + \mathop \sum \limits_{n = 1}^\infty {b_n}\sin nx\)
where
\({a_o} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)dx;\;{a_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\cos nxdx;\;{b_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\sin nxdx\)
An even function is any function f such that f(-x) = f(x)
Example: cos x, sec x, x2, x4, x6 …….., x-2, x-4 ……..
An odd function is any function f such that f(-x) = -f(x)
Example: sin x, tan x, cosec x, cot x, n, x3 ……., x-1, x-3 ……..
\(\mathop \smallint \limits_{ - L}^L f\left( x \right)dx = \left\{ {\begin{array}{*{20}{c}} {2\mathop \smallint \limits_0^L f\left( x \right)dx,\;\;when\;f\left( x \right)\;is\;an\;even\;function}\\ {0,\;\;when\;f\left( x \right)\;is\;an\;odd\;function} \end{array}} \right.\)
When f is an even periodic function of period 2L, then its Fourier series contains only cosine (include possibly, the constant term) terms.
\(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\frac{{\cos n\pi x}}{L}\)
\({a_o} = \frac{1}{L}\mathop \smallint \limits_{ - L}^L f\left( x \right)dx = \frac{2}{L}\mathop \smallint \limits_0^L f\left( x \right)dx\)
\({a_n} = \frac{1}{L}\mathop \smallint \limits_{ - L}^L f\left( x \right)\cos \frac{{n\pi x}}{L}dx = \frac{2}{L}\mathop \smallint \limits_0^L f\left( x \right)\cos \frac{{n\pi x}}{L}dx\)
When f is an odd periodic function of period 2L, then its Fourier series contains only sine terms.
\(f\left( x \right) = \mathop \sum \limits_{n = 1}^\infty {b_n}\sin \frac{{n\pi x}}{L}\)
\({b_n} = \frac{1}{L}\mathop \smallint \limits_{ - L}^L f\left( x \right)\sin \frac{{n\pi x}}{L}dx = \frac{2}{L}\mathop \smallint \limits_0^L f\left( x \right)\sin \frac{{n\pi x}}{L}dx\)
- Based on the symmetry of the periodic signal given we can conclude the Fourier series coefficients and the type of harmonics that are present in the signal given.
- With the Fourier series, the non-sinusoidal periodic waveform can be converted into the sinusoidal wave.
The below table shows the type of Fourier coefficient terms corresponding to the symmetry.
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Symmetry
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Condition
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a0
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an
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bn
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Terms
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Even
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x(t) = x(-t)
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Non zero
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Non zero
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Zero
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DC and Cosine
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|
Odd
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x(t) = - x(-t)
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Zero
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Zero
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Non zero
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Only Sine
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|
Half wave
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\(x\left( t \right) = - x\left( {t \pm \frac{T}{2}} \right)\)
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Zero
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Zero ; n even
Non zero; n odd
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Zero; n even
Non zero; n odd
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Only odd harmonics
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