Fourier Series for Symmetric Functions
The question states that the function $f(x)$ is symmetric, meaning $f(x) = f(-x)$. This is the definition of an **even function**.
For a periodic function $f(x)$ defined over an interval $[-L, L]$ (or any interval of length $2L$), its Fourier series is generally given by:
$ f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos\left(\frac{n\pi x}{L}\right) + b_n \sin\left(\frac{n\pi x}{L}\right) \right) $Here, $k$ in the options typically represents $\frac{\pi}{L}$.
Properties of Fourier Series for Even Functions
If $f(x)$ is an even function ($f(x) = f(-x)$):
- The odd harmonics (sine terms) vanish, meaning all $b_n = 0$.
- The Fourier series simplifies to include only a constant term and cosine terms.
Therefore, the Fourier series of a symmetric (even) function takes the form:
$ f(x) = a_0 + \sum_{n=1}^{\infty} a_n \cos(nkx) $where $a_0 = \frac{1}{2L} \int_{-L}^{L} f(x) dx$ and $a_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos(nkx) dx$. Since $f(x)$ and $\cos(nkx)$ are both even, their product is even, leading to non-zero $a_n$ values.
Analysis of Options
- Option 1: $f(x) = \sum_{n=1}^{\infty}(a_n \cos(nkx) + b_n \sin(nkx))$. This is the general form for any periodic function, not specific to symmetric (even) functions.
- Option 2: $f(x) = a_0 + \sum_{n=1}^{\infty}(a_n \cos(nkx))$. This matches the derived form for an even function, as $b_n = 0$.
- Option 3: $f(x) = \sum_{n=1}^{\infty}(b_n \sin(nkx))$. This is the form for an odd function ($f(x) = -f(-x)$), where $a_0 = 0$ and $a_n = 0$.
- Option 4: $f(x) = a_0 + \sum_{n=1}^{\infty}(b_n \sin(nkx))$. This form is incorrect as it combines a constant term with sine terms, which doesn't align with the properties of even or odd functions.
Thus, the correct form for a symmetric (even) periodic function is Option 2.
Important Questions from Fourier Series
If we use the Fourier transform ϕ(x, y) = \(\int {{{\rm{e}}^{{\rm{ikx}}}}} {ϕ _{\rm{k}}}\left( {\rm{y}} \right){\rm{dk}}\) to solve the partial differential equation \({\rm{ - }}\frac{{{\partial ^2}ϕ \left( {x,y} \right)}}{{\partial {y^2}}}\, - \,\frac{1}{{{y^2}}}\frac{{{\partial ^2}ϕ \left( {x,y} \right)}}{{\partial {x^2}}} + \frac{{{m^2}}}{{{y^2}}}ϕ \left( {x,y} \right) = 0\) in the half-plane {(x, y) : -∞ < x < ∞, 0 < y < ∞} the Fourier modes ϕ k(y) depend on y as y α and y β . The values of α and β are
When a time-domain signal is converted into its Fourier representation, which of the following is/are conserved?
I. Energy
II. Power
The trigonometric Fourier series of a periodic time function can have
The Fourier series expansion of x3 in the interval −1 ≤ x < 1 with periodic continuation has
The Fourier series to represent x-x2 for –π ≤ x ≤ π is given by \(x - {x^2} = \frac{{{a_0}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}cosnx + \mathop \sum \limits_{n = 1}^\infty {b_n}sinnx\)
The value of a0 (round off to two decimal places), is
