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Question

What is the value of the Fourier coefficient $a_0$ for the function given below?

 $f(x) = \begin{cases} \sin 2x, & -\pi < x < -\frac{\pi}{2} \\ 0, & -\frac{\pi}{2} \le x \le 0 \\ \sin 2x, & 0 < x \le \pi \end{cases}$

The correct answer is
$1/(2\pi)$

Fourier Coefficient $a_0$ Calculation

The Fourier coefficient $a_0$ represents the average value of the function $f(x)$ over the interval $[-\pi, \pi]$. It is calculated using the formula:

$a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) dx$

Integration Steps

The function $f(x)$ is defined piecewise. To calculate the integral $\int_{-\pi}^{\pi} f(x) dx$, we split the integral according to the function's definition:

  • From $-\pi$ to $-\frac{\pi}{2}$, $f(x) = \sin 2x$.
  • From $-\frac{\pi}{2}$ to $0$, $f(x) = 0$.
  • From $0$ to $\pi$, $f(x) = \sin 2x$.

The total integral becomes:

$ \int_{-\pi}^{\pi} f(x) dx = \int_{-\pi}^{-\pi/2} \sin(2x) \, dx + \int_{-\pi/2}^{0} 0 \, dx + \int_{0}^{\pi} \sin(2x) \, dx $

Integral Calculation Details

  1. Integral over $(-\pi, -\pi/2)$:

    $ \int_{-\pi}^{-\pi/2} \sin(2x) \, dx = \left[ -\frac{\cos(2x)}{2} \right]_{-\pi}^{-\pi/2} $

    Evaluate at the limits:

    $ = \left(-\frac{\cos(2 \times (-\pi/2))}{2}\right) - \left(-\frac{\cos(2 \times (-\pi))}{2}\right) $

    $ = \left(-\frac{\cos(-\pi)}{2}\right) - \left(-\frac{\cos(-2\pi)}{2}\right) = \left(-\frac{-1}{2}\right) - \left(-\frac{1}{2}\right) = \frac{1}{2} + \frac{1}{2} = 1 $

  2. Integral over $[-\pi/2, 0]$:

    The function is $0$ in this interval:

    $ \int_{-\pi/2}^{0} 0 \, dx = 0 $

  3. Integral over $(0, \pi]$:

    $ \int_{0}^{\pi} \sin(2x) \, dx = \left[ -\frac{\cos(2x)}{2} \right]_{0}^{\pi} $

    Evaluate at the limits:

    $ = \left(-\frac{\cos(2\pi)}{2}\right) - \left(-\frac{\cos(2 \times 0)}{2}\right) $

    $ = \left(-\frac{1}{2}\right) - \left(-\frac{1}{2}\right) = -\frac{1}{2} + \frac{1}{2} = 0 $

Integral Value

Sum the results from the individual intervals:

$ \int_{-\pi}^{\pi} f(x) dx = 1 + 0 + 0 = 1 $

$a_0$ Calculation

Now substitute the value of the total integral back into the formula for $a_0$:

$ a_0 = \frac{1}{2\pi} \times (1) = \frac{1}{2\pi} $

Result

The value of the Fourier coefficient $a_0$ is $\frac{1}{2\pi}$.

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Important Questions from Fourier Series

  1. 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  

  2. When a time-domain signal is converted into its Fourier representation, which of the following is/are conserved?

    I. Energy

    II. Power

  3. The trigonometric Fourier series of a periodic time function can have

  4. The Fourier series expansion of x3 in the interval −1 ≤ x < 1 with periodic continuation has

  5. 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
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