Chapter 7: Q4P (page 355)

Sketch several periods of the corresponding periodic function of period $2 π$ . Expand the periodic function in a sine-cosine Fourier series.
$f (x) = {\begin{cases} - 1, & - π & < & x & < & \frac{π}{2} \\ 1, & \frac{π}{2} & < & x & < & π, \end{cases}$

Short Answer

Expert verified

The answer of the given function is:

$f (x) = \frac{1}{4} + \frac{1}{π} \{\sum_{n = 1 + 4}^{\infty} \frac{\cos n x}{n} - \sum_{n = 3 + 4}^{\infty} \frac{\cos n x}{n}\} + \frac{2}{π} \{- 2 \sum_{n = 2 + 4 m}^{\infty} \frac{\sin n x}{n} + 2 \sum_{n = 1 + 2 m}^{\infty} \frac{\sin n x}{n}\} w h e r e m = 0, 1, 2, - - .$

Step by step solution

Given

The given function is $f (x) = {\begin{cases} - 1, & - π & < & x & < & \frac{π}{2} \\ 1, & \frac{π}{2} & < & x & < & π, \end{cases}$ .

The concept of the Fourier series for the function f(x)

The Fourier series for the function :

$f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} (a_{n} \cos nx + b_{n} \sin nx) a_{0} = \frac{1}{π} \int_{- π}^{π} f (x) d x a_{n} = \frac{1}{π} \int_{- π}^{π} f (x) c o s n x d x b_{n} = \frac{1}{π} \int_{- π}^{π} f (x) s i n n x d x$

If is an even function:

$b_{n} = 0 f (x) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} \cos nx$

If is an odd function:

$a_{0} = a_{a} = 0 f (x) = \sum_{n = 1}^{\infty} b_{n} \sin nx$

From the given information

The sketch of the given function is shown below.

Coefficients of $a_{0}$ :

$a_{0} = \frac{1}{π} \int_{- π}^{π} f (x) d x = \frac{1}{π} \{- \int_{- π}^{\frac{π}{2}} d x + \int_{\frac{π}{2}}^{π} d x\} = \frac{1}{π} [{[x]}_{π}^{\frac{π}{2}} + {[x]}_{\frac{π}{2}}^{π}] = - 1$

Coefficients of $a_{n}$ :

$a_{n} = \frac{1}{π} \int_{- π}^{π} f (x) \cos n x d x = \frac{1}{π} [\int_{- π}^{\frac{π}{2}} \cos n x d x + \int_{\frac{π}{2}}^{π} \cos n x d x] = \frac{1}{π} {[\sin n x]}_{π}^{\frac{π}{2}} + {[\sin n x]}_{\frac{π}{2}}^{π} = \frac{2}{nπ} \sin (\frac{n π}{2})$

Calculate Coefficients of bn

Coefficients of $b_{n}$ are $\frac{1}{π}, \frac{2}{2 π}, \frac{1}{3 π}, 0, \frac{1}{5 π}, \frac{2}{6 π}, \frac{1}{7 π}$ .

Hence the expansion is $f (x) = \frac{1}{2} + \frac{2}{π} \{\sum_{n = 1 + 4}^{\infty} \frac{\cos n x}{n} - \sum_{n = 3 + 4}^{\infty} \frac{\cos n x}{n}\} + \frac{2}{π} \{- 2 \sum_{n = 2 + 4 m}^{\infty} \frac{\sin n x}{n} + 2 \sum_{n = 1 + 2 m}^{\infty} \frac{\sin n x}{n}\}$ where $m = 0, 1, 2, - - .$ .

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Sketch several periods of the corresponding periodic function of period $2 π$ . Expand the periodic function in a sine-cosine Fourier series.
$f (x) = {\begin{cases} - 1, & - π & < & x & < & \frac{π}{2} \\ 1, & \frac{π}{2} & < & x & < & π, \end{cases}$

Short Answer

Step by step solution

Given

The concept of the Fourier series for the function f(x)

From the given information

Calculate Coefficients of bn

One App. One Place for Learning.

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Sketch several periods of the corresponding periodic function of period 2π . Expand the periodic function in a sine-cosine Fourier series.f(x)={-1,-π<x<π21,π2<x<π,

Short Answer

Step by step solution

Given

The concept of the Fourier series for the function f(x)

From the given information

Calculate Coefficients of bn

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Geometrical and Physical Optics

Scientific Method Physics

Rotational Dynamics

Kinematics Physics

Space Physics

Study anywhere. Anytime. Across all devices.

Sketch several periods of the corresponding periodic function of period $2 π$ . Expand the periodic function in a sine-cosine Fourier series.
$f (x) = {\begin{cases} - 1, & - π & < & x & < & \frac{π}{2} \\ 1, & \frac{π}{2} & < & x & < & π, \end{cases}$