Answer
Verified
442.5k+ views
Hint: A Fourier series is a means of representing a periodic function as a sum of sine and cosine functions (possibly infinite).In such problems, finding zero coefficients is time consuming and can be prevented. With understanding of even and odd functions, without implementing the integration, the zero coefficient can be predicted.
Complete step by step answer:
A function $ y = f(t) $ is said to be odd if $ f( - t) = - f(t) $ for all values of $ t $ . The graph of an odd function is always symmetrical about the origin.
Let the odd function be $ f(t) = \sin t $
Then the graph of the following function will be represented as:
The above graph has amplitude $ 1 $ and period $ 2\pi $
For an odd function $ f(t) $ defined over the range $ - L $ to $ L $ (period $ = 2L) $
We can observe that $ {a_n} = 0 $ for all $ n $
Then we have
$ {a_n} = \dfrac{1}{L}\int_{ - L}^L f (t)\cos \dfrac{{n\pi t}}{L}dt $
So, the zero coefficients in this case are: $ {a_0} = 0 $ and $ {a_n} = 0 $
The coefficients of $ {b_n} $ is given by
$ {b_n} = \dfrac{1}{L}\int_{ - L}^L f (t)\sin \dfrac{{n\pi t}}{L}dt $
Therefore, the Fourier series of the following odd function is given by
$ f(t) = \sum\limits_{n = 1}^\infty {{b_n}} \sin \dfrac{{n\pi t}}{L} $
Hence, the Fourier series of an odd periodic function contains only sine terms.
Hence the correct option is (D).
Note:
Using the Fourier series in various questions makes our task easy, fast, and more productive. We can easily find out the zero coefficient of vast problems by the help of Fourier series even without performing the actual integration. Baron Jean Baptiste Joseph Fourier introduced the idea that a series of harmonically related sines and cosines can represent any periodic function.
Complete step by step answer:
A function $ y = f(t) $ is said to be odd if $ f( - t) = - f(t) $ for all values of $ t $ . The graph of an odd function is always symmetrical about the origin.
Let the odd function be $ f(t) = \sin t $
Then the graph of the following function will be represented as:
The above graph has amplitude $ 1 $ and period $ 2\pi $
For an odd function $ f(t) $ defined over the range $ - L $ to $ L $ (period $ = 2L) $
We can observe that $ {a_n} = 0 $ for all $ n $
Then we have
$ {a_n} = \dfrac{1}{L}\int_{ - L}^L f (t)\cos \dfrac{{n\pi t}}{L}dt $
So, the zero coefficients in this case are: $ {a_0} = 0 $ and $ {a_n} = 0 $
The coefficients of $ {b_n} $ is given by
$ {b_n} = \dfrac{1}{L}\int_{ - L}^L f (t)\sin \dfrac{{n\pi t}}{L}dt $
Therefore, the Fourier series of the following odd function is given by
$ f(t) = \sum\limits_{n = 1}^\infty {{b_n}} \sin \dfrac{{n\pi t}}{L} $
Hence, the Fourier series of an odd periodic function contains only sine terms.
Hence the correct option is (D).
Note:
Using the Fourier series in various questions makes our task easy, fast, and more productive. We can easily find out the zero coefficient of vast problems by the help of Fourier series even without performing the actual integration. Baron Jean Baptiste Joseph Fourier introduced the idea that a series of harmonically related sines and cosines can represent any periodic function.
Recently Updated Pages
Class 10 Question and Answer - Your Ultimate Solutions Guide
Master Class 10 Science: Engaging Questions & Answers for Success
Master Class 10 Maths: Engaging Questions & Answers for Success
Master Class 9 Social Science: Engaging Questions & Answers for Success
Master Class 10 General Knowledge: Engaging Questions & Answers for Success
Master Class 10 Social Science: Engaging Questions & Answers for Success
Trending doubts
Pigmented layer in the eye is called as a Cornea b class 11 biology CBSE
Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE
What organs are located on the left side of your body class 11 biology CBSE
The lightest gas is A nitrogen B helium C oxygen D class 11 chemistry CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE
What are ekaboron ekaaluminium and ekasilicon class 11 chemistry CBSE