Solved

Consider the Following Periodic Function with Period :
F bn=1(2n1)π,n=1,2,3, b_{n}=\frac{1}{(2 n-1) \pi}, n=1,2,3, \ldots

Question 22

Multiple Choice

Consider the following periodic function with period  Consider the following periodic function with period   : F (t)  =   F   = f (t)  The Fourier representation has the form   Which of these are the coefficients   ? A)    b_{n}=\frac{1}{(2 n-1)  \pi}, n=1,2,3, \ldots   B)    b_{n}=-\frac{2}{\pi} \cdot \frac{1}{2 n-1}, n=1,2,3, \ldots   C)    b_{2 n}=0, b_{2 n-1}=-\frac{2}{\left(4 n^{2}-1\right)  \pi}, n=1,2,3, \ldots   D)    b_{2 n}=-\frac{2}{\left(4 n^{2}-1\right)  \pi}, b_{2 n-1}=0, n=1,2,3, \ldots   E)    b_{n}=0, n=1,2,3, \ldots :
F (t) =  Consider the following periodic function with period   : F (t)  =   F   = f (t)  The Fourier representation has the form   Which of these are the coefficients   ? A)    b_{n}=\frac{1}{(2 n-1)  \pi}, n=1,2,3, \ldots   B)    b_{n}=-\frac{2}{\pi} \cdot \frac{1}{2 n-1}, n=1,2,3, \ldots   C)    b_{2 n}=0, b_{2 n-1}=-\frac{2}{\left(4 n^{2}-1\right)  \pi}, n=1,2,3, \ldots   D)    b_{2 n}=-\frac{2}{\left(4 n^{2}-1\right)  \pi}, b_{2 n-1}=0, n=1,2,3, \ldots   E)    b_{n}=0, n=1,2,3, \ldots
F  Consider the following periodic function with period   : F (t)  =   F   = f (t)  The Fourier representation has the form   Which of these are the coefficients   ? A)    b_{n}=\frac{1}{(2 n-1)  \pi}, n=1,2,3, \ldots   B)    b_{n}=-\frac{2}{\pi} \cdot \frac{1}{2 n-1}, n=1,2,3, \ldots   C)    b_{2 n}=0, b_{2 n-1}=-\frac{2}{\left(4 n^{2}-1\right)  \pi}, n=1,2,3, \ldots   D)    b_{2 n}=-\frac{2}{\left(4 n^{2}-1\right)  \pi}, b_{2 n-1}=0, n=1,2,3, \ldots   E)    b_{n}=0, n=1,2,3, \ldots = f (t)
The Fourier representation has the form  Consider the following periodic function with period   : F (t)  =   F   = f (t)  The Fourier representation has the form   Which of these are the coefficients   ? A)    b_{n}=\frac{1}{(2 n-1)  \pi}, n=1,2,3, \ldots   B)    b_{n}=-\frac{2}{\pi} \cdot \frac{1}{2 n-1}, n=1,2,3, \ldots   C)    b_{2 n}=0, b_{2 n-1}=-\frac{2}{\left(4 n^{2}-1\right)  \pi}, n=1,2,3, \ldots   D)    b_{2 n}=-\frac{2}{\left(4 n^{2}-1\right)  \pi}, b_{2 n-1}=0, n=1,2,3, \ldots   E)    b_{n}=0, n=1,2,3, \ldots
Which of these are the coefficients  Consider the following periodic function with period   : F (t)  =   F   = f (t)  The Fourier representation has the form   Which of these are the coefficients   ? A)    b_{n}=\frac{1}{(2 n-1)  \pi}, n=1,2,3, \ldots   B)    b_{n}=-\frac{2}{\pi} \cdot \frac{1}{2 n-1}, n=1,2,3, \ldots   C)    b_{2 n}=0, b_{2 n-1}=-\frac{2}{\left(4 n^{2}-1\right)  \pi}, n=1,2,3, \ldots   D)    b_{2 n}=-\frac{2}{\left(4 n^{2}-1\right)  \pi}, b_{2 n-1}=0, n=1,2,3, \ldots   E)    b_{n}=0, n=1,2,3, \ldots ?


A) bn=1(2n1) π,n=1,2,3, b_{n}=\frac{1}{(2 n-1) \pi}, n=1,2,3, \ldots
B) bn=2π12n1,n=1,2,3, b_{n}=-\frac{2}{\pi} \cdot \frac{1}{2 n-1}, n=1,2,3, \ldots
C) b2n=0,b2n1=2(4n21) π,n=1,2,3, b_{2 n}=0, b_{2 n-1}=-\frac{2}{\left(4 n^{2}-1\right) \pi}, n=1,2,3, \ldots
D) b2n=2(4n21) π,b2n1=0,n=1,2,3, b_{2 n}=-\frac{2}{\left(4 n^{2}-1\right) \pi}, b_{2 n-1}=0, n=1,2,3, \ldots
E) bn=0,n=1,2,3, b_{n}=0, n=1,2,3, \ldots

Correct Answer:

verifed

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Related Questions

Unlock this Answer For Free Now!

View this answer and more for free by performing one of the following actions

qr-code

Scan the QR code to install the App and get 2 free unlocks

upload documents

Unlock quizzes for free by uploading documents