Question

. Let x(t) be a periodic function with period T = 10. The Fourier series coefficients for this series are denoted by $a_k$, that is $x\left(t\right)={\sum }_{k=-\infty }^{\infty }{a}_k{e}^{jk\frac{2\pi }{T}t}$ The same function x(t) can also be considered as a periodic function with period T' = 40. Let $b_k$ be the Fourier series coefficients when period is taken as T'. If ${\sum }_{k=-\infty }^{\infty }|a_k|=16,then{\sum }_{k=-\infty }^{\infty }|b_k|$ is equal to

• A. 256
• B. 64
• C. 16
• D. 4

Answer 1

The correct option is C 16

Given, x(t) is periodic function with period T = 10, and fourier series coefficinet is $a_k$
$x\left(t\right)={\sum }_{k=-\infty }^{\infty }{a}_k{e}^{jk\frac{2\pi }{T}t}$
$x\left(0\right)={\sum }_{k=-\infty }^{\infty }{a}_k=16$
If time period T' = 40 then fourier series coefficient is $b_k$ so
$x\left(t\right)={\sum }_{k=-\infty }^{\infty }{b}_k{e}^{jk\frac{2\pi }{T^{\prime }}t}$
$x\left(0\right)={\sum }_{k=-\infty }^{\infty }{b}_k$
But x(0) = 16
${\sum }_{k=-\infty }^{\infty }{b}_k=16$