Question

A periodic signal with period $T_0$ is real. The signal x(t) has even and odd part shown as $x_e\left(t\right)$ and $x_0\left(t\right)$ , also
$C_k$ is the $k^{th}$ exponential Fourier series coefficient of x(t) .
If $C_k=\{\begin{array}{cc}(\frac{10}{3|k|}+\frac{3}{k}j)k\ne 0and|K|<3& \\ 10,K=0& \end{array}$

Then which of the following is correct?

• A. The power of odd part of x(t) is 25.50 W
• B. The power of even part of x(t) is 120.46 W
• C. The power of even part of x(t) is 127.76 W
• D. None of the above
• E. The power of odd part of x(t) is 28.50 W

Answer 1

The correct option is C The power of even part of x(t) is 127.76 W

If x(t) is real then Fourier coefficient of even part of x(t) i.e.,
$x_e\left(t\right)$ is Re {$C_K$}

$so,B_k=\{\begin{array}{cc}\frac{10}{3|k|}|k|<3& \\ 10,k=0& \end{array}$

$so,B_k=\{\begin{array}{ccc}10,k=0& & \\ \frac{10}{3},k=1,-1& & \\ \frac{10}{6},k=2,-2& & \end{array}$

so power of $x_e\left(t\right)=\sum _{k=-\infty }^{\infty }|B_k{|}^2$

$=100+2(\frac{100}{9}+\frac{100}{36})$
$=127.76W$

If x(t) is real, the Fourier coefficient of odd part of x(t) i.e.,
$x_0\left(t\right)$ is img {$c_K$ }

$So,D_K=\{\begin{array}{cc}\frac{3}{K},forK\ne 0and|K|<3& \\ 0,forK=0& \end{array}$

So , Power of $x_0\left(t\right)$ = $\sum _{k=-\infty }^{\infty }|D_K{|}^2$

$=[{\left(\frac{3}{-2}\right)}^2+{\left(\frac{3}{-1}\right)}^2+(3{)}^3+\left(\frac{3}{2}{)}^2\right]$

$=\frac{9}{4}+9+9+\frac{9}{4}=22.5W$