Question

Consider the signal,

$x\left(t\right)=\sum _{m=-\infty }^{\infty }\delta (t-\frac{m}{3})+\delta (t-\frac{2m}{3})$
Where $C_n$ is the exponential Fourier series coefficient of signal x(t). Which of the following is/are correct?

• A. $C_1=C_3=C_5=1.5$
• B. x(t) is a periodic signal with time period 2.
• C. $C_2=C_4=C_6=0$
• D. x(t) is a periodic signal with time period $\frac{2}{3}$

Answer 1

Answer: (A), (D)

x(t) is a periodic signal with time period $\frac{2}{3}$

(a,d)
$\left(xt\right)=\sum _{m=-\infty }^{\infty }\delta (t-\frac{m}{3})+\delta (t-\frac{2m}{3})$

$\sum _{m=-\infty }^{\infty }\delta (t-\frac{m}{3})$ can be plotted as

$\sum _{m=-\infty }^{\infty }\delta (t-\frac{2m}{3})$ can be plotted
$x\left(t\right)=\sum _{m=-\infty }^{\infty }\delta (t-\frac{m}{3})+\delta (t-\frac{2m}{3})$

x(t) is periodic signal with,

$T=\frac{2}{3}$

${\omega }_0=\frac{2\pi }{\frac{2}{3}}=3\pi$

$C_n=\frac{1}{T}\underset{0}{\overset{T}{\int }}x\left(t\right)e^{-jn{\omega }_{0}tdt}$

$C_n=\frac{3}{2}\underset{0}{\overset{2/3}{\int }}\left[2\delta \right(t)+\delta (t-\frac{1}{3})]e^{-jn\left(3\pi \right)t}dt$

$\underset{t_1}{\overset{t_2}{\int }}f\left(t\right)\delta (t-t_0)dt=f\left(t_0\right)\forall to\in (t_1,t_2)$

$C_n=\frac{3}{2}[2+e^{-jn\pi }]=\frac{3}{2}[2+{(-1)}^n]$

$C_n=4.5forniseven$

$C_n=1.5fornisodd$

$x\left(t\right)=\sum _{m=-\infty }^{\infty }\delta (t-\frac{m}{3})+\delta (t-\frac{2m}{3})$

$T=\frac{2}{3}$

${\omega }_0=\frac{2\pi }{\frac{2}{3}}=3\pi$

$C_n=\frac{1}{T}\underset{0}{\overset{T}{\int }}x\left(t\right)e^{-jn{\omega }_{0}t}dt$

$C_n=\frac{3}{2}\underset{0}{\overset{2/3}{\int }}\left[2\delta \right(t)+\delta (t-\frac{1}{3})]e^{-jn\left(3\pi \right)t}dt$

$C_n=\frac{3}{2}[2+e^{-jn\pi }]=\frac{3}{2}[2+{(-1)}^n]$