Question

Determine whether the following signals are periodic or non-periodic.If the signal is periodic find the fundamental period (i) x(t)=6 cos(5)/(4)t+2 cos(9)/(7)t (ii) x(t)=sin sqrt(3t)+cos sqrt(3)t (iii) x(t)=e^(j((5)/(3)t+(pi)/(4)))

Answer 1

i)$x(t)=\cos{\dfrac{5}{4}t}+2\cos{\dfrac{9}{7}t}\\ \omega_1=\dfrac{5}{4}\\ \omega_2=\dfrac{9}{7}\\ T_1=\dfrac{2\pi}{\omega_1}\\ T_1=\dfrac{8\pi}{5}\\ T_2=\dfrac{14\pi}{9}\\ \dfrac{T_1}{T_2}=\dfrac{36}{35}\\$

since the ratio of the periods is a rational number, the signal is periodic

$T_o=35T_1=36T_2=56\pi$

Fundamental period is 56$\pi$

ii)

$x(t)=\sin{\sqrt{3t}}+\cos{\sqrt{3}t}\\ \omega_2=\sqrt{3}$

Not periodic because the first expressiin cannot be expressed as a signal. Hence, no fundamental frequency. However, if $x(t)=\sin{\sqrt{3}t}+\cos{\sqrt{3}t}\\$. Then, the signal is periodic with fundamental period of 1

iii)

$x(t)=e^{j(\dfrac{5}{3}t+\dfrac{\pi}{4})}\\ x(t)=\cos({\dfrac{5}{3}t+\dfrac{\pi}{4}})+j\sin({\dfrac{5}{3}t+\dfrac{\pi}{4}})\\$

The signal is periodic with a fundamental period of 1s