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# Answer to Question #226768 in Electrical Engineering for Mohitur rahaman

Question #226768
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)))
1
2021-08-19T04:19:01-0400

i)"x(t)=\\cos{\\dfrac{5}{4}t}+2\\cos{\\dfrac{9}{7}t}\\\\\n\\omega_1=\\dfrac{5}{4}\\\\\n\\omega_2=\\dfrac{9}{7}\\\\\nT_1=\\dfrac{2\\pi}{\\omega_1}\\\\\nT_1=\\dfrac{8\\pi}{5}\\\\\nT_2=\\dfrac{14\\pi}{9}\\\\\n\\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}\\\\\n\\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})}\\\\\nx(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

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