Answer:
amplitudes : 1 , 0.05, 0.05
frequencies : 50/[tex]\pi[/tex], 105/[tex]2\pi[/tex], 95/2[tex]\pi[/tex]
phases : [tex]\pi /2 , \pi /2 , \pi /2[/tex]
Explanation:
signal s(t) = ( 1 + 0.1 cos 5t )cos 100t
signal s(t) = cos100t + 0.1cos100tcos5t . using the identity for cosacosb
s(t) = cos100t + [tex]\frac{0.1}{2}[/tex] [cos(100+5)t + cos (100-5)t]
s(t) = cos 100t + 0.05cos ( 100+5)t + 0.05cos (100-5)t
= cos100t + 0.05cos(105)t + 0.05cos 95t
= cos 2 [tex](\frac{50}{\pi } )t + 0.05cos2 (\frac{105}{2\pi } )t + 0.05cos2 (\frac{95}{2\pi } )t[/tex] [ ∵cos (∅) = sin(/2 +∅ ]
= sin ( 2 [tex](\frac{50}{\pi } ) t[/tex] + /2 ) + 0.05sin ( 2 [tex](\frac{105}{2\pi } ) t + /2 )[/tex] + 0.05sin ( 2 [tex](\frac{95}{2\pi } )t + /2[/tex] )
attached is the remaining part of the solution
