Answer:
Given in explanation.
Explanation:
First seperate the signal into components,
[tex](1 + 0.1 \cos5t)\cos100t=\cos100t+0.1 \cos5t\cos100t[/tex]
Now, we will use the identity given in the hint,
[tex]\cos A \cos B = \frac{1}{2} (\cos(A+B) + \cos(A-B))\\\\\cos100t+0.1 \cos5t\cos100t=\cos100t + 0.05\cos105t + 0.05\cos95t[/tex]
Now the signal is decomposed into its different components.
Accordingly,
[tex]at\:\:100/(2*\pi) = 50/\pi \:Hz \:(16\:Hz)\:\:The \:\:amplitude\:\: is\:\: 1 \\at\:\:105/(2*\pi) = 52.5/\pi \:Hz \:(17\:Hz)\:\:The \:\:amplitude\:\: is\:\: 0.05 \\at\:\:95/(2*\pi) = 47.5/\pi \:Hz \:(15\:Hz)\:\:The \:\:amplitude\:\: is\:\: 0.05[/tex]
Since there is no other term within the cos function, the phase differences in all of the above cases are zero.
The amplitude, frequency, and phase of each component are respectively;
Amplitude; 1, 0.05, 0.05
Frequency; 50/π, 52.5/π, 47.5/π
Phase; All are zero
We are given the signal;
(1 + 0.1 cos5t) cos100t
Expanding by multiplying out the bracket gives;
cos100t + 0.1 cos5t cos100t
From trigonometric identity, we know that;
cosA cosB = ¹/₂[cos(A + B) + cos(A - B)]
⇒ 0.5cos(A + B) + 0.5cos(A - B)
Thus;
cos100t + 0.1 cos5t cos100t = cos100t + 0.1[0.5cos(5t + 100t) + 0.5cos(5t - 100t)]
⇒ cos100t + 0.1[0.5cos(105t) - 0.5cos(95t)]
⇒ cos100t + 0.05cos(105t) - 0.05cos(95t)
We know that formula for simple harmonic motion is;
x(t) = Acos(ωt + φ)
where;
A is amplitude
ω is angular frequency
φ is phase angle
decomposing the signal equation gives;
cos100t
0.05cos(105t)
0.05cos(95t)
Comparing each with x(t) = Acos(ωt + φ), we have;
For cos100t;
A = 1
ω = 100
phase = 0
Frequency; f = ω/2π
f = 100/2π
f = 50/π
For 0.05cos(105t);
A = 0.05
ω = 105
phase = 0
Frequency; f = ω/2π
f = 105/2π
f = 52.5/π
For 0.05cos(95t);
A = 0.05
ω = 95
phase = 0
Frequency; f = ω/2π
f = 95/2π
f = 47.5/π
Read more about sinusoidal wave motion at; https://brainly.com/question/13226564
