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noname [10]
3 years ago
3

Decompose the signal (1+0.1 cos5t) cos100t into a linear combination of sinusoidal functions, and find the amplitude, frequency,

and phase of each component. Hint: use the identity for cosacosb.

Engineering
1 answer:
qaws [65]3 years ago
5 0

Answer:

amplitudes : 1 , 0.05, 0.05

frequencies : 50/\pi,   105/2\pi,  95/2\pi

phases : \pi /2   ,   \pi /2   , \pi /2

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 + \frac{0.1}{2} [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 (\frac{50}{\pi } )t + 0.05cos2 (\frac{105}{2\pi } )t + 0.05cos2 (\frac{95}{2\pi } )t [ ∵cos (∅) = sin(/2 +∅ ]

= sin ( 2 (\frac{50}{\pi } ) t  + /2 ) + 0.05sin ( 2 (\frac{105}{2\pi } ) t + /2 ) + 0.05sin ( 2 (\frac{95}{2\pi } )t + /2 )

attached is the remaining part of the solution

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Let <em>f(z)</em> = (4<em>z </em>² + 2<em>z</em>) / (2<em>z </em>² - 3<em>z</em> + 1).

First, carry out the division:

<em>f(z)</em> = 2 + (8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1)

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2<em>z </em>² - 3<em>z</em> + 1 = (2<em>z</em> - 1) (<em>z</em> - 1)

so you can separate the rational part of <em>f(z)</em> into partial fractions. We have

(8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1) = <em>a</em> / (2<em>z</em> - 1) + <em>b</em> / (<em>z</em> - 1)

8<em>z</em> - 2 = <em>a</em> (<em>z</em> - 1) + <em>b</em> (2<em>z</em> - 1)

8<em>z</em> - 2 = (<em>a</em> + 2<em>b</em>) <em>z</em> - (<em>a</em> + <em>b</em>)

so that <em>a</em> + 2<em>b</em> = 8 and <em>a</em> + <em>b</em> = 2, yielding <em>a</em> = -4 and <em>b</em> = 6.

So we have

<em>f(z)</em> = 2 - 4 / (2<em>z</em> - 1) + 6 / (<em>z</em> - 1)

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Now condense <em>f(z)</em> into one series:

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