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The input to a certain filter is given by vin(t)=2cos(104πt−25∘) and the steady-state output is given by vout(t)=6cos(104πt+20∘). Part A Determine the (complex) value of the transfer function of the filter for f = 5000 Hz. Enter your answer to three significant figures using polar notation. Express argument in degrees.

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Answer:

[tex]H(5000) = 3 \angle45\°[/tex]

Explanation:

Easy, we know that

[tex]v_{in}(t)= 2cos(104\pi t -25\°)\\v_{out}(t)= 6cos(10 \pi t +20\°)[/tex]

Transforming to polar:

[tex]V_{in} = 2 \angle -25\° V\\V_{out} = 6 \angle 20\°V[/tex]

For [tex]f=5000Hz[/tex]

[tex]H(5000) = \frac{V_{out}}{V_{in}}[/tex]

[tex]H(5000) = \frac{6 \angle 20\°V}{2 \angle -25\° V}[/tex]

[tex]H(5000) = 3 \angle45\°[/tex]

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