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ololo11 [35]
3 years ago
7

a. Determine R for a series RC high-pass filter with a cutoff frequency (fc) of 8 kHz. Use a 100 nF capacitor. b. Draw the schem

atic (label Vin and Vout). c. Determine the frequency response H(f) in dB at f = 20 kHz given fc = 8 kHz.

Engineering
1 answer:
Readme [11.4K]3 years ago
5 0

Answer:

a) 199.04 ohms

b) attached in image

c) -0.696dB

Explanation:

We are given:

Fc = 8Khz = 8000hz

C = 100nF = 100*10^-^9F

a)Using the formula:

F_c = \frac{1}{2pie*Rc}

8000= \frac{1}{2*3.14*R*100*10^-^9}

R =\frac{1}{2*3.14*100*10^-^9*8000}

R = 199.04 ohms

b) diagram is attached

c) H(w) = \frac{V_out(w)}{Vin(w)} = \frac{1}{1-j\frac{wc}{w}}

H(F) = \frac{1}{1-j\frac{fc}{f}}

At F = 20KHz and Fc= 8KHz we have:

H(F)= \frac{1}{1-j\frac{8}{20}} = \frac{1}{1-j(0.4)}

|H(F)|= \frac{1}{\sqrt{1^2+(0.4)^2}}

=0.923

|H(F)| in dB = 20log |H(F)|

=20log0.923

= -0.696dB

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Answer:

A.) 62.5 ft

B.) 3.58 seconds

C.) 8.58 seconds

Explanation:

A.) Given that a ball is thrown vertically upward from the top of a building of 60ft with an initial velocity of vA=35 ft/s

To determine how high above the top of the building the ball will go before it stops at B, let us use the third equation of motion.

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H = 1225/19.6

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(B) The time tAB it takes to reach its maximum height will be achieved by using second equation of motion

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Substitutes all the parameters into the formula

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Divide all by 4.9

t^2 - 7.143t + 12.755 = 0

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Ignore the negative one.

(C) the total time tAC needed for it to reach the ground at C from the instant it is released.

When the object is falling back from B, the initial velocity = 0. And the height h will be 60 + 62.5 = 122.5 ft

Using equation 2 of equations of motion again.

h = 1/2gt^2

122.5 = 1/2 × 9.8 × t^2

122.5 = 4.9t^2

t^2 = 122.5/4.9

t^2 = 25

t = 5

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