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dmitriy555 [2]
3 years ago
5

In this design problem, your input signal is a mono audio signal that is the output of anelectret condenser microphone. An elect

ret condenser microphone output alwaysincludes a DC offset in addition to the signal itself. Suppose that this microphoneoutputs a signal with a voltage swing of 400 mVpp plus a DC offset of 0.8 V. Thismicrophone output signal needs to be changed before being input to a sensitive audiocircuit. Specifically, it needs to be inverted, amplified, and the DC offset needs to beremoved. Design and build a circuit that will cancel out the DC offset and achievemaximum signal amplification without exceeding the 20 Vpp input limit of the audioequipment. Hint: use superposition.
Engineering
1 answer:
Free_Kalibri [48]3 years ago
8 0

Answer:

DC offset is usually undesirable when it causes saturation or change in the operating point of an amplifier. An electrical DC bias will not pass through a transformer; thus a simple isolation transformer can be used to block or remove it, leaving only the AC component on the other side. In signal processing terms, DC offset can be reduced in real-time by a high-pass filter. When one already has the entire waveform, subtracting the mean amplitude from each sample will remove the offset. Often, very low frequencies are called "slowly changing DC" or "baseline wander"

Explanation:

You might be interested in
The Gill Art Gallery wishes to maintain data on their customers, artists and paintings. They may have several paintings by each
sashaice [31]

Answer:

a. $79,122.50

b. 314. 84 miles.

c. 700 ways.

Explanation:

a. Compound interest = A = P (1 + r/n)nt ;

where P = 7500 (principal amount)

r = 15% (interest rate)

t = 16 (no of years)

n = 4 (no of times compounded annually.)

and finally A is the Amount (present value after 16 years)

Upon calculation A = $79,122.50.

b. Answer: 314. 84 miles.

Explanation:

Given data. 160 miles / 7.75 liters of gasoline

Therefore 1 liter of gasoline = 160/7.75 = 20.645 miles/ liter.

15.25 liters of gasoline = 20.645 * 15.25 = 314.84 miles.

c. Answer: 700 ways.

Explanation:

This is an application of generalized principle of counting.

4 Americans can be choosen from 7 Americans = 7C4 = 35

3 Russians can be choosen from 6 Russians = 6C3 = 20.

Total number of ways = (7C4)* (6C3) = 35 * 20 = 700 ways.

3 0
3 years ago
The ignition temperature of 87-octane gasoline vapor is about 430 ∘C and, assuming that the working gas is diatomic and enters t
Flura [38]

Answer:

I have solved the problem below. I hope it will let you clear the concept.

For any inquiries ask me in the comments.

Explanation:

6 0
4 years ago
Engineers please help im not good when it comes to drawing​
kakasveta [241]

Answer:

I'm good at drawing and computer-animated design

Explanation:

3 0
3 years ago
A thin-walled tube with a diameter of 6 mm and length of 20 m is used to carry exhaust gas from a smoke stack to the laboratory
Molodets [167]

Answer:

Explanation:

Mean temperature is given by

T_mean = \frac{T_i + T_ \infinity}{2}\\\\T_mean = \frac{200 + 15}{2}

Tmean = (Ti + T∞)/2

T_mean = 107.5^{0}

Tmean = 107.5⁰C

Tmean = 107.5 + 273 = 380.5K

Properties of air at mean temperature

v = 24.2689 × 10⁻⁶m²/s

α = 35.024 × 10⁻⁶m²/s

\mu = 221.6 × 10⁻⁷N.s/m²

\kappa = 0.0323 W/m.K

Cp = 1012 J/kg.K

Pr = v/α  = 24.2689 × 10⁻⁶/35.024 × 10⁻⁶

              = 0.693

Reynold's number, Re

Pv = 4m/πD²

where Re = (Pv * D)/\mu

Substituting for Pv

Re = 4m/(πD\mu)

     = (4 x 0.003)/( π × 6 ×10⁻³ × 221.6 × 10⁻⁷)

     = 28728.3

Since Re > 2000, the flow is turbulent

For turbulent flows, Use

Dittus - Doeltr correlation with n = 0.03

Nu = 0.023Re⁰⁸Pr⁰³ = (h₁D)/k

(h₁ × 0.006)/0.0323 = 0.023(28728.3)⁰⁸(0.693)⁰³

(h₁ × 0.006)/0.0323 = 75.962

h₁ = (75.962 × 0.0323)/0.006

h₁ = 408.93 W/m².K

4 0
3 years ago
A rectangular weir is in a rectangular channel 2.9 m wide. The length of the weir is 1.9 m and is centered in the channel. If th
WINSTONCH [101]

Answer:

discharge = 0.310976 m³/s

Explanation:

given data

rectangular channel  wide = 2.9 m

length of weir L = 1.9 m

water level H = 0.2 m

solution

we get here discharge that is express as

discharge = \frac{2}{3} * C_d * L* \sqrt{2g} * H^{\frac{3}{2} }    ............................1

we consider here Coefficient of discharge Cd = 0.62

put here value we get

discharge = \frac{2}{3} * 0.62 * 1.9* \sqrt{2*9.8} * 0.2^{\frac{3}{2} }

discharge = 0.310976 m³/s

3 0
4 years ago
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