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nordsb [41]
3 years ago
6

Consider a sinusoidal oscillator consisting of an amplifier having a frequency-independent gain A (where A is positive) and a se

cond-order bandpass filter with a pole frequency ω0, a pole Q denoted Q, and a positive center-frequency gain K.
a) Find the frequency of oscillation, and the condition that A and K must satisfy for sustained oscillation.
Engineering
1 answer:
mafiozo [28]3 years ago
4 0

Sinusoidal oscillator frequency of oscillation is given below.

Explanation:

The criterion for a stable oscillator is given in the equation

l A(jw)β(jw) l ≥ 1

In this task A represents the gain of the amplifier , and

β represents gain/attenuation of the second-order bandpass filter.

This sinusoidal oscillation is a special edge case where the product is equal to one.

So the condition is A-K=1

to obtain the sustained oscillations at the desired frequency of oscillations, the product of the voltage gain A and the feedback gain β must be one or greater than one. In this case, the amplifier gain A must be 3. Hence, to satisfy the product condition, feedback gain β must be 1/3.

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ryzh [129]

Answer:

The velocity at R/2 (midway between the wall surface and the centerline) is given by (3/4)(Vmax) provided that Vmax is the maximum velocity in the tube.

Explanation:

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v = (ΔPR²/4μL) [1 - (r²/R²)]

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So, velocity at any point in the radial direction is

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At the point r = (R/2)

r² = (R²/4)

(r²/R²) = r² ÷ R² = (R²/4) ÷ (R²) = (1/4)

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1 - (r²/R²) = 1 - (1/4) = (3/4)

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4 0
3 years ago
A venturi meter is to be installed in a 63 mm bore section of a piping system to measure the flow rate of water in it. From spac
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Answer:

Throat diameter d_2=28.60 mm

Explanation:

 Bore diameter d_1=63mm  ⇒A_1=3.09\times 10^{-3} m^2

Manometric deflection x=235 mm

Flow rate Q=240 Lt/min⇒ Q=.004\frac{m^3}{s}

Coefficient of discharge C_d=0.8

We know that discharge through venturi meter

 Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}

h=x(\dfrac{S_m}{S_w}-1)

S_m=13.6 for Hg and S_w=1 for water.

h=0.235(\dfrac{13.6}{1}-1)

h=2.961 m

Now by putting the all value in

Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}

0.004=0.8\times \dfrac{3.09\times 10^{-3} A_2\sqrt{2\times 9.81\times 2.961}}{\sqrt{(3.09\times 10^{-3})^2-A_2^2}}

A_2=6.42\times 10^{-4} m^2

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4 0
3 years ago
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8 0
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Answer:

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