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STALIN [3.7K]
2 years ago
11

A) If a given directional antenna can receive 15 times the power of an isotropic antenna, what is

Engineering
1 answer:
Sauron [17]2 years ago
4 0

The gain of the directional antenna is calculated based on the comparison

with an hypothetical isotropic antenna.

  • a) The gain in dBi of the isotropic antenna is approximately <u>11.76 dBi</u>
  • b) The power received by the isotropic antenna is approximately <u>0.61 mW</u>

Reasons:

a) The number of times the power of the directional antenna is stronger than the  isotropic antenna, G = 15 times

The gain in dBi is given as follows;

\displaystyle G = \mathbf{10^{\dfrac{G(dBi)}{10} }}

Which gives;

\displaystyle 15 = 10^{\dfrac{G(dBi)}{10} }

\displaystyle \dfrac{G(dBi)}{10}  = \frac{ln(15)}{ln(10)}

\displaystyle G(dBi)} = 10 \times \frac{ln(15)}{ln(10)} \approx 11.76

The power gain of the isotropic antenna, G(dBi) ≈ <u>11.76 dBi</u>

b) When the gain of the directional antenna is 2.15 dBi and it received 1 mW of power, we have;

The number of times stronger the directional antenna is, is found as follows;

\displaystyle G  = 10^{\dfrac{2.15}{10} } \approx \mathbf{1.64}

The 2.15 dBi directional antenna receives the signal approximately 1.64 times stronger than the isotropic antenna, therefore;

  • \displaystyle The \ power \ received \ by \ the \ isotropic \ antenna  = \frac{1 \, mW}{1.64}  \approx \underline {0.61 \, mW}

Learn more here:

brainly.com/question/17587029

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At the instant shown, slider block B is moving with a constant acceleration, and its speed is 150 mm/s. Knowing that after slide
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Answer:

a) aA = - 13.33 mm/s²

aB = - 20 mm/s²

b) aD = - 13.33 mm/s²

c) vB = 70 mm/s

d) xB = 440 mm

Explanation:

Given

The initial speed of B is: v₀B = 150 mm/s

Distance moved by A is: xA = 240 mm

Velocity of A is: vA = 60 mm/s

Assuming:

Displacement of blocks are denoted by:

A = xA

B = xB

C = xC

D = xD

From the pic shown, the total length of the cable is:

xB + (xB - xA) + 2*(d - xA) = L

⇒ 2*xB - 3*xA = L - 2*d

where L - 2*d is constant. Differentiating the above equation with respect to time:

d(2*xB)/dt - d(3*xA)/dt = 0

⇒ 2*vB - 3*vA = 0    (i)

Substituting in equation (i)

2*(150 mm/s) - 3*vA = 0

⇒ v₀A = 100 mm/s  (initial speed of A)

Then, we use the equation

vA² = v₀A² + 2*aA*xA

Substituting the values in above equation:

(60 mm/s)² = (100 mm/s)² + 2*aA*(240 mm)

⇒ aA = - 13.33 mm/s²

If  2*vB - 3*vA = 0

Differentiating the above equation with respect to time:

d(2*vB)/dt - d(3*vA)/dt = 0

⇒ 2*aB - 3*aA = 0    (ii)

Substituting in equation (ii)

2*aB - 3*(- 13.33 mm/s²) = 0

⇒ aB = - 20 mm/s²

b) From the pic shown,

xD - xA = constant

If we apply

d(xD)/dt - d(xA)/dt = 0

⇒ vD - vA = 0

then

d(vD)/dt - d(vA)/dt = 0

⇒ aD - aA = 0

⇒ aD = aA = - 13.33 mm/s²

c) We use the formula

vB = v₀B + aB*t

Substituting the values in above equation:

vB = 150 mm/s + (- 20 mm/s²)*(4 s)

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d) We apply the equation

xB = v₀B*t + 0.5*aB*t²

Substituting the values in above equation:

xB = (150 mm/s)*(4 s) + 0.5*(- 20 mm/s²)*(4 s)²

⇒ xB = 440 mm

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