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2 years ago

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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A) The amount of space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 31.0 °C is; 0.6048 cm

B) The stress in the rails on a summer day when their temperature is 31.0 °C is; 86.4 × 10⁶ Pa

<h3>Linear Thermal Expansion</h3>

We are given;

Length; L = 14 m

Initial Temperature; T_i = −5 °C

Final Temperature; T_f = 31 °C

The formula for Linear Thermal Expansion is;

ΔL = L_i * α * ΔT

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L_i is initial length

α is thermal expansion

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ΔT is change in temperature

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Thus;

ΔL = 14 * 1.2 × 10⁻⁵ * (31 - (-5))

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The formula to get the stress is;

σ = Y * α * ΔT

where;

Y is young's modulus of steel = 20 × 10¹⁰ Pa

α is thermal expansion

ΔT is change in temperature

Thus;

σ = 20 × 10¹⁰ × 1.2 × 10⁻⁵ × (31 - (-5))

σ = 86.4 × 10⁶ Pa

The complete question is;

Steel train rails are laid in 14.0-m long segments placed end to end. The rails are laid on a winter day when their temperature is −5 °C.

(a) How much space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 31.0 °C?

(b) If the rails are originally laid in contact, what is the stress in them on a summer day when their temperature is 31.0 °C?

Read more about Linear Thermal Expansion at; brainly.com/question/6985348

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