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1 month ago

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For each of the following combinations of parameters, determine if the material is a low-loss dielectric, a quasi-conductor, or

a good conductor, and then calculate α,β,λ,up, and ηc:(a)Glass with μr=1,εr=5, andσ=10−12S/m at 10 GHz.(b)Animal tissue with μr=1,εr=12, and σ=0.3 S/m at 100 MHz.(c)Wood with μr=1,εr=3, and σ=10−4S/m at 1 kHz

1 answer:

choli [298]1 month ago

8 0

Answer:

Glass: Low-Loss dielectric

α = 8.42*10^-11 Np/m

β = 468.3 rad/m

λ = 1.34 cm

up = 1.34*10^8 m/s

ηc = 168.5 Ω

Tissue: Quasi-Conductor

α = 9.75 Np/m

β = 12.16 rad/m

λ = 51.69 cm

up = 0.52*10^8 m/s

ηc = 39.54 + j 31.72 Ω

Wood: Good conductor

α = 6.3*10^-4 Np/m

β = 6.3*10^-4 Np/m

λ = 10 km

up = 0.1*10^8 m/s

ηc = 6.28*( 1 + j )

Explanation:

Given:

Glass with µr = 1, εr = 5, and σ = 10−12 S/m at 10 GHz

Animal tissue with µr = 1, εr = 12, and σ = 0.3 S/m at 100 MHz.

Wood with µr = 1, εr = 3, and σ = 10−4 S/m at 1 kHz

Find:

Determine if the material is a low-loss dielectric, a quasi-conductor, or a good conductor, and then calculate α, β, λ, up, and ηc:

Solution:

- We need to determine the loss tangent to classify the medium as follows:

σ / w*εr*εo

Where, w is the wave's angular frequency

εo is the permittivity of free space = 10^-9 / 36*pi

- Now we categorize according to the following guidelines:

$Glass = \frac{10^-^1^2 }{2*\pi * 10*10^9 * \frac{5*10^-^9}{36\pi } } = 3.6*10^-^1^3\\\\Tissue = \frac{0.3 }{2*\pi * 100*10^6 * \frac{12*10^-^9}{36\pi } } = 4.5\\\\Wood = \frac{10^-^4 }{2*\pi * 1*10^3 * \frac{3*10^-^9}{36\pi } } = 600\\$

- For σ / w*εr*εo < 0.01 --- Low-Loss dielectric and σ / w*εr*εo > 100 --- Good conductor.

Glass: Low-Loss dielectric

Tissue: Quasi-Conductor

Wood: Good conductor

- Next, we will apply the categorized material equations specified in Table 17-1 as follows:

Glass: Low-Loss dielectric

α = (σ / 2)*sqrt(u / εr*εo) = (10^-12 / 2)*sqrt( 4*pi*10^-7/5*8.85*10^-12)

α = 8.42*10^-11 Np/m

β = w*sqrt (u*εr*εo) = 2pi*10^10*sqrt (4*pi*10^-7*5*8.85*10^-12)

β = 468.3 rad/m

λ = 2*pi / β = 2*pi / 468.3

λ = 1.34 cm

up = λ*f = 0.0134*10^10

up = 1.34*10^8 m/s

ηc = sqrt ( u / εr*εo ) = sqrt( 4*pi*10^-7/12*8.85*10^-12)

ηc = 168.5 Ω

Tissue: Quasi-Conductor

α = (σ / 2)*sqrt(u / εr*εo) = (0.3 / 2)*sqrt( 4*pi*10^-7/12*8.85*10^-12)

α = 9.75 Np/m

β = w*sqrt (u*εr*εo) = 2pi*100*10^6*sqrt (4*pi*10^-7*12*8.85*10^-12)

β = 12.16 rad/m

λ = 2*pi / β = 2*pi / 12.16

λ = 51.69 cm

up = λ*f = 0.5169*100*10^6

up = 0.52*10^8 m/s

ηc = sqrt ( u / εr*εo )*( 1 - j (σ / w*εr*εo))^-0.5

ηc = sqrt (4*pi*10^-7*12*8.85*10^-12)*( 1 - j 4.5)^-0.5

ηc = 39.54 + j 31.72 Ω

Wood: Good conductor

α = sqrt (pi*f*σ u) = sqrt( pi* 10^3 *4*pi* 10^-7 * 10^-4 )

β = α = 6.3*10^-4 Np/m

λ = 2*pi / β = 2*pi / 6.3*10^-4

λ = 10 km

up = λ*f = 10,000*1*10^3

up = 0.1*10^8 m/s

ηc = α*( 1 + j ) / б = 6.3*10^-4*( 1 + j ) / 10^-4

ηc = 6.28*( 1 + j )

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