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aleksandr82 [10.1K]
3 years ago
10

A 200 Ω transmission line is to be matched to a computer terminal with ZL = (50 − j25) Ω by inserting an appropriate reactance i

n parallel with the line. If f = 800 MHz and r = 4, determine the location nearest to the load at which inserting:
A) A capacitor can achieve the required matching, and the value of the capacitor. PROBLEMS 131 B) An inductor can achieve the required matching, and the value of the inductor.
Engineering
1 answer:
snow_lady [41]3 years ago
7 0

Answer:

The answer is below

Explanation:

a) To solve this problem, we are going to use the smith chart. After entering the value of Zo = 200 ohm and ZL = (50 − j25) Ω, this give us z_L\ and\ y_L, the intersection between y_L and the SWR line gives:

y(d)=1.026-j1.54\\\\We\ are\ using\ the\ imaginary\ part\ to calculate\ the\ capacitance,hence:\\\\wC=1.54*Y_o\\\\C=\frac{1.54}{Z_o*w}=\frac{1.54}{200*2\pi*800*10^6}=1.53*10^{-12}\\  \\C=1.53*10^{-12}F

b) Also, we get:

y(d)=1.0-j1.52\\\\We\ are\ using\ the\ imaginary\ part\ to\ calculate\ the\ inductance,hence:\\\\1/wL=1.52*Y_o\\\\L=\frac{Z_o}{1.52*w}=\frac{200}{1.52*2\pi*800*10^6}=2.6*10^{-8}\\  \\L=2.6*10^{-8}H

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5. A typical paper clip weighs 0.59 g and consists of BCC iron. Calculate (a) the number of
marta [7]

Answer:

(a) 3.185*10^{21} cells

(b) 6.37*10^{21} atoms

Explanation:

(a)

Volume, V of unit cell

V=(2.866*10^{-8})^{3}=2.354*10^{-23}

Number of unit cells, N

N=\frac {W_{mat}}{V\rho_{mat}} Where W_{mat} is weight of material and \rho_{mat} is density of material

N=\frac{0.59}{7.87*(2.354*10^{-23}}=3.185*10^{21} cells

(b)

Number of atoms in paper clip

This is a product of number of unit cells and number of atoms per cell

Since iron has 2 atoms per cell

Number of atoms of iron=3.185*10^{21} cells*2 atoms/cell=6.37*10^{21} atoms

8 0
3 years ago
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user100 [1]

Answer: vehicles with a front engine and FWD or a rear engine and RWD.

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3 years ago
What is the function maintenance? List some important steps for vibration monitoring based maintenance.
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3.balance

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4 0
3 years ago
On what frequency can you expect to monitor air traffic in and around<br> Lincoln Airport?
Phantasy [73]

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118.5

Explanation:

Hope this helps!

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3 years ago
If we have silicon at 300K with 10 microns of p-type doping of 4.48*10^18/cc and 10 microns of n-type doping at 1000 times less
liq [111]

Answer:

The resistance is 24.9 Ω

Explanation:

The resistivity is equal to:

R=\frac{1}{N_{o}*u*V } =\frac{1}{4.48x10^{15}*1500*106x10^{-19}  } =0.93ohm*cm

The area is:

A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²

w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{A} }+\frac{1}{N_{D} })

If NA is greater, then, the term 1/NA can be neglected, thus the equation:

w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{D} })

Where

V = 0.44 V

E = 11.68*8.85x10¹⁴ f/cm

V_{o} =\frac{KT}{p} ln(\frac{N_{A}*N_{D}}{n_{i}^{2}  } , if n_{i}=1.5x10^{10}cm^{-3}  \\V_{o}=0.02585ln(\frac{4.48x10^{18}*4.48x10^{15}  }{(1.5x10^{10})^{2}  } )=0.83V

w=\sqrt{\frac{2*11.68*8.85x10^{-14}*(0.83-0.44) }{1.6x10^{-19}*4.48x10^{15}  } } =3.35x10^{-5} cm=0.335um

The length is:

L = 10 - 0.335 = 9.665 um

The resistance is:

Re=\frac{pL}{A} =\frac{0.93*9.665x10^{-4} }{0.36x10^{-4} } =24.9ohm

7 0
3 years ago
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