Assume that we have a BS with a 6-dB antenna gain and an MS with antenna gain of 2 dB, at heights 10 m and 1.5 m, respectively,
operating in an environment where the ground plane can be treated as perfectly conducting. The lengths of the two antennas are 0.5 m and 15 cm, respectively. The BS transmits with a maximum power of 40 W and the mobile with a power of 0.1 W. The center frequency of the links (duplex) are both at 900 MHz, even if in practice they are separated by a small duplex distance (frequency difference).(a) Assuming that holds, calculate how much received power is available at the output of the receive antenna (BS antenna and MS antenna, respectively), as a function of distance d?(b) Plot the received powers for all valid distances d holds and the far field condition of the antennas is fulfilled.
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Answer:
The answers are as follow:
a) 10 mm
b) 12.730 N/
c) 127.307 N/
d) 0.25
Explanation:
d1 = 10mm , L1 = 40 mm, L2 = 50 mm, reduction in area = 90% = 0.9
Force = F =1000 N
let us find initial area first, A1 = pi* = 78.55
using reduction in area formula : 0.9 = (A1 - A2 ) / A1
solving it will give, A2 = 0.1 A1 = 7.855
a) The specimen elongation is final length - initial length
50 - 40 = 10 mm
b) Engineering stress uses the original area for all stress calculations,
Engineering stress = force / original area = F / A1 = 1000 / 78.55
Engineering stress = 12.730
c) True stress uses instantaneous area during stress calculations,
True fracture stress = force / final area = F / A2 = 1000 / 7.855
True Fracture stress = 127.30
e) strain = change in length / original length
strain = 10 / 40 = 0.25
Answer:
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As we infer from the above equation If the velocity of the escalator is doubled then the Power required will also be doubled and become 31.215kW