b) There is a sending and receiving system on the same floor at a distance of 50 m. This f has several characteristics: The exponent value is (1.8), the floor attenuation facto (10dB), and the partition attenuation factor is (16dB). The boundary between the n field and the far field is (d. =1 m). Calculate the attenuation constant if: i- The receiving is fixed. ii- The receiving moves to another location on the same floor and the same distance.. iii- Comment on your results? F(d)[dB] = [(d)[dB] + 10nSF Log () + FAF[dB] + Σ PAF[dB] FL(d)[dB] = [(do)[dB] + 10nMpLog() + Σ PAF[dB] P(d)[dB]= F(do) [dB]+20Log +x+d+ FAF[dB]+PAF[dB]

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Note that f=900MHz

Subject (mobile communication)

b) There is a sending and receiving system on the same floor at a distance of 50 m. This f
has several characteristics: The exponent value is (1.8), the floor attenuation facto
(10dB), and the partition attenuation factor is (16dB). The boundary between the n
field and the far field is (d. =1 m). Calculate the attenuation constant if:
i- The receiving is fixed.
ii- The receiving moves to another location on the same floor and the same distance..
iii- Comment on your results?
F(d)[dB] = [(d)[dB] + 10nSF Log () + FAF[dB] + Σ PAF[dB]
FL(d)[dB] = [(do)[dB] + 10nMpLog() + Σ PAF[dB]
P(d)[dB]= F(do) [dB]+20Log
+x+d+ FAF[dB]+PAF[dB]
Transcribed Image Text:b) There is a sending and receiving system on the same floor at a distance of 50 m. This f has several characteristics: The exponent value is (1.8), the floor attenuation facto (10dB), and the partition attenuation factor is (16dB). The boundary between the n field and the far field is (d. =1 m). Calculate the attenuation constant if: i- The receiving is fixed. ii- The receiving moves to another location on the same floor and the same distance.. iii- Comment on your results? F(d)[dB] = [(d)[dB] + 10nSF Log () + FAF[dB] + Σ PAF[dB] FL(d)[dB] = [(do)[dB] + 10nMpLog() + Σ PAF[dB] P(d)[dB]= F(do) [dB]+20Log +x+d+ FAF[dB]+PAF[dB]
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