# LAB-4 Wireless Networks Course Instructor: Dr.Amina Saleem Lab

### LAB-4 Wireless Networks Course Instructor: Dr.Amina Saleem Lab Conducted by Miss Moonerah Al-Eidi

Question No 1:

The room temperature is usually taken to be π = 17 π πΆ or 290πΎ . At this temperature find the thermal noise power density.

Solution:

π΅ π= π²π» ( πΎππππ π―π )

Where

π² = π. ππππ βππ

is the Boltzman constant. And

π΅ π

is the thermal noise power density per Hertz or bandwidth.

π΅ π= π²π» ( πΎππππ π―π ) = (π. ππ Γ ππ βππ Γ πππ = π Γ ππ βππ ( πΎππππ π―π ) = βπππππ©πΎ/π―π

The expression for

ππ© πΎ = πππππ ππ (π·ππππ ππ πππππ/ππΎπππ)

.

is an absolute level of power relative to 1W which selected as a reference and defined to be

The thermal noise is called white noise as it is constant over the entire bandwidth. So given a receiver with the above noise temperature and a 10MHz bandwidth, the thermal noise at the receiverβs output is given by:

π΅ π = π²π»π© π΅ ππ© = βπππ. π + πππππ(πππ) + πππππππ π π΅ = βπππ. πππ©πΎ

Question No 2: Consider a voice channel being used via modem to transmit digital data. Assuming a bandwidth of 3100 Hz. Then the Nyquist capacity is given by the following considering M=2

If we use M-Ary Signalling where the total number of levels are M=8 then the channel capacity is eaual to 18600 b/s over a bandwidth of 3100Hz.

πͺ = ππππ Γ π = πππππ ππππ/ππππππ

Question No 3 Prove the following expression for the Signal energy/bit to Noise power per Hz or the ratio

π¬ π π΅ π The relationship between the π¬ π π΅ π and the SNR can be represented as The relationship between the π¬ π π΅ π and the spectral efficiency can be found out as

Question No 4:

Given π¬ π π΅ π = 8.4ππ΅ is required to achieve a bit error rate of 10 β4 .If the effective noise temperature is 290π and the data rate is 2400bps, what received signal is required. 8.4 = π(ππ΅π) β 10 log(2400) + 228.6ππ΅π β 10πππ290 = β161.8ππ΅π

Question No 5:

Find the minimum

π¬ π /π΅ π

required to achieve a spectral efficiency of 6bps/Hz.

π¬ π π΅ π = π π(π π β π) = ππ. π = ππ. ππππ©

Question No 6: If the maximum distance between two antennas for LOS transmission is 41Km. Find the length of the transmitting antenna if the length of the receiving antenna is 10 m.

ππ = π. ππ(βπ²π π + βπ/π Γ ππ) π π = ππ. ππ

Question No 7: Following Figure gives us the relationship between the free space loss with distance and the frequency. What can you interpret between the relationship of the loss with increasing distance and increasing frequencies. Solution: We can see from the following curve, that the free space loss increases with increasing distance from the transmitter and it is higher for higher frequencies.