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**Question No 1: **

The room temperature is usually taken to be π = 17 π πΆ or 290πΎ . At this temperature find the thermal noise power density.

**Solution: **

π΅ π= π²π» ( πΎππππ π―π )

**Where **

π² = π. ππππ βππ

** is the Boltzman constant. And **

π΅ π

** is the thermal noise power density per Hertz or bandwidth. **

π΅ π= π²π» ( πΎππππ π―π ) = (π. ππ Γ ππ βππ Γ πππ = π Γ ππ βππ ( πΎππππ π―π ) = βππππ π©πΎ/π―π

**The expression for **

π π© πΎ = πππππ ππ (π·ππππ ππ πππππ/ππΎπππ)

**.**

π π© πΎ

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

ππ π© π

**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 receiverβs output is given by: **

π΅ π = π²π»π© π΅ π π© = βπππ. π + πππππ(πππ) + πππππππ π π΅ = βπππ. ππ π©πΎ

**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 **

πͺ = ππ© = πππππππ

**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. **

πͺ = ππππ Γ π = πππππ ππππ/ππππππ

**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 = 8.4ππ΅ ** **is required to achieve a bit error rate of 10 β4 .If the effective noise temperature is 290π and the data rate is 2400bps, what received signal is required. 8.4 = π(ππ΅π) β 10 log(2400) + 228.6ππ΅π β 10πππ290 = β161.8ππ΅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) = ππ. π = ππ. πππ 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. **

ππ = π. ππ(βπ²π π + βπ/π Γ ππ) π π = ππ. ππ

**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. **