Question: Q3.16) We can do this by using Shannon's equation: C-2Blog2M C = 9600bps a) b) Q3.19) C- Blog2 (1+ SNR) SNR = ? Q3.20) a)
Q3.16) We can do this by using Shannon's equation: C-2Blog2M C = 9600bps a) b) Q3.19) C- Blog2 (1+ SNR) SNR = ? Q3.20) a) In figure 3.7c, Output wave form: [sin (2xft) +1/3 sin(2n(3fi)t)+ 1/5 sin (2m(5fi)1)+ 1/7 sin (2n (7fi)t)] where fi = 1/T= 1kHz Output power-? watt b) where f= 8kHz, N.-0.1 watt Hz Output noise power -f No Output noise power? mWatt SNR-output signal power/ output noise power SNR-? Therefore: (SNR)dB = 10 log 10 (signal poweroise power) (SNR)all = ? Gains GdB = 10log 10 (Pout/Pin) Losses LdB-10logio (Pin/Poat) Q3.22) Decibels 1 Losses- Gains
Step by Step Solution
There are 3 Steps involved in it
1 Expert Approved Answer
Step: 1 Unlock
Question Has Been Solved by an Expert!
Get step-by-step solutions from verified subject matter experts
Step: 2 Unlock
Step: 3 Unlock
