Question

(a) Consider a digital communication system with a frequency band ranging from 3 MHz to 4 MHz. It is required to operate without unacceptable bit errors for which let the required SNR min = 24dB. Compute the channel transmission capacity for this minimum value of SNR. (b) Let us consider a Nyquist channel capacity formula C = 2B log2 (M) (in bits/second) where M is the number of signal levels per second, and B is the bandwidth assumed in part (a). (i) Compute the Nyquist channel capacity for 4 levels per symbol. (ii) Compute the Nyquist channel capacity for 8 levels per symbol. (c) Now, using the two Nyquist channel capacities found in part (b), and the bandwidth assumed in part (a), what is the minimum possible SNR required for both cases in part (b) to achieve error-free transmission?

Answer 1

sol: The Data given in the question is as follows: frequency Band of a Digital Communication System = 3MHZ-4MHZ Bandwidth = 4 MHz-3MH2 = 1MHz Bu= 1MHZ SNR min = 24 dB a) the know that formula for Carculating Channel Capacity is given by. 3 - 0 C = Bloga 1+ () = 2408 lolog ( ) = 240B () Now Subsitating 102.4 - 251.18, BW= B = 1 MHz the Abore Values in equation ① Meget - c = lxlob luge S 4 + 251,18} bits/sec C= 1x lol loga (252.133 - 7.97X106 bits/sec = 7.97 Mbps Hence, the Channel Transmission Copacity of a Above Digital System is 7.97X106 bis.'

(b) The Myrnist Channed copacity formula is given by C = Blog (m) Бs/see (in kenen M= 4 (number of lerees are 4) В. Дмн2 - Ios - Hence, the Channed Capacity will be C: \ \о ) с. 16 То е с. 2 - 2 Ельск [c. 26 bes | (ii) kinen M= al & levees Per Symbul) B= 1 MH2 = 1x lob Hz Hence, the Channes Copacity killbe. с 1xu love (83 с 1xts (04, 2° С- 3х1,6 bes (с. 3 нь 1

@ Aste hare Calculate c=2x106 bis keinen (M= 4, B= 1MHz) Hence, using C - Bloga [ 1 + 3 - ② Subsitating the obtained Data in equation ② heget 2x108 4x108 loga ( 14 3 log, Sa+S/3= 2 () = 41 Toking logrithinic on both sides tolog () = lolog (3) ( dB = 4.77 dB Mininimum Possible (Sthildr = 4.77 dB. Hence, In this

As the hare Calculated- c= sinbes= 3x108 bps [M=8, B = JMHz] Now C= Bloga Subsituating the obtained Date in abore equation veget- 3x106 = 1x16 loge 1 + x3 → loga [ 14 4 3 = 3 1+(84) = 23 (s) = 7 Taking logrithmic on both sides me get lolog ( 4 ) = lo log(71) ( de = 84508 Hence, In this case Minimum (SHR) Required for enter free is 8.45 dB, Transmission