CISC 7332X Spring 2022 Midterm Exam

1) a
2) d
3) c
4) d
5) c
6) c
7) d
8) b
9) c
10) 96Mbps; 24*log(16)/log(2)
11) 23.4 kHz;  96Kbps = 23486 * log( (64+4)/4 ) /log(2)
12) 4;  96000/(2*23486) = log(M)/log(2) 
13) 4;  5*log(15+1)/log(2) = 20 = 2*B Log(M); ((20/2)/5) = log(m)/log(2)
14) 13;  10 * log(20)/log(10)
15) 26Mbps;  SNR=100; 4 * log(100+1)/log(2) = 26.6
16) 757.71bps;  1000 - (1000.0 * -(0.04*log(0.04)/log(2) + 0.96*log(0.96)/log(2)))
17) -0.967 (or -1); 10 * log ( 20 / 25 ) /log(10)
18) d
19) d
20) a