Want this question answered?
43600Hz (43.6KHz)
no of sources: 5 bandwidth required for each source= 400 Hz no of guard times= 5 bandwidth of each guard time = 200 Hz minimum bandwidth = 5 *400 + 5*200 Hz
ssb modulation scheme
answer in www.ent.mrt.ac.lk/~ekulasek/cni/cni4-eck.ppt last slide
600 Hz
C = 9600 = 2B*3 = 2B * 3 W = 1600 Hz
600
7000 Hz
The answer depends on the accuracy desired. a. The minimum bandwidth, a rough approximation, is B = bit rate /2, or 500 kHz. We need a low-pass channel with frequencies between 0 and 500 kHz. b. A better result can be achieved by using the first and the third harmonics with the required bandwidth B = 3 × 500 kHz = 1.5 MHz. c. A still better result can be achieved by using the first, third, and fifth harmonics with B = 5 × 500 kHz = 2.5 MHz
38,400 bps... But how old is this question that the speed is so insanely slow?
7000 Hz
150 khz