Question

new defined PCM system is going to be employed to send 2 MByte of data. Samp rate to be used is 50% higher than Nyquist rate. Referring to the frame structure given calculate the following. Frame Structure Number of channel = 32 Number of bits in one time slot = 8 32 channels = 1 frame Number of bits in a frame = 32 x 8 = 256 bits This frame must be transmitted within the sampling period i. Sampling frequency. (CO2 - 2 mark ii. Transmission rate. (CO2 - 1.5 mark iii. Duration of a frame. (CO2 - 1.5 mar! iv. Number of frame needed. (CO2 - 3 marl v. Time taken to send the data. (CO2 - 2 mar

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soln let anume pom model is used for voice . fransmission where standard sampling frequency fó' = 6000 Samples/s As perproblem, new sampling frequency is fs = fit ft fs = 12000 Samplests. Sampling duration. To = 1/2 = 12000 ** each sample contains & bit, and there are 32 Channel each producing one sample in each time slot so & Trang mission rated is $b = 8X12000x32 "R5 = 3.0 72 mbps.. So C) Duration of frame is Ts = Number of bits in frame TE = 256 = I. 1 g Transmission rate. 2000 32X8 X 12000 IF = 33.33 AS do let N number of frame required to transmitt 2 m Byte data, 256XN = 2 M Bite (As Size of one frame is 256 bits N = 1024x1024x6x2 = 3X32X1024 => N= 3276882 256 N = 65536. 1 e) Since trans mission rate is Rb=3.072 mbps. let of time required to send 2 m Byte data, ot=18x1024X1024 At= 5, 46588 Rb