8/13/2019 Assignment 2 Analog and Digital Comm.

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a) Plot the transmitted signal (the sum of the individual pulses) corresponding to the datapattern. Be sure to include sufficient detail in your plot. Explain in detail how youobtained the answer.

Graph of sum if the individual pulses could be observed on next page.

Data taken from the question:

Bit period = 2/rb When data 1 amplitude = 1V,but when data 0 amplitude = -1 V Bitrate@rb = 100k bits/s Period = 2/rb = 20u seconds. Triangle pulses

This graph were obtained by plotting pulse of first bit .During first bit the amplitude is 1 Vbecaused the data is 1.The plot for the first bit could be observed in diagram next page.

Now for pulse of second bit, the data is 0 thus the amplitude is -1 V.

Now for pulse of third bit and forth bit, the data is 1 thus the ampltiude of is 1 V.

Finally for pulse of fifth bit the data is 0 thus the amplitude is -1V.

All the pulse pulses could be observed diagram next page.

Sum of the invidual pulses in the sum of the pulses. When there is a same data the amplitudebecome larger because of combination of that pulses.

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(b) Suppose that the transmitted signal is sent across an ideal baseband channel.Explain and indicate how the receiver extracts the digital information from the receivedsignal.

When the data delivered is a 1,the receiver received 1 volt but if the data delivered is 0 thereceiver received -1 volt.

This process could be observed in each of pulses diagram on previous page. The receiver willsample the received signal at exactly at the center of each bit period.

Lets take first bit as an example, the center of bit period which is at 10usec the received signalis 1 volt.

c) Suppose that the receiver is perfectly synchronized with the transmitter. Analyze theISI inflicted on the received signal.

No ISI when the receiver is perfectly synchronized with the transmitter. It will sample at exactlythe middle of the bit period.

d) Suppose that the channel attenuates the signal amplitude by 80%. Calculate the noisemargin for the second bit if perfect synchronization is assumed.

0.8 =

0.8 from exact amplitude. Exact amplitude is -1 V for second bit

0.8 x -1 = 0.8 thus noise margin is 0.8

e) Suppose that the receiver is out of synchronization and it samples the received signal2sec after the optimum time. Using the attenuation values from Part (d), determine the noise margin for the second bit.Optimum sampling time is in the middle of bit period for second bit = 30uS

0.8 =

New sampling time 30+2 = 32uS

Equation for the line the time is between 30us to 40us is: y= 0.08x-3.2

Now let us check what is the slope when it received the signal after 2usec.

Y=(0.08)(32)-3.2 = 0.64 thus the margin of error is 0.64

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