Question

Consider additive white Gaussian noise with a double-sided noise power spectral density (PSD) 12-90 dBm/Hz 1E-12 W/Hz. This noise corrupts a baseband polar NRZ signal with rectangular pulses, like that 1 mV and the pulse duration is Tb such that the symbol rate is Rb 1/Tb. Since this is binary signaling, the symbol rate equals the bit rate of the incoming baseband pulse train: B Rt. The signal is then sampled at the center of the noisy shown in the figure below. The pulse amplitude The noisy signal is applied to an ideal low pass filter with bandwidth B that equals the effective (first-null) bandwidth pulse and a decision circuit determines that a 1 was sent if the sampled voltage is greater than 0 V, and a 0 was sent if the sampled voltage error Pe-0 2066E-4 Hint: Note that Q(4.1) 0.2066E-4 less than 0 V Calculate the bit rate that corresponds to a probability of bit 29.744 kbps 39.744 kbps 9.744 kbps 19.744 kbps

Answer 1

Probability of error is given by

Pe QIVEd/2N]

Where Ed =difference in energy of signal that is

Ed S2(t)2 dt S1 (t)

Here S1(t) is the amplitude when 1 is transmitted and

S2(t) is the amplitude when 0 is transmitted.

For NRZ baseband signal given, we have

S1(t) = 1

S2(t) = -1

So

-1)12 dt Ed 1 -

Ed [2d

Ed=4T

Hence

Pe QIV4Th/2No]

PeQV2Tb/No]

In the question, we have Pe =0.2066 E-4 and No = 2 ×1E-12W/Hz(multiplied by 2 because of being double sided signal)

Substituting values

0.2066 x e 12 QlV2Tb/No]

Given is Q[4.1] =0.2066E-12

Hence $\sqrt{2T_{b}/N_{0}} =4.1$

27h/No 16.81

T No X 16.81/2

$T_{b}=2 \times e^{-12}\times 16.81/2$

$T_{b}=1.033\times 10^{-4}$

Hence bit rate is

Ri 1/T 1/1.033 x 10-4

$R_{b}=9.68kbps$

Hence from option c, we can say bit rate $R_{b}=9.744kbps$