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A BPSK scheme operating over an AWGN channel with noise power spectral density of $$\frac{{{N_0}}}{2},$$  uses equiprobable signals
$${s_1}\left( t \right) = \sqrt {\frac{{2E}}{T}} \sin \left( {{\omega _c}t} \right)$$     and $${s_2}\left( t \right) = \sqrt {\frac{{2E}}{T}} \sin \left( {{\omega _c}t} \right)$$
over the symbol internal (0, T). If the local oscillator in a coherent receiver is ahead in phase by 45° with respect to the received signal, the probability of error in the resulting system is

A BPSK scheme operating over an AWGN channel with noise power spectral density of $$\frac{{{N_0}}}{2},$$  uses equiprobable signals
$${s_1}\left( t \right) = \sqrt {\frac{{2E}}{T}} \sin \left( {{\omega _c}t} \right)$$     and $${s_2}\left( t \right) = \sqrt {\frac{{2E}}{T}} \sin \left( {{\omega _c}t} \right)$$
over the symbol internal (0, T). If the local oscillator in a coherent receiver is ahead in phase by 45° with respect to the received signal, the probability of error in the resulting system is Correct Answer $$Q\left( {\sqrt {\frac{E}{{{N_0}}}} } \right)$$

Related Questions

Coherent orthogonal binary FSK modulation is used to transmit two equiprobable symbol waveforms s1(t) = αcos2πf1t & s2(t) = αcos2πf2t1 where α = 4 mV. Assume an AWGN channel with two-sided noise power spectral density $$\frac{{{N_0}}}{2} = 0.5 \times {10^{ - 12}}W/Hz.$$     Using an optimal receiver and the relation $$Q\left( v \right) = \frac{1}{{\sqrt {2\pi } }}\int\limits_v^\infty {{\theta ^{\frac{{{u^2}}}{2}}}du,} $$     the bit error probability for a data rate of 500 kbps is
The capacity of a band-limited additive white Gaussian noise (AWGN) channel is given by $$C = W\,{\log _2}\left( {1 + \frac{P}{{{\sigma ^2}W}}} \right)$$     bits per second (bps), where W is the channel bandwidth, P is the average power received and σ2 is the one-sided power spectral density of the AWGN. For a fixed $$\frac{P}{{{\sigma ^2}}} = 1000,$$   the channel capacity (in kbps) with infinite bandwidth (W → ∞) is approximately
$$\frac{1}{{\left( {\sqrt 9 - \sqrt 8 } \right)}} \, - $$   $$\frac{1}{{\left( {\sqrt 8 - \sqrt 7 } \right)}} \, + $$   $$\frac{1}{{\left( {\sqrt 7 - \sqrt 6 } \right)}} \, - $$   $$\frac{1}{{\left( {\sqrt 6 - \sqrt 5 } \right)}} \, + $$   $$\frac{1}{{\left( {\sqrt 5 - \sqrt 4 } \right)}}$$   is equal to ?
Which of the following is TRUE? \({\rm{I}}.{\rm{\;}}\sqrt[3]{{11}} > \sqrt 7 > \sqrt[4]{{45}}\) \({\rm{II}}.{\rm{\;}}\sqrt 7 > \sqrt[3]{{11}} > \sqrt[4]{{45}}\) \({\rm{III}}.{\rm{\;}}\sqrt 7 > \sqrt[4]{{45}} > \sqrt[3]{{11}}\) \({\rm{IV}}.{\rm{\;}}\sqrt[4]{{45}} > \sqrt 7 > \sqrt[3]{{11}}\)
Which of the following is TRUE? \({\rm{I}}.{\rm{\;}}\frac{1}{{\sqrt[3]{{12}}}} > \frac{1}{{\sqrt[4]{{29}}}} > \frac{1}{{\sqrt 5 }}\) \({\rm{II}}.\;\frac{1}{{\sqrt[4]{{29}}}} > \frac{1}{{\sqrt[3]{{12}}}} > \frac{1}{{\sqrt 5 }}\) \({\rm{III}}.\;\frac{1}{{\sqrt 5 }} > \frac{1}{{\sqrt[3]{{12}}}} > \frac{1}{{\sqrt[4]{{29}}}}\) \({\rm{IV}}.{\rm{\;}}\frac{1}{{\sqrt 5 }} > \frac{1}{{\sqrt[4]{{29}}}} > \frac{1}{{\sqrt[3]{{12}}}}\)
Simplify : $$\frac{1}{{\sqrt 3 + \sqrt 4 }} \,+ $$   $$\frac{1}{{\sqrt 4 + \sqrt 5 }} \,+ $$   $$\frac{1}{{\sqrt 5 + \sqrt 6 }} \,+ $$   $$\frac{1}{{\sqrt 6 + \sqrt 7 }} \,+ $$   $$\frac{1}{{\sqrt 7 + \sqrt 8 }}\, + $$   $$\frac{1}{{\sqrt 8 + \sqrt 9 }} = ?$$
In a fast FH spread spectrum system, the information is transmitted via FSK with non coherent detection. Suppose there are N = 3 hops/bit with hard decision decoding of the signal in each hop. The channel is AWGN with power spectral density 0.5No and an SNR 20 ~13 dB (total SNR over the three hops). The probability of error for this system is
If the value of \(\frac{{3x\sqrt y + 2y\sqrt x }}{{3x\sqrt y - 2y\sqrt x }} - \frac{{3x\sqrt y - 2y\sqrt x }}{{3x\sqrt y + 2y\sqrt x }}\) is same as that of \(\sqrt x\sqrt y\), then which of the following relations between x and y is correct?
Which of the following statement(s) is/are TRUE? I. \(\sqrt{1}+ \sqrt{2} + \sqrt{3} + \sqrt{4}+ \sqrt{5} + \sqrt{6} > 10\) II. \(\sqrt{(10)} + \sqrt{(12)} + \sqrt{(14)} > 3 \sqrt{(12)}\)
The Fourier transform of a signal
h(t) is $$H\left( {j\omega } \right) = {{\left( {2\cos \omega } \right)\left( {\sin \omega } \right)} \over \omega }$$
The value of h(0) is