in [DSP]

Prev: Draft paper submission deadline is extended: IVPCV-10, Orlando, USA
Next: coff file download from Host processor
From: rbb on 15 Feb 2010 11:06 BPSK has a 3dB better SNR than coherent OOK(FSK) for a given BER, say 10^-3. Is this due to the DSB-SC nature of BPSK, such that the sidebands add coherently, while the noise adds randomly. Thus in the BPSK demodulator the signal is increased by 2x, while the noise remains the same, which decreases the SNR by 3dB? Thanks.
From: dvsarwate on 15 Feb 2010 11:56 On Feb 15, 10:06 am, "rbb" <Rory.Bucha... (a)onsemi.com> wrote:> BPSK has a 3dB better SNR than coherent OOK(FSK) for a given BER, say > 10^-3. Is this due to the DSB-SC nature of BPSK, such that the sidebands > add coherently, while the noise adds randomly. No, the improvement is not due to the DSB-SC nature of BPSK. > Thus in the BPSK > demodulator the signal is increased by 2x, while the noise remains the > same, which decreases the SNR by 3dB? I don't understand how increasing signal by 2x with noise remaining the same ***decreases*** SNR by 3 dB.
From: Vladimir Vassilevsky on 15 Feb 2010 12:15 rbb wrote: > BPSK has a 3dB better SNR than coherent OOK(FSK) for a given BER, say > 10^-3. Uhm, no. Ideally, for the same average transmit power, BPSK has 3dB advantage over OOK, and 1.7dB over FSK. Is this due to the DSB-SC nature of BPSK, such that the sidebands > add coherently, while the noise adds randomly. Thus in the BPSK > demodulator the signal is increased by 2x, while the noise remains the > same, which decreases the SNR by 3dB? Think constellations and euclidean distances. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com
From: rbb on 15 Feb 2010 12:16 > >I don't understand how increasing signal by 2x with noise >remaining the same ***decreases*** SNR by 3 dB. > Sorry. Increases the SNR by 3dB.
From: dvsarwate on 15 Feb 2010 17:37
Suppose that s0(t) and s1(t) are the two signals in a binary communication system. Define sc(t) = [s0(t) + s1(t)]/2 as the *common* signal and sd(t) = [s0(t) - s1(t)]/2 as the *difference* signal. Then, s0(t) = sc(t) + sd(t) and s1(t) = sc(t) - sd(t) meaning that the common signal is always transmitted while the data is being transmitted by the *antipodal* signal set {+sd(t), -sd(t)}. The energy in the common signal does not contribute to the data transmission at all, and in fact is wasted in the sense that the BER is Q(sqrt{2Ed/N0}) where Q is the complementary unit Gaussian CDF and Ed is the energy in the difference signal. In BPSK, the common signal is 0 and no energy is wasted. In OOK or orthogonal FSK, some energy is wasted in the common signal. Let's be specific: BPSK uses the signal set {+s(t), -s(t)} with no common signal and achieves BER of Q(sqrt{2E/N0}) where E is the energy in s(t). The energy per bit is also E. To get the same BER with OOK, we would need to use signal set {0, 2s(t)} so that the common signal is sc(t) = s(t) and the difference signal is +- s(t) giving rise to the same BER. But the energy being transmitted is either 0 or 4E (for an average energy of 2E per bit). Thus, OOK needs twice as much energy per bit to get the same BER which accounts for the usual statement of a 3 dB difference between OOK and BPSK. One should also keep in mind possible peak power limitations since OOK needs 4 times the peak power as BPSK. Hope this helps. Dilip Sarwate |