I understand that the BER performance of BPSK and QPSK are the same. What does this mean? Does this mean that if I have a fixed bit rate (eg. 1Mbps), I can use the same transmit power and half the bandwidth (0.5MHz), and achieve the same performance? This is counter-intuitive, as I would then be able to transmit using a lower bandwidth with no tradeoff in transmit power.
BER performance of BPSK and QPSK
Started by ●November 3, 2008
Reply by ●November 3, 20082008-11-03
anthonychua wrote:> I understand that the BER performance of BPSK and QPSK are the same. What > does this mean?This means you don't understand that the BER performance of BPSK and QPSK is the same.> Does this mean that if I have a fixed bit rate (eg. > 1Mbps), I can use the same transmit power and half the bandwidth (0.5MHz), > and achieve the same performance?In the ideal case, yes. Think of QPSK as of a combination of two BPSK signals.> This is counter-intuitive, as I would then be able to transmit using a > lower bandwidth with no tradeoff in transmit power.BTW, ternary PSK provides for even better performance. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com
Reply by ●November 3, 20082008-11-03
anthonychua wrote:> I understand that the BER performance of BPSK and QPSK are the same. What > does this mean? Does this mean that if I have a fixed bit rate (eg. > 1Mbps), I can use the same transmit power and half the bandwidth (0.5MHz), > and achieve the same performance?Yes.> This is counter-intuitive, as I would then be able to transmit using a > lower bandwidth with no tradeoff in transmit power.It's no more counter-intuitive than the notion that you can spend less money on fuel (and take less time) to load your possessions in a small pickup truck and drive from Missouri to Oregon than you would if you loaded those same possessions into a Conestoga wagon, hitched up a pair of mules, and walked (and bought feed) all the way. Some things are just more efficient than others, and QPSK is more efficient at using the available power and bandwidth than BPSK is, at least in a linear Gaussian white-noise channel (there may be channels where BPSK does better, but I don't know if there are any such that you'll run into in the real world). As Vladimir mentions, there are modulation methods that do even better than QPSK. In fact, you can find out what the best possible performance is using Shannon's channel capacity theorem; then you have the line in the sand that you can strive for with your modulation methods. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" gives you just what it says. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●November 3, 20082008-11-03
anthonychua <anthony.chua@gmail.com> wrote:>I understand that the BER performance of BPSK and QPSK are the same. What >does this mean? Does this mean that if I have a fixed bit rate (eg. >1Mbps), I can use the same transmit power and half the bandwidth (0.5MHz), >and achieve the same performance? > >This is counter-intuitive, as I would then be able to transmit using a >lower bandwidth with no tradeoff in transmit power.It is simplest to look at antipodal performance and not worry about bandwidth. Then it becomes clear the two perform the same, since QPSK can be described as two BPSK signals running orthogonally to each other. Steve
Reply by ●November 3, 20082008-11-03
On Mon, 03 Nov 2008 09:07:54 -0600, "anthonychua" <anthony.chua@gmail.com> wrote:>I understand that the BER performance of BPSK and QPSK are the same. What >does this mean? Does this mean that if I have a fixed bit rate (eg. >1Mbps), I can use the same transmit power and half the bandwidth (0.5MHz), >and achieve the same performance? > >This is counter-intuitive, as I would then be able to transmit using a >lower bandwidth with no tradeoff in transmit power. >I'll add a little bit to what's been said. Realize that transmitting at half the bandwidth with the same signal power means that the power concentration results in an SNR increase of 3dB. So while BPSK and QPSK have the same BER vs Eb/No performance, the SNR will differ by 3dB for the same BER. This is why BPSK is still useful: it provides a 3dB improvement in link margin compared to QPSK. Eric Jacobsen Minister of Algorithms Abineau Communications http://www.ericjacobsen.org Blog: http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php
Reply by ●November 3, 20082008-11-03
Eric Jacobsen <eric.jacobsen@ieee.org> writes:> On Mon, 03 Nov 2008 09:07:54 -0600, "anthonychua" > <anthony.chua@gmail.com> wrote: > >>I understand that the BER performance of BPSK and QPSK are the same. What >>does this mean? Does this mean that if I have a fixed bit rate (eg. >>1Mbps), I can use the same transmit power and half the bandwidth (0.5MHz), >>and achieve the same performance? >> >>This is counter-intuitive, as I would then be able to transmit using a >>lower bandwidth with no tradeoff in transmit power. >> > > I'll add a little bit to what's been said. > > Realize that transmitting at half the bandwidth with the same signal > power means that the power concentration results in an SNR increase of > 3dB. So while BPSK and QPSK have the same BER vs Eb/No performance, > the SNR will differ by 3dB for the same BER.Hi Eric, This conflicts with my understanding. Let R = bit rate [bits / second] W = bandwidth [Hz] Eb = Energy per bit [Joules / bit] Ps = signal power [Watts] Pn = noise power [Watts] No = noise power spectral spectral density [Joules] r = spectral efficiency = R / W [bits] Then Ps = R * Eb Pn = W * No and SNR = Ps / Pn = (R * Eb) / (W * No) = (R / W) * (Eb / No) = r * Eb / No. According to the table on page 282 of Proakis [proakiscomm], BPSK (2 PAM) and QPSK (4 PSK) have the same Eb / No AND the same spectral efficiency r = R / W for a BER of 10^(-5). So at any given bitrate, both BPSK and QPSK have the SNR for the same BER performance. Am I making a mistake? --Randy @BOOK{proakiscomm, title = "{Digital Communications}", author = "John~G.~Proakis", publisher = "McGraw-Hill", edition = "fourth", year = "2001"} -- % Randy Yates % "Midnight, on the water... %% Fuquay-Varina, NC % I saw... the ocean's daughter." %%% 919-577-9882 % 'Can't Get It Out Of My Head' %%%% <yates@ieee.org> % *El Dorado*, Electric Light Orchestra http://www.digitalsignallabs.com
Reply by ●November 3, 20082008-11-03
Randy Yates <yates@ieee.org> wrote:>Eric Jacobsen <eric.jacobsen@ieee.org> writes:>> Realize that transmitting at half the bandwidth with the same signal >> power means that the power concentration results in an SNR increase of >> 3dB. So while BPSK and QPSK have the same BER vs Eb/No performance, >> the SNR will differ by 3dB for the same BER.I agree with this.>This conflicts with my understanding.>Let > > R = bit rate [bits / second] > W = bandwidth [Hz] > Eb = Energy per bit [Joules / bit] > Ps = signal power [Watts] > Pn = noise power [Watts] > No = noise power spectral spectral density [Joules] > r = spectral efficiency > = R / W [bits]>Then > > Ps = R * Eb > Pn = W * No > >and > > SNR = Ps / Pn > = (R * Eb) / (W * No) > = (R / W) * (Eb / No) > = r * Eb / No. > >According to the table on page 282 of Proakis [proakiscomm], BPSK (2 >PAM) and QPSK (4 PSK) have the same Eb / No AND the same spectral >efficiency r = R / W for a BER of 10^(-5).Is this particular definition of spectral efficiency the same one that Proakis is using? In my experience spectral efficiency is a figure of merit that is the same for BPSK and QPSK, but becomes gradually worse for higher-order QAM modulation. Whereas R / W defined as above seems to be simply the number of bits per Hz. Steve
Reply by ●November 3, 20082008-11-03
spope33@speedymail.org (Steve Pope) writes:> Randy Yates <yates@ieee.org> wrote: > >>Eric Jacobsen <eric.jacobsen@ieee.org> writes: > >>> Realize that transmitting at half the bandwidth with the same signal >>> power means that the power concentration results in an SNR increase of >>> 3dB. So while BPSK and QPSK have the same BER vs Eb/No performance, >>> the SNR will differ by 3dB for the same BER. > > I agree with this. > >>This conflicts with my understanding. > >>Let >> >> R = bit rate [bits / second] >> W = bandwidth [Hz] >> Eb = Energy per bit [Joules / bit] >> Ps = signal power [Watts] >> Pn = noise power [Watts] >> No = noise power spectral spectral density [Joules] >> r = spectral efficiency >> = R / W [bits] > >>Then >> >> Ps = R * Eb >> Pn = W * No >> >>and >> >> SNR = Ps / Pn >> = (R * Eb) / (W * No) >> = (R / W) * (Eb / No) >> = r * Eb / No. >> >>According to the table on page 282 of Proakis [proakiscomm], BPSK (2 >>PAM) and QPSK (4 PSK) have the same Eb / No AND the same spectral >>efficiency r = R / W for a BER of 10^(-5). > > Is this particular definition of spectral efficiency the same > one that Proakis is using?Hi Steve, Actually I couldn't find a definition or usage of the term "spectral efficiency" at all in Proakis. I thought that's the way spectral efficiency (r) was defined and I put it in those terms since it simplified the result. However, the chart on p.282 is in terms of R/W and doesn't even mention spectral efficiency explicitly: http://www.digitalsignallabs.com/efficiency.jpg> In my experience spectral efficiency is a figure of merit > that is the same for BPSK and QPSK, but becomes gradually worse > for higher-order QAM modulation. Whereas R / W defined as above > seems to be simply the number of bits per Hz.Actually it's bits per second per Hz. In any case, I still don't see the error in my understanding (if there is one). Whether you use "r" and call it spectral efficiency, or you use R/W, the point is still that the SNR of one modulation is equivalent to the other for the same bit rate and BER. -- % Randy Yates % "Midnight, on the water... %% Fuquay-Varina, NC % I saw... the ocean's daughter." %%% 919-577-9882 % 'Can't Get It Out Of My Head' %%%% <yates@ieee.org> % *El Dorado*, Electric Light Orchestra http://www.digitalsignallabs.com
Reply by ●November 3, 20082008-11-03
Randy Yates <yates@ieee.org> writes:> Eric Jacobsen <eric.jacobsen@ieee.org> writes: > >> On Mon, 03 Nov 2008 09:07:54 -0600, "anthonychua" >> <anthony.chua@gmail.com> wrote: >> >>>I understand that the BER performance of BPSK and QPSK are the same. What >>>does this mean? Does this mean that if I have a fixed bit rate (eg. >>>1Mbps), I can use the same transmit power and half the bandwidth (0.5MHz), >>>and achieve the same performance? >>> >>>This is counter-intuitive, as I would then be able to transmit using a >>>lower bandwidth with no tradeoff in transmit power. >>> >> >> I'll add a little bit to what's been said. >> >> Realize that transmitting at half the bandwidth with the same signal >> power means that the power concentration results in an SNR increase of >> 3dB. So while BPSK and QPSK have the same BER vs Eb/No performance, >> the SNR will differ by 3dB for the same BER. > > Hi Eric, > > This conflicts with my understanding. > > Let > > R = bit rate [bits / second] > W = bandwidth [Hz] > Eb = Energy per bit [Joules / bit] > Ps = signal power [Watts] > Pn = noise power [Watts] > No = noise power spectral spectral density [Joules] > r = spectral efficiency > = R / W [bits] > > Then > > Ps = R * Eb > Pn = W * No > > and > > SNR = Ps / Pn > = (R * Eb) / (W * No) > = (R / W) * (Eb / No) > = r * Eb / No. > > According to the table on page 282 of Proakis [proakiscomm], BPSK (2 > PAM) and QPSK (4 PSK) have the same Eb / No AND the same spectral > efficiency r = R / W for a BER of 10^(-5). > > So at any given bitrate, both BPSK and QPSK have the SNR for the > same BER performance. > > Am I making a mistake?OK, I see the problem. The point on the chart I'm referring to is for SSB 2 PAM. I guess you're referring to DSB 2 PAM, which does have half the spectral efficiency of QPSK. But now I have a new disgreement with your point, or perhaps a "clarification" is a better term. Yes, the SNR required for QPSK is 3 dB higher than that for DSB 2 PAM. BUT..., since the bandwidth is twice as large, IT REQUIRES THE SAME TRANSMIT POWER. And that (transmit power) is usually the issue when it comes to link margin, not SNR. -- % Randy Yates % "She tells me that she likes me very much, %% Fuquay-Varina, NC % but when I try to touch, she makes it %%% 919-577-9882 % all too clear." %%%% <yates@ieee.org> % 'Yours Truly, 2095', *Time*, ELO http://www.digitalsignallabs.com
Reply by ●November 3, 20082008-11-03
On Nov 3, 8:16�pm, Randy Yates <ya...@ieee.org> wrote:> Randy Yates <ya...@ieee.org> writes: > > Eric Jacobsen <eric.jacob...@ieee.org> writes: > > >> On Mon, 03 Nov 2008 09:07:54 -0600, "anthonychua" > >> <anthony.c...@gmail.com> wrote: > > >>>I understand that the BER performance of BPSK and QPSK are the same. �What > >>>does this mean? �Does this mean that if I have a fixed bit rate (eg. > >>>1Mbps), I can use the same transmit power and half the bandwidth (0.5MHz), > >>>and achieve the same performance? > > >>>This is counter-intuitive, as I would then be able to transmit using a > >>>lower bandwidth with no tradeoff in transmit power. > > >> I'll add a little bit to what's been said. > > >> Realize that transmitting at half the bandwidth with the same signal > >> power means that the power concentration results in an SNR increase of > >> 3dB. � So while BPSK and QPSK have the same BER vs Eb/No performance, > >> the SNR will differ by 3dB for the same BER. � > > > Hi Eric, > > > This conflicts with my understanding. > > > Let > > > � R �= bit rate [bits / second] > > � W �= bandwidth [Hz] > > � Eb = Energy per bit [Joules / bit] > > � Ps = signal power [Watts] > > � Pn = noise power [Watts] > > � No = noise power spectral spectral density [Joules] > > � r �= spectral efficiency > > � � �= R / W [bits] > > > Then > > > � Ps = R * Eb > > � Pn = W * No > > > and > > > � SNR = Ps / Pn > > � � � = (R * Eb) / (W * No) > > � � � = (R / W) * (Eb / No) > > � � � = r * Eb / No. > > > According to the table on page 282 of Proakis [proakiscomm], BPSK (2 > > PAM) and QPSK (4 PSK) have the same Eb / No AND the same spectral > > efficiency r = R / W for a BER of 10^(-5). > > > So at any given bitrate, both BPSK and QPSK have the SNR for the > > same BER performance. > > > Am I making a mistake? > > OK, I see the problem. The point on the chart I'm referring to is for > SSB 2 PAM. I guess you're referring to DSB 2 PAM, which does have half > the spectral efficiency of QPSK. > > But now I have a new disgreement with your point, or perhaps a > "clarification" is a better term. > > Yes, the SNR required for QPSK is 3 dB higher than that for DSB 2 > PAM. BUT..., since the bandwidth is twice as large, IT REQUIRES THE SAME > TRANSMIT POWER. And that (transmit power) is usually the issue when it > comes to link margin, not SNR. > -- > % �Randy Yates � � � � � � � � �% "She tells me that she likes me very much, > %% Fuquay-Varina, NC � � � � � �% � � but when I try to touch, she makes it > %%% 919-577-9882 � � � � � � � �% � � � � � � � � � � � � � �all too clear." > %%%% <ya...@ieee.org> � � � � � % � � � �'Yours Truly, 2095', *Time*, ELO �http://www.digitalsignallabs.com- Hide quoted text - > > - Show quoted textI understand the confusion, but remember that there's no such thing as a free lunch. QPSK requires twice the power to achieve the same distance between symbols. John
