One Question of my note is like this.

Consider a communication channel with a bandwidth of 2400Hz. If QPSK technique is
used, what is the possible transmission rate? (Assume that rectangular
pulses are used in the baseband signals, and that 90% energy preservation is

For a baseband system, a bipolar signal of bandwidth B can support a rate of B
with 90% energy preservation (since bandwidth =1/τ in this case).
For a bandpass system, a BPSK signal of bandwidth of B can support a rate of B/2.
This is because, after modulation, the bandwidth is 2/τ.
For QPSK, rate can be doubled using the same bandwidth. Therefore the
supported rate is 2400bps.

Why the BPSK signal of bandwidth 2/τ can support a datarate of B/2? and what is the relationship between the bandwidth and the datarate? and why in the QPSK, the rate can be doubled using the same bandwidth?

  • $\begingroup$ Channel capacity is a function of SNR. With no noise, the capacity of any non-zero-bandwidth channel is infinite. $\endgroup$
    – Jason R
    Commented Dec 7, 2012 at 23:20
  • $\begingroup$ Where did you get the quoted section from? I'm not sure what they mean by "90% energy preservation." Perhaps they're suggesting that 90% bandwidth of a communication signal is a meaningful measure of channel capacity, which is dubious to me. $\endgroup$
    – Jason R
    Commented Dec 8, 2012 at 0:54
  • $\begingroup$ 90% energy preservation means the bandwidth that contains 90% Signal Power. and I dont know where this question come form, may be my teacher made this up. $\endgroup$
    – Samuel
    Commented Dec 8, 2012 at 5:24
  • 1
    $\begingroup$ Although it is true that QPSK and BPSK have identical BER vs Eb/No, BPSK will have 3 dB more link margin than QPSK. There is no free lunch. $\endgroup$
    – John
    Commented Dec 10, 2012 at 13:08

3 Answers 3


Ideal symbol pulses are sequences of rectangular pulses. Rectangular pulses, unfortunately, have horrible frequency profiles, as shown below. FFT of rectangular pulse

The main lobe is where most of the pulse power is, and the other lobes are called side lobes. Though they do diminish in power the farther out they go, they don't diminish very quickly. The main purpose of pulse shaping is to dramatically reduce the power of those side lobes.

The width of the primary lobe is $\frac{2}{T}$, where $T$ is the rectangular pulse width, which is equivalent to the symbol period. Thus, once pulse shaping is employed, the signal bandwidth is $\frac{2}{T}$.

Since BPSK transmits one bit per symbol, and the symbol rate is $\frac{1}{T}$, BPSK can transmit $\frac{1}{T}$ bits per second. Therefore, the bit rate divided by the bandwidth is $\frac{\frac{1}{T}}{\frac{2}{T}}$, which reduces to $\frac{1}{2}$. Thus, the bit rate is $\frac{1}{2}$ the bandwidth.

When the signal is at baseband half of the mainlobe is in the positive frequencies and half is in the negative frequencies. Given that it is a real signal, the two halves mirror each other so some people consider it the bandwidth to be halved, or $\frac{1}{T}$. Thus, at baseband the bit rate would be equal to the bandwidth.

QPSK transmits two bits per symbol, so the bit rate for QPSK is $\frac{2}{T}$. It follows that QPSK can transmit $2$ bits per Hz of bandwidth at baseband, and $1$ bit per Hz at passband.

  • $\begingroup$ @@ thanks Jim, very clear explanation and graph $\endgroup$
    – Samuel
    Commented Dec 11, 2012 at 13:14
  • $\begingroup$ @Jim I'm not completely clear on the baseband/passband distinction: if a QPSK-modulated signal of bandwidth B comes in as a passband signal, its maximum bitrate is B, but if it comes in at baseband, it can be 2B? What if the signal is shifted to baseband? When I use a bandwidth calculator like this one (jaunty-electronics.com/blog/2012/05/…), I get a bandwidth:bitrate ratio of 0.6 (0.5, theoretically). In short, can a QPSK demodulator capable of demodulating a signal of bandwidth B receive a bandpass signal with a maximum information rate of B or 2B? $\endgroup$ Commented Jan 8, 2016 at 14:44
  • 1
    $\begingroup$ @TomislavNakic-Alfirevic There is a bit of ambiguity about what people mean when they talk about bandwidth at baseband. Usually they mean the one-sided bandwidth, i.e. 0 Hz to whatever their maximum frequency is. They might mean the two-sided bandwidth though. If the maximum frequency is $f_b$, then the two-sided bandwidth is from $-f_b$ to $f_b$, or $2f_b$. When a signal is modulated its bandwidth is always $2f_b$. $\endgroup$
    – Jim Clay
    Commented Jan 8, 2016 at 19:48
  • $\begingroup$ @JimClay Thank you for the clarification. Opens up more questions than it answers for me, but they all seem to point to "go read a good book on signal processing". $\endgroup$ Commented Jan 17, 2016 at 15:05

Keep in mind, QPSK has the same bit error rate as BPSK when both schemes are used to transmit underlying data at the same rate. If you are trying to use QPSK to double your data rate, then you get half as much power per bit, hence half as much Eb/N0, hence a higher bit error rate. So you have to choose: maintain your data rate and keep your BER, or use QPSK to double your data rate (more symbols), but your BER goes up. So Dave's statement above is true that when using QPSK to double your data rate over BPSK, you have to increase SNR to maintain your BER. Jason's statement above is also true, QPSK and BPSK, for the same data rate, have the same Eb/N0, and hence the same BER.


QPSK sends two bits per baud, where a baud is defined as one change in state of the signal. The tradeoff is that QPSK requires a better SNR, because it is more difficult to discriminate among 4 states per baud than 2.

  • 2
    $\begingroup$ Not true, if you properly use a normalized measure for SNE, like Eb/No. QPSK has the same bit-error rate performance as BPSK. $\endgroup$
    – Jason R
    Commented Dec 7, 2012 at 23:19
  • $\begingroup$ That should have been SNR; typo. $\endgroup$
    – Jason R
    Commented Dec 8, 2012 at 0:23
  • $\begingroup$ @JasonR I think that's because when you introduce imaginary signal with the same magnitude, Es/N0 is double, but there are double the bits in the symbol so they both have the same Eb/N0. $\endgroup$ Commented Oct 19, 2021 at 23:34

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