Why does QASK with a same bitrate as BASK use only half of the bandwidth?

Question

I think the title is pretty self explanatory, what else could I add.

Why does QASK with a same bitrate as BASK use only half of the bandwidth? Does anybody have an idea about why this is the case?

Thank you

Answer 1

The efficiency comes from the fact that a QASK modulation scheme has 4 levels and therefore transmits 2 bits per symbol whilst a BASK scheme you will only get 1 bit per symbol because you only have 2 levels. So for every symbol you send with QASK you are transmitting twice as many bits. The price you pay for this gain in bit-rate is an reduction in amplitude immunity i.e your scheme is a lot less resilient to noise and will thus have a higher bit error rate (BER).

EDIT:

I will try and simplify it as much as possible. What you have to think about is how efficiently you can send bits over a fixed bandwidth with both schemes given the facts given above. With QASK you have a efficiency of 2 bits/s/Hz (bit per second per hertz) and with BASK you have an efficiency of 1 bit/s/Hz.

Now assume you have an fixed bandwidth $B$, which scheme would send the most bits across the channel in that case? The QASK scheme would because it will be sending 2 bits with each symbol. In order to get the two bit rates to equate you can to double the symbol duration ($T_s$) of the QASK scheme (or in other words slow it down by a factor of 2) and this would half the Bandwidth you occupy and also half your carrier frequency.

Let me give you a way to think about it. Let's say at first you have a centre frequency of 150Hz and a bandwidth of 100Hz for both schemes, you are transmitting a signal of form

$$s(t) = a_1 \sin(2 \pi 100 t) + ...+a_i \sin(2 \pi 150 t) + + a_n \sin(2 \pi 200 t)$$

for both schemes, if you double the symbol period for the QASK scheme, you are effectively stretching out the signal (or in other words halving the frequency of each component), leaving you with something of form

$$s_1(t) = a_1 \sin(2 \pi 50 t) + ...+a_i \sin(2 \pi 75 t) + + a_n \sin(2 \pi 100 t)$$

now $s_1(t)$ has a bandwidth that is half of the bandwidth of $s(t)$. You can just shift $s_1(t)$ to the previous centre frequency and have the same centre frequency whilst using half the bandwidth. Hope the point is a bit clearer now.

Answer 2

I assume that QASK stands for Quaternary Amplitude-Shift-Keying and BASK for Binary Amplitude-Shift-Keying.

The standard baseband signal used in an amplitude-shift-keyed digital communication system operating at $T^{-1}$ baud $$s(t) = \sum_{n=-\infty}^\infty A_n g(t-nT) \tag 1$$ where $g(t)$ is the baseband pulse which can be a square pulse of duration $T$ or a raised-cosine pulse or a root-raised-cosine pulse or whatever, and $A_n$ is a parameter that takes on values $\pm A$ in a BASK system and values $\pm A, \pm 3A$ in a QASK system, depending on the bit(s) to be transmitted during that signaling interval. Note that one bit is transmitted every $T$ seconds in the BASK system while two bits are transmitted every $T$ seconds in the QASK system. (As all Americans know, two bits make a quarter and hence the name quaternary). Note also that the four values in the PASK system are spaced $2A$ apart, the same as the two values in the BASK system.

Now, the power spectral density (PSD) of $s(t)$, treated as a random process, is proportional of $|G(f)|^2$ where $G(f)$ is the Fourier transform of the baseband pulse $g(t)$. (The constant of proportionality is different for BASK and QASK). Define the bandwidth $B$ of the baseband pulse using whichever definition of bandwidth you prefer. Then, we can say that

In the bandwidth $B$, the QASK communication system using $s(t)$ transmits at a data rate twice that of the BASK system using $s(t)$.

This is not quite what we want; we want to be able to say that the two systems have the same data rate but the QASK system uses only bandwidth $\frac B2$, half the bandwidth of the BASK system. Not to worry. Define the baseband pulse $\hat{g}(t)$ by $$\hat{g}(t) = g\left(\frac t2\right)$$ which is a stretching of the pulse $g(t)$ along the time axis so that the signal lasts twice as long, and define the QASK baseband signal as $$\hat s(t) = \sum_{n=-\infty}^\infty A_n \hat g(t-2nT)\tag 2$$ which transmits two bits in $2T$ seconds for a data rate of $T^{-1}$ bps, the same as the BASK system using $(1)$. But, the Fourier transform of $\hat{g}(t)$ is related to the Fourier transform of $G(f)$ as $$\hat{G}(f) = 2G(2f)$$ which is a contraction of $G(f)$ along the frequency axis by a factor of $2$. Thus, the bandwidth of $\hat{g}(t)$ is half the bandwidth of $g(t)$, and we have that

$\hat{s}(t)$ is a QASK signal with bandwidth $\frac B2$ that transmits data at rate $T^{-1}$ bps. This is the same data rate as that of the BASK signal $s(t)$ which occupies bandwidth $B$.

Note that $A_n$ in $(1)$ is restricted to two values in the BASK signal while the $A_n$'s in $(2)$ are four-valued. The passband signals corresponding to these baseband signals are $$s(t)\cos(2\pi f_ct) ~~\text{and}~~ \hat s(t)\cos(2\pi f_ct)$$ where $f_c$ is the carrier frequency. Applying the modulation theorem of Fourier theory tells us that the passband QASK signal has half the passband bandwidth of the passband BASK signal if we keep the data rate fixed at $T^{-1}$ bps.