Choosing a suitable modulation format

Question

I'm trying to deal with this question about multiplexing and choosing an appropriate modulation format. It's in two parts.

(i) You need to design a system for the transmission of surround audio channels (7+1). The maximum frequency of each channel is 22 KHz, however, you need to oversample the signal at twice the minimum sampling rate. Calculate the total bandwidth required to transmit the 7+1 channels, if they are time division multiplexed, quantised at 16 bits per sample and use a 16‐QAM modulation with carrier frequency centred at 100 MHz. ii.

(ii) Then assume you need to transmit 10 such multi‐channel surround systems. The 10 systems are multiplexed using frequency division multiplexing, with guard bands of 60 KHz. However, your maximum available bandwidth is 20 MHz. What modulation format do you need to use to fit the 10 systems in the given 20 MHz bandwidth?

I've tried working it out myself, but I just don't feel confident about my answer.

For the first part, I calculated the sampling frequency as $f_s = 2(2(22000))=88\text{ KHz}$. The data rate of the 8 channels is $8\times(88000\times16)=11,264,000$ bit/s. Then, for modulation, letting $d=1$, $B_m=2\times S=2\times\frac{R}{log_2(L)}=2\times\frac{11264000}{log_2(16)}=5.632\text{ MHz}$.

Then, moving on to trying to decide on a suitable modulation format, I did the following. The remaining bandwidth after guard bands is $B_r=20,000,000-9\times 60000=19,460,000\text{ Hz}$. Thus, each system of 10 channels can have a bandwidth of $\frac{19,460,000}{10}=1,946,000\text{ Hz}$. Assuming we're not picking M-PSK, $B_m=2\times S$, so $log_2(L)=2\times\frac{R}{B_m}$. After computing this as $log_2(L)=2\times\frac{11,264,000}{1,946,000}\approx11.58$, I get $L=3055$, which is a very unsatisfactory answer. Assuming this is QAM, this goes far beyond what's commercially used.

I'm 99% sure I've misunderstood something about how TDM and FDM works, I was wondering if anyone knew what the issue was here?

Answer 1

In (i), indeed 11.264 Mb/s are required. However, there are a few steps required to go from this to a bandwidth. First, you need to calculate the baud rate (or symbols per second). In 16-QAM there are four bits per symbol, so the symbol rate is $$R_s = \frac{11.264 \times 10^6}{4} = 2.816\text{ MBd},$$ where $\text{Bd}$ is the abbreviation for the "baud".

To calculate the required bandwidth, first you need to choose a pulse shape. In an ideal system, sinc pulses can be used. With sync pulses, one can transmit two symbols per second (two baud) per Hz of available baseband bandwidth. In other words, with sinc pulses, $B = R_s / 2$, so in your case: $$B_\text{BB} = \frac{Rs}{2} = 1.408\text{ MHz},$$ where the subindex $\text{BB}$ stands for "baseband".

When upconverting to passband, the bandwidth doubles, so the actual passband bandwidth required is: $$B_\text{PB} = 2.816\text{ MHz}.$$

When using other pulses (such as a square-root raised cosine pulse), a bit more bandwidth will be required. If not specified in the problem, then I'd suggest going with sinc pulses.

Regarding question (ii), we can see that 10 channels do not fit in 20 MHz. However, they almost fit, so intuitively increasing the modulation order a little should be enough. Try running the calculations again with 5, 6 or more bits per symbol until you get a low-enough bandwidth.