Relationship between signal rate, data rate, minimum bandwidth

Question

I need to understand the relationship between bit rate, signal rate and minimum bandwidth. I read in a textbook that

We define three cases for relationship between bit rate and signal rate for this, they are: the worst, best and average. Worst case is when we need maximum signal rate, best case is when we need the minimum. In data communication, we usually prefer the average case and the relationship between data rate and signal rate is

$$S = c \times N \times \frac{1}{r} \quad\text{baud}$$

where $N$ is data rate, $c$ is case factor, $S$ is no number of signal elements and $r$ is previously defined ratio.

I don't understand what the above formula signifies.

(i) $S$ is said to be the number of signal elements and not the signal rate, i.e number of signal elements per unit time. How does this then tell the relationship between signal rate and data rate?

(ii) What exactly is $c$ and $r$ in the formula?

(iii) Secondly, there's a formula given for calculating the minimum bandwidth of a digital signal which is also exactly the same as the formula given above for the relationship between data rate and signal rate, i.e, $B_\text{min} = S = c \times N \times \frac{1}{r}$ . How does this same formula calculates the minimum bandwidth?

Answer 1

Here's an attempt at an answer, by comparing your equations with the more commonly used ones.

We can write the "pulse rate" or the "signaling rate" $R_p$, measured in baud, or equivalently in pulses per second or symbols per second, as: $$R_p = \frac{1}{k} R_b,$$ where $R_b$ is the bit rate, and $k$ is the number of bits carried by a pulse.

To elaborate: a digital communications signal can be written as a sequence of pulses of different amplitudes, each pulse time-shifted an amount $T_p = 1/R_p$, and with the information carried in the pulses' amplitude: $$s(t) = \sum_j a_j p(t-jT_p).$$

Note that:

• For each value of $j$, the pulse $a_j p(t-T_j)$ carries the information in its amplitude $a_j$. The amount of information depends on how many distinct values $a_j$ can take. Usually this number is represented by $M = 2^k$ for integer $k$, and we say that each pulse carries $k$ bits.

• Since one pulse is transmitted every $T_p$ seconds, the pulse rate is $R_p = 1/T_p$ baud.

Now, (ignoring some more advanced considerations) the bandwidth of $s(t)$ is the same as the bandwidth of the pulse $p(t)$, denoted $B_p$. The pulse that has the minimum possible bandwidth for a given $R_p$ is a sinc pulse, and in that case the bandwidth of $s(t)$ is given by $$B_{min} = \frac{R_p}{2} = \frac{1}{2}\frac{1}{k}R_b.$$ Practical pulses achieve a fraction larger than $1/2$, or in other words, they require a larger bandwidth. If we define this fraction by $c$, we have $$B = c\frac{1}{k}R_b.$$

In conclusion, it seems to me that:

• In your equations, $N$ is the bit rate, $r$ represents the number of bits transmitted per pulse, and $c$ represents the bandwidth efficiency of the chosen pulse. The variable name $k$ is more commonly used than $r$.

• The first equation in your book is wrong; it should be $S = N \times \frac{1}{r}$.

• The equation for the bandwidth is correct.

Answer 2

The value of the case factor (c) is not limited to specific numbers; it can take a wide range of values depending on the design and operational requirements of the communication system. The specific conditions and constraints of the system will determine the actual value of the case factor. It would take a value greater than 1 in a worst-case scenario, close to 1 in the ideal case, and less than in a practical scenario. r is the data elements per signal element. The ratio of how many bits are represented by how many signal levels. r is 1 when 1 bit of the data is represented by 1 signal element and k/n when k bits are represented by n signal levels.