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When using for example, polar signaling of full width rectangular pulses, the bandwidth is equal to the pulse rate $Rb$, which is also the bit rate.

$B=R_b$ for full width rectangular pulses.

But also we can describe the minimum required bandwidth for transmission of this data.

$B_{min}=\frac{R_b}{2}$

If I am sending data over a channel using polar signaling of the full width rectangular pulses, surely I cannot achieve a bandwidth of $B_{min}$?

How would I go about converting my data to achieve this theoretical minimum bandwidth? I'm looking for an intuitive understanding of minimum bandwidth versus the actual bandwidth of my data.

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    $\begingroup$ What is your source for the claim that $B_{min} = \frac{R_b}{2}$? $\endgroup$ – Jason R Oct 20 '18 at 2:21
  • $\begingroup$ Personally, I think $B=R_b$ for rectangular pulses is awfully optimistic... I prefer to use $B=5R_b$ myself. $\endgroup$ – MBaz Oct 20 '18 at 15:14
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We must be careful in how we define bandwidth. To maintain truly rectangular pulses in the time domain, infinite bandwidth is required, as in the frequency domain the spectrum would be a Sinc function that extends to infinity (as the Fourier Transform of the rectangular pulse). The requirement for infinite bandwidth is intuitively consistent with a rectangular pulse that can change instantaneously from a low level to a high level (infinite bandwidth is required to transition that fast!). Note for now that the Sinc function has nulls at 1/T and integer multiples of that, where T is the width of the rectangular pulse.

As we constrain the bandwidth in frequency, the edges of the signal will likewise slow in transition from low to high. A good rule of thumb related to this (specific to first order systems but generally a good estimate) is:

$$t_r = 0.35/BW$$

Where $t_r$ is the 10% to 90% rise time in seconds, and BW is the bandwidth in Hz.

Ultimately for a digital signal the fastest bandwidth is required when we send the sequence 1 0 1 0 1 0 1 0 .... We note that for rectangular pulses this would be a 50% duty cycle square wave which according to the Fourier Series Expansion is decomposed in frequency to the odd harmonics of the fundamental sinusoidal frequency that fits in the 1 0 1 0 pattern. This is still the Sinc function we described above, just when the time domain signal repeats the only frequency components that can exist are at integer multiples of the repetition rate, and therefore those frequencies become samples of the Sinc function. As described above, the nulls would exist at 1/T, where T in this case is $1/R_b$ which is also the location of the 2nd harmonic that we already reasoned is nulled (only odd harmonics exist for the 50% square wave). The first harmonic is at $R_b/(2)$, and if we were to represent it as a double sideband signal (positive and negative frequencies) then the minimum bandwidth required to contain both would be $R_b$. This would result in a sinusoidal waveform (sine or cosine) but would be sufficient to represent our 1 0 1 0 pattern without data loss (we recover with a simple threshold at y=0).

However, we can also represent the same sequence as a single sided signal (positive or negative frequencies only) which would then only require half that bandwidth, which is how you came to $R_b/2$. This would be consistent with a use of Single Sided Modulation. (Note we don't require any bandwidth if the pattern was always 1 0 1 0 as it would be represented by a single frequency, but this is the case of the highest frequency needed, and we would also need all the lower frequencies to represent any other patterns).

Depending on your background this may be a lot to swallow, and if so I also point you to the links below that also help explain these fundamental concepts:

How do I modulate complex time domain signal to a carrier frequency?

Frequency shifting of a quadrature mixed signal

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