Bandwidth Theorem

Figure 8.5: A digital bit transmitted over AM radio.

Let us now consider how we might transmit a digital signal over AM radio. Suppose that each data “bit” in the signal takes the form of a Gaussian envelope, of characteristic duration $\sigma_t$, superimposed on a carrier wave whose frequency is $\omega_0$; that is,

$$\psi(t) = \exp\left(-\frac{t^{\,2}}{2\,\sigma_t^{\,2}}\right)\,\cos(\omega_0\,t).$$ (8.68)

Let us assume that $\omega_0\,\sigma_t\gg 1$. In other words, the period of the carrier wave is much less than the duration of the bit. Figure 8.5 illustrates a digital bit calculated for $\omega_0\,\sigma_t=20$.

The sine Fourier transform of the signal (8.68) is zero by symmetry. However, its cosine Fourier transform takes the form

$$C(\omega) =\frac{1}{2\pi}\int_{-\infty}^{\infty} \exp\left(-\frac{t^{\,2}}{2\,\sigma_t^{\,2}}\right)\cos(\omega_0\,t)\,\cos(\omega\,t)\,dt,$$

$$=\frac{1}{4\pi}\int_{-\infty}^\infty \exp\left(-\frac{t^{\,2}}{2\... ...ight) \left\{\cos[(\omega-\omega_0)\,t] + \cos[(\omega+\omega_0)\,t]\right\}dt.$$ (8.69)

A comparison with Equations (8.15)–(8.18) reveals that

$$C(\omega>0) = \frac{1}{2\,(2\pi\,\sigma_\omega^{\,2})^{1/2}}\,\exp\left[-\frac{(\omega-\omega_0)^{\,2}}{2\,\sigma_\omega^{\,2}}\right],$$ (8.70)

where

$$\sigma_\omega = \frac{1}{\sigma_t}.$$ (8.71)

In other words, the Fourier transform of the signal takes the form of a Gaussian in $\omega$-space that is centered on the carrier frequency, $\omega_0$, and is of characteristic width $\sigma_\omega=1/\sigma_t$. Thus, the bandwidth of the signal is of order $\sigma_\omega$. The shorter the signal duration, the higher the bandwidth. This is a general rule. A signal of full width at half maximum temporal duration ${\mit\Delta} t\simeq \sqrt{2\pi}\,\sigma_t$ generally has a Fourier transform of full width at half maximum bandwidth ${\mit\Delta} \omega\simeq \sqrt{2\pi}\,\sigma_\omega$, so that

$${\mit\Delta}\omega\,{\mit\Delta} t\simeq 2\pi.$$ (8.72)

This can also be written

$${\mit\Delta} f\,{\mit\Delta} t\simeq 1,$$ (8.73)

where ${\mit\Delta} f={\mit\Delta}\omega/2\pi$ is the bandwidth in hertz. The previous result is known as the bandwidth theorem. The duration of a digital bit is closely related to the maximum rate at which information can be transmitted by a digital signal. Because the individual bits cannot overlap in time, the maximum number of bits per second that can be transmitted by a digital signal is of order $1/{\mit\Delta} t$. In other words, it is of order the bandwidth of the signal. Thus, digital signals that transmit information at a rapid rate require large bandwidths, and, consequently occupy a wide range of frequency space.

An old-fashioned black and white TV screen consists of a rectangular grid of black and white spots scanned by an electron beam that can be switched on or off rapidly. A given spot is “white” if the phosphorescent TV screen was recently (i.e., within about $1/50$ th of a second) struck by the energized electron beam at that location. The spot separation is about $1$ mm. A typical screen is $50\,{\rm cm}\times 50\,{\rm cm}$, and thus has 500 lines with 500 spots per line, or $2.5\times 10^{\,5}$ spots. Each spot is renewed every $1/30$ th of a second. (Every other horizontal line is skipped during a given traversal of the electron beam over the screen. The skipped lines are renewed on the next traversal. This technique is known as interlacing. Consequently, a given region of the screen, that includes many horizontal lines, has a flicker rate of 60 Hz.) Thus, the rate at which the instructions “turn on" and “turn off” must be sent to the electron beam is $30\times 2.5\times 10^{\,5}$ or $8\times 10^{\,6}$ times a second. The transmitted TV signal must therefore have about $10^{\,7}$ on-off instruction blips per second. If temporal overlap is to be avoided, each blip can be no longer than ${\mit\Delta} t \sim 10^{-7}$ seconds in duration. Thus, the required bandwidth is ${\mit\Delta} f\sim 1/{\mit\Delta} t\sim 10^{\,7}\,{\rm Hz}=10\,{\rm MHz}$. The carrier wave frequencies used for conventional broadcast TV lie in the so-called VHF band, and range from about 55 to 210 MHz. Our previous discussion of AM radio might lead us to think that the 10 MHz bandwidth represents the combined extents of an upper and a lower sideband of modulation frequencies. In practice, the carrier wave and one of the sidebands are suppressed. That is, they are filtered out, and never applied to the antenna. However, they are regenerated in the receiver from the information contained in the single sideband that is broadcast. This technique, which is called single sideband transmission, halves the bandwidth requirement to about $5\,{\rm MHz}$. (Incidentally, the lower sideband carries the same information as the upper one, and thus can be used to completely regenerate the upper sideband, and vice versa.) Thus, between $55$ and $210\,{\rm MHz}$ there is room for about 30 TV channels, each using a 5 MHz bandwidth. (Actually, there are far fewer TV channels than this in the VHF band, because part of this band is reserved for FM radio, air traffic control, air navigation beacons, marine communications, etc.)