AM Coverage: Frequency vs. Conductivity
     

It is common knowledge and experience that the groundwave coverage areas of AM broadcast stations can vary significantly, even for identical transmitter powers and antenna system parameters.

AM stations at the lower end of the AM broadcast band have greater coverage areas than those at the upper end, all other things being equal.

This variation in groundwave coverage area with frequency is related to the conductivity of the earth along the propagation path, and the differing loss that this conductivity produces for different AM broadcast frequencies.

Earth conductivity along a groundwave path is dependent on the basic composition, moisture and mineral content present at, and below, the surface of the earth. Groundwave radiation in the AM broadcast band travels along the surface of the earth, and also penetrates the earth to a depth of several tens of meters.

The portion of the radiated groundwave that penetrates the earth encounters losses, causing the wavefront to tip forward slightly in the direction of its travel, which is useful in providing coverage beyond the radio horizon. Both poorer earth conductivities and higher broadcast frequencies result in greater losses.

The table lists the descriptions, dielectric constants and conductivities for various land types common in the United States, in descending order. For reference, sea water has a dielectric constant of 80, and conductivity of 5,000 mS/m.

Table 1

Best for coverage?

The amount of AM broadcast coverage difference that results from earth conductivity and frequency can be difficult to appreciate fully, without a detailed analysis. For this reason, a set of graphs is included as Figs. 1–4 to show these effects visually.

The graphs lead to some interesting observations. For example, Figs. 1–3 show that a 1 kW AM station on the lowest AM frequencies can have a greater radius to its 0.5 mV/m daytime contour than a 50 kW station on the highest AM frequencies. However this does not mean that a high-power, high-frequency AM station has no advantage in comparison to a low-power, low-frequency station. The high-power, high-frequency station can have significantly more field intensity, ranging to 40 or more miles from the transmit site, which serves a large portion of the city of license and surrounding area.

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Fig. 1
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Fig. 2

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Fig. 3
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Fig. 4

This effect is shown in Fig. 4, where the distances to equal field intensities for a radius of at least 26 miles and conductivities of 8 mS/m and 15 mS/m are greater for a 50 kW station on 1600 kHz than for a 1 kW station on 600 kHz. Probably this advantage to high-power, higher frequency stations is more important to them and their local advertisers than the distances to their 0.5 mV/m contours.

Richard J. Fry, CPBE, was an RF systems applications engineer with Harris Broadcast. He is retired. Comment on this or any article. E-mail rwee@nbmedia.com.


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Of course, the optimum would be 50 kilowatts on 540 kHz. That would guarantee monster coverage of the 0.5 mv/m contour, along with massive signal intensities for quite some distance. It is however also of note that SKYWAVE is better on the HIGHER frequencies: a 50 kW station on 1600 located in, say, Maine, would probably be heard in California more often at night than a 50 kWer on 540 in the same location.
By Phil Boersma on 6/18/2011

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