1-26Figure 1-20.—Transmitter moving relative to an observer.In 1 second the transmitter moves 360 feet and transmits 60 hertz. At the end of 1 second, the firstcycle of the transmitted signal reaches the observer, just as the sixtieth cycle is leaving the transmitter atpoint P1. Under these conditions the 60 hertz emitted is located between the observer and point P1.Notice that this distance is only 720 feet (1,080 minus 360). The 60 hertz is spread over the distance frompoint P1 to the observer and has a wavelength of just 12 feet (720 divided by 60). To find the newfrequency, use the following formula:The original frequency, 60 hertz, has changed to an apparent frequency of 90 hertz. This newfrequency only applies to the observer. Notice that the Doppler frequency variation is directlyproportional to the velocity of the approaching transmitter. The faster the transmitter moves toward theobserver, the greater the number of waves that will be crowded into the space between the transmitter andthe observer.Suppose the transmitter were stationary and the observer moving. When approaching the transmitter,the observer would encounter waves per unit of time. As a result, the observer would hear a higher pitchthan the transmitter would actually emit.If the transmitter were traveling away from the observer, as shown in view B of figure 1-20, the firstcycle would leave the transmitter at point P and the sixtieth at point P2. The first cycle would reach theobserver when the transmitter reached P2. You would then have 60 cycles stretched out over 1,080 plus360 feet, a total of 1,440 feet. The wavelength of these 60 hertz is 1,440/60, or 24 feet. The apparentfrequency is 1,080 divided by 24, or 45 hertz.Uses of CW Doppler SystemThe continuous-wave, or Doppler, system is used in several ways. In one radar application, the radarset differentiates between the transmitted and reflected wave to determine the speed of the moving object.

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