The bandwidth of a signal is the width of frequencies between the highest and the lowest frequency.
So 500Hz - 50Hz = 450Hz bandwidth.
AE7HD
-450
What is the bit rate of a signal in which 10 bit lasts 20 microseconds?
Physically the two fiber types differ in the diameter of their cores, the light-carrying region of the fiber. This is signified by the numeric nomenclature. In 62.5/125 fiber, for example, the core has a diameter of 62.5 microns and the cladding diameter is 125 microns. In terms of performance, the difference lies in the fibers' bandwidth, or information-carrying capacity. Bandwidth is actually specified as a bandwidth-distance product with units of MHz·km. The bandwidth needed to support an application depends on the data rate. As the data rate goes up [MHz], the distance that rate can be transmitted [km], goes down. Thus, a higher fiber bandwidth enables you to transmit at higher data rates or for longer distances. 50 micron multimode fiber offers nearly three times more bandwidth (500 MHz·km) than FDDI-grade 62.5 micron fiber (160 MHz·km) at a wavelength of 850 nm [nanometers]. Network planners often choose 50 micron fiber when they know the network will need to carry high bandwidth applications over longer link distances, or when they anticipate running higher speed protocols in the future.
how can Technology be harmful? Explain in 500 words
$500-$1000
There are over 500 designs and fabrics
The Nyquist frequency for a signal with a maximum bandwidth of 1 KHz is 500 Hz, however that will lead to aliasing unless perfect filters are available. The Nyquist rate for a signal with a maximum bandwidth of 1 KHz is 2 KHz, so the answer to the question is 2 KHz, or 500 microseconds.
If three pure signals with frequencies of 100, 200, and 500 Hz are added, their sum is a signal with afrequency of 100 Hz, which has additional "wiggles" superimposed on its shape. The wiggles appear ingroups of 2 per cycle and 5 per cycle on the main signal; their exact shape and depth is determined bythe relative amplitudes of the three individual component signals.
The answer depends on the accuracy desired. a. The minimum bandwidth, a rough approximation, is B = bit rate /2, or 500 kHz. We need a low-pass channel with frequencies between 0 and 500 kHz. b. A better result can be achieved by using the first and the third harmonics with the required bandwidth B = 3 × 500 kHz = 1.5 MHz. c. A still better result can be achieved by using the first, third, and fifth harmonics with B = 5 × 500 kHz = 2.5 MHz
The Nyquist Therorem states that the lowest sampling rate has to be equil to or greather than 2 times the highest frequency. Therefore the sampling rate should be 400Hz or more.
500 ft
Low frequency waves are waves with long wavelengths, by definition, since their cycle is slowly repeated. One example is AM radio waves compared to visible light waves. AM radio waves have a frequency of around 1 MegaHertz (1,000,000 s-1) and are around 500 meters long. In contrast, visible light waves have a frequency of around 500 TeraHertz (300,000,000,000,000 s-1) and are around 500 nanometers long (.000000500 meters).
500 feet
500 feet
500 feet
10 feet.
You must buy a complete signal mirror cost around 200.00
a 1 bits/second b 500 bits per second c 500 bits per second. I assume you meant 20 msec for c.