sky waves
KS-15676 horns used were broadband 3.6-12 GHz. According to the actual specs; manual., they were designed to work between 3700 MHz to 4200 MHz. 5925 MHz to 6425 Mhz and 10,700 to 11,700 MHz on common carrier bands.
38 MHz. digital bit-rates vary depending on compression, modulation, typically QPSK. Expect at least 30 Mbps.
To determine the theoretical minimum system bandwidth needed for a 10 Mbps signal using 16-level Pulse Amplitude Modulation (PAM), we first calculate the symbol rate. Since 16-level PAM represents 4 bits per symbol (as (2^4 = 16)), the symbol rate is (10 \text{ Mbps} / 4 \text{ bits/symbol} = 2.5 \text{ Msymbols/s}). According to the Nyquist formula, the minimum bandwidth required is half the symbol rate, which leads to a theoretical minimum bandwidth of (2.5 \text{ MHz} / 2 = 1.25 \text{ MHz}) to avoid inter-symbol interference (ISI).
Since the 8085 has a maximum clock frequency of 6 MHz, increasing the crystal frequency from 5 MHz to 20 MHz, a corresponding clock frequency change of 2.5 MHz to 10 MHz, the chip would malfunction.
6 MHz
sky waves
The minimum bandwidth required for a multiplexed channel depends on the number of channels being multiplexed and the bandwidth of each individual channel. In general, the total minimum bandwidth needed is the sum of the bandwidths of all the channels being combined. For example, if you are multiplexing four channels, each requiring 1 MHz, a minimum bandwidth of 4 MHz would be necessary. Additionally, some multiplexing techniques may require guard bands to prevent interference, which would increase the total bandwidth requirement.
88 MHz-108MHz
6 mhz.
he rating for RIMM memory is based on the maximum theoretical bandwidth (in MHz) and included speed ratings of 800 MHz, 1066 MHz, 1200 MHz, 1333 MHz, and 1600 MHz.
1.95 MHz 1,950 KHz 1,950,000 Hz
Solution Let fh is the highest frequency and fl is the lowest frequency. Bandwidth = fh - fl = 4000 - 40 KHz = 3960 KHz = 3.96 MHz
Channel bonding
Channel bonding
22 MHz Each channel is a contiguous band of frequencies 22 MHz wide
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