Homework #3 S421
- A cable system is used to transmit a signal, s(t), that occupies frequencies between 67-73 MHz. That is, the signal is 6MHz wide, and centered at 70MHz. The cable has a loss of 2.0 dB/100 feet at 100 MHz. The cable’s thermal noise Ncable(f) = 2kT Watts/Hz. The receiver adds an equivalent noise at it’s input of an additional 2kT Watts/Hz (i.e., the receiver’s noise figure is 3dB). The transmitter output signal is 5 millivolts peak. The receiver must admit all frequencies within the transmitted bandwidth of 6MHz. The cable’s characteristic impedance is 75 ohms, and this is matched to the input impedance of the receiver. The noise temperature is 300K.
The receiver requires a peak signal - to- rms noise ratio at its input of 46 dB in order to properly demodulate the signal and extract the underlying information with the desired fidelity.
Assuming that no line amplifiers are used, (i.e., a single cable section between the transmitter and the receiver)…what is the maximum allowable cable length?
Solution:
The formula for signal-to-noise ratio is:
SNR = [{cable input voltage x cable attenuation factor}**2 / R] / [4kT x Bandwidth]
Where cable attenuation factor = ratio of cable output/cable input at the highest signal frequency.
SNR= 46 dB => 40,000
Cable input voltage = .005 volts
Bandwidth = 6,000,000 Hz
kT=1.38 x 10**-23 x 300 Watts/Hz = 4.14 x 10**-21 Watts/Hz
R=75 ohms
Therefore we can solve for the cable attenuation factor, which is 0.109. This means that the ratio of cable output to cable input at the highest frequency of the signal must be 0.109 or more if we are going to satisfy the signal-to-noise ratio requirement.
Since the cable attenuation factor is a ratio of voltages, the allowable cable attenuation in decibels is 20 log 0.109 = -19.25 dB.
Since the loss of the cable is 2 dB/100 feet at 100MHz, it follows that the loss of the cable at the highest frequency of the signal is 2 x [(73/100)**0.5] = 1.71 dB/100 feet. Therefore, the maximum length of the cable is 19.25 dB / [1.71 dB/100 feet] = 1126 feet.
2. A 100Base-T Ethernet link operates over Category 5 twisted pair cable. The cable
loss at 10 MHz is 3dB/100 feet. The maximum allowable rolloff of H(f), the cable
frequency response, is a factor of 1.414 at 100 MHz. What is the maximum length of
cable that can be used?
Solution:
If the attenuation factor of the cable at 100 MHz can be no larger than a factor of 1.414, i.e., H(f) = [voltage out / voltage in] must be greater than (1/1.414), then the attenuation of the cable, in decibels, can be no more than 20 log (0.707) = -3dB. If the loss of the cable is 3dB/100 ft at 10MHz, it follows that the loss of the cable at 100MHz is 3 x [(100/10)**0.5] = 9.49dB/100 ft. Therefore, we can allow no more than 3 dB / [9.49 dB/100 feet] = 31.6 feet of cable.