Access Networks Exercises 2008 / 09 Examples On Calculating Properties Of Metalic Lines - Mainly Twisted Pairs

Example 1 We have measured the value of twisted pair longitudinal ballance (LB) 40,92 dB. This parameter was measured with the signal source with signal level 1 Veff at 1kHz. Calculate the value of differential voltage measured between these both wires. Result: 9 mV

Example 2 At the near end, there were found the line-to-ground impedances modules 90,48 MΩ and 84,63 MΩ respectively. Calculate the longitudinal balance of this pair and decide, whether this pair satisfies to use of broadband services (the condition is: LB>40 dB!) Result: 29,52 dB  the pair does not satisfy

Example 3 We know, that the twisted pair line (Φ= 0,4 mm) should have characteristic impedance 1100Ω at 1 kHz. The open ended line has impedance Zoc= 24,8 kΩ. Determine the input impedance of the short circuited line (short circuit impedance). Result: about 49 Ω

Example 4 We have a coaxial line by which is fed the device with asymmetric impedance 75Ω (analogue TV receiver). This device needs minimal input voltage 696 μVeff for good operation. We send into line the signal power 1dBm. What must be the maximal line attenuation for correct operation of the device? Solution: The device input power must be P2 = U.I=U2/R=6,46 .10-9 W, after calculating to dBm it is -51,9 dBm. For the line input signal power P1=P2+A; therefore attenuation is A=P1-P2 =1-(-51,9) = 52,9 dB

Example 5 There is utilised the twisted pairs cable with declared impedance 190 Ω at 1 kHz. One of pairs is ended by device with input impedance of 300 Ω. Calculate both the reflection coefficient and the reflection loss in such case. Results: r=0,224, RL=12,986 dB

Example 6 We have a signal source and send out the signal with power level of 1 dBm. In the neighbour pair in the same end of cable there was measured crosstalk with level -50dBm at frequency 600 Hz. State the parameter ANEXT in [dB] (NEXT attenuation) and also power level of crosstalk in [mW]. Results ANEXT=10log(P1N/P2N)=P1N[dBm]-P2N[dBm]=1+50=51dB rosstalk level PNEXT= P2N = logarithm inversion of (-50dBm)  =10-5 mW

Example 7 What was the reflection time from fault place, if our measurement device evaluates distance of this place as 113 ft? (1 ft = 0.3048 m, propagation velocity in media is v = 0.65 c) Result: 0,353 μs (don’t forget the distance must be taken 2 times!)

Example 8 a) By measuring there was found out the line capacity 52,1 nF. If we know the specific line capacity (52pF/m), state the line length. Result: 1001,92 m b) There may be an similar example with loop resistivity and loop length…

Example 9 The line has 1 km length, its input power is 0.35 W and output power for some electronic device must be minimal 0.1 mW. State, how must be maximal specific attenuation of transport line (in [dB/100m]) to feed the device correctly. Result: 3,5 dB /100 m

Example 10 We have measured longitudinal balance of twisted pair by means of voltage method. The common voltage was 1Veff and between the wires there was measured the differential voltage of 0.0025 Veff. Calculate the parameter LB. Result: 52 dB

Example 11 Calculate effective value of noise voltage on the input of receiver in the 8 MHz frequency band, at the temperature 40oC, if input resistance is 75 Ω. How must be the minimal value of signal (effective voltage) on the input of receiver, if minimal SNR must be 43 dB ? SNR is the parameter “signal-to-noise ratio”, which is in [dB] defined as follows: [dB] where PS is signal power, PN is noise power and VS eff, VN eff are effective voltages of signal and noise respectively, measured on the same impedance. Results: UN = 3.22.10-6 V, Usig min = 0,455 mV