Pulses in transmission lines

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1 Pulses in transmission lines Physics 401, Fall 013 Eugene V. Colla

2 Definition Distributed parameters networ Pulses in transmission line Wave equation and wave propagation eflections. esistive load Thévenin's theorem eflection. Non resistive load Appendix. Error propagation

3 Transmission line is a specialized cable designed to carry alternating current of radio frequency, that is, currents with a frequency high enough that its wave nature must be taen into account. Courtesy Wiipedia 3

4 Simplified equalent circuit i-1 i i+1 Ideal case C i-1 C i C i+1 i i eal situation C i G i+1 4

5 Specification: Impedance: 53 Ω Capacitance: 83 pf/m Conductor: Bare Copper Wire (1/1.0mm) 5

6 reflected r out i(x,t) V(t) V(x,t) Z x V 0 forward 6

7 Wavete 81 Tetronix 301B Sync output Triggering input Signal output G8U oad 7

8 V(t) r out i(x,t) V(x,t) x dx C = capacitance per unit length = inductance per unit length CdxV dq; V q C t t i V = -C x t i; dv V x di ( dx) ; dt i dt 8

9 i x = -C V t V x i dt t x i tx C V t V x i xt (1) () Combining (1) and () i x C i t V x V t C 9

10 i x V x i C t i dt V x V C t Now substituting V(x,t) and i(x,t) in We can find V i 0 0 i x C V = -C t or V ( x, t) i( x, t) Z i( x, t) C ooing for solution V ( x, t) V0 sin t i( x, t) i0 sin t v 1 C Speed of wave propagation Z - characteristic Impedance Equivalent to Ohm s law equation x v x v 10

11 C = capacitance per unit length Z = C Cross-section of the coaxial cable = inductance per unitlength πε0εr C= (F/m) D ln d d D e r dielectric permittivity m r -magnetic permeability 1 ε = m mm 0 r D ln d -1 (F/m) -7 0 =4 10 (H/m) (H/m) Finally for coaxial cable: Z 138 D log ( Ohms ) e d r 11

12 1 = C Speed of wave propagation = 1 c c m m e e m e e 0 r 0 r r r r 1 Delay time G-8/U, G58U: 1 (s/m) 9 For polyethylene e r ~.5(up to 1GHz) e ( s / m) e ( ns / m) Inner Insulation Materials: Polyethylene Nominal Impedance: 5 ohm Delay time ~5ns/m r r 1

13 x reflected esistive load Z = Z forward V i V x solution for the traveling in opposite direction V C t For reflected wave V r =-Z i r At any point of the transmission line: V V V V Z i r i r or V r V ( x, t) V0 sin t i( x, t) i0 sin t Z V V V Z r i i i i r i Z V i i V x v x v V Z r 13

14 V V V V Z i r esistive load Z = i r or V r Z Z V i Open line = V r = V i Incident pulse eflected pulse End of the line 14

15 Attenuation (db per 100 feet) Theory: = V r = V i MHz G-58U Experiment G 58U incident reflected V i V r V i ATTN ( db) 0log Vr Important parameter for cable is attenuation per length 15

16 In our case: 4.18 Attn(00 ft) 0 log 1.46dB 3.54 Where it is coming from? i i C i G i+1 16

17 G-58U 4.18 Attn(00 ft ) 0 log 1.46dB 3.54 G-8U > Core ø=0.81 mm Core ø=.17 mm Dielectric ø=.9 mm 9/16/ Attn(00 ft ) 0 log 0.335dB 3.78 Dielectric ø=7. mm Fall

18 eflected pulse does not follow the shape of the incident pulse G-58U Frequency dependence of the attenuation G-58U cable 18

19 FFT Spectrum correction Incident pulse spectrum reflected pulse spectrum IFFT 19

20 V V V V Z i r esistive load Z = i r or V r Z Z V i Shorted line =0 V r = - V i Incident pulse V(V) eflected pulse time (ns) 0

21 esistive load Z = Vi Vr V V Z i r or V r Z Z V i V (V) Matching the load impedance Z ; V r 0 Incident pulse V(V) Incident pulse time (ns) -0. V(V) time (ns) eflected pulse time (ns) 1 End of the line

22 Any combination of batteries and resistances with two terminals can be replaced by a single voltage source e and a single series resistor r Hermann udwig Ferdinand von Helmholtz ( ) 3 éon Charles Thévenin ( ) r E 1 E e + -

23 V V V i i i i r i V i r i Z V Z r i V i Z Z Vi From this equivalent equation we can find the maximum possible power delivered to V i P i Z P=P max if =Z (no reflection) 3

24 G 8U Pulse at the end of the line Vi Z incident reflected This experiment better to perform on G 8U cable because of lower attenuation =, amplitude of the pulse at the end of line is expected to be Vi, where Vi is the amplitude of the incident pulse 4

25 i V 0 V i Z Z Z di V i iz ; dt t i i0 1exp ; Z V i V V r = V - V i time 5

26 i V i Z Z di V i iz ; dt t i i0 1exp ; V 0 Z Z 50ns, =Z~. 5mH 6

27 i V i Z Z Z V 0 C Z C C 3. nf Z 7

28 8

29 1. The reports should be uploaded to the proper folder and only to the proper folder For example folder C ab eport_1 should used by students from 1 section only I would recommend the file name style as: 1_lab_student1 ab section ab number Your name. Origin template for this wee ab: \\engr-file-03\phyinst\ap Courses\PHYCS401\Common\Origin templates\transmission line\time trace.otp 9

30 limit s a serious to C circuit resistor Fitting parameters: (1) limit, () ω 0 = 1 C, (3) Q = 1 C Initial values can be estimated from circuit parameters. H 1 j limit j Q 0 limit 1 j Q j C 1 1 j Q1 0 Q 0 limit Q For fitting results and actual fitting function equation go to: \\engr-file-03\phyinst\ap Courses\PHYCS401\Common\Simple Examples\ ab 3 Frequency Domain Analysis_example.opj 30

31 Now to simplify the equation we can introduce the reduced frequency γ = ω ω 0 and transfer function components can be presented as: Q j 1 Q 1 H( ) Q ; 0 Q 1 H E Q Q 0 Q 1 ; H IM Q j 1 Q 1 0 Q 1 31

32 Two other components the modulus and the phase shift: H Q Q 0 Q 1 Q 1 1 ; arctan H H IM E arctan 1 Q 1 Q 3

33 X Y Data courtesy Tsung-in Hsieh, Physics 401, Fall

34 y = f(x1, x... xn) n f i i i i1 xi f ( x, x ) x 1.15 f(x i ) f±fx 1.10 x i ± x i x i 34

35 Derive resonance frequency f from measured inductance ± and capacitance C± C f (, C) C 10 1mH, C 10 μf 1 1 f f f (, C,, C) C C f f C 1 C C ; esults: f( 1,C 1 )= Hz f= hz f( 1,C 1 )=503±56Hz 35

36 10 1mH, C 10 1μF Where these numbers are coming from? Using commercial resistors, capacitors, inductances C=500pF±5% =35mH±10%. Measuring the parameters using standard equipment SENCOE Z meter model C53 Capacitance measuring accuracy ±5% Inductance measuring accuracy ±% Agilent E4980A Precision C Meter Basic accuracy ±0.05% 36

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