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Reflection & Refraction
 At Discontinuity (change of Z0):
1-Adjustment to keep the
proportionality of V and I
2-in form of initiation of 2 new waves
 The new waves : Reflected &
Transmitted
 Satisfying portionality & continuity
 The energy conservation Auto. Satisfied
 α: reflection coeff. β: refraction coeff.
 α=(ZB-ZA)/(ZB+ZA), β=2ZB/(ZB+ZA)
Energy Conservation
 Assuming ZA>ZB,
I3=V3/ZB,
I1=V1/ZA,
V1+V2=V3,
I2=-V2/ZA
I1+I2=I3
 Solving for V2,V3 :
V2=[ZB-ZA]/[ZA+ZB] V1
V3=2ZB/[ZA+ZB] V1
I1V1=V1 /ZA
V2 /ZA +V3 /ZB=V1 /ZA{[(ZAZB)/(ZA+ZB)] +4ZAZB/(ZA+ZB)
=V1
/ZA
}
Traveling on multiple joint
 i.e. a line connected to n other lines
I3B=V3B/ZB, I3C=V3C/ZC, …. I3N=V3N/ZN
I2A=-V2A/ZA
 For Continuity of Voltage:
V1A+V2A=V3B=V3C=….=V3N
I1A+I2A=I3B+I3C+…+I3N
 These sufficient for Analysis
Line Termination




Open CCT voltage coeff.s :α=1, β=2
Sh. CCT Voltage coeff.s : α=-1,β=0
A real surge:
V=V0(e^-αt - e^-βt)
For a C termination:
ZB(s)=1/C1s
 a=[(1/C1s)-ZA]/[1/(C1s)+ZA],
 b=2/C1s/[1/(C1s)+ZA]
 v2(s)=av1(s)
Time response of C termination
 if α=1/(C1ZA):
v2(s)=V1/s{[(1/C1s –ZA)/(1/C1s+ZA)]}
 V2(t)=V1(1-2e^-αt),V3(t)=V1(2-2e^-αt)
 Interpretation of V3 response:
1- At step application, sh. CCT. : O/P zero
2- Finally open CCT.:
O/P 2V1
 Similarly For a termination Inductance L1 :
 v2(s)=v1/s [(L1s-ZA)/(L1s+ZA)]
 Assuming 1/β=L1/ZA

 v2(s)=v1{1/(s+β)-β/[s(s+β)]},
v3=v1[2/(s+β)]
Time response of L termination
 V2(t)=-V1(1-2e^-βt)
 V3(t)=2V1e^-βt
 Application of Thevenin theorem to:
calculation of refl. & refr. at Termination
 Steps : to calculate current in ZB
1- branch 1st removed,& V0 across it
2-all sources sh.& replaced by int. Imp.s
3-looking to its terminals x,y ; ZA determined
 I=V0/(ZA+ZB),VB=IZB=V0ZB/(ZA+ZB)
Attenuation and Distortion
rate of Electrical energy supplied:
1/2CV ν watts, dissipated rate: GV ν
both ~ V
result in an exp. Form voltage wave: V0exp(-G/C t)
 current wave supplies:
1/2LI , dissipate I R ;
both ~ I
: 
I=I0 exp(-R/L t)
 However to preserve the relation of V/I=Z0
 requires: R/L=G/C or R/G=L/C=Z0 =V /I
 says: I R=V G rate. loss. LR=rate. loss. Line
Leak.
 In power trans. res. losses>> leakance losses

Switching Operations and
Transmission Lines
 Source Impedance
 Voltage on Line:
V(0)/s x Z0/(Ls+Z0)
 V=V(0)[1-exp(- Z0t/L)]
 complicated source
 The source impedance
shown
 When study energization
of single line
Closing Resistor
stiff source impress
100% voltage
on line
 closing resistor reduce
percentage impressed
by factor: Z0/(Z0+R)

 S2 close 1st , S1 short
some time later
 Comparison of
reclosing transient
voltage
Lattice Diagram
 Example of Line:
Voltage: at instant t,
and at point M
Add incident &
reflected up to that
instant
 A general Method
 voltage&current at
any location vs
time
Example(Lattice Diagram Appl.)
 A sys. of O/H line &
Cable
O/H parameters:
Zc=270Ω,T=100μs
Cable parameters:
Zc’=30Ω,T’=50μs
 Unit step, Zs=0
 C open CCT
 VB, IB ?
Refl. & Refr coefficients
 αA=-1 , βA=0
 for B junction if O/H ~1, cable~2
αB1-2=(30-270)/300=-0.8
βB1-2=600/300=0.2
αB2-1=0.8 βB2-1=540/300=1.8
 αC=1 β=2
 Consider the Lattice Diagram
Lattice Diagram of Example
 T=2T’
t=0 eB=0
t=T eB=1-.8=.2
t=2T eB=1-.8+.36=.56
t=3T eB=.56+.8-.352=1
t=4T eB=1-.36+.288+.51
=1.454
t=5T eB=1.454+.352+.285.28=1.8
Voltage Variation at B
 The voltage at B
1-rising continuously
2-increasing to 2 pu
3- since C open
 What current is
expected?
 Any possible
response?
Current Refl. & Refr. coefficients
 αA=1 βA=0
 αB12=0.8 βB12=1.8
 αB21=-.8 βB21=0.2
 αC=-1
βc=0
 draw a similar Lattice Diagram
Lattice Diagram for Current
 t=0
iB=0
 t=T
iB=(1+.8)/Zc=1.8/Zc
 t=2T iB=1.44/Zc
 t=3T iB=2.59/Zc
 T=4T iB=1.425/Zc
 T=5T iB=1.774/Zc
 Next the IB curve
Variation of IB
 variation different
 no similarity
 there is some
similarity in single
line propagation
 Method capable of
application in a
software
 High memory size
Characteristic Method
Wave Equations:
L ∂i/dt+Ri+∂e/∂x=0
(1)
C ∂e/dt+Ge+∂i/∂x=0
(2)
Difference of a function of 2 variables:
de=∂e/∂t dt+ ∂e/∂x dx
di=∂i/∂t dt + ∂i/∂x dx
From these if ∂e/∂x,∂i/∂t as follows
∂e/∂x=[de-∂e/∂t dt]/dx
∂i/∂t=[di-∂i/∂x dx]/dt
be substituted in EQ 1:
L{ [di-∂i/∂x dx]/dt} +RI+ {[de-∂e/∂t dt]/dx}=0
(3)
∂e/∂t=- 1/C ∂i/∂x – G/C e [from (2) substituted in (3)]

Reforming the Equations
 Ldi/dt+Ri+de/dx+G/C e dt/dx+
(-Ldx/dt+1/C dt/dx) ∂i/∂x=0
 term in ()=0 to cancel the partial
derivatives; then
 2 resultant ODEs:
Ldi/dt+Ri+de/dx+1/C Ge dt/dx=0
(dx/dt) =1/LC
or in form of:
 LdI dx/dt+Ridx+de+1/C Ge dt/dx=0
dx/dt=+(-)1/√LC
(4)
(5)
(6)
Solution based on Characteristic
Method
 if dx/dt=1/√LC:
√L/C di+(RI+√L/C Ge)dx+de=0 (7)
 If dx/dt=-1/√LC:
-√L/C di +(Ri-√L/C Ge)dx+de=0 (8)
 The characteristics are straight Lines
called Forward & Backward
 e & i are found from above EQs
Finding lossless line solution
 dx/dt=1/√LC=v,
de=-√L/C di=-Zc di
(9)
 dx/dt=-1/√LC=-v
de=√L/C di=Zc di
(10)
 1st method employed by Bergeron
1928 in Hydraulic
 Application to single phase
transmission line
Integration of ODEs 7 & 8
 integrating EQ set (7):
e=-Zci+c1 (9), x=vt+c2
 where c1 & c2 are constants:
(10)
found from initial conditions
 X=0 line terminal, if point (d, t) satisfy EQ (10)
then satisfy EQ(9) and:
e(d,t)=-Zc i(d,t)+c1 , d=vt+c2
(11)
 Similarly for
point (0,t’):
e(0,t’)=-Zc i(0,t’)+c1, 0=vt’+c2
(12)
 Subtracting EQs 11 & 12 respectively:
 e(d,t)-e(0,t’)=-Zc[i(d,t)-i(0,t’)], d=v(t-t’)
Solution Continued
 t’=t-d/v=t-τ,
where: τ=d/v
 e(d,t)-e(0,t-τ)=-Zc [i(d,t)-i(0,t-τ)]
(13)
and:
 e(0,t)-e(d,t-τ)=Zc[i(0,t)-i(d,t-τ)]
(14)
 Rearranging (13)&(14):
e(d,t)=-Zc i(d,t)+[e(0,t-τ)+Zc i(0,t-τ)]
(15)
e(0,t)= Zc i(0,t)+[e(d,t-τ)- Zc i(d,t-τ)]
(16)
 Defining, 2 terms in right brackets as History
dependent voltage sources;
Ef(0,t-τ)=-[e(0,t-τ)+Zc i(0,t-τ)]
Eb(d,t-τ)=-[e(d,t-τ)-Zc i(d,t-τ)]
Lossless line Equivalent CCTs
 Substituting in
(15)&(16)
e(d,t)=-Zc i(d,t)-Ef(0,t-τ)
(17)
e(0,t)=Zc i(0,t)–Eb(d,t-τ)
(18)
 Equiv. CCT. ,

 The Norton Eq. CCT
more useful
Line Norton Eq. CCT.

rewriting (17)&(18):
i(d,t)=-1/Zc e(d,t)-If(0,t-τ) (19)
i(0,t)=1/Zc e(0,t) + Ib(d,t-τ)(20)
If & Ib Hist. depend. Cur. Sources:
If(0,t-τ)=-1/Zc e(0,t-τ)-i(0,t-τ)
Ib(d,t-τ)=-1/Zc e(d,t-τ)–i(d,t-τ)
Simple H.D.S. evaluation:
Ef(0,t)=-[2e(0,t)+Eb(d,t-τ)]
Eb(d,t-τ)=-[2e(d,t)-Ef(0,t-τ)]
Eq. CCT. Of Lumped Elements

Inductance:
 ea-eb=L(dia,b/dt)
 Trapezoidal Rule:
ia,b(t)-ia,b(t-∆t)=
1/L∫(ea-eb)dt=
1/L{[ea(t)-eb(t)]+ [ea(t-∆t)+
eb(t-∆t)]}/2 . ∆t
 ia,b(t)=∆t/2L [ea(t)eb(t)] +Ia,b(t-∆t)
 Ia,b(t-∆t)=ia,b(t-∆t)+
∆t/2L[ea(t-∆t)-eb(t-∆t)]
Eq. CCT for Lumped Capacitor
 Similar derivation:
 ia,b(t)=2C/∆t[ea(t)eb(t)]+Ia,b(t-∆t)
 Where:
Ia,b(t-∆t)=-ia,b(t-∆t)-2C/∆t
[ea(t-∆t)-eb(t-∆t)]
 all in form of:
algebraic EQs
Distributed Line Model in 3ph
network
 for a 3ph lossless line in general:
[-∂eph/∂x]=[L][∂iph/∂t]
[-∂iph/∂x]=[C][∂eph/∂t]
 wave EQs similarly for 3ph is:
[∂ eph/∂x ]=[L][C][∂ eph/∂t ]
[∂ iph/∂x ]=[C][L][∂ iph/∂t ]
[L],[C] inductance & capacitance matrices
of 3 ph line with mutuals
Similarity Transformation






to solve the complexity of EQs
instead of 3ph Domain, Modal Domain
solved for 3 independent voltages
Results of Modal Domain Transferred to 3ph
[eph]=[M][eM] and [iph]=[N][iM]
[∂ eM/∂x ]=
[M][L][C][M][∂ eM/∂t ]=[Λ][∂ eM/∂t ]
[∂ iM/∂x ]=
[N][L][C][N][∂ iM/∂t ]=[Λ][∂ iM/∂t ]
Similarity Transformation
[Λ] is diagonal matrix
Diagonal elements are
eigen values of:
[L][C] or [C][L]
 EQ of λn is independent of
other modes:


∂ eM/∂x
∂ eM/∂t
1
1
1
1
-2
1
1
1
-2
=λn
λn≈LC of single phase
 A case where:
[M]=[N] is shown 
 Vn=√1/λn,τn=l/vn
Bergeron EQs for 3ph network
 Eq. Modal Domains of
3ph.
 i1a-2a(t)=
-1/Za e1a(t)-Ifa(t-τa)
 i1b-2b(t)=
-1/Zb e1b(t)-Ifb(t-τb)
 i1c-2c(t)=
-1/Zc e1c(t)-Ifc(t-τc)
The 3ph Eq CCT Equations
 In matrix form:
[iM(1-2)]=-[Λ / ]- [eM1]-[IMf(t-τ)]
[IMf(t-τ)]=-{[Λ / ][eM2(t-τ)]-[iM(2-1)(tτ)]}
Then :
[N]- [i1-2(t)]=[Λ / ][M][e1(t)]+[IMf(t-τ)]
Or:
[i1-2(t)]=[G][e1(t)] + [I]
1st Mid Term Exam
 Question 1:

Xc1=20 /20=20Ω,
Xc2=40Ω
C1=1/(3.14x20)=1
59.1μF,C2=79.6μF
 Vp=20√2/√3
Ceq=C1C2/[C1+C2]
 Z0=√[(40x238.68)
/12661]=0.868Ω
Q1 continued
 d I/dt +1/τs dI/dt +I/T =0
 i(s)=(s+1/Ts)/[s +s/Ts+1/T ]I(0)+I’(0)/[
s +s/Ts+1/T ]
 I(0)=0, I’(0)=Vc(0)/L
 i(s)=Vc(0)/L x 1/[s +s/Ts+1/T ]
 i(s)=Vc(0)/L x 1/[s +1/T ] undamped
 I(t)=Vc(0)/Z0 sinω0t
 Ip=Vp/0.868=18.81 KA
 Ip%=13.5/18.81=0.715fig4.4:λ=2.0
 λ=Z0/R=0.868/R=2  R=0.434 Ω
Q1 solution
 Vcf=Vpx159.1/[159.1+79.6]=10.88KV
 in undamp, C2swing to 21.76 KV
 with damping:
C1V1=C1V1(0)-C2V2 V1=V1(0)-C2/C1V2
V1=IR+L dI/dt +V2, I=C2dV2/dt
d V2/dt +L/R dV2/dt+V2/LC=V1(0)/LC2
V2(0)=V’2(0)=0
V2(s)=V1(0)/T 1/[s(s +s/Ts+1/T )]
xC1/[C1+C2]
2x15/21.76=1.38fig 4.7: λ=1.8, R=Z0/λ
R=Z0/λ=0.868/1.8=0.482Ω
Question 2
 30/[20√3]=0.866 KA
 I XL /[20/√3]=0.12

XL=0.12x20/√3/0.86
8=1.6 Ω
 L=1.6/314.15=5.1
mH

R=0.05x20√3/0.868
=0.666Ω
Q2 continued
Z=√0.666 +1.6 =1.73Ω
Φ=tan- 1.6/0.666=tan- 2.4=67.38◦
Z0=√0.0051/(1.2x10^-8)=651.92Ω
λ=651.92/0.666=978.8
TRV :Almost
undamped:2Vp=2x16.33=32.66 KV
 t=Π√LC=3.1415√0.0051x1.2x10^8=24.47 μs
 k=20/32.66=0.6125 fig 4.7:η=1.2





Q2 continued




R/651.9=1.2 R=782.3Ω
RRRV=32.65/24.57=1.327 KV/μs
t’=3.6  t=3.6x 7.82=28.15μs
RRRV=20/28.15=0.71 KV/μs