ulaby equations

Circuitsby Fawwaz T. Ulaby and Michel M. Maharbiz

Equations

Fawwaz T. Ulaby and Michel M. Maharbiz, Circuits c 2013 National Technology Press

Chapter 1: Circuit Terminology

Chapter 2: Resisitive Circuits

Chapter 3: Analysis Techniques

Chapter 4: Operational Amplifiers

Chapter 5: RC and RL First-Order Circuits

Chapter 6: Circuit Analysis by Laplace Transform

Chapter 7: ac Analysis

Chapter 8: ac Power

Chapter 9: Frequency Response of Circuits and Filters

Chapter 10: Three-Phase Circuits

Chapter 11: Magnetically Coupled Circuits

Chapter 12: Fourier Analysis Techniques


I =VR

(1.2)

i=dqdt

(A) (1.3)

q(t) = t

i dt (C) (1.6)

p= i (W) (1.9)

n

k=1

pk = 0 (1.10)


R=`

A=

`

A() (2.2)

= iR (2.3)

p= i = i2R=2

R(W) (2.4)

G=1R

(S) (2.5)

p= i = G2 (W) (2.7)

N

n=1

in = 0 (KCL) (2.8)

N

n=1

n = 0 (KVL) (2.11)

Req =N

i=1

Ri (resistors in series) (2.22a)

i =(

RiReq

)s (2.22b)

1Req

=N

i=1

1Ri

(resistors in parallel) (2.31)

Geq =N

i=1

Gi (conductances in parallel) (2.34)


R1 = R2 (2.38a)

is =sR1

(2.38b)

R1 =RbRc

Ra+Rb+Rc(2.42a)

R2 =RaRc

Ra+Rb+Rc(2.42b)

R3 =RaRb

Ra+Rb+Rc(2.42c)

Ra =R1R2+R2R3+R1R3

R1(2.43a)

Rb =R1R2+R2R3+R1R3

R2(2.43b)

Rc =R1R2+R2R3+R1R3

R3(2.43c)

R1 = R2 = R3 =Ra3

(if Ra = Rb = Rc) (2.44a)

Ra = Rb = Rc = 3R1 (if R1 = R2 = R3) (2.44b)

Rx =(R2R1

)R3 (balanced condition) (2.47)

Vout ' V04(RR

)(2.48)


GV = It (3.26)

RI = Vt (3.27)

iN =ThRTh

(3.37a)

RN = RTh (3.37b)

RL Rs (maximum current transfer) (3.39)

RL Rs (maximum voltage transfer) (3.40)

RL = Rs (maximum power transfer) (3.42)

pmax =2s RL

(RL+RL)2=

2s4RL

(3.43)


ip = in = 0 (ideal op-amp model) (4.16)

p = n (ideal op-amp model) (4.17)

G=os

=(RfRs

)(4.24)

o = G11+G22 (4.31)

o =(RfR

)[1+2] (equal gain) (4.32)

o =(1+2) (inverted adder) (4.33)

o =(RfR1

)1+

(RfR2

)2+ +

(RfRn

)n (4.34)

o =[(

R4R3+R4

)(R1+R2R1

)]2

(R2R1

)1 (4.40)

o =(R2R1

)(21) (equal gain) (4.44)

o =(

1+2RR2

)(21) (4.56)


u(t) =

{0 for t < 01 for t > 0

(5.2)

r(tT ) ={

0 for t T(tT ) for t T (5.4)

r(t) = t

u(t) dt = t u(t) (5.6)

rect(tT

)

=

0 for t < (T /2)1 for (T /2) t (T + /2)0 for t > (T + /2)

(5.8)

C =q

(F) (any capacitor) (5.20)

C =Ad

(parallel-plate capacitor) (5.21)

C =2pi`

ln(b/a)(coaxial capacitor) (5.22)

(t) = (t0)+1C

tt0i dt (5.24)

(t) =1C

t0i dt

(capacitor uncharged before t = 0)

(5.25)

w(t) =12C 2(t) (J) (5.28)


1Ceq

=N

i=1

1Ci

=1C1

+1C2

+ + 1CN

(capacitors in series)

(5.35)

Ceq =N

i=1

Ci (capacitors in parallel) (5.40)

C11 =C22 (5.46)

L=N2S`

(solenoid) (5.51)

i(t) = i(t0)+1L

tt0 dt (5.55)

Leq =N

i=1

Li = L1+L2+ +LN

(inductors in series) (5.62)

1Leq

=N

i=1

1Li=

1L1

+1L2

+ + 1LN

(inductors in parallel) (5.65)

ddt

+a = 0 (source-free) (5.69)

(t) = (0) et/ (natural response) (5.77)

= RC (s) (5.78)


i(t) =VsR

et/ u(t) (for t 0)(natural response) (5.82)

ddt

+a = b (5.87)

(t) = ()+ [(0)()]et/ (for t 0)(switch action at t = 0) (5.95)

(t) = ()+ [(T0)()]e(tT0)/(for t T0) (5.97)

(switch action at t = T0)

i(t) = i(0) et/ (for t 0)(natural response) (5.102)

=1a=

LR

(5.103)

i(t) = i()+ [i(0) i()]et/ (for t 0)(switch action at t = 0) (5.106)

i(t) = i()+ [i(T0) i()]e(tT0)/(for t T0)

(switch action at t = T0)

(5.107)

out(t) = 1RC tt0i(t ) dt +out(t0) (5.128)

out(t) = 1RC t

0i(t ) dt (if out(0) = 0) (5.129)


out =RC didt (5.130)

tfall =CnD+C

pD

g(5.155)


damping coefficient =R2L

(Np/s) (6.1a)

resonant frequency 0 =1LC

(rad/s) (6.1b)

(series RLC)

iL+a2iL+b2iL = c2 (6.12)

=1

2RC(parallel RLC) (6.14)

(tT ) = 0 for t 6= T

(tT ) dt = 1(6.15a)

(6.15b)

u(tT ) = t

(T ) dddt

[u(tT )] = (tT )

(6.19a)

(6.19b)

x(t) (tT ) dt = x(T )

(sampling property)

(6.23)

e j = cos + j sin (6.27)

x= |z|cos y= |z|sin|z|=

x2+ y2 = tan1(y/x) (6.30)

z = (x+ jy) = x jy= |z|e j = |z| (6.31)

|z|=

zz (6.32)


F(s) =LLL[ f (t)] =

0f (t) est dt (6.40)

(t) 1 (6.46)

[cos(t)] u(t)s

s2+2(6.47)

f (at)1a

F( sa

)a> 0

(time-scaling property)

(6.49)

f (tT ) u(tT ) eT s F(s)T 0

(time-shift property)

(6.53)

f =d fdt

s F(s) f (0)(time-differentiation property)

(6.58)

f =d2 fdt2

s2 F(s) s f (0) f (0)(second-derivative property) (6.61)

t0

f (t ) dt 1s

F(s)

(time-integration property)

(6.62)

F(s) =A1

s+ p1+

A2s+ p2

+ + Ans+ pn

=n

i=1

Ais+ pi

(6.83)


A1 = (s+ pi) F(s)s=pi

i= 1,2, . . . ,n(6.84)

B j ={

1(m j)!

dm j

dsm j[(s+ p)m F(s)]

}s=p

j = 1,2, . . . ,m (6.92)

LLL1[(n1)!(s+a)n

]= tn1eat u(t) (6.94)

= Ri V = RI (6.107)

= Ldidt

V = sLIL i(0) (6.110)

i=Cddt

I = sCVC (0) (6.111)

ZR = R, ZL = sL and ZC =1

sC(6.112)


= 2pi f (rad/s) (7.3)

T =1f

(s) (7.4)

didt

jI (7.21)

i dt

Ij

(7.23)

Z =VI

() (7.29)

ZR =VRIR

= R (7.30)

ZL =VLIL

= jL (7.35)

ZC =VCIC

=1

jC(7.38)

Zeq =N

i=1

Zi (impedances in series) (7.64)

Yeq =N

i=1

Yi (admittances in parallel) (7.66)

Z1 =ZbZc

Za+Zb+Zc(7.69a)


Z2 =ZaZc

Za+Zb+Zc(7.69b)

Z3 =ZaZb

Za+Zb+Zc(7.69c)

Za =Z1Z2+Z2Z3+Z1Z3

Z1(7.70a)

Zb =Z1Z2+Z2Z3+Z1Z3

Z2(7.70b)

Zc =Z1Z2+Z2Z3+Z1Z3

Z3(7.70c)

Z1 = Z2 = Z3 =Za3, if Za = Zb = Zc (7.71a)

Za = Zb = Zc = 3Z1 if Z1 = Z2 = Z3 (7.71b)

21

=N2N1

= n (7.120)

i2i1=

N1N2

(7.121)


Xav =1T

T0

x(t) dt (8.5)

cos2 x=12+

12

cos2x

1T

T0

cos2(

2pintT

+1)

dt =12

and

1T

T0

sin2(

2pintT

+2)

dt =12

(8.10)

Ieff =

1T

T0

i2(t) dt (8.13)

Xrms = Xeff =

1T

T0

x2(t) dt (8.14)

1T

T0

cos(nt+) dt = 0 (n= 1,2, . . .) (8.22)

Pav =VmIm

2cos( i) (W) (8.23)

Pav =VrmsIrms cos( i) (W) (8.24)

Pav =VrmsIrms =V 2rmsR

(purely resistive load)

(8.25)

Pav =VrmsIrms cos90 = 0

(purely reactive load)

(8.26)


S =12

VI (VA) (8.29)

S = VrmsIrms (VA) (8.32)

Q=VrmsIrms sin( i) (VAR) (8.34)

Pav =Re[S] (average absorbed power) (8.36a)

Q= Im[S] (peak exchanged power) (8.36b)

Pav =Re[S] =12|I|2R= I2rmsR (W) (8.39a)

Q= Im[S] =12|I|2X = I2rmsX (VAR) (8.39b)

n

i=1

Pavi = 0 andn

i=1

Qi = 0 (8.40)

S= |S|=P2av+Q2 =VrmsIrms (8.43)

pf=PavS

= cos( i) (8.44)

pf= cosz (8.49)

pf=

{cosZL for the RL circuit alonecosnew for the compensated circuit

(8.53)

XL =Xs (8.63)


RL = Rs (8.64)

ZL = Zs (maximum power transfer) (8.65)

Pav(max) =18|Vs|2RL

(8.66)


M(c) =M0

2= 0.707M0 (9.5)

0 =1LC

(RLC circuit) (9.11)

R = KmR,

L = KmL,

C =CKm

,

and

=

(magnitude scaling only)

(9.23)

R = R,

L =LKf,

C =CKf,

and

= Kf

(frequency scaling only)

(9.25)

R = KmR,

L =KmKf

L,

C =1

KmKfC,

and

= Kf

(magnitude and frequency scaling)

(9.26)

G= XY G [dB] = X [dB]+Y [dB] (9.31)


G=XY

G [dB] = X [dB]Y [dB] (9.32)

0 =1LC

(9.48)

c1 =R2L

+

(R2L

)2+

1LC

(9.50a)

c2 =R2L

+

(R2L

)2+

1LC

(9.50b)

B= c2c1 =RL

(9.51)

0 =c1c2 (9.52)

Q= 2pi(WstorWdiss

)=0

(9.53)

Q=0LR

(bandpass filter) (9.61)

Q=0B

(bandpass filter) (9.62)

c1 =1RC

(RC filter) (9.72)


Y-Source Configuration

V1 =VYs0

V2 =VYs120

V3 =VYs240(10.1)

-Source Configuration

V12 = V1V2=VYs0VYs120

=

3VYs30 =Vs30

V23 = V2V3 =Vs90V31 = V3V1 =Vs150

with Vs =

3VYs

(10.3)

VN = 0 (balanced network) (10.8)

Z = 3ZY (10.12)

PT = 3VYLIYL cosYQT = 3VYLIYL sinY

(balanced network)

(10.18a)

(10.18)

ST = PT+ jQT =

3VLILY

(balanced Y-load)

(10.20)

PT(t) = 3VYLIYL cosY (10.27)

PT = P1+P2(any 3-phase load)

(10.41)

QT = 3V 2LZ

sin (10.43)


QT =

3 (P2P1) (balanced load) (10.45)


2 =M21 di1dt (11.6)

1 = L1di1dt

+Mdi2dt

and

2 = L2di2dt

+Mdi1dt

(11.8a)

(11.8b)

1 = L1di1dtM di2

dtand

2 = L2di2dtM di1

dt

(11.9a)

(11.9b)

k =ML1L2

(11.21)

M(max) =L1L2 (11.22)

(perfectly coupled transformer with k = 1)

ZR =2M2

R2+ jL2+ZL(11.25)

[V1V2

]=[jL1 jMjM jL2

][I1I2

](transformer)

(11.27c)

[V1V2

]=[j(Lx+Lz) jLz

jLz j(Ly+Lz)

][I1I2

](T-equivalent circuit) (11.28)


Lx = L1MLy = L2M

and

Lz =M

(transformer dots on same ends)

(11.29a)

(11.29b)

(11.29c)

Lx = L1+M

Ly = L2+M

and

Lz =M(transformer dots on opposite ends)

(11.30a)

(11.30b)

(11.30c)

La =L1L2M2L1M

Lb =L1L2M2L2M

and

Lc =L1L2M2

M(transformer with dots on same ends)

(11.31a)

(11.31b)

(11.31c)

M(max) =L1L2

(ideal transformer)

(11.34)

L2L1

=N22N21

= n2 (11.35)

V2V1

= n (ideal transformerwith dots on same side) (11.36)

I2I1=

1n

(ideal transformerdots on same ends) (11.39)


V2V1

=N2N

=N2

N1+N2and

I2I1=

V1V2

=N1+N2

N2(step-down autotransformer)

(11.42)

V2V1

=NN2

=N1+N2

N2and

I2I1=

V1V2

=N2

N1+N2(step-up autotransformer)

(11.43)

VLsVLp

=ILpILs

= n (Y-Y and -) (11.44)

ST =

3VLIL (Y and ) (11.45)

VLsVLp

=ILpILs

=n3

(Y-)

and

VLsVLp

=ILpILs

=

3 n (-Y)

(11.46)


f (t) = a0+

n=1

(an cosn0t+bn sinn0t)

(sine/cosine representation) (12.15)

a0 =1T

T0

f (t) dt

an =2T

T0

f (t) cosn0t dt

bn =2T

T0

f (t) sinn0t dt

(12.17a)

(12.17b)

(12.17c)

An =a2n+b2n

and

n =

tan1

(bnan

)an > 0

pi tan1(bnan

)an < 0

(12.26)

Ann = an jbn (12.27)

f (t) = a0+

n=1

An cos(n0t+n)

(amplitude/phase representation)

(12.28)

Even Symmetry: f (t) = f (t)

a0 =2T

T/20

f (t) dt,

an =4T

T/20

f (t) cos(n0t) dt, (12.31)

bn = 0,

An = |an|, and n ={

0 if an > 0180 if an < 0


Odd Symmetry: f (t) = f (t)a0 = 0, an = 0,

bn =4T

T/20

f (t) sin(n0t) dt, (12.32)

An = |bn| and n ={90 if bn > 090 if bn < 0

Solution Procedure:Fourier Series Analysis Procedure

Step 1: Express s(t) in terms of an amplitude/phase Fourier series as

s(t) = a0+

n=1

An cos(n0t+n) (12.33)

with Ann = an jbn.

Step 2: Establish the generic transfer function ofthe circuit at frequency as

H() = Vout when s = 1cost. (12.34)

Step 3: Write down the time-domain solution as

out(t) = a0 H( = 0)

+

n=1

AnRe{H( = n0) e j(n0t+n)}.(12.35)

Pav =VdcIdc+12

n=1

VnIn cos(nin) (12.43)


cn =an jbn

2and

cn =an+ jbn

2= cn

(12.47)

f (t) =

n=

cne jn0t

(exponential representation)

(12.48)

cn =1T

T/2T/2

f (t) e jn0t dt (12.50)

sinc(x) =sinxx

(12.54)

f (t) =

n=

cne jn0t (12.58a)

cn =1T

T/2T/2

f (t) e jn0t dt (12.58b)

F() =F [ f (t)] =

f (t) e jt dt (12.62a)

f (t) =F1[F()] =1

2pi

F() e jt d (12.62b)

K1 f1(t)+K2 f2(t) K1 F1()+K2 F2()

(linearity property) (12.65)


(t t0) e jt0and

(t) 1

(12.67a)

(12.67b)

e j0t 2pi (0)and

1 2pi ()

(12.68a)

(12.68b)

e j0t f (t) F(0)(frequency-shift property)

(12.69)

f (t t0) e jt0 F()(time-shift property)

(12.70)

cos0t pi[ (0)+ (+0)] (12.71)

sin0t jpi[ (+0) (0)] (12.72)

Aeat u(t)A

a+ j, for a> 0 (12.73)

sgn(t) = u(t)u(t) (12.74)

u(t) pi ()+1j

(12.79)

f (t) j F() (12.81)

cos0t f (t)12[F(0)+F(+0)] (12.82)


F() = F()(reversal property)

(12.85)

f 2(t) dt =1

2pi

|F()|2 d

(Parsevals theorem)

(12.86)


Chapter 1Chapter 2Chapter 3Chapter 4Chapter 5Chapter 6Chapter 7Chapter 8Chapter 9Chapter 10Chapter 11Chapter 12

ulaby equations

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