Download Circuit Theory

Distributed constants
• Lumped constants are inadequate models of
extended circuit elements at high frequency.
• Examples are telephone lines, guitar strings,
and organ pipes.
• We shall develop the model for an electrical
transmission line.
• Also a model for sound traveling in a pipe,
and show how they are equivalent.
Ideal transmission line model
x
L
L
+
v
-
L
i+
C
C
C
i-
A voltage v is applied between two wires comprising a transmission line.
A current i+ enters one wire. An equal current i- returns from the other.
This voltage and current generate electric and magnetic fields around the
wires, shown respectively as dashed and dot-dashed lines in a cross
sectional view on the right. In turn, these fields manifest themselves as a
series inductance L and a parallel capacitance C per length x of the
transmission line, depicted by the circuit component overlay on the left.
An infinitesimal length of line
Consider a short segment dx of an infinite
dL
Z
line. Its measurable input impedance Z1 is
dC
Z
identical to Z2, that of the next segment.
Thus we write, Z1  jdL  1/(1/ Z2  jdC),
then let Z2 = Z1 = Z0 and simplify:
jdL
dL
2
2
Z 0  jZ 0 dL 
 0  Z0 
 jZ 0 dL
jdC
dC
As dx0, the second term goes to zero, while the constant
first term is simply L/C, the inductance and capacitance per
unit length of the line. Thus, Z0  L / C , which is called
the characteristic impedance of the transmission line.
1
2
Transmission lines with loss
• Wires (except superconductors) have some
resistance R per unit length.
• Likewise, most insulators have some
conductance G per unit length.
Infinitesimal line model
R
L
G
C
Characteristic impedance
R  jL
Z0 
G  j C
Signal propagation on a line (1)
dv
 i1 ( R  jL)
dx
di
(2)
 v2 (G  jC )
dx
(1)
v1
i1
dv/dx = v2 - v1
R
di/dx = i 1 - i 2
v2
i2
L
G
C
d 2v
di
Take the first derivative of (1),
 ( R  jL) ,
2
dx
dx
2
d
v
then substitute (2) to get,
 ( R  jL)(G  jC )v.
2
dx
d 2v
2
Let propagation constant   ( R  jL)(G  jC ) , so


v.
2
dx
The solutions are, v( x)  VAe x  VB ex . Similarly, i ( x)  I Ae x  I B ex .
Signal propagation on a line (2)
IA and IB are related to VA and VB as follows. From eq(1),
dv
 VAe x  VB ex  i ( R  jL), so
dx
i

( R  jL)
(VAe x  VB ex )
( R  jL)(G  jC )
1
x
x

(VAe  VB e ) 
(VAe x  VB ex ).
( R  jL)
Z0
The transmission line equations for sinusoidal signals are,
jt
e
v( x, t )  e jt (VAe x  VB ex ), i ( x, t ) 
(VAe x  VB ex ),
Z0
where the explicit time dependence (usually ignored) is e jt.
Traveling waves
The propagation constant g = a + ib
where a is the attenuation constant
and b is the phase constant, provides
a complete description of a wave on a
transmission line. If VB = 0, we have
a pure (forward) traveling wave. The
amplitude of such a wave is plotted
here over a 1 [m] length of transmission line for a = 2
[nepers/m] and b = 16p [radians/m]. The neper is a
dimensionless natural logarithmic unit of measure. Thus, a
specifies the exponential decay rate of a wave, while b
specifies its spatial angular frequency.
Boundary conditions
Vo
VA is the amplitude of the forward
traveling wave so if Rs = Z0, we can
write VA  Vs / 2 . At the load, the
voltage and current are related by
the load impedance,
Vl
(VAe x  VB ex )
Zl   Z0
Il
(VAe x  VB ex )

Vl
Rs
Vs
Zl
Transmission line, Zo, 
x=0
VB  VA e 2l
x=l
Zl  Z0
.
Zl  Z0
Notice that the reverse traveling wave vanishes iff Z l  Z 0 ,
that is if the transmission line and the load impedances match.
Standing waves
Zl  Z0
The voltage reflection coefficient  v 
; clearly,  v  1.
Zl  Z0
If Zl = Z0 at the load, i.e., rv = 0, we have seen that no signal
energy is reflected. Conversely, if Zl =  (open circuit) or 0
(short circuit), rv = ±1, respectively; in either case, all the
energy is reflected, resulting in a pure standing wave, which
over time appears not to move, rather just to oscillate in place.
For intermediate values of rv, the
voltage standing wave ratio
VSWR = (1 + |rv|)/(1 - |rv|) is the
ratio between the max and min
of the voltage envelope. A plot
for rv = 0.5 is shown.
Figure 2-3 from Matick, Transmission Lines for Digital
and Communication Networks, Mcgraw Hill, 1969.
The wave equation
2
k




k

 
  
2
   
, or in one dimension,
 
.
2
2
2
x
 s  t
 s  t
Solutions are of the form   e ( kx  st )  e ( kx ) e ( st ) . Factoring
2
2
d  x  k  2 ( st )
   s e  x , or
2
dx
s
2
out the time dependence , e
2
2
( st )
2
d 2 x
2
( kx )

k

,
where


e
is just the spatial function.
x
x
2
dx
d 2v
2
The transmiss ion line equation


v is of this form.
2
dx
Acoustic wave in a pipe (1)
Let a piston in a tube of area A move
through a distance d . The volume V of
air in the cylinder increases by dV , and
the pressure accordingl y drops by dp from
A
dV
V
its intial value, say atmospheri c pressure pa .
dp
The bulk modulus K  V
is the " stiffness" of the air.
dV
dV
Thus, dp   K
. For sound, the variation dp  pa ,
V
and represents the acoustic portion p of the total pressure
dV
( pa is constant). Therefore we can write, p   K
.
V
d
Acoustic wave in a pipe (2)
Consider a wave traveling down a tube of area A.
As it passes through a short column of air dx, it
displaces one end of the column by a distance  ,

and the other end by  
dx. Thus the displaced
x
  
column has a length dx1 
, and therefore a
 x 

  dx
dx
x
  
volume V  dV  Adx1 
,
 x 
A
dV 
dV
or

. But p   K
,
V
x
V

  
dx1 


so p   K
.
 x 
x
Acoustic wave in a pipe (3)
The column of air also has a mass m  V ,
dx
where  is the density of air. A force of
f i  pi A is exerted on each end of the
column by the adjacent air. The sum of
m
f2
p f1 A
these forces f  A( p1  p2 )   Adx .
 2
 2 x
p
 2
Now f  ma  V 2  Adx 2 , therefor e 
 2.
t
t
x
t

 2 K  2
Substituti ng p   K
, we get 2 
, which is the
2
x
t
 x
one dimensiona l wave equation. Differenti ating this with

p
2 p K 2 p
respect to x and substituti ng

gives

.
2
2
x
K
t
 x
Acoustic plane wave propagation
2
2 p

p
2
c
.
• Same as wave in a pipe:
2
2
t
x
2
• c = K/r; where c is the speed of sound.
– c = 332 [m/s] at 0 [degC] at 50% RH.
– dc/dT = 0.551 [m/s/degC].
• Solutions are of the form p  Aest kx  Be st  kx ,
– where s = jw for temporal sine waves.
 kx
kx
– Factoring out time, p  Ae  Be ,
just as for transmission lines.
Reflection of sound
p
• Wave impedance of a medium Z i  .

• Reflection of a wave normal to an interface
between media is,
B Z 2  Z1

,
A Z 2  Z1
which is the same as rv for transmission lines
Homework problem
100 [Ohms]
Vin
1 [V]
10 [m]
318 [pF]
10 [MHz]
A 10 [m] length of transmission line with 100 [pF/m]
capacitance and 250 [nH/m] inductance in series with a
1 [W/m] wire resistance is driven by a 10 [MHz] sine
wave from a 1 [V] 100 [Ohm] Thevenin source, and
loaded by a 318 [pF] capacitor. Calculate the complex
input voltage Vin of the transmission line.
Extra credit: plot the magnitude and phase of the voltage
as a function of position on the transmission line.