Semi-infinite Lossless Transmission Line

Above discussion elaborates on semi-differentiation and semi- integration obtained for driving point impedance of semi-infinite lossy line. A lossless transmission line constitutes of L and C distributed throughout its length. In Fig. 3.4, the element L will replace R. The line considered here is the semi-infinite lossless line whose impedance is constant or an operator of zero order. The problem is written as

∂²v(x,t)

∂t² = 1 LC

∂²v(x,t)

∂x² , v(0,t)=v1(t), v(∞,t)=0, v(x,0)&v(x,0) given with

∂i (x,t)

∂x = −C∂v(x,t)

∂t &∂v(x,t)

∂x = −L∂i (x,t)

∂t ,

wherevis the voltage, i is the current,vI(t) is time-dependent input variable, L is inductance per unit length, and C is capacitance per unit length. A classical solution to this problem is obtained through iterated Laplace transforms as done for semi- infinite lossy line in Sect. 3.5. The main results are given below:

V (0,s)= −V^∗(0,s)

√LC s + 1

√LC [V ( p,0)]_p=s^√_LC+ 1

√LC s

V( p,0)

p=s√ LC

V^∗(0,s)= d V (0,s)

d x &V= dv(x,0) dt

This contains transfer function of driving point (not impedance) as V (0,s)

V^∗(0,s) = − 1

√LC s and in time domain

v(0,t)= − 1

√LC

dv(0,t)

d x dt+ϕ1(t)

ϕ1(t) is the time-dependent initial condition. The transfer function consists of two parts: the forced response due to V^∗(0,s) and the initial condition response due to the initial voltage distribution in the lossless line. Using current expression as given,

3.6 Semi-infinite Lossless Transmission Line 55 the driving point impedance is obtained as follows:

V (0,s)= L

CI (0,s)− L

C

[I (0,0)]

s + 1

√LC [V ( p,0)]_p=s^√_LC + 1

√LC s

V( p,0)

p=s√ LC

Notice that the voltage is composed of two parts: the forced response due to I (0,s) and the initial condition response due to the initial voltage distribution in the lossless line. Considering only the first term, it can be seen that the impedance looking into this line is thus

Z (s)= V (0,s) I (0,s) =

L C

which is simply a constant. Mathematically, the impedance expressed in time domain as

v(0,t)= L

Ci (0,t)+ϕ2(t)

has a time-dependent initial condition response due to initial voltage and current dis- tribution and can be obtained by Laplace inverse of the last three terms of equation showing V (0,s),I (0,s) relationship, i.e.,

V (0,s)= L

CI (0,s)− L

C

[I (0,0)]

s + 1

√LC [V ( p,0)]_p=s^√_LC + 1

√LC s

V( p,0)

p=s√ LC

Thus it can be seen that a simple constant gain operator (zero- order operator) can also have time-varying initial condition terms. Figure 3.10 gives the diagram of a lossless semi-infinite transmission line (a zero-order element). Though the order of operation is zero, i.e., it returns the input function (variable) unaltered (except gain or attenuation), yet in the theory of generalized calculus, the initial distributed charges and voltage stored will be returned to the output. This initial function is time varying into future. The initial conditions on the distributed L and C along the infinite line gives rise to initialization functions (of time). Note that this particular element (of zero order) does not call for differintegrations, but the initial conditions ϕ associated with this distributed characteristics is very important to generalized theory of initialized (fractional) calculus. Operational amplifier circuit realized with zero-order distributed element will give practical understanding for generalized (ini- tialized) calculus; it is dealt in detail in Chaps. 6 and 7.

… i(t)

v(t) c c c

l l

l l

Fig. 3.10 Semi-infinite lossless transmission line

In Fig. 3.6, the input element is a lumped resistor R and the feedback element is a lumped capacitor C. Then this circuit configuration gives lumped integrator circuit.

Let the input voltagevi(t) to the circuit be switched on at some time a, before the time t = a the voltage is zero, and we start the circuit process (of integration) at time t=c>a. This implies that the capacitor is pre-charged with q(c) Coulombs, from time ato c with initial voltagevo(c). This constant is thus the initial condition for this lumped element integrator circuit.

The describing equations for this configuration is as follows:

vi(t)−0=ifR 0−vo(t)= 1

C t t=c

if(t)dt+ 1 C

t=c

t=a

if(t)dt = 1 C

t c

if(t)dt+q(c) C

= 1 C

t c

if(t)dt+[0−vo(c)]= 1

C^cD_t⁻¹if(t).

cD⁻¹_t is integer order one integration process starting from time t = c which includes the initialization, process that is charging of the capacitor C from time t =a to t=c which is represented as

ψ^/(t)= c a

ifdt = 1

Cq(c)= −vo(c).

Therefore, the total process is un-initialized integration starting from time t = c, that is:

cd_t⁻¹if = t

c

ifdt

plus initialization integration process from a to c, that is:

ad_c⁻¹if = c

a

ifdt =ψ^/(t)

3.6 Semi-infinite Lossless Transmission Line 57 These equations yield the final result by putting ii(t)=if(t)

vo(t)= − 1 RC

t c

vi(t)dt+vo(c)= − 1

RC^cD_t⁻¹vi(t),

withψ(t) = −RCvo(t). This is classical integer order calculus, with initialization as constant.

In the circuit of Fig. 3.6, we now replace the input element with semi-infinite lossless (LC) transmission line, a zero-order element, and the feedback element with lumped capacitor Cf. The transmission line terminal equation is re-written as

i (t)= C

Lv(t)+ϕ(t),

withϕ(t) as initial charge distribution on the distributed element.

The defining equations of this circuit are ii(t)=

C

L [vi(t)−0]+ϕ(t) 0−vo(t)= 1

Cf

t t=c

if(t)dt+[−vo(c)],

as done for the lumped integrator case above ii(t)=if(t) Therefore solving forvo(t), we obtain

vo(t)=−

# 1 Cf

C L

$t t=c

vi(t)dt− 1 Cf

t t=c

ϕ(t)dt+vo(c)=−

# 1 Cf

C L

$

cD_t⁻¹vi(t) where

ψ(t)= L

C t

c

ϕ(t)dt+Cf

L Cvo(c)

Here the initialization function is not a constant, but a function of time.

This expression is similar to the classical integer order integrator with lumped parameters as obtained earlier. Here the integrator is realized with distributed element.

The important difference is the values of initialization function. For a distributed element integrator, the effect of past history is not only contained in a constantvo(c),

which is charge on the capacitor Cf, but also carried in the remainder of theψ(t) function, which accounts for the distributed charge along the semi-infinite line. It is also observed here that the zero-order input element, as it is a wave equation

∂²v(x,t)

∂t² = 1 LC

∂²v(x,t)

∂x² ,

will simply propagate any perturbations invi(t) along the semi-infinite line, never returning, thus never seen again, the only effect being proportional variations in the ii(t). This behavior is true for terminal charging. However for side charging (arbitrary charging with voltages on the distributed line), an additional time function may return to the circuit output, which is dependent on initial voltage distribution on the line.

The circuit of Fig. 3.6 when configured with input element as lumped capaci- tor C and the feedback element as lumped resistance R behaves as integer order differentiator. The constituent equations are

vi(t)−0= 1 C

t t=c

ii(t)dt+ 1 C

t=c

t=a

ii(t)dt= 1 C

t c

ii(t)dt+q(c) C

= 1 C

t c

ii(t)dt+vi(c)= 1

C^cD⁻¹_t ii(t) 0−vo(t)=ifR

ii(t)=if(t) This gives

vo(t)= −RC d

dt(vi(t)−vi(c))

= −RCcD_t¹vi(t)= −RC

cd_t¹vi(t)+ψ(t) The initialization term

ψ(t)= d dtvi(c)

is taken normally as zero. However, the presence of initial charge in the input capac- itor gives an impulse output at the start of differentiation process at t =c.

Modifying the circuit of Fig. 3.6 with input element as capacitor Ci and the feedback element with distributed LC zero-order element gives the integer order differentiator transfer character, with the concept of initialization function and gen- eralized calculus. The defining equations are

vi(t)−0= 1 Ci

t c

ii(t)dt+vi(c)= 1 Ci

cD⁻¹_t ii(t)

3.6 Semi-infinite Lossless Transmission Line 59 For the distributed feedback zero-order elements, the expression in the circuit is

0−vo(t)= L

Cif(t)+ϕ(t)= L

Cif(t)+ L

Cψ(t) Putting ii(t)=if(t), yields the final result as

vo(t)= − L

C

Ci

d

dt(vi(t)−vi(c))+ψ(t)

= −Ci

L C

d

dtvi(t)+ 1 Ciψ(t)

= −Ci

L

C^cD¹_tvi(t)

The generalized differentiation requires an initialization function. However for ter- minal charging case for integer order differentiation this initialization is zero but for side charged transmission line, an additional time function will be returned to the circuit output.

A simple gain (memory) less zero-order operator is realized by configuring the circuit of Fig. 3.6 with Ri as lumped resistor at input leg and Rf as lumped resistor at the feedback. The transfer characteristics will then be

vo(t)= −Rf

Rivi(t)= −Rf

Ri

cD⁰_tvi(t)= −Rf

Ri

cd_t⁰vi(t)+ψ(t)

,withψ(t)=0,

clearly this circuit has no memory.

Zero-order circuit may be realized by employing semi-infinite distributed lossless transmission lines at input leg and one lumped resistor R at feedback, of circuit of Fig. 3.6.

The input leg equation with LC line is

ii(t)= C

L [vi(t)−0]+ϕi(t)= C

L^cD_t⁰[vi(t)−0]

The feedback leg equation is

0−vo(t)=Rif(t) and ii(t)=if(t) gives

vo(t)= −Rif(t)= −R C

L^cD_t⁰vi(t)= −R C

L '

cd_t⁰vi(t)+ψi(t)( whereψi(t)=

L Cϕi(t)

This zero-order operation in general returns the input functionvi(t) (with ampli- fication or attenuation), also provides the extra time function (associated with the memorized charges on the distributed element). This zero-order circuit has memory.