5.3 Reimann–Liouville Fractional Differintegral
5.3.2 Convolution
5.3 Reimann–Liouville Fractional Differintegral 85
0 1 2 3 4
0 1 2 3 4
5
5
6 7
Fig. 5.3 Homogeneous and heterogeneous time
(of the moving object). Fractional integration in time means transformation of the local time to cosmic time.
2
a 1 h(τ)
0-10 -5 0 5 10
1
b 0
-1-10 -5 0 5 10
f(τ)
2
c 1
0-10 -5 0 5 10
2
d 1
0-10 -5 0 5 10
2
e 0
-2-10 -5 0 5 10
1
f 0
-1-10 -5
TIME
0 5 10
h(−τ)
h(5−τ)
f(τ).h(5−τ) Y=f(t)
X
d–1/2f(t)
Fig. 5.4 RL integral interpreted as convolution
the order of fractional integration is half q =1/2. Figure 5.4a shows h(τ) versusτ; Fig. 5.4b shows that f (t),t >0. Figure 5.4c shows h(−τ). The curve is obtained for the value t =5, and Fig. 5.4d shows the plot of h(t−τ) at t=5, i.e., h(5−τ) versusτ. Figure 5.4e shows the full integrand for t =5. Now moving this h(t−τ) for several continuous values of t from 0 to 10, repeating the graphs d and e and obtaining the value of the integral of the product (for several values of t) the final graph f is obtained. For t =5, the graph e shows full integrand as h(5−τ) f (τ).
The integral of this product becomes t =5, value of the semi-integral of f (t).
5.3 Reimann–Liouville Fractional Differintegral 87 Fig. 5.5 Convoluting
function h (t) for several t
2 1
0 1
t=1 t=2 t=3 t=4 t=5 t=6
2 3 4 5 6
In Fig. 5.4f, the point X is 5 0
f (τ)h(5−τ)dτ,
definite value of the integration.
Figure 5.5 demonstrates the several h(t−τ) for t=1,2,3,4,5,6, . . . Figure 5.6 demonstrates the semi-derivative of
f (t)=cos 2π 5 t
,
which is obtained from differentiating once the RL semi-integral graph.
5.3.3 Practical Example of RL Differintegration in Electrical Circuit Element Description
These types of intermediate devises are becoming reality in electrical circuits as evident from patent US 20060-267595 of November 2006. We shall start with a resistoductance alone, which is a linear circuit element whose behavior is interme- diate between that of an inductor element and ohmic resistor element. The term
Fig. 5.6 Semi-derivative of function. Differentiating once the RL semi-integral of f (t)
1.0 0.5 0 -0.5 -1 2 0 d–1/2 f(t)
d1/2 f(t) -2
0 1 2 3 4 5 6
0 1 2 3 4 5 6
resistoductance is combination of pure resistance and pure inductor. As integer order equations, the fractional order requires fractional derivatives (or integrals) as initial conditions. How does then one relate to initial condition expressed in terms of fractional differintegrals? The constitutive equation of such an element is
v(t)=K0Dtαi (t) or i (t)= 1
K0Di−αv(t)
Herev(t) is across variable, i.e., the voltage across the circuit element, and i (t) is the through variable, i.e., current through the circuit element. Ifα=0, the circuit is purely resistive and K = R ohms, and ifα=1, the circuit is purely inductive and K =L henrys.
For a step input of voltagev(t) = V0 applied at t = 0, the current is described as fractional integration of the forcing function, i.e., i (t)= K 01 Dt−αv(t). In terms of convolution definition (as forcing function being causal) and h(t)= Γtα−(α)1, the power function, the current is obtained as follows:
i (t)= 1 K
h(t)∗v(t)
= t 0
(t−τ)α−1
KΓ(α) V0dτ = V0 KΓ(α)
0 t
−(x )α−1d x
= V0
KΓ(α)
− xα−1+1 α−1+1
0 t
= V0
KΓ(α)
−xα α
0 t
= V0
KαΓ(α)tα UsingΓ(α+1)=αΓ(α), we obtain the current to step input as
i (t)= V0 KΓ(α+1)tα
for pure resistanceα =0, and i (t) = V0/K and for pure inductanceα = 1, and i (t)=(V0/K )t.
The initial value of the current vanishes, i.e., there is no instantaneous current, only retarded response. However, the first ordinary derivative of i (t) = K1tα is unbounded, so that a finite though undefined current can be reached in arbitrary small time interval. The change of i (t) is described by the fractional differential equation as0Dtαi (t)=V0/K .
In accordance with the theory of fractional differential in terms of RL derivatives, an initial condition involving0Dtα−1i (t) is thus required. Physically this initial con- dition has no representation and cannot be directly obtained from measurement. This condition can be found by taking the first-order integral of the constitutive equation.
This process relates the fractional (immeasurable) initial condition to something of reality and measurable as
0Dtα−1i (t)
t→0=
0D−1t (V0/K )
t→0
5.3 Reimann–Liouville Fractional Differintegral 89 In the case under consideration, voltage stress is finite at all times hence
0D−1t V0
t→0 = 0, which leads to the condition of zero initial condition involv- ing fractional differintegral, namely,
0Dtα−1i (t)
t→0 = 0. The same consid- eration applies to general finite voltage v(t), and the equation to be solved is
0Dtαi (t)=v(t)/K , and same zero initial condition be attached.
Now for the impulse input voltage at time 0, i.e., v(t) = Bδ(t) at t = 0, the current expression is again i (t)= K 01 D−αt v(t) using the convolution definition and h(t)= Γtα−1(α), the current expression is
i (t)= 1 K
h(t)∗v(t)
= 1 K
h(t)∗Bδ(t)
= B
Kh(t)=B tα−1 KΓ(α)
This is obtained as the convolution of function with impulse at t =0, this returns the function itself. The power function with gain B is retuned. This is property of the convolution. As observed from the derived current expression for the impulse voltage, the initial voltage–stress singularity gives rise to lower order current singu- larity, since resistoductance cannot respond instantaneously.
The impulse response is mathematical convenience to evaluate transfer charac- teristics and is seldom used in practice because it is even more problematic to apply homogeneous impulse voltage on circuit element than to apply step. However inves- tigating the impulse response will follow the same reasoning as for the step.
For the impulse voltage excitation for t >0, the fractional differential equation is0Dαti (t) = 0. In accordance with the theory of fractional differential equations with RL derivatives, an initial condition involving [0Dα−1t i (t)]t→0is required. This can be found through integration of constitutive equation as
0Dtα−1i (t)
t→0 =
0Dt−1(v(t)/K )
t→0 =B/K
which gives the initial condition in terms of fractional differintegral as
0Dα−t 1i (t)
t→0 =B/K . This fractional differintegral initial condition is non-zero, well defined, and bounded, whereas both current and its integer order derivatives are unbounded, and its first-order integral is zero so that a meaningful initial condition expressing the loading conditions cannot be obtained using integer order derivatives.
In the above example of resistoductance, it is possible to attribute physical mean- ing to initial condition expressed in terms of fractional differintegral. Expressing initial condition in terms of fractional derivative of a function u(t) is not a prob- lem because it does not require a direct experimental evaluation of these fractional derivatives. Instead, one should consider its counterpart (in separable twin),v(t) via basic physical law, and measure (or consider) its initial values.
Similarly, other intermediate models can be considered as resistocaptance. Resis- tocaptance will be similar in nature to resistoductance, where the circuit element will be intermediate between pure resistance and pure capacitance. The constitutive part will be
v(t)= 1
K0Dt−αi (t) or i (t)=K0Dαtv(t),
where forα=1 the element is pure capacitor with K =C farads, and forα=0 the element will be pure conductance K =G mho.
These intermediate models can explain the behavior of “time-constant” disper- sion effects in the circuit behavior when the relaxation observations cannot be explained by singleτ =R/L or RC time constant.