This section presents a procedure to construct a solution of the NP problem when it is solvable. The procedure is developed inductively: First, the casen= 1 is solved; then the case ofnpoints is reduced to the case of n−1 points.
Let us begin by lettingDdenote the open unit disk,|z|<1, andDthe closed unit disk, |z| ≤1. A M¨obius functionhas the form
Mb(z) =
z−b
1−zb,
where |b|<1. Following is a list of some properties of M¨obius functions (you should check these): 1. Mb has a zero atz=b and a pole atz= 1/b. ThusMb is analytic inD.
2. The magnitude of Mb equals 1 on the unit circle.
3. Mb maps Donto Dand the unit circle onto the unit circle.
4. The inverse map is
Mb−1(z) = z+b 1 +zb
(i.e.,Mb−1 =M−b). So the inverse map is a M¨obius function too.
We will also need the all-pass function
Aa(s) :=
s−a
s+a, Rea >0.
With the aid of these functions we can solve the NP problem for the data
a1
b1
There are two cases.
Case 1 |b1| = 1 A solution is G(s) = b1. By the maximum modulus theorem this solution is unique.
Case 2 |b1|<1 There are an infinite number of solutions:
Lemma 4 The set of all solutions is
{G:G(s) =M−b1[G1(s)Aa1(s)], G1 ∈ Sc,kG1k∞≤1}.
9.3. NEVANLINNA’S ALGORITHM 155
Proof LetG1 ∈ Sc,kG1k∞≤1, and defineGas
G(s) =M−b1[G1(s)Aa1(s)].
Thus Gequals the composition of the two functions
s 7→ G1(s)Aa1(s), z 7→ M−b1(z).
The first is analytic in the closed right half-plane and maps it into the closed disk D; the second is analytic inD and maps it back intoD. It follows thatG∈ Sc and kGk∞≤1. Also,Ginterpolates
b1 at a1:
G(a1) =M−b1[G1(a1)Aa1(a1)] =M−b1(0) =b1.
Thus Gsolves the NP problem. Moreover, if G1 is an all-pass function, then so is G1Aa1, hence so
is G(because M−b1 maps the unit circle onto itself).
Conversely, suppose thatGsolves the NP problem. DefineG1 so that
G(s) =M−b1[G1(s)Aa1(s)], that is, G1(s) = Mb1[G(s)] Aa1(s) .
The function Mb1[G(s)] belongs to Sc, has ∞-norm ≤ 1, and has a zero at s = a1. Therefore, G1 ∈ Sc and kG1k∞≤1.
Example 1 For the interpolation data 2
0.6
the formula in the lemma gives
G(s) = G1(s) s−2 s+ 2 + 0.6 1 + 0.6G1(s) s−2 s+ 2 .
The all-pass functionG1(s) = (s−1)/(s+ 1) results in
G(s) = s
2−0.75s+ 2
s2+ 0.75s+ 2.
Now we turn to the NP problem with ndata points, the problem being assumed solvable, and see how to reduce it to the case of n−1 points. Again, there are two cases.
Case 1 |b1|= 1 Since the problem is solvable, by the maximum modulus theorem it must be that
Case 2 |b1|<1 Pose a new problem, labeled the NP′ problem, with the n−1 data points
a2 · · · an
b′
2 · · · b′n
where b′i:=Mb1(bi)/Aa1(ai).
Lemma 5 The set of all solutions to the NP problem is given by the formula
G(s) =M−b1[G1(s)Aa1(s)],
where G1 ranges over all solutions to the NP′ problem. If G1 is all-pass, so isG.
Proof Gsolves the NP problem iff
G∈ Sc, kGk∞≤1, G(a1) =b1, and G(ai) =bi, i= 2, . . . , n.
From Lemma 4, the set of all Gs satisfying the first three conditions is
{G:G(s) =M−b1[G1(s)Aa1(s)], G1 ∈ Sc,kG1k∞≤1}.
Then Gsatisfies the fourth condition iff
G1(ai) =
Mb1(bi) Aa1(ai)
, i= 2, . . . , n
(i.e., G1 solves the NP′ problem).
It follows by induction that the NP problem always has an all-pass solution.
Example 2 Consider the NP problem for the data
a1 a2 a3
b1 b2 b3 =
1 2 3
1 2 13 14 The Pick matrix is
0.3750 0.2778 0.2188 0.2778 0.2222 0.1833 0.2188 0.1833 0.1563 .
The smallest eigenvalue equals 0.0004. Since this is positive, the NP problem is solvable.
A solution can be obtained by reduction to one interpolation point by applying Lemma 5 twice. First, reduce to the NP′ problem with two points:
a2 a3
b′
2 b′3
= 2 3
−0.6 −0.5714
Here b′i :=Mb1(bi)/Aa1(ai),i= 2,3. Second, reduce to the NP
′′ problem with only one point:
a3
b′′3 =
3 0.2174
9.3. NEVANLINNA’S ALGORITHM 157
Here b′′3 :=Mb′
2(b
′
3)/Aa2(a3).
Now solve the problems in reverse order. By Lemma 4 the solution of the NP′′problem is
G2(s) =M−b′′
3[G3(s)Aa3(s)],
where G3 is an arbitrary function inSc of ∞-norm≤1. Let’s takeG3(s) = 1, the simplest all-pass function. Then
G2(s) =
1.2174s−2.3478 1.2174s+ 2.3478.
The induced solution to the NP′ problem is
G1(s) =M−b′
2[G2(s)Aa2(s)] =
0.4870s2−7.6522s+ 1.8783 0.4870s2+ 7.6522s+ 1.8783. Finally, the solution to the NP problem is
G(s) =M−b1[G1(s)Aa1(s)] =
0.7304s3−4.0696s2+ 14.2957s−0.9391 0.7304s3+ 4.0696s2+ 14.2957s+ 0.9391.
Notice that the degree of the numerator and denominator ofGequals 3, the number of data points. In general, there always exists an all-pass solution of degree ≤n.
In our application of NP theory to the model-matching problem, the data
a1 · · · an
b1 · · · bn
will have conjugate symmetry; that is, if (ai, bi) appears, so will the conjugate pair (ai, bi). Then
we will want the solution Gto be real-rational instead of complex. Suppose that the data do have conjugate symmetry and that G is a solution inSc. It can be written uniquely as
G(s) =GR(s) +jGI(s),
where GR and GI both are real-rational. Then GR belongs to S and is also a solution to the NP
problem (the proof is left as an exercise).
Example 3 For the data 5 + 2j 5−2j
0.1−0.1j 0.1 + 0.1j
the NP problem is solvable (the smallest eigenvalue of the Pick matrix equals 0.0051). Starting from the all-pass function 1, Nevanlinna’s algorithm produces the all-pass solution
G(s) = (0.5268 + 0.1213j)s
2−(9 +j)s+ (47.1073 + 1.1410j) (0.5268−0.1213j)s2+ (9−j)s+ (47.1073−1.1410j).
LetGdenote the function obtained fromGby conjugating all coefficients. The functionGRis then
GR= 1 2(G+G), that is, GR(s) = 0.2628s4−30.6418s2 + 2217.7975 0.2923s4+ 9.7255s3 + 131.9119s2+ 850.2137s+ 2220.4012. This solution is not all-pass.