Single-input single-output transceiver optimization - Signal Processing and Optimization for Transceiver Systems - page 337

for all ω, where

ω s = 2 π
T
See the review of sampling theory given in Appendix G.

10.2.2.A Formulating the optimization problem
The integral in Eq. (10.7) can be written as A ( ω ) B ( ω ) dω/ 2 π, where

A ( ω )= S 1 / 2

qq ( jω )

( ω )= |

H c ( jω ) |

, (10 . 15)

| H ( jω ) | T

are non-negative quantities. Given A ( ω ) ≥ 0, find a function B ( ω ) ≥ 0such

that

ψ = ∞
−∞

A ( ω ) B ( ω ) dω
2 π

(10 . 16)

is minimized under the constraint
∞

B ( ω + kω s )=1 (10 . 17)

k = −∞

for all ω. Since the left-hand side is periodic, it is su cient to ensure the condition
for 0

≤ ω<ω s . For convenience we rewrite the objective function as

ψ = ω s
0

∞

A ( ω + kω s ) B ( ω + kω s ) dω
2 π

(10 . 18)

k = −∞

Since A ( ω ) is non-negative, the non-negative B ( ω ) which minimizes Eq. (10.18)
should be such that, for each frequency ω 0 in 0 ≤ ω 0 <ω s , the sum

∞

A ( ω 0 + kω s ) B ( ω 0 + kω s ) (10 . 19)

k = −∞

is minimized under the constraint (10.17). Let k 0 be an integer such that

A ( ω 0 + k 0 ω s ) ≤ A ( ω 0 + kω s )

for all integer k.

♠ Lemma 10.1 .If B ( jω ) is chosen such that

B ( ω 0 + kω s )= 1for k = k 0
0

(10 . 20)

otherwise,

then Eq. (10.19) is minimized, and the constraint (10.17) is satisfied as well. ♦

Proof. Let a k be a fixed set of non-negative numbers arranged such that

a 0 ≤ a 1 ≤ a 2 ... (10 . 21)