Computationally Efficient Waveform Design in Spectrally Dense - PowerPoint PPT Presentation

Computationally Efficient Waveform Design in Spectrally Dense Environment Markus Yli-Niemi & Sergiy A. Vorobyov markus.yli-niemi@aalto.fi sergiy.vorobyov@aalto.fi July 5, 2018

Introduction ◮ Recently in radar systems waveform design in spectrally dense environment [1] has aroused noticeable interest ◮ Solution methods exist for the problem (see e.g. [2], [3]) but they are computationally inefficient ◮ When radar system operates at GHz level radar code dimension becomes large, need for computationally efficient solution methods ◮ Here we develop new computationally efficient method to design transmitter waveform in spectrally dense environment ◮ New method is based on ADMM algorithm [4] alongside Majorization-Minimization step [5] July 5, 2018 2/23

Problem formulation ◮ Similarly to [3], denote transmitted fast-time radar code vector by c and fast-time observation signal by v : c = ( c [ 1 ] , c [ 2 ] , ..., c [ N ]) T , v = α c + n , c , v ∈ C N , α ∈ C (1) ◮ Matched filtering v with filter h ∈ C N yields y = h H v . Write y = y s + y n , where y s = α h H c and y n = h H n . SINR is given as: SINR = | y s | 2 | y n | 2 = | α | 2 | h H c | 2 = | α | 2 | h H c | 2 (2) h H nn H | h H n | 2 h ���� = M ◮ To maximize SINR w.r.t. h , we choose h = M − 1 c , which yields SINR = | α | 2 c H M − 1 c July 5, 2018 3/23

Problem formulation ◮ Introduce constrained bandwidths { Ω k } k ∈{ 1 , 2 ,..., K } , where � � f k 1 , f k Ω k = . The energy c radiates to constrained 2 bandwidths is (see e.g. [3]): � K � |F N { c }| 2 df = c H R I c , w k (3) Ω k k = 1 where { w k } K k = 1 are non-negative weights, F N { c } stands for the discrete-time Fourier transform of c given as F N { c } � � N k = 1 c [ k ] e − j 2 π kf , and R I � � K k = 1 w k R k I with 2 ( m − l ) − e j 2 π f k [ R k I ] m , l = ( e j 2 π f k 1 ( m − l ) ) / e j 2 π ( m − l ) , if m � = l , and [ R k I ] m , l = f k 2 − f k 1 , if m = l . July 5, 2018 4/23

Problem formulation ◮ If radar code energy � c � 2 is unit constrained and required to be in similarity region with reference code c 0 alongside radiation energy constraint c H R I c ≤ E I , SINR maximization problem can be written:  | α | 2 c H M − 1 c max (4a)    c   � c � 2 = 1 s.t. : (4b) P 1 :  c H R I c ≤ E I  (4c)    � c − c 0 � 2 ≤ ǫ (4d) July 5, 2018 5/23

Problem formulation ◮ P 1 is equal to:  − c H Rc min (5a)    c   � c � 2 = 1 s.t. : (5b) P ( 1 ) : 1   c H R I c ≤ E I (5c)    � c − c 0 � 2 ≤ ǫ (5d) where c , c 0 ∈ C N and R I , R = M − 1 ∈ C N × N July 5, 2018 6/23

Majorization-Minimization step ◮ Due to independence of real and imaginary components we can write c , c 0 , R and R I as: � Re { R } � � Re { c } � � Re { c 0 } � − Im { R } R = , c = and c 0 = . Im { R } Re { R } Im { c } Im { c 0 } ◮ Let us use use surrogate Q = µ I − R � 0, µ > 0 to upper-bound objective. We get real-valued optimization problem P 2 :  c T Qc min (6a)    c   � c � 2 = 1 s.t. : (6b) P 2 :  c T R I c ≤ E I  (6c)    � c − c 0 � ≤ ǫ (6d) where c , c 0 ∈ R 2 N and Q , R I ∈ R 2 N × 2 N July 5, 2018 7/23

Apply ADMM to P 2 ◮ To allow separability of c T Qc , let us introduce slack variable z with constraint c = z . Augmented Lagrangian L ρ ( c , z , λ ) for minimization problem min c c T Qc s.t.: c = z : L ρ ( c , z , λ ) = c T Qc + λ T ( c − z ) + ρ 2 � c − z � 2 . (7) ◮ ADMM-steps for P 2 :  c k + 1 = arg min L ρ ( c , z k , λ k ) (8a)    c z k + 1 = arg min L ρ ( c k + 1 , z , λ k ) (8b)   z  λ k + 1 = λ k + ρ ( c k + 1 − z k + 1 ) , (8c) ◮ Next c -variable update and z -variable update are solved. July 5, 2018 8/23

c -variable update ◮ c -variable update (8a) can be written as: � � c T Qc + ( λ − ρ z ) T c c k + 1 = arg min L ρ ( c , z k , λ k ) = arg min c c | s.t. � c � 2 = 1, � c − c 0 � 2 ≤ ǫ. = arg min h ( c ) (9) c ◮ Objective function h ( c ) is continuously differentiable and ∇ c h is L -Lipschitz continuous. To minimize h ( c ) we use gradient descent: �� Q + Q T � � c k + 1 = c k − 1 c k + ( λ − ρ z ) , (10) L where Lipschitz constant can be found by noticing: � � � Q + Q T � � � |∇ c h ( κ ) − ∇ c h ( c ) | = ( κ − c ) � ≤ L | κ − c | � � � � � � � 2 N � � � Q [ i , p ] + Q T � � ⇒ ≤ L , ∀ i = 1 , · · · , 2 N � � [ i , p ] � � p = 1 July 5, 2018 9/23

c -variable update ◮ Gradient descent yields updated c that has � c � 2 2 � = 1 and possibly � c − c 0 � ≥ ǫ . ◮ Denote Θ = { c ∈ R 2 N | � c � 2 = 1 and � c − c 0 � 2 ≤ ǫ , for some c 0 ∈ R 2 N } ◮ Cheap way to project c back to unitary region is to divide updated c by its L 2 -norm: ˆ c k + 1 = c k + 1 / � c k + 1 � (11) July 5, 2018 10/23

c -variable update Rotation of c towards c 0 α * e ✲ c 0 〛 2 ≤ ϵ 〚 c ◮ Next ˆ c k + 1 is rotated to region Θ with α * e c 0 � c 〚 c 〛 2 ✯ steps introduced in Algorithm 1. ❂ ✶ Algorithm 1: Rotate c toward c 0 as long as region � c − c 0 � ≤ ǫ is reached 1 function RotateVector ( c , c 0 , α ′ , ǫ ) ; c c c 0 : c , c 0 , α ′ and ǫ ✁ Input Output : c 2 while � c − c 0 � > ǫ do c = c 0 − proj c ( c 0 ) = c 0 − c H 0 , c � � c � 2 c ; 3 c � , c ∗ = c + α ′ e , � c e = 4 � � c ∗ c = � c ∗ � ; 5 end July 5, 2018 11/23

c -variable update ◮ The combination of steps (10), (11) and Algorithm 1 can be shown to be solution steps to projected gradient step for problem min c h ( c ) subject to c ∈ Θ :  y k + 1 = c k − 1   L ∇ h ( c k ) (12a) � �  � . � y k + 1 − c  c k + 1 = min (12b) c ∈ Θ ◮ By using angular coordinates φ ∈ R 2 N − 1 step (12b) can be written as: � φ k + 1 = arg min � φ ∗ − φ � (13a) φ ∈ Ω c k + 1 = c ( φ k + 1 ) . (13b) � � φ ∈ R 2 N − 1 | � c ( φ ) − c 0 ( φ ) � 2 ≤ ǫ where Ω = and φ ∗ = arg min φ h ( c ( φ )) . July 5, 2018 12/23

z -variable update ◮ z -variable update (8b) can be written as: z k + 1 = arg min L ρ ( c k + 1 , z , λ k ) z � 2 � c − z � 2 � λ T ( c − z ) + ρ = arg min z �� 2 � � � � � z − ( c + 1 � � | s.t. z T R I z ≤ E I . (14) = arg min ρ λ ) � z ◮ Lagrangian for (14) is given as: � � 2 � � � z − ( c + 1 � � + γ ( z T R I z − E I ) . L ( z , γ ) = ρ λ ) (15) � July 5, 2018 13/23

z -variable update ◮ Karush-Kuhn-Tucker (KKT) conditions for the minimization problem (14):  ∇ z L ( z ∗ , γ ∗ ) = 0 (16a)     γ ∗ ≥ 0  (16b)   γ ∗ (( z ∗ ) T R I z ∗ − E I ) = 0 (16c)    ( z T R I z − E I ) ≤ 0  (16d)    ∇ zz L ( z ∗ , γ ∗ ) � 0 , (16e) ◮ By (16a) and (16c): ∇ z L ( z ∗ , γ ∗ ) = 0 ⇒ ( I + γ ∗ R I ) z ∗ = c + 1 ρ λ , (17) ( z ∗ ) T R I z ∗ − E I = 0 , (18) where z ∗ and γ ∗ denotes critical points of Lagrangian L ( z , γ ) . July 5, 2018 14/23

z -variable update ◮ Now (17) can be written as iteration step (19): � � c + 1 z k + 1 = ( I + γ k + 1 R I ) − 1 ρ λ � � � � 2 N � γ k + 1 σ i c + 1 p i p T = I + . (19) ρ λ i 1 + γ k + 1 σ i i = 1 ◮ γ k + 1 > 0 can be found as the solution to (18): 2 N � a i σ i z T k + 1 R I z k + 1 = E I ⇔ ( 1 + γσ i ) 2 − E I = 0 (20) i = 1 where a i = ( p T i ( c + 1 ρ λ )) 2 , σ i is i ’th eigenvalue and p i corresponding eigenvector of R I . Equation (20) can be efficiently solved by using Newton’s method. July 5, 2018 15/23

Proposed algorithm ◮ Collect c and z -variable updates to get final algorithm: Algorithm 2: MM-algorithm 1 function MM ( Q , c 0 , R I , E I , ǫ, K ′ ) ; : Q = µ I − R � 0, c 0 , R I , E I , ǫ and K ′ Input Output : c 2 Initialize c , z and λ ; 3 for k = 1 , k ≤ K ′ , k ++ do �� Q + Q T � � c k + 1 = c k − 1 ˆ c k + ( λ − ρ z ) ; 4 L ˆ c k + 1 � c k + 1 = c k + 1 � ; 5 � ˆ c k + 1 = RotateVector ( � c k + 1 , c 0 , α, ǫ ) ; 6 � 2 N a i σ i Solve ( 1 + γσ i ) 2 − E I = 0 for γ k + 1 > 0; 7 i = 1 � � � � 2 N � 1 + γ k + 1 σ i p i p T c + 1 z k + 1 = I + γ k + 1 σ i ; ρ λ 8 i i = 1 λ k + 1 = λ k + ρ ( c k + 1 − z k + 1 ) ; 9 10 end July 5, 2018 16/23

Time-Complexity graph 10 15 n 2 n 3.5 Algorithm 2 n n*log(n) 10 10 Runtime (s) 10 5 10 0 10 -5 10 0 10 1 10 2 10 3 10 4 10 5 Problem dimension July 5, 2018 17/23