5

Can someone point me to a webpage or other resource that shows how to analytically solve the beamformer Unconstrained Array Gain expression in Henry Cox's 1987 IEEE paper "Robust Adaptive Beamforming"?

$$ \max_{\mathbf{w}} \frac{|\mathbf{w}^H\mathbf{d}|^2}{\mathbf{w}^H\mathbf{Q}\mathbf{w}} $$

Cox says:

The well-known solution is $\mathbf{w} = \alpha\mathbf{Q}^{-1}\mathbf{d}$

I'd just like to better understand this by learning how to derive this myself.

Matt L.
  • 89,963
  • 9
  • 79
  • 179
j03y_
  • 111
  • 5
  • Not 100% sure, but perhaps they calculated the derivative of the expression and validated where the derivative equals 0. – Ben Dec 29 '20 at 19:34
  • The solution of this equality constrained quadratic programming problem is possible that uses a pseudoinverse, rather than simple inverse, of matrix Q. For general equality constrained QP problems, this extended solution has a benefit of relaxing the requirement for matrix Q to be strictly positive definite. Don't know if this generalization may be useful for beamformers. – V.V.T Jan 01 '21 at 10:27

3 Answers3

5

A common way is to make use of the Schwarz inequality. First note that:

$$\frac{|w^Hd|^2}{w^HQw} = \frac{|w^HQ^{1/2}Q^{-1/2}d|^2}{w^HQw}$$

Using the Schwarz inequality on the numerator: $$\frac{|w^HQ^{1/2}Q^{-1/2}d|^2}{w^HQw} \leq \frac{(w^HQw)(d^HQ^{-1}d)}{w^HQw} = d^HQ^{-1}d$$

Thus, $$\frac{|w^HQ^{1/2}Q^{-1/2}d|^2}{w^HQw} \leq d^HQ^{-1}d$$

From this, it can easily be seen that the beamformer that achieves equality is $w = \alpha Q^{-1}d$.

Sources:

Gillespie
  • 1,767
  • 4
  • 26
  • 1
    To show that this $w$ maximizes SNR, simply substitute $w = Q^{-1}d$ into the inequality above. First, expand the numerator of the left side of the inequality, noting that $Q$ is Hermitian:

    $$\frac{|w^HQ^{1/2}Q^{-1/2}d|^2}{w^HQw} = \frac{(w^HQ^{1/2}Q^{-1/2}d)^H(w^HQ^{1/2}Q^{-1/2}d)}{w^HQw} = \frac{d^Hww^Hd}{w^HQw} $$

    Then, substituting $w = Q^{-1}d$: $$ \frac{d^Hww^Hd}{w^HQw} = \frac{d^HQ^{-1}dd^HQ^{-1}d}{d^HQ^{-1}QQ^{-1}d} = \frac{d^HQ^{-1}dd^HQ^{-1}d}{d^HQ^{-1}d} = d^HQ^{-1}d $$

    Thus, this definition of $w$ achieves the maximum possible SNR indicated by the Schwartz inequality.

    – Gillespie Feb 12 '21 at 18:37
3

You can solve such a problem using the method of Lagrange multipliers. First note that maximizing the expression in your question is equivalent to minimizing the inverse function:

$$\min_{\mathbf{w}}\frac{\mathbf{w}^H\mathbf{Q}\mathbf{w}}{|\mathbf{w}^H\mathbf{d}|^2}\tag{1}$$

Next note that the solution of $(1)$ is invariant to scaling of $\mathbf{w}$, i.e., replacing $\mathbf{w}$ by $c\cdot\mathbf{w}$ in $(1)$ with an arbitrary scalar constant $c$ will not change the value of the function. So we may as well use a scaling such that $\mathbf{w}^H\mathbf{d}=1$ is satisfied. This scaling corresponds to a unity response for the desired signal. With this constraint, problem $(1)$ can be reformulated as

$$\min_{\mathbf{w}}\mathbf{w}^H\mathbf{Q}\mathbf{w}\qquad\textrm{s.t.}\qquad \mathbf{w}^H\mathbf{d}=1\tag{2}$$

We can solve $(2)$ using the method of Lagrange multipliers by minimizing

$$\mathbf{w}^H\mathbf{Q}\mathbf{w}-\lambda(\mathbf{w}^H\mathbf{d}-1)\tag{3}$$

Formally taking the derivative of $(3)$ with respect to $\mathbf{w}^H$ and setting it to zero gives

$$\mathbf{w}=\lambda\mathbf{Q}^{-1}\mathbf{d}\tag{4}$$

The constraint in $(2)$ is satisfied for

$$\lambda=\frac{1}{\mathbf{d}^H\mathbf{Q}^{-1}\mathbf{d}}\tag{5}$$

From $(4)$ and $(5)$ we finally obtain

$$\mathbf{w}=\frac{\mathbf{Q}^{-1}\mathbf{d}}{\mathbf{d}^H\mathbf{Q}^{-1}\mathbf{d}}\tag{6}$$

Note that the scaling in $(6)$ is optional and the general solution is given by $(4)$.

Matt L.
  • 89,963
  • 9
  • 79
  • 179
  • Lagrange multipliers are a great trick! Years ago, I had always thought they were a little sketchy, but I've used them to solve several constrained optimization problems. And they work well. – Peter K. Dec 29 '20 at 23:17
  • "Formally taking the derivative of (3) with respect of $w^H$": do you suggest that the Lagrangian (3) is the function of three independent variables, $w, w^H,$ and $λ$? If so, you can likewise take the derivative with respect to $w$ and, setting it to zero, arrive at another minimization equation $w^HQ=0$, can't you? – V.V.T Jan 01 '21 at 10:33
  • @V.V.T: No, this wouldn't make sense, but I understand that the version I used also looks like it didn't make sense ... This is related to the Wirtinger derivative. Please take a look at this related answer, and the other answer on math.SE that it links to. – Matt L. Jan 01 '21 at 12:11
  • I'm quite familiar with "intricacies" of complex differentiation. Struggling to connect the derivation of formula for MOE beamformer weights and the equality constrained QP problems (like, e.g., in https://www.cis.upenn.edu/~cis515/cis515-20-sl15.pdf, Def. / Prop. 18.3), I wonder if a factor 1/2 in definition of a real quadratic form matrix (page #821, or 3/39) may be offset for complex-valued matrices rightly as manifestation of these intricacies. Would you please comment on relevance of this explanation? – V.V.T Jan 01 '21 at 13:36
  • @V.V.T: Maybe it would be a good idea to formulate this as a full question, either here or maybe even better on math.SE. – Matt L. Jan 01 '21 at 14:24
2

First, a sketch of the solution for the maximum SINR beamformer problem $$ \text{max}_{\mathbf{w}} \frac{|\mathbf{w}^H\mathbf{d}|^2}{\mathbf{w}^H\mathbf{Q}\mathbf{w}} $$ Start with writing down a functional $$ \mathbf{w}^H\mathbf{Q}\mathbf{w} $$ to be minimized, and a set of constraints. Indeed, the weight vectors w and wH are considered the two independent set of variables when taking derivates with respect to these variables; therefore, the output signal energy, typically written as a squared modulus of the weights-signals coproduct, has to be written down as an analytic function, without computing the norm that takes the square root: $$ |\mathbf{w}^H\mathbf{d}|^2 = \mathbf{w}^H\mathbf{d}·\mathbf{d}^H\mathbf{w} $$ The resulting set of linear constraints is $$ \mathbf{w}^H\mathbf{d} = c \\ \mathbf{d}^H\mathbf{w} = c^* $$ and we have to write down a Lagrangian with two Lagrange multipliers, λ and μ: $$ \mathbf{w}^H\mathbf{Q}\mathbf{w}-λ(\mathbf{w}^H\mathbf{d}-c)-μ(\mathbf{d}^H\mathbf{w}-c^*) $$ Taking the two derivatives of the Lagrangian -- the first, with respect to w, and the second, with respect to wH -- we obtain the expressions for λ and μ, and, substituting these to the constraint expressions, finally arrive at the formula for weights: $$ \mathbf{w}=c\frac{\mathbf{Q}^{-1}\mathbf{d}}{\mathbf{d}^H\mathbf{Q}^{-1}\mathbf{d}} $$ To my surprize, searching a web for "a webpage or other resource that shows how to analytically solve the beamformer" per OP's request, I could find only curtailed, flawed versions of this formula's derivation, a typical document being the course notes Optimal Beamforming, a detailed and useful introduction into the subject in all the other aspects. I even suspect that the OP posted the question with the purpose to broadcast this omission of the learning resource (excuse my awkward attempt to joke).

For now, I can only recommend the learning material on general linear constraint quadratic programming to students interested in the optimal beamforming. For example, refs. https://www.math.uh.edu/~rohop/fall_06/Chapter3.pdf and https://www.cis.upenn.edu/~cis515/cis515-20-sl15.pdf . Only real-valued quadratic forms are considered in these documents, but the main results can be generalized to the complex domain.

V.V.T
  • 1,729
  • 8
  • 8
  • Thank you to all who responded. Very helpful. I had no idea that and could be treated as independent variables. Many of the details here seem to be above my standard math background from electrical engineering. – j03y_ Jan 02 '21 at 05:52
  • Something else I just noticed - isn't the Gain expression required to be a real number, yet the denominator is complex? – j03y_ Jan 02 '21 at 15:11
  • Oh - I just figured it out - When Q is Hermitian, the denominator does reduce to a real number. I proved this to myself both by hand with a 2x2 Q and 2x1 w, and in Octave as well. – j03y_ Jan 02 '21 at 22:03
  • "seem to be above my standard math background from electrical engineering": Wait, your questions on SE reveal, least to say, your intuitive sense of crucial issues arisen from application of math tools (SVD vs. Colesky, remember?). – V.V.T Jan 02 '21 at 23:48
  • I know a few radically-minded (and successful) engineers declaring that EE is "a branch of applied mathematics". Certainly it is not, being an autonomous discipline with a distinct field of study, uses, and tooling; but it certainly does use advanced mathematics and, in reciprocal motion, contributes to a development of math theories (e.g., see https://math.stackexchange.com/questions/226818/is-there-any-abstract-theory-of-electrical-networks). – V.V.T Jan 02 '21 at 23:48
  • "I proved this ... both by hand ... and in Octave": an efficient, although not universal, way of learning and research. "had no idea that and could be treated as independent variables" notice that this treatment also requires two sets of Lagrange multipliers in Lagrange equations, this is the gist of my answer. – V.V.T Jan 02 '21 at 23:49