% Tri-state bus sizing and topology design % Section 5.4, L. Vandenberghe, S. Boyd, and A. El Gamal % "Optimizing dominant time constant in RC circuits" % Original by Lieven Vandenberghe % Adapted for CVX by Joelle Skaf - 11/27/05 % Modified by Michael Grant - 3/8/06 % % We optimize a tri-state bus connecting six nodes ( The model for the bus % is shown in the paper, fig.16). The total wire area is sum_{i>j} lij*xij % The bus can be driven from any node. When node i drives the bus, the ith % switch is closed and the others are all open. Thus we really have six % different circuits, each corresponding to a given node driving the bus. % we require that the dominant time constant of each of the six drive % configuration circuits has dominant time constant less than Tmax. % The problem can be formulated with the following SDP: % minimize sum_{i>j}(x_ij*l_ij) % s.t. 0 <= xij <= wmax % Tmax*(G(x) + GE_kk) - C(x) >= 0 , 1 <=k<= 6 % The matrix E_kk is zero except for the kth diagonal element, which is 1. % % Circuit parameters % n=6; % number of nodes m=15; % number of wires beta = 0.5; % capacitance per segment is twice beta times xi*li alpha = 1; % conductance per segment is alpha times xi/li G0 = 1; % source conductance C0 = 10; % load capacitor wmax = 1; % upper bound on x % % Node positions % xpos = [ 0 1 6 8 -4 -1 ; 0 -1 4 -2 1 4 ] ; X11 = repmat(xpos(1,:),n,1); X12 = repmat(xpos(1,:)',1,n); X21 = repmat(xpos(2,:),n,1); X22 = repmat(xpos(2,:)',1,n); LL = abs(X11-X12) + abs(X21-X22); L = tril(LL); L = L(L>0); % % Construct the capacitance and conductance matrices % C(x) = C0 + w11 * C1 + w21 * C2 + ... % G(x) = G0 + w11 * G1 + w21 * G2 + ... % and we assemble the coefficient matrices together as follows: % CC = [ C0(:) C1(:) C2(:) ... ] % GG = [ G0(:) G1(:) G2(:) ... ] % CC = zeros(n,n,m+1); GG = zeros(n,n,m+1); CC(:,:,1) = C0 * eye(n); % segment capacitances and conductances k3 = 1; for k1 = 1 : 5, for k2 = k1 + 1 : 6, CC([k1,k2],[k1,k2],k3) = beta *[1, 0; 0,1]*L(k3); GG([k1,k2],[k1,k2],k3) = alpha*[1,-1;-1,1]/L(k3); k3 = k3 + 1; end end GG = reshape( GG, n*n, m+1 ); CC = reshape( CC, n*n, m+1 ); % % Compute points the tradeoff curve and the two desired points % % points on the tradeoff curve npts = 50; delays = linspace( 410, 2000, npts ); xdelays = [ 410, 2000 ]; xnpts = length(xdelays); areas = zeros(1,npts); xareas = zeros(1,xnpts); sizes = zeros(m,xnpts); for i = 1 : npts + xnpts, if i > npts, xi = i - npts; delay = xdelays(xi); disp( sprintf( 'Particular solution %d of %d (Tmax = %g)', xi, xnpts, delay ) ); else delay = delays(i); disp( sprintf( 'Point %d of %d on the tradeoff curve (Tmax = %g)', i, npts, delay ) ); end % % Construct and solve the convex model % cvx_begin sdp quiet variable x(m) variable G(n,n) symmetric variable C(n,n) symmetric minimize( L'*x ) G == reshape( GG * [ 1 ; x ], n, n ); C == reshape( CC * [ 1 ; x ], n, n ); for k = 1 : n, delay * G - C + sparse(k,k,delay,n,n) >= 0; end 0 <= x <= wmax; cvx_end if i <= npts, areas(i) = cvx_optval; else xareas(xi) = cvx_optval; sizes(:,xi) = x; % % Plot the step response % T = linspace(0,2*delay,1000); for inp = 1 : 6, figure(6*xi-5+inp); GQ = G + sparse(inp,inp,delay,n,n); A = -inv(C)*GQ; B = -A*ones(n,1); Y = simple_step(A,B,T(2),length(T)); hold off; plot(T,Y,'-'); hold on; ind=0; for j=1:size(Y,1), ind = max(min(find(Y(j,:)>=0.5)),ind); end tdom = max(eig(inv(GQ)*C)); elmore = max(sum((inv(GQ)*C)')); tthres = T(ind); plot( tdom * [1;1], [0;1], '--', ... elmore * [1;1], [0;1], '--', ... tthres * [1;1], [0;1], '--'); text(tdom, 0,'d'); text(elmore,0,'e'); text(tthres,0,'t'); ylabel('Voltage'); title(sprintf('Step response for solution %d, Tmax=%d, with switch %d is closed',xi,delay,inp)); end end end; % % Plot the tradeoff curve % figure(1) ind = isfinite(areas); plot(areas(ind), delays(ind)); xlabel('Area'); ylabel('Tdom'); title('Area-delay tradeoff curve'); hold on for k = 1 : xnpts, text( xareas(k), xdelays(k), sprintf( '(%d)', k ) ); end