Acoustic Source Localization using Hough Transform

This is rather ugly solution out of conversation with MFTI students. Warm up for "Programming Kata". Scene: cheap hardwire assempled into robot with 3 mics separated 10 cm each. On "Clap" robot is running towards clapper. Simulation: .. code:: julia # Scene c_speed_m = 340.29 # m / s c_speed_cm= 34029 # cm/sec # Grid size - area of interest, my assumption our mics would not pickup sound of clap (chirp) further than 10 m. x_max = 1000 # cm y_max = 1000 # cm # Target position x_t = 751 y_t = 657 # Sensor position x1 = 035 y1 = 036 x2 = 045 y2 = 036 x3 = 035 y3 = 026 sigma_time=7.4E-9 # this is not correct value for this system - 7.4E-7 is GPS timing error synchronisation, assuming all mics connected by wire real_distance1=sqrt((x_t-x1)^2+(y_t-y1)^2) real_distance2=sqrt((x_t-x2)^2+(y_t-y2)^2) real_distance3=sqrt((x_t-x3)^2+(y_t-y3)^2) # time_2=real_distance2*10^3/c+(sigma_time*rand); # TIME of Arrival time_2=real_distance2/c_speed_cm; time_1=real_distance1/c_speed_cm; time_3=real_distance3/c_speed_cm; TDOA1 = time_2 - time_1 + (sigma_time*randn()); # sensor 1 is a reference TDOA2 = time_3 - time_1 + (sigma_time*randn()); TDOA3 = time_3 - time_2 + (sigma_time*randn()); TDOA4 = time_2 - time_3 + (sigma_time*randn()); TDOA can be obtained using ROOT-MUSIC or cross-correlation techniques. In the past I saw XOR based TDOA estimator. Applying Hough Transform approach (non linear, non-parametric etc estimator) with principle "throw everything on the wall, see what sticks", very bruteforce, not julia-like example, but very robust algorithm. .. code:: julia A_tdoa=zeros(x_max,y_max); sigma_range=c_speed_cm*sigma_time/(sqrt(2)) tic(); for x = 1:x_max for y = 1:y_max sigma_range=c_speed_cm*sigma_time/(sqrt(2)); r2=sqrt((x-x2)^2+(y-y2)^2); r1=sqrt((x-x1)^2+(y-y1)^2); delta_r=(r2-r1); p=(delta_r)-(c_speed_cm*TDOA1); l=exp(-0.5*p^2/sigma_range^2)/2; A_tdoa[x,y]=A_tdoa[x,y]+(l); # very ugly second tdoa measurement TDOA2 r3=sqrt((x-x3)^2+(y-y3)^2); r1=sqrt((x-x1)^2+(y-y1)^2); delta_r=(r3-r1); p=(delta_r)-(c_speed_cm*TDOA2); l=exp(-0.5*p^2/sigma_range^2)/2; A_tdoa[x,y]=A_tdoa[x,y]+(l); # very ugly third tdoa measurement TDOA3 r3=sqrt((x-x3)^2+(y-y3)^2); r2=sqrt((x-x2)^2+(y-y2)^2); delta_r=(r3-r2); p=(delta_r)-(c_speed_cm*TDOA3); l=exp(-0.5*p^2/sigma_range^2)/2; A_tdoa[x,y]=A_tdoa[x,y]+(l); r3=sqrt((x-x3)^2+(y-y3)^2); r2=sqrt((x-x2)^2+(y-y2)^2); delta_r=(r2-r3); p=(delta_r)-(c_speed_cm*TDOA4); l=exp(-0.5*p^2/sigma_range^2)/2; A_tdoa[x,y]=A_tdoa[x,y]+(l); end end toc(); using PyPlot; imshow(A_tdoa) maxp_value = maximum(A_tdoa,1) (value,y_est)= findmax(maxp_value) maxp_value_x = maximum(A_tdoa,2) (value_x,x_est)= findmax(maxp_value_x) rmse=sqrt((x_t-x_est)^2+(y_t-y_est)^2) Out of one measurement I manage to get fairly good results - 30 cm, super precision! Feel free to use or ask questions, I applied some "heuristic" - sqrt(2) in equation should be something like `sqrt(1-cos(sqrt(angle between sensor and point on hyperbolae)))`