+| % Files: | PlotFresnelInfiniteWedgeRectangular...PlotFresnelInfiniteWedgeRectangular...
+ | PlotFresnelInfiniteWedgeRectangular...
+ | PlotFresnelInfiniteWedgeRectangular...
+ | PlotFresnelInfiniteWedgeRectangul...
+ | PlotFresnelInfiniteWedgeRe...
+ | PlotFresnelInfiniteWed...
+ | PlotFresne...
+ | PlotFr...
+ | Fraunho...
+ | Baffe...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
+ | Baffl...
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+ | Baffl...
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+
+
+
+
+| % Files: |
+PlotFresnelInfiniteWedgeRectangul...a) |
+PlotFresnelInfiniteWedgeRectangul...b) |
+PlotFresnelInfiniteWedgeRectangul...c) |
+PlotFresnelInfiniteWedgeRectangul...d) |
+PlotFresnelInfiniteWedgeRectangul...e) |
+PlotFresnelInfiniteWedgeRectangul...f) |
+PlotFresnelInfiniteWedgeRectangul...g) |
+PlotFresnelInfiniteWedgeRectangul...i) |
+PlotFresnelInfiniteWedgeRectangul...j) |
+PlotFresnelInfiniteWedge Rectangul...k) |
+PlotFresnelInfiniteWedge Rectangul...l) |
+PlotFresnelInfiniteWedge Rectangul...m) |
+PlotFresnel Infini...n) |
+FRENSEL INTEGRATION
% Files: |
+PlotFresnelInfiniteWedge Rectangul...a) |
+PlotFresnel Infini...n) |
+FRENSEL INTEGRATION
% Files: |
+PlotFr...a) |
+Fraunho...b) |
+Fraunho...c) |
+Fraunho...d) |
+Fraunho...e) |
+Fraunho...f) |
+Fraunho...i) |
+Fraunho...j) |
+Fraunho...k) |
+Fraunho...l) |
+Fraunho...m) |
+FRAUNHOFER INTEGRATION
% Files: |
+PlotFr...a) |
+Fraunho...b) |
+Fraunho...c) |
+Fraunho...d) |
+Fraunho...e) |
+Fraunho...f) |
+Fraunho...i) |
+Fraunho...j) |
+Fraunho...k) |
+Fraunho...l) |
+Fraunho...m) |
+FRAUNHOFER INTEGRATION
% Files: |
+
+
+
+
+| % Files: |
+PlotFr...a) |
+Fraunho...b) |
+Fraunho...c) |
+Fraunho...d) |
+Fraunho...e) |
+Fraunho...f) |
+Fraunho...i) |
+Fraunho...j) |
+Fraunho...k) |
+Fraunho...l) |
+
+
+
+
+| % Files: |
+PlotFr...a)(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=inf.)t=inf.(t=INFINE) (THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS, SO THE RESULTS ARE COPIOUS). (THE FIRST MODEL IS USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THE TWO MODELS. THE RESULTS OF THE FIRST MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE SECOND MODEL. THE RESULTS OF THE SECOND MODEL ARE USED AS EXAMPLE TO BE COMPARED WITH THE FIRST MODEL. THERE ARE NO DIFFERENCES BETWEEN THI
+```matlab
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+end
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab
+
+% %% Modello con direzione elettronica e direzione radioattiva
+
+```matlab}
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzione elettronica e direzione radioattiva.
+
+%% Modello con direzioni...
+
+```
+
+### Modelli di diffrerenza per due modi di misurare la diffrazione:
+
+$$ \text{Modelli di diffrerenza per due modi di misurare la diffrazione} $$
+---PAGE_BREAK---
+
+% Execution time: max 3 min.
+
+UseModel = 0;
+
+% UseModel = 0 Discard models.
+% UseModel = 1 Fresnel semi-infinite baffle.
+% UseModel = 2 Fresnel rectangular baffle.
+% UseModel = 3 Fraunhofer semi-infinite baffle.
+
+if (UseModel ==1)
+
+% === FRESNEL - SEMI-INFINITE BAFFLE ===============================
+
+MainTitle = 'Fresnel integration - Loudspeaker baffle at 0 deg';
+DataFile='4-RectangularBaffle\FreqRespCentre000.dat';
+
+A = 0.10; % Horisontal dimension (m); Width = 2A
+D = 1.30; % Observation distance (m).
+th = 0; % Observation angle (degrees).
+
+% --- PARAMETERS
+
+ds = A*cos(pi/2-pi*th/180); % Distance to source (m).
+dp = D-ds; % Distance to observer (m).
+y1 = -A*sin(pi/2-pi*th/180); % Wedge height (m).
+
+% --- MODEL
+
+HA = 0.5; % Aperture constant (large aperture).
+L = ds*dp/(ds+dp); % Fresnel length (m).
+h = sqrt(2*f/(c*L)); % Scaling factor (1/m).
+
+disp('...1/2');
+Cy = mfun('FresnelC',h*y1);
+disp('...2/2');
+Sy = mfun('FresnelS',h*y1);
+p0 = HA*(1-i)*(0.5-Cy-i*(0.5-Sy));
+
+elif (UseModel == 2)
+
+% === FRESNEL - RECTANGULAR BAFFLE ===============================
+
+MainTitle = 'Fresnel integration - Rectangular baffle at 0 deg, ds = 20 cm';
+DataFile='4-RectangularBaffle\FreqRespCentre000.dat';
+
+A = 0.10; % Horisontal dimension (m); Width = 2A
+B = 0.17; % Vertical dimension (m); height = 2B.
+r0 = 1.30; % Observation distance (m).
+th = 0; % Observation angle (degrees).
+
+% --- PARAMETERS
+
+ds = 0.20; % Source to aperture (m).
+dp = r0*cos(pi*th/180); % Distance to observer (m).
+x1 = -A+r0*sin(pi*th/180); % Aperture coordinates (m).
+x2 = A+r0*sin(pi*th/180);
+y1 = -B;
+y2 = B;
+
+% --- MODEL
+
+s = sqrt(A*B)/ds; % Aperture size.
+HA = (1/2)*sqrt((1+s^2)/(sqrt(1+s^2)+1)); % Aperture correction.
+L = ds*dp/(ds+dp); % Fresnel length (m).
+h = sqrt(2*f/(c*L)); % Scaling factor (1/m).
+disp('...1/8'); % Display step number.
+
+Cx1 = mfun('FresnelC',h*x1);
+disp('...2/8');
+Cx2 = mfun('FresnelC',h*x2);
+disp('...3/8');
+Sx1 = mfun('FresnelS',h*x1);
+disp('...4/8');
+Sx2 = mfun('FresnelS',h*x2);
+disp('...5/8');
+Cy1 = mfun('FresnelC',h*y1);
+disp('...6/8');
+Cy2 = mfun('FresnelC',h*y2);
+disp('...7/8');
+Sy1 = mfun('FresnelS',h*y1);
+disp('...8/8');
+Sy2 = mfun('FresnelS',h*y2);
+p0 = HA*(Cx1-Cx2-i*(Sx1-Sx2)).*(Cy1-Cy2-i*(Sy1-Sy2));
+
+elif (UseModel == 3)
+
+% === FRAUNHOFER - RECTANGULAR BAFFLE ===============================
+---PAGE_BREAK---
+
+MainTitle = 'Fraunhofer integration - Rectangular baffle at 30 deg';
+DataFile='4-RectangularBaffle\FreqRespCentre030.dat';
+
+% --- INPUT PARAMETERS
+
+th = 030; % Observation angle (deg).
+r0 = 0.41; % Obsevation distance (m).
+B = 0.10; % Baffle half-width (m).
+
+% --- INTERNAL PARAMETERS
+
+h = -B*sin(pi/2-th*pi/180) % Edge height (observer = 0).
+zP = r0-B*cos(pi/2-th*pi/180) % Edge to observer distance.
+
+amax = 20; % Horisontal lintegration limits.
+bmax = 20; % Vertical upper integration limit.
+bmin = h/zP % Vertical lower integration limit.
+M = 1200; % Number points for alfa.
+N = 1200; % Number points for beta.
+da = amax/M; % Increment in alfa (note: amin = -amax).
+db = (bmax-bmin)/N; % Increment in beta (note: asymmetric).
+FP = 100; % Number of frequency points calculated.
+
+% --- INTEGRAL
+alfa = -amax:da:(amax*(M-1)/M); % Vector containing alfa-values.
+
+% --- Frequency loop
+
+for g=1:6401 % Index for the frequency vector.
+f(g)=2*(g-1); % Frequency (Hz).
+if (mod(g,FP) == 1) % Calculate for 1, FP+1, 2FP+1, ...
+disp(g); % Output the loop-counter value.
+ikz = i*2*pi*f(g)*zP/c; % Frequency-distance.
+% --- Summation loop
+for n=1:N % Index for beta with N elements.
+beta = (n-1)*db+bmin; % Beta-value.
+abv = sqrt(1+alfa.^2+beta^2); % Vector.
+wf = sqrt(1+(alfa/3).^2+(beta/3)^2); % Weight.
+% --- ALFA summation (2M elements)
+alfasum(n) = sum(exp(-ikz*abv)./(abv.*wf));
+end
+% --- BETA summation (N elements)
+p0(g) = abs(ikz*amax*(bmax-bmin)*sum(alfasum)/(M*N*2*pi));
+if (g > FP)
+ for h=(g-FP+1):(g-1)
+ p0(h) = ((g-h)*p0(g-FP) + (h+FP-g)*p0(g))/FP;
+ end
+end
+end
+end
+
+% =====================================================================
+
+% PLOT THE MODEL
+
+% =====================================================================
+
+% Model data is stored in "p0" as amplitude values from 0 Hz to the
+% upper limit at 12.8 kHz with 2 Hz spacing. Data is output with a
+% logarithmic frequency axis showing the range from 60 Hz to 10 kHz,
+% set through the parameters fmin and fmax, and with max and min
+% sound pressure levels set through the parameters pmin and pmax.
+
+disp('Plotting ...');
+FreqSpec = 20*log10(abs(p0)); % Change format and make decibel.
+figure(1); % Prepare output picture #1.
+semilogx(f,FreqSpec,'b-',); % Plot versus frequency.
+axis([fmin, fmax, pmin, pmax]); % Set freq-axis and decibel-axis.
+title(MainTitle); % Set selected title.
+xlabel('Frequency (Hz)'); % Set freq-axis label.
+ylabel('Amplitude (dB)'); % Set decibel-axis.
+hold on; % Keep figure ready for more plots.
+
+% =====================================================================
+
+% PLOT THE MEASUREMENT
+
+% =====================================================================
+
+% The average of four files are defined as the reference for output.
+% The files are read, data is converted from real and imaginary to
+% magnitude and the average is computed.
+---PAGE_BREAK---
+
+% The "fopen" command assumes data files containing:
+
+% (1) Line number, (2) Frequency, (3) Real value, (4) Imag value.
+
+% DataFile is converted to format 6401 rows with the above 4 columns.
+
+% Variable "MeasRefX" contains 6401 rows with the amplitude value.
+
+% Variable "MeasRef" contains the resulting reference spectrum.
+
+% --------------------------------------------------------------------
+
+% --- Get measurements source file.
+fid=fopen(DataFile); d=fscprintf fid, '%e %e %e %e', [4 inf]); fclose(fid);
+d=d'; MeasSource=sqrt(d(:,3).^2 + d(:,4).^2);
+
+% --- Offset measurement for rectangular baffle to 1.41 m distance.
+if (DataFile == '4-RectangularBaffle\FreqRespCentre000.dat')
+ corr = -0.71;
+elseif (DataFile == '4-RectangularBaffle\FreqRespCentre030.dat')
+ corr = -0.57;
+elseif (DataFile == '4-RectangularBaffle\FreqRespCentre060.dat')
+ corr = -0.22;
+elseif (DataFile == '4-RectangularBaffle\FreqRespCentre090.dat')
+ corr = 0.24;
+elseif (DataFile == '4-RectangularBaffle\FreqRespCentre120.dat')
+ corr = 0.68;
+elseif (DataFile == '4-RectangularBaffle\FreqRespCentre150.dat')
+ corr = 0.99;
+elseif (DataFile == '4-RectangularBaffle\FreqRespCentre180.dat')
+ corr = 1.19;
+elseif (DataFile == '4-RectangularBaffle\FreqRespOffset.dat')
+ corr = -0.71;
+elseif (DataFile == '4-RectangularBaffle\FreqRespAsym.dat')
+ corr = -0.71;
+elseif (DataFile == '3-CircularBaffle\FreqRespFront00cm.dat')
+ corr = -0.71;
+elseif (DataFile == '3-CircularBaffle\FreqRespFront03cm.dat')
+ corr = -0.71;
+elseif (DataFile == '3-CircularBaffle\FreqRespFront06cm.dat')
+ corr = -0.71;
+elseif (DataFile == '3-CircularBaffle\FreqRespFront09cm.dat')
+ corr = -0.71;
+elseif (DataFile == '3-CircularBaffle\FreqRespFront12cm.dat')
+ corr = -0.71;
+elseif (DataFile == '3-CircularBaffle\FreqRespFront15cm.dat')
+ corr = -0.71;
+elseif (DataFile == '3-CircularBaffle\FreqRespFront18cm.dat')
+ corr = -0.71;
+elseif (DataFile == '3-CircularBaffle\FreqRespFront25cm.dat')
+ corr = -0.71;
+else
+ corr = 0.00;
+
+end
+
+corr
+
+% --- Get measurements reference files.
+ref1='5-CircularBaffle\FreqRespInitial3.dat';
+ref2='5-CircularBaffle\FreqRespInitial4.dat';
+ref3='5-CircularBaffle\FreqRespSource000.dat';
+ref4='5-CircularBaffle\FreqRespSource360.dat';
+fid=fopen(ref1); r1=fscprintf fid, '%e %e %e %e', [4 inf]); fclose(fid);
+r1=r1'; MeasRef1=sqrt(r1(:,3).^2 + r1(:,4).^2);
+fid=fopen(ref1); r2=fscprintf fid, '%e %e %e %e', [4 inf]); fclose(fid);
+r2=r2'; MeasRef2=sqrt(r2(:,3).^2 + r2(:,4).^2);
+fid=fopen(ref1); r3=fscprintf fid, '%e %e %e %e', [4 inf]); fclose(fid);
+r3=r3'; MeasRef3=sqrt(r3(:,3).^2 + r3(:,4).^2);
+fid=fopen(ref1); r4=fscprintf fid, '%e %e %e %e', [4 inf]); fclose(fid);
+r4=r4'; MeasRef4=sqrt(r4(:,3).^2 + r4(:,4).^2);
+MeasRef=(MeasRef1+MeasRef2+MeasRef3+MeasRef4)/4;
+
+% --- Calculate and plot magnitude of the measured spectrum.
+MeasSpec=20*log10(MeasSource)-20*log10(MeasRef)+offs+corr;
+semilogx(f,MeasSpec,'r-',);
+legend('Model','Measurement','Location','SouthWest');
+hold off;
+
+% =====================================================================
+% PLOT THE DIFFERENCE
+---PAGE_BREAK---
+
+% =====================================================================
+% The difference between data and reference is computed and output.
+% =====================================================================
+
+UseDif = 1; % Set UseDif = 0 to discard the plot.
+ % Set UseFif = 1 to use the plot option.
+
+if (UseDif == 1)
+ figure(2); % Prepare output picture #2.
+ Dif = FreqSpec-MeasSpec; % Calculate difference.
+ semilogx(f,Dif,'g-');
+ axis([fmin, fmax, pmin, pmax]); % Set freq-axis and decibel-axis.
+ title(MainTitle); % Set selected title.
+ xlabel('Frequency (Hz)'); % Set freq-axis label.
+ ylabel('Amplitude (dB)'); % Set decibel-axis.
+ legend('Difference','Location','SouthWest');
+ hold off;
+end
+
+### 6.1.4 Fraunhofer far-field model
+
+This section presents the implementation of the algorithm for calculating the transfer function using the assumption of plane waves approaching an infinite edge. The input is edge height *h* above the origin of the coordinate system. Top of the edge is at *y* = *h* with the edge along the *x*-axis and the observer located at *y* = 0. The height *h* is negative for an edge obstructing less than 50 % of the plane wave. The observer is located at *x* = *y* = 0 and *z* = *z*P. See Figure 112 for details. The implementation includes a weight function for improved convergence and a correction factor for 0 dB when the edge is removed (*h* << -1).
+
+The main equation is:
+
+$$L_w = 20 \log_{10} \left| \frac{ik\bar{\zeta}_p}{2\pi} \frac{\alpha_{MAX}(\beta_{MAX} - \beta_{MIN})}{MN} \sum_{n=1}^{N} \sum_{m=-M}^{M-1} \frac{\exp(-ik\bar{\zeta}_p \sqrt{1+\alpha_m^2 + \beta_n^2})}{\sqrt{(1+\alpha_m^2+\beta_n^2)(1+\frac{1}{9}\alpha_m^2+\frac{1}{9}\beta_n^2)}} \right| \text{dB}$$
+
+The normalised variables are defined by:
+
+$$\alpha = \frac{x}{z_p}, \quad \beta = \frac{y}{z_p}, \quad \beta_{MIN} = \frac{h}{z_p}$$
+
+$$\alpha_m = m\Delta\alpha, \quad m = -MK \text{ 0K M}-1$$
+
+$$\beta_n = (n-1)\Delta\beta + \beta_{MIN}, \quad n = 1K N$$
+
+The increments (step sizes) are defined by:
+
+$$\Delta\alpha = \frac{\alpha_{MAX} - \alpha_{MIN}}{2M}, \quad \Delta\beta = \frac{\beta_{MAX} - \beta_{MIN}}{N}$$
+
+Integration limits $\alpha_{MIN}$, $\alpha_{MAX}$, and $\beta_{MAX}$ and number of steps *M* and *N* are input parameters.
+
+The MATLAB dot-operators for efficient multiplication and division of vector elements are used to build a vector containing the $\alpha$-elements. The vector size is 2*M*. A for-loop is built for controlling the frequency axe and another for-loop is used to step through the $\beta$-elements. The counter must start from 1 due to the rule for indexing within MATLAB so the current $\beta$-value is calculated as shown above. The *abv*-vector contains the calculations over all $\alpha$-elements for the current $\beta$-value corresponding to the first square root in the denominator above and the *wf*-vector contains the calculations for the weight function corresponding to the second square root. The weight function values are scaled by 1/3 before squaring to widen the
+---PAGE_BREAK---
+
+weight function according to the previous discussion. The complex calculation is stored in `alfasum(n)` where *n* is the current index for the $\beta$-element. After the for-loop are all elements within the vector summed and scaled with the frequency, integration range and number of steps. The output is expressed in decibels. A warning is output from MATLAB since the value is zero for the first frequency point.
+
+```matlab
+% ---------------------------------------------------------------------
+% FRAUNHOFER INTEGRATION - INFINITE WEDGE
+% ---------------------------------------------------------------------
+% Compute the calculated frequency response.
+%
+% Written by Tore Skogberg, 2006.
+% ---------------------------------------------------------------------
+% Infinite wedge at far-field conditions.
+% Includes weight function and scaling for 0 dB at no wedge.
+
+clear
+format compact
+
+% Files: PlotInfiniteWedge1.png Height 0 and -10.
+% PlotInfiniteWedge2.png Weight 1, 1/3 and 1/10.
+% PlotInfiniteWedge3.png Weight 1/10 and 1/30.
+% PlotInfiniteWedge4.png No weighting.
+% PlotInfiniteWedge5.png H = -1, 0 and 1.
+% PlotInfiniteWedge6.png H = -1, 0 and 1 with grid.
+
+% --- CONSTANTS
+c = 345; % Speed of sound (m/s).
+
+% --- INPUT PARAMETERS
+zP = 1; % Observer horizontal distance from wedge (m).
+h = 1; % Observer vertical distance above wedge (m).
+f0 = 10; % Frequency increment (Hz).
+
+% --- INTERNAL PARAMETERS
+amax = 100; % Horizontal limits, alfa maximum.
+bmax = 100; % Vertical limit, beta maximum.
+bmin = h/zP; % Vertical limit, beta minimum.
+M = 1000; % Number of horizontal points for alfa.
+N = 1000; % Number of vertical points for beta.
+da = amax/M; % Increment in alfa (note: amin = -amax).
+db = (bmax-bmin)/N; % Increment in beta (note: asymmetric).
+
+% --- INTEGRAL
+alfa = -amax:da:(amax*(M-1)/M); % Vector containing alfa-values.
+% --- Frequency loop
+for g=1:201 % Index for the frequency vector.
+ g % Output step number to screen.
+ f(g)=(g-1)*f0; % Frequency (Hz).
+ ikz = i*2*pi*f(g)*zP/c; % Frequency-distance.
+ % --- Summation loop
+ for n=1:N % Index for beta with N elements.
+ beta = (n-1)*db+bmin; % Beta-value.
+ abv = sqrt(1+alfa.^2+beta^2); % Alfa-beta vector.
+ wf = sqrt(1+(alfa/3).^2+(beta/3)^2); % Weight function.
+ % --- ALFA summation (2M elements)
+ alfasum(n) = sum(exp(-ikz*abv)./(abv.*wf));
+ end
+ % --- BETA summation (N elements)
+ H(g) = abs(ikz*amax*(bmax-bmin)*sum(alfasum)/(M*N*2*pi));
+end
+
+% --- PLOT RESULT
+plot(f,20*log10(H),'r-',);
+title('Fraunhofer integral with the height as parameter');
+axis([0 2000 -25 5]);
+xlabel('Frequency (Hz)');
+ylabel('Transfer function, L (dB)');
+legend ('Above wedge, h/z_P = -1', ... % b
+ 'At wedge top, h/z_P = 0', ... % g
+ 'Below wedge, h/z_P = 1', ... % r
+```
+---PAGE_BREAK---
+
+'Location', 'SouthEast');
+
+grid on
+
+hold on
+---PAGE_BREAK---
+
+# 7 Appendix
+
+This chapter presents the mathematics required for derivations within the document.
+
+## 7.1 Mathematical definitions
+
+This section contains miscellaneous derivations used within this document.
+
+### 7.1.1 Partial integration
+
+Partial integration (integration by parts) is defined as follows for definite integrals, which is used here, but the procedure applies to indefinite integrals as well (Westergren, 141):
+
+$$ \int_a^b f(x)g(x)dx = [F(x)g(x)]_a^b - \int_a^b F(x)g'(x)dx $$
+
+$$ F(x) = \int f(x)dx $$
+
+This relation will be used to transform $p_r$ and $p_{rr}$.
+
+### 7.1.2 First-order differentiation
+
+First-order differentiation of a product is defined as follows (Westergren, 137), where $f(x)$ and $g'(x)$ denotes the first-order derivatives of $f(x)$ and $g(x)$ respectively:
+
+$$ \frac{d}{dx} \{f(x)g(x)\} = f'(x)g(x) + f(x)g'(x) $$
+
+### 7.1.3 Second-order differentiation
+
+Second-order differentiation of a product is defined as follows, where $f''(x)$ and $g'''(x)$ denote the second-order derivatives of $f(x)$ and $g(x)$ respectively:
+
+$$
+\begin{aligned}
+\frac{d^2}{dx^2} \{f(x)g(x)\} &= \frac{d}{dx} \left\{ \frac{d}{dx} \{f(x)g(x)\} \right\} \\
+&= \frac{d}{dx} \{f'(x)g(x) + f(x)g'(x)\} \\
+&= \frac{d}{dx} \{f'(x)\}g(x) + f'(x)\frac{d}{dx}\{g(x)\} + \frac{d}{dx}\{f(x)\}g'(x) + f(x)\frac{d}{dx}\{g'(x)\} \\
+&= f''(x)g(x) + f'(x)g'(x) + f'(x)g'(x) + f(x)g''(x)
+\end{aligned}
+$$
+
+Hence the rule for second-order differentiation:
+
+$$ \frac{d^2}{dx^2} \{f(x)g(x)\} = f''(x)g(x) + 2f'(x)g'(x) + f(x)g''(x) $$
+---PAGE_BREAK---
+
+## 7.2 Blurring
+
+A circular baffle delays the diffracted signal by a fixed amount of time while any other baffle shape distributes the delay over time, thus blurring the diffracted signal, which has the important consequence of reducing the over-all amplitude of the ripples. The baffle shape is thus an important parameter and it is the purpose of this section to derive a method for computing the degree of blurring of the diffracted signal. The model is suited for analytical work using the expressions as well as numerical simulations using a computer.
+
+### 7.2.1 Overview
+
+Expression will be derived for the below baffle shapes. All expressions are approximations based upon the method to be presented.
+
+$$p_1 = p_0 \exp(-ikB)$$
+
+$$p_1 = p_0 \exp\left(-ikB\left[\frac{3}{4} + \frac{\cos^2(\theta)}{4}\right]\right) \text{sinc}\left(kB\left[\frac{1}{4} - \frac{\cos^2(\theta)}{4}\right]\right)$$
+
+$$p_1 = \frac{p_0}{2} \left[ \exp\left(-ikB \frac{1-d_{B1}}{2} + \sqrt{d_{B1}^2 + \cos^2(\theta)}\right) \text{sinc}\left(kB \frac{1-d_{B1}}{2} - \sqrt{d_{B1}^2 + \cos^2(\theta)}\right) + \exp\left(-ikB \frac{\sqrt{d_{B1}^2 + \cos^2(\theta)} + 1}{2}\right) \text{sinc}\left(kB \frac{\sqrt{d_{B1}^2 + \cos^2(\theta)} - 1}{2}\right) \right]$$
+
+$$d_B = \frac{D}{B}$$
+
+$$p_1 = \frac{p_0}{2} [\exp(-1.04ikB)\text{sinc}(0.04kB) + \exp(-1.25ikB)\text{sinc}(0.17kB)]$$
+
+$$p_1 = \frac{p_0}{3} \exp(-ikB) \left[ \exp(-ikD) + 2 \cdot \cos\left(\frac{kD}{2}\right) \right] \text{sinc}\left(\frac{kD}{2}\right)$$
+
+Figure 140 – Sound pressure of diffraction (first reflection) for different baffle shapes.
+
+Dimension *B* represents the size of the baffle and *D* an offset from the centre along one of the axes. All baffles are assumed rigid and the loudspeaker is represented by a point source.
+---PAGE_BREAK---
+
+### 7.2.2 Theory
+
+The sound pressure generated by the diffracted sound at the edge of the baffle is calculated from integration of infinitesimal sound sources located on an infinite plane at positions corresponding to the edges of the baffle. The assumption of large observation distance and on-axis listening generates the below expression (from equation 53), which describes the sound pressure of the diffracted sound at the observation point as a transfer function (the integral) multiplied by the direct signal ($p_0$).
+
+$$p_1 = \frac{p_0}{2\pi} \int_{0}^{2\pi} \exp(-ikb) d\varphi$$
+
+Distance $b$ is function of angle $\varphi$ and will be written as $b = Bf(\varphi)$, where $B$ is a fixed length, characterising the baffle size, and $f(\varphi)$ is a dimensionless function, characterising the baffle shape and where $f(\varphi) > 0$ for all $\varphi$ since the distance between source and edge must be positive. Thus, we have:
+
+$$p_1 = \frac{p_0}{2\pi} \int_{0}^{2\pi} \exp(-ikBf(\varphi))d\varphi$$
+
+The function $f(\varphi)$ will be approximated by linear sections of fixed length. The shape will thus be described by a straight line from $f(\varphi_0)$ to $f(\varphi_1)$, another straight line from $f(\varphi_1)$ to $f(\varphi_2)$ and so forth, with each section being defined by the starting point $f(\varphi_n)$ and the slope $s_n$, where $n$ is the section number (0 to $N-1$) and the total number of sections is $N = 2\pi/\Delta\varphi$.
+
+Two examples are shown in the picture below with the angle $\varphi$ ranging from 0 to $2\pi$, but the starting and ending points can be at any angle as long as the total range is $2\pi$. Since the range is closed by one full revolution of $\varphi$, the starting function value $f(\varphi_0)$ must equal the limiting function value $f(\varphi_N)$ since they describe the same point. The first drawing represents a baffle with circular shape whose radius is constant so $b = B$ and $f(\varphi) = 1$. The second drawing represents a square baffle where $b$ is at the maximum value at the corner points.
+
+Figure 141 – The shape function for two baffle shapes, circular (left) and square (right) both assuming that the loudspeaker is placed symmetrically at the centre.
+
+Assuming that the sections are of identical length, each with the step size $\Delta\varphi$, and that the number of line segments is $N$, the function can be approximated by:
+
+$$f(\varphi) = s_n \cdot (\varphi - \varphi_n) + f(\varphi_n), \quad \varphi_n < \varphi < \varphi_n + \Delta\varphi, \quad n = 1, 2, 3, \dots, N-1$$
+
+$$s_n = \frac{f(\varphi_n + \Delta\varphi) - f(\varphi_n)}{\Delta\varphi}, \quad \Delta\varphi = \frac{2\pi}{N}$$
+---PAGE_BREAK---
+
+Insertion of this definition into the integral produces:
+
+$$
+\begin{align*}
+P_1 &= P_0 \frac{1}{2\pi} \sum_{n=0}^{N-1} \int_{\varphi_n}^{\varphi_{n+\Delta\varphi}} \exp(-ikB[s_n \cdot (\varphi - \varphi_n) + f(\varphi_n)]) d\varphi \\
+&= P_0 \frac{1}{2\pi} \sum_{n=0}^{N-1} \exp(-ikBf(\varphi_n)) \int_{\varphi_n}^{\varphi_{n+\Delta\varphi}} \exp(-ikBs_n \cdot (\varphi - \varphi_n)) d\varphi
+\end{align*}
+$$
+
+The first exponential corresponds to a delay defined by baffle size Bf(φn) at the considered
+point while the second exponential represents the effect of blurring. Computing the integral
+generates the following expression:
+
+$$
+\begin{align*}
+P_1 &= P_0 \frac{1}{2\pi} \sum_{n=0}^{N-1} \exp(-ikBf(\varphi_n)) \left[ \frac{\exp(-ikBs_n \cdot (\varphi-\varphi_n))}{-ikBs_n} \right]_{\varphi_n}^{\varphi_n+\Delta\varphi} \\
+&= P_0 \frac{1}{2\pi} \sum_{n=0}^{N-1} \exp(-ikBf(\varphi_n)) \frac{\exp(-ikBs_n\Delta\varphi) - 1}{-ikBs_n}
+\end{align*}
+$$
+
+The last term represents a sine function, which can be seen by moving half the value of the
+exponential function outside the brackets and using Euler's formula for the sine function on
+the brackets (Westergren, 62). Introducing the step size Δφ shows that the sine function is
+divided by its argument, and it is thus recognised as the sinc function.
+
+$$
+\begin{align*}
+P_1 &= P_0 \frac{1}{2\pi} \sum_{n=0}^{N-1} \exp(-ikBf(\varphi_n)) \exp\left(-\frac{ikBs_n\Delta\varphi}{2}\right) \frac{2}{kBs_n} \frac{\exp\left(\frac{ikBs_n\Delta\varphi}{2}\right) - \exp\left(-\frac{ikBs_n\Delta\varphi}{2}\right)}{2i} \\
+&= P_0 \frac{1}{2\pi} \sum_{n=0}^{N-1} \exp\left(-ikB\left(f(\varphi_n) + \frac{s_n\Delta\varphi}{2}\right)\right) \frac{2\Delta\varphi}{kBs_n\Delta\varphi} \sin\left(\frac{kBs_n\Delta\varphi}{2}\right) \\
+&= P_0 \frac{\Delta\varphi}{2\pi} \sum_{n=0}^{N-1} \exp\left(-ikB\left(f(\varphi_n) + \frac{s_n\Delta\varphi}{2}\right)\right) \text{sinc}\left(\frac{kBs_n\Delta\varphi}{2}\right)
+\end{align*}
+$$
+
+Using the definition of $s_n$ enables the delay function being expressed by the average function
+value for both ends of the current segment and the same relation can be used to express the
+sinc function as the change in function value. The constant in front of the sum is equal to 1/N
+and the result is the required expression:
+
+$$
+P_1 = P_0 \frac{1}{N} \sum_{n=1}^{N-1} \exp\left(-ikB \frac{f(\varphi_n + \Delta\varphi) + f(\varphi_n)}{2}\right) \text{sinc}\left(kB \frac{f(\varphi_n + \Delta\varphi) - f(\varphi_n)}{2}\right) \quad 130
+$$
+
+This equation will be addressed in the following sections but before generating a library of
+baffle shapes, a method shall be introduced for calculating the expression using the least
+possible number of segments.
+
+7.2.3 Optimisation
+
+A quite large number of points are required for description of a typical baffle but the number
+of segments required for calculation can be reduced for most baffles using symmetry of the
+baffle as well as mathematical properties of the expression. Using the square baffle as an
+example, the below drawing shows that although the shape function is defined within the
+---PAGE_BREAK---
+
+closed range from 0 to $2\pi$, the function itself is periodic within the range from 0 to $\pi/2$, so
+only this range needs consideration. This reduces the number of segments to one-fourth.
+
+Figure 142 – The shape function for a square baffle shape oscillates between function values $f_1$ and $f_2$ (left). The periodicity of the shape function can be used to remove most of the calculation (middle) and the symmetry of the exponential and sinc functions can be used to swap the sign of the linear sections (right).
+
+The exponential function within equation 130 is defined by the average value of two consecutive points, $f(\phi_n)$ and $f(\phi_{n+1})$, from now on written as $f_1$ and $f_2$, but the actual order of the points is unimportant; the function values can be interchanged without affecting the result. A similar conclusion applies to the sinc function due to the symmetry of the function to negative arguments: $\text{sinc}(-x) = \text{sinc}(x)$. As a consequence, one can swap the sign of the slope to the sections, thus calculating the remaining two sections as one section. The physical explanation is that the effective delay is calculated at the midpoint of each linear section.
+
+The above problem can be calculated from the function values $f_1$ and $f_2$ as:
+
+$$p_1 = p_0 \exp\left(-ikB \frac{f_1 + f_2}{2}\right) \text{sinc}\left(kB \frac{f_1 - f_2}{2}\right) \qquad 131$$
+
+One possible implementation for this equation is an elliptic baffle where the radius is increased gradually between the minimum value and the maximum value. The shape function may not exactly model the elliptic baffle but it is at least an approximation.
+
+Figure 143 – The shape function for one section of the baffle.
+
+This can be generalised, approximating the baffle by a finite number of sections with the delay defined by the mean distance and the ripple defined by the difference between the distances to the endpoints of the section. The baffle is thus described by the sum of the above contributions and divided by the number of segments. Using *N* segments to describe the baffle we have *N* + 1 endpoints and the following equation approximates the diffracted signal.
+
+$$p_1 = \frac{p_0}{N} \sum_{n=1}^{N} \exp\left(-ikB \frac{f_n + f_{n+1}}{2}\right) \text{sinc}\left(kB \frac{f_n - f_{n+1}}{2}\right), \quad f_{n+1} = f_1 \qquad 132$$
+
+With two segments ($N = 2$) we have $f_1$, $f_2$ and $f_3$, with $f_3 = f_1$ since the loop is closed.
+---PAGE_BREAK---
+
+The sections do not need to be of the same size; $\varphi$ can step with different values for each section. The simplification of the problem opens the possibility of improving the precision of the approximation by increasing the number of segments.
+
+Figure 144 – The shape function for a square baffle shape can be reduced from eight segments to one segment as shown previously (left) thus allowing an improvement in the accuracy by using two segments to represent the shape function (right).
+
+Using two straight lines for each segment, one for function values $f_1$ and $f_2$, and another segment for function values $f_2$ and $f_3$, results in the following expression:
+
+$$p_1 = \frac{p_0}{2} \left[ \exp\left(-ikB \frac{f_1+f_2}{2}\right) \text{sinc}\left(kB \frac{f_1-f_2}{2}\right) + \exp\left(-ikB \frac{f_2+f_3}{2}\right) \text{sinc}\left(kB \frac{f_2-f_3}{2}\right) \right] \quad 133$$
+
+The function values $f_1$ and $f_2$ are closer together than the function values $f_2$ and $f_3$, so the first sinc function will attenuate the signal at a higher frequency than the second sinc function.
+
+### 7.2.4 Circular baffle
+
+Radius of the circular baffle is $B$, which is constant so the shape function $f(\varphi)$ is unity for all $\varphi$ and it is thus not required dividing the function into sections, one section from $\varphi_0$ to $\varphi_1$ using a step size of $\Delta\varphi = 2\pi$ will do. The result for the circular baffle becomes a delay function without the sinc function.
+
+$$p_1 = p_0 \exp(-ikB) \qquad 134$$
+
+The circular baffle does not affect the amplitude of the diffracted signal; only the phase is changed, so the circular baffle causes serious interference throughout the frequency range.
+
+Figure 145 – Derivation of function $f(\varphi)$ for an elliptic baffle.
+---PAGE_BREAK---
+
+### 7.2.5 Elliptic baffle
+
+An ellipse is defined similarly to the circle as $x = B_x\cos(\varphi)$ and $y = B_y\cos(\varphi)$, where the shape is specified through the coefficients $B_x$ and $B_y$ and becomes a circle for $B_x = B_y$. With the major axis of the ellipse along the x-axis, the equation for the ellipse can be written:
+
+$$x = B \cos(\varphi), \quad y = B \cos(\theta)\sin(\varphi), \quad 0 < \varphi < 2\pi, \quad 0 < \theta < \frac{\pi}{2}$$
+
+This represents a circle for the projection angle $\theta = 0^\circ$, and the ellipse collapses to a straight line for $\theta = 90^\circ$, with the elliptic shapes within this range. The ellipse may alternatively be constructed as the projection of a circle onto an angled plane, which can be considered as a circular baffle observed at angle $\theta$, so the following theory applies equally well to a circle observed off-axis at observation angle $\theta$. The elliptic baffle and its shape function is shown below.
+
+Figure 146 – Derivation of function $f(\varphi)$ for an elliptic baffle.
+
+The distance from the sound source to the edge is:
+
+$$b = \sqrt{x^2 + y^2} = B\sqrt{\cos^2(\varphi) + \cos^2(\theta)\sin^2(\varphi)}$$
+
+The shape function, i.e. the function describing $b$, becomes:
+
+$$b = Bf(\varphi) \Rightarrow f(\varphi) = \sqrt{\cos^2(\varphi) + \cos^2(\theta)\sin^2(\varphi)}$$
+
+Using the trigonometric identity for the sum of squared sine and cosine functions, the cosine can be rewritten as a squared sine.
+
+$$ \begin{aligned} f(\varphi) &= \sqrt{1 - \sin^2(\varphi) + \cos^2(\theta)\sin^2(\varphi)} \\ &= \sqrt{1 + (\cos^2(\theta) - 1)\sin^2(\varphi)} \end{aligned} $$
+
+A Taylor series expansion will be used to simplify the expression by removing the square root (Westergren, 196). Only the two first terms of the series expansion are used, which results in an error around 6% for an argument approaching ±1.
+
+$$\sqrt{1+x} \approx 1 + \frac{x}{2} \Rightarrow f(\varphi) \approx 1 + \frac{\cos^2(\theta) - 1}{2}\sin^2(\varphi)$$
+---PAGE_BREAK---
+
+The function is $\pi$-periodic, so only the range from 0 to $\pi$ is needed for calculation. The function will be approximated by two sections within this range, so $\Delta\varphi = \pi/2$. The maximum and minimum values are:
+
+$$
+\begin{aligned}
+f_1 &= 1 && \varphi = 0 \\
+f_2 &= \left[ \frac{1}{2} + \frac{\cos^2(\theta)}{2} \right] && \varphi = \frac{\pi}{2}
+\end{aligned}
+ $$
+
+The expression for the sound pressure from the diffracted signal will use equation 131:
+
+$$ p_1 = p_0 \exp\left(-ikB \frac{f_1+f_2}{2}\right) \text{sinc}\left(kB \frac{f_1-f_2}{2}\right) $$
+
+Inserting the function values into the exponential results in the following expression:
+
+$$
+\begin{aligned}
+\exp\left(-\frac{ikB}{2}[f_1 + f_2]\right) &= \exp\left(-\frac{ikB}{2}\left[1 + \frac{1}{2} + \frac{\cos^2(\theta)}{2}\right]\right) \\
+&= \exp\left(-ikB\left[\frac{3}{4} + \frac{\cos^2(\theta)}{4}\right]\right)
+\end{aligned}
+ $$
+
+The delay becomes identical to the circular baffle for $\theta = 0^\circ$, as it should. The sinc function can be written as:
+
+$$
+\begin{aligned}
+\text{sinc}\left(kB \frac{f_1 - f_2}{2}\right) &= \text{sinc}\left(\frac{kB}{2}\left[1 - \left[\frac{1}{2} + \frac{\cos^2(\theta)}{2}\right]\right]\right) \\
+&= \text{sinc}\left(kB\left[\frac{1}{4} - \frac{\cos^2(\theta)}{4}\right]\right)
+\end{aligned}
+ $$
+
+The sound pressure of the diffracted signal for the elliptic baffle becomes:
+
+$$ p_1 = p_0 \exp\left(-ikB\left[\frac{3}{4} + \frac{\cos^2(\theta)}{4}\right]\right) \text{sinc}\left(kB\left[\frac{1}{4} - \frac{\cos^2(\theta)}{4}\right]\right) \quad 135 $$
+
+The delay changes from the maximum value $kB$ obtained for $\theta = 0^\circ$, where the elliptic baffle has become a circle, and to the minimum value of $0.75kB$ obtained at $\theta = 90^\circ$, where the baffle is a straight line. The latter value is obviously too large but this is a result of the approximation. The sinc function changes from unity at $\theta = 0^\circ$ to $kB/4$ at the projection angle of $\theta = 90^\circ$.
+
+For an elliptic baffle with height equal to 50 % of the width, the projection angle is $\theta = 60^\circ$, for a cosine of 0.50, and the equation becomes:
+
+$$ p_1 = p_0 \exp(-0.81ikB) \text{sinc}(0.19kB) $$
+
+The elliptic baffle introduces the sinc function into the diffracted signal thus attenuating the level of the diffracted signal at the high-frequency end.
+---PAGE_BREAK---
+
+The baffle cut-off frequency can be defined as the first zero in the frequency response, which is determined from equation 135 as:
+
+$$ \text{sinc}\left(\frac{kB}{4}\left[\frac{1 - \cos^2(\theta)}{4}\right]\right) = 0 \Rightarrow \frac{kB}{4}[1 - \cos^2(\theta)] = \pi \Rightarrow f_{\text{CUT}} = \frac{2c}{(1 - \cos^2(\theta))B} $$
+
+For $B = 170$ mm and $\theta = 60^\circ$ will the first zero occur at 5.4 kHz so the elliptic baffle is an improvement to the circular baffle only at high frequencies.
+
+### 7.2.6 Off-axis correction
+
+A circular baffle looks like an elliptic baffle when it is viewed from an angle different from on-axis and similar for a square baffle, which looks like a rectangle when it is tilted. This is shown below for the circular baffle.
+
+Figure 147 – The observer may see the circle turned into an ellipse but the sound must travel through a distance, which is increased for one side and decreased for the other side.
+
+It is possible modelling the tilt using the above tools if the change in distance to the observer is taken into account; distance $d_{30A}$ is shorter than the nominal distance $d_0$ while distance $d_{30B}$ is larger. Hence, the model approximates a circle with offset source. Similar applies to a tilted square, which is turned into a rectangle, and a rectangle is turned into another rectangle.
+
+The change in distance for a tilted baffle is:
+
+$$ d = B \sin(\theta) \qquad 137 $$
+
+The change in distance is reflected within the shape function, which becomes:
+
+$$ f_1 = 1 + \sin(\theta), \quad f_2 = 1 - \sin(\theta) $$
+
+Insertion into equation 130 gives:
+
+$$ p_1 = p_0 \exp(-ikB) \text{sinc}(kB \sin(\theta)) \qquad 138 $$
+
+For $\theta=30^\circ$ the equation becomes:
+
+$$ p_1 = p_0 \exp(-ikB) \text{sinc}\left(\frac{kB}{2}\right) \qquad 139 $$
+---PAGE_BREAK---
+
+### 7.2.7 Elliptic baffle with offset source
+
+A model will be derived for an elliptic baffle with offset source.
+
+Figure 148 – Derivation of function f(φ) for an elliptic baffle with offset source.
+
+Offsetting the source to $x = D$ does not affect the definition of the ellipse but it changes the distance from the sound source to the edge.
+
+$$
+\begin{aligned}
+b &= \sqrt{(x-D)^2 + y^2} \\
+&= B\sqrt{\left(\cos(\varphi)-D\right)^2 + \left(\cos(\theta)\sin(\varphi)\right)^2}
+\end{aligned}
+ $$
+
+The offset distance will be normalised to $d_B = D/B$ and the shape function becomes:
+
+$$ f(\varphi) = \sqrt{(\cos(\varphi) - d_B)^2 + \cos^2(\theta)\sin^2(\varphi)} $$
+
+The function is $2\pi$-periodic, but symmetry can halve the range needed for calculation to $\pi$. The function will be approximated by two sections within this range, so $\Delta\varphi = \pi/2$ and the function values must be calculated at $\varphi=0, \pi/2$ and $\varphi=\pi$. The function values are:
+
+$$
+\begin{align*}
+f_1 &= f(0) = \sqrt{(1-d_B)^2} = 1-d_B \\
+f_2 &= f(\pi/2) = \sqrt{d_B^2 + \cos^2(\theta)} \\
+f_3 &= f(\pi) = \sqrt{(1+d_B)^2} = 1+d_B
+\end{align*}
+ $$
+
+Insertion into equation 133 generates the following expression:
+
+$$ P_1 = \frac{p_0}{2} \left[ \exp\left(-ikB \frac{1-d_B+\sqrt{d_B^2+\cos^2(\theta)}}{2}\right) \sinc\left(kB \frac{1-d_B-\sqrt{d_B^2+\cos^2(\theta)}}{2}\right) + \exp\left(-ikB \frac{\sqrt{d_B^2+\cos^2(\theta)}+1+d_B}{2}\right) \sinc\left(kB \frac{\sqrt{d_B^2+\cos^2(\theta)}-1-d_B}{2}\right) \right] \quad d_B = \frac{D}{B} \quad 140 $$
+
+It is seen, that the model reduces to the simple exponential for a projection angle of $\theta = 0^\circ$ and an offset of $D = 0$; as it is expected to do.
+---PAGE_BREAK---
+
+For an offset different from zero the cosine functions become unity and we have an equation for a circular baffle with offset source:
+
+$$p_1 = \frac{p_0}{2} \left[ \exp\left(-ikB \frac{1-d_B+\sqrt{d_B^2+1}}{2}\right) \text{sinc}\left(kB \frac{1-d_B-\sqrt{d_B^2+1}}{2}\right) + \exp\left(-ikB \frac{\sqrt{d_B^2+1}+1+d_B}{2}\right) \text{sinc}\left(kB \frac{\sqrt{d_B^2+1}-1-d_B}{2}\right) \right]$$
+
+Approximating the square roots by the Taylor series-expansion and ignoring the squaring of the $d_B$ term within the square root we get the approximation:
+
+$$p_1 = \frac{p_0}{2} \left[ \exp\left(-ikB\left(1-\frac{d_B}{4}\right)\right) \text{sinc}\left(kB\frac{3d_B}{4}\right) + \exp\left(-ikB\left(1+\frac{3d_B}{4}\right)\right) \text{sinc}\left(kB\frac{d_B}{4}\right) \right] \quad 141$$
+
+The errors introduced by the approximations are negligible; the model is an approximation anyway. For $D = B/2$ we get $d_B = 1/2$ and the equation becomes:
+
+$$p_1 = \frac{p_0}{2} \left[ \exp\left(-\frac{7}{8}ikB\right) \text{sinc}\left(\frac{3}{8}kB\right) + \exp\left(-\frac{11}{8}ikB\right) \text{sinc}\left(\frac{1}{8}kB\right) \right]$$
+
+Using a projection angle of $\theta = 60^\circ$ and an offset of $D = B/2$, the result becomes:
+
+$$p_1 = \frac{p_0}{2} [\exp(-0.60 \cdot ikB) \text{sinc}(0.10 \cdot kB) + \exp(-1.10 \cdot ikB) \text{sinc}(0.79 \cdot kB)] \quad 142$$
+
+The elliptic baffle with offset source attenuates the diffracted signal above a frequency defined by the baffle dimension *B* and the offset. The attenuation starts where the first of the sinc function has its zero and accelerates where the second sinc function has its zero. The two zero frequencies are:
+
+$$\text{sinc}(0.10 \cdot kB) = 0, \quad \text{sinc}(0.79 \cdot kB) = 0$$
+
+The frequencies are located at different frequencies so the cut-off frequency is defined here as the geometrical midpoint of the two frequencies:
+
+$$\begin{cases} 0.10 \cdot kB = \pi & \Rightarrow f_{CUT1} = 5 \frac{c}{B} \\ 0.79 \cdot kB = \pi & \Rightarrow f_{CUT2} = 0.63 \frac{c}{B} \end{cases} \quad \Rightarrow \quad f_{CUT} = \sqrt{f_{CUT1} f_{CUT2}} = 1.8 \frac{c}{B}$$
+
+The cut-off frequency becomes 3.6 kHz with $B = 170$ mm, which is in reasonable agreement with the plot.
+
+### 7.2.8 Square baffle
+
+Using a square baffle with the loudspeaker located symmetrically at the centre produces the following relation exists for the x and y coordinates of the vertical right-hand side:
+
+$$x = B, \quad y = B \tan(\varphi), \quad -\frac{\pi}{4} < \varphi < \frac{\pi}{4}$$
+---PAGE_BREAK---
+
+Similar expressions can be stated for the other sides, but this is not required due to symmetry, as explained in section 7.2.3.
+
+Figure 149 – Derivation of function f(φ) for a square baffle. Only the first quarter of the range is of interest due to periodicity and even this range can be halved due to symmetry.
+
+The relation for determination of the shape function for the range from 0 to π/4 is:
+
+$$
+b_B f(\varphi) = \sqrt{x^2 + y^2} = B\sqrt{1 + \tan^2(\varphi)} \Rightarrow f(\varphi) = \frac{B}{b_B}\sqrt{1 + \tan^2(\varphi)}
+$$
+
+One of the possible definitions is shown in the figure above where $b_B = B$, for which $f(\varphi)$ becomes equal to the square root and this definition will be used below, but any other definition is equally valid. Using the symmetry, the method needs only two function values, $f(\varphi_0) = 1.00$ and $f(\varphi_1) = 1.41$, for computation of the expression.
+
+$$
+p_1 = p_0 \exp\left(-ikB \frac{1+1.41}{2}\right) \text{sinc}\left(kB \frac{1.41-1}{2}\right)
+$$
+
+The expression becomes:
+
+$$
+p_1 = p_0 \exp(-1.21ikB) \text{sinc}(0.21kB) \tag{143}
+$$
+
+This is almost identical to the expression of an elliptic baffle; the main difference is the delay, which 1.5 times larger thus corresponding to the larger baffle. The expression can be improved by using four two linear sections with $\varphi = 0, \pi/8$ and $\pi/4$. The function values are: $f(\varphi_0) = 1.00$, $f(\varphi_1) = 1.08$ and $f(\varphi_3) = 1.41$ and insertion into equation 133 generates the following expression:
+
+$$
+p_1 = P_0 \frac{1}{2} \left[ \exp(-1.04ikB) \sinc(0.04kB) + \exp(-1.25ikB) \sinc(0.17kB) \right] \quad 144
+$$
+
+The expression can be re-arranged by moving the first term of the sum outside the brackets:
+
+$$
+p_1 = P_0 \frac{1}{2} \exp(-1.04ikB) \text{sinc}(0.04kB) \left[ 1 + \exp(-0.21ikB) \frac{0.04 \sin(0.17kB)}{0.17 \sin(0.04kB)} \right]
+$$
+
+The term in front of the brackets is – almost – the expression of a circular baffle; the decay is close to exp(-ikB) and the sinc function is almost independent upon frequency, the first zero is located around 50 kHz. The term inside the bracket represents the change due to the four corners of the square.
+
+The right-hand term includes division with a sine function but the equation does not blow up when the function approaches zero since the sinc function in front of the brackets approaches
+---PAGE_BREAK---
+
+zero at the same frequency. The net result at the limit becomes equal to the scaling factor within the brackets, i.e. the peak will be 0.04/0.17 or around -8 dB.
+
+The cut-off frequency will here be defined as the zero frequency of the sinc function within the square brackets, since the sinc function in front of the brackets has limited influence within the audible range.
+
+$$ \text{sinc}(0.17kB) = 0 \Rightarrow 0.17kB = \pi \Rightarrow f_{\text{CUT}} = 2.9 \frac{c}{B} $$
+
+The cut-off frequency becomes 5.9 kHz with $B = 170$ mm, which is in reasonable agreement with the plot although it is close to the local peak. Interference due to diffraction is reduced above the cut-off frequency but serious rippling within the useful frequency range can be expected since the cut-off is relatively high.
+
+### 7.2.9 Rectangular baffle
+
+A rectangular baffle is not easily modelled within the polar coordinate system, so the baffle to be analysed will be fixed in the high to width ratio in order to survive the calculation. The description will use twelve segments ($Δφ = 30°$), fixing the baffle size at 1.15B high by 2B wide (1 high by 1.73 wide), which is intended to represent a typical rectangular baffle.
+
+The shape function will be approximated by three segments from 0 to $π$ using the symmetry, and ignoring the minor variations within the function, which corresponds to rounding the corners of the baffle, as shown dashed in the figure below.
+
+Figure 150 – Derivation of function $f(φ)$ for a rectangular baffle with offset sound source. The shape function is shown for $d = B/2$ as the offset and the effect of the simplification is shown as dotted lines; the edges are rounded.
+
+Key values of the shape function are shown below.
+
+$$ f_1 = \frac{B-D}{B} = 1 - \frac{D}{B}, \quad f_2 = \frac{f_1 + f_3}{2} = 1, \quad f_3 = \frac{B+D}{B} = 1 + \frac{D}{B} $$
+---PAGE_BREAK---
+
+Insertion into expression 130 (with N = 3) the equation becomes:
+
+$$
+\begin{aligned}
+p_1 = \frac{p_0}{3} & \left[ \exp\left(-\frac{ikB}{2}\left(1-\frac{D}{B}+1\right)\right) \sinc\left(\frac{kB}{2}\left(\left(1+\frac{D}{B}-1\right)\right)\right) \right. \\
+& + \exp\left(-\frac{ikB}{2}\left(1+1+\frac{D}{B}\right)\right) \sinc\left(\frac{kB}{2}\left(1-1-\frac{D}{B}\right)\right) \\
+& \left. + \exp\left(-\frac{ikB}{2}\left(1+\frac{D}{B}+1\right)\right) \sinc\left(\frac{kB}{2}\left(1+\frac{D}{B}-1\right)\right) \right]
+\end{aligned}
+$$
+
+The sinc functions are identical despite the different signs and reduction of the exponentials results in:
+
+$$
+\begin{align*}
+p_1 &= \frac{p_0}{3} \left[ 2 \cdot \exp\left(-ikB\left(1+\frac{D}{2B}\right)\right) + \exp\left(-ikB\left(1-\frac{D}{2B}\right)\right) \right] \sinc\left(\frac{kD}{2B}\right) \\
+&= \frac{p_0}{3} \exp(-ikB) \left[ \exp\left(-\frac{ikD}{2}\right) + \exp\left(-\frac{ikD}{2}\right) + \exp\left(\frac{ikD}{2}\right) \right] \sinc\left(\frac{kD}{2}\right)
+\end{align*}
+$$
+
+The square bracket is recognised as the sum of a delay and a cosine function according to the Euler formula for the cosine function (Westergren, 62). Hence:
+
+$$
+p_1 = \frac{p_0}{3} \exp(-i k B) \left[ \exp(-i k D) + 2 \cos\left(\frac{k D}{2}\right) \right] \sin c\left(\frac{k D}{2}\right) \quad 145
+$$
+
+For offset *D* = *B*/2 the expression becomes:
+
+$$
+p_1 = \frac{p_0}{3} \exp(-i k B) \left[ \exp\left(-\frac{i k B}{2}\right) + 2 \cos\left(\frac{k B}{4}\right) \right] \sin c\left(\frac{k B}{4}\right) \quad 146
+$$
+
+Two zero-frequencies are possible; the first is due to the terms within the square bracket and the second is generated by the sinc function. The expression can be solved analytically but MATLAB is better suited for this purpose so a file was generated representing the terms within the square bracket and the amplitude is plotted below.
+
+The first minimum is found for $kB/2 = 2.58$ so the cut-off frequency becomes $f_{CUT} = 0.82c/B$, which is at 1.7 kHz for $B = 170$ mm. The sinc function generates a null at $kB/4 = \pi$, so the
+---PAGE_BREAK---
+
+next cut-off is at $f_{CUT} = 2c/B = 4.1$ kHz for $B = 170$ mm. The geometric mean of the two values is 2.6 kHz, which is approximately at the peak before the level finally drops off.
+
+## 7.3 Blurring of the second reflection
+
+It is the purpose of this section to derive an equation for the blurring of the diffracted signal due to the second reflection. The reflected signal is assumed to propagate as spherical waves within the $2\pi$ solid angle in front of or behind the baffle generating new reflections whenever an edge is reached. This is illustrated below where reflection is represented by a monopole sound source located at the edge. Hence, the model may also be used to calculate the amplitude response of a loudspeaker mounted at the edge of a circular baffle.
+
+Figure 151 – Estimation of distance d as function of the observation point expressed by $\varphi$ for a circular baffle with radius b. The monopole sound source of the figure represents each of the infinitesimal sound sources along the periphery of the circle.
+
+The model will be presented for the circular baffle only, due to the complexity of the problem.
+
+### 7.3.1 Theory
+
+An infinitesimal sound source is defined on the circle periphery with the x and y coordinates determined by:
+
+$$x = B \cos(\varphi), \quad y = B \sin(\varphi)$$
+
+The source is assumed located at $(B,0)$ so the length of distance $d$ to a point on the circle is changing from zero to $2B$ according to the following relation:
+
+$$
+\begin{aligned}
+d &= \sqrt{(B-x)^2 + y^2} \\
+&= B\sqrt{(1-\cos(\varphi))^2 + \sin^2(\varphi)} \\
+&= B\sqrt{2(1-\cos(\varphi))}
+\end{aligned}
+ $$
+
+The square root can be removed using a trigonometric identity (Westergren, 127):
+
+$$ \sin^2\left(\frac{\varphi}{2}\right) = \frac{1 - \cos(\varphi)}{2} \Rightarrow d = 2B \sin\left(\frac{\varphi}{2}\right) $$
+---PAGE_BREAK---
+
+The sound pressure at distance $r_1$ from each of the infinitesimal sound sources, located at the periphery of the circle and assuming an infinite plane, is:
+
+$$dp_2 = \frac{i\omega\rho q}{2\pi(r_1 + B + d)} \exp(i\omega t - ikr_1 - ikB - ikd) \frac{d\varphi}{2\pi}$$
+
+Ignoring the change in amplitude due to $B$ and $d$, but not the change in phase, simplifies the expression by removing $B$ and $d$ from the dominator and setting $r_1$ equal to $r_0$. This requires the dimension $B + d$ being less than $er_0$, where $e$ is the accepted error and $r_0$ is the distance to the observation point. Using this information the equation simplifies to:
+
+$$dp_2 = \frac{i\omega\rho q}{2\pi r_0} \exp(i\omega t - ikr_1) \exp(-ikB) \exp(-ikd) \frac{d\varphi}{2\pi}$$
+
+The first two terms are recognised as the original monopole sound source $p_0$ and the third term is recognised as the delay of a circular baffle, $p_1/p_0$, so the equation can be written:
+
+$$dp_2 = p_0 \frac{p_1}{2\pi} \frac{d\varphi}{2\pi} = p_1 \frac{d\varphi}{2\pi}$$
+
+The resultant sound pressure from the sound source becomes:
+
+$$p_2 = p_1 \frac{1}{2\pi} \int_{0}^{2\pi} \exp(-ikd) d\varphi$$
+
+The expression for $d$ is inserted:
+
+$$p_2 = p_1 \frac{1}{2\pi} \int_{0}^{2\pi} \exp\left(-2ikB\sin\left(\frac{\varphi}{2}\right)\right) d\varphi$$
+
+This is a half-sine for $\varphi$ varying from $\pi$ to $2\pi$ and will be approximated by three segments as shown below.
+
+Figure 152 – Shape function for the blurring due to a circular disk.
+
+The function values are $f(0) = f(2\pi) = 0$ and $f(2\pi/3) = f(4\pi/3) = 2$ and insertion these values directly into expression 130 (with $N=3$) yields:
+
+$$p_1 = \frac{p_0}{3} [\exp(-ikB)\text{sinc}(kB) + \exp(-2ikB) + \exp(-ikB)\text{sinc}(kB)]$$
+---PAGE_BREAK---
+
+After reduction the expression for the source at the edge of the baffle becomes:
+
+$$p_1 = \frac{P_0}{3} \exp(-ikB) [2 \cdot \text{sinc}(kB) + \exp(-2ikB)] \quad 147$$
+
+A plot of the transfer function is shown below with a monopole sound source at the edge of the baffle.
+
+Figure 153 – Frequency response with a circular baffle and the source located at the edge of the baffle.
+
+The effect of blurring is to attenuate the high-frequencies.
+
+## 7.4 Wave equation
+
+It was the objective modelling the wave front through solving the wave equation for a circular baffle with the sound source at the centre of the baffle. The problem was considered spherical due to the point source and the circular baffle but the wave front on the rear side of the baffle is not spherical (it is like an inflating bicycle tube) so I decided after some work to ignore the rear side radiation and concentrate on the front side only. The model is thus assuming that the propagating wave front is propagating as a spherical wave within the half-space at the front side and it was the idea to model the baffle through the boundary conditions.
+
+Figure 154 – Graphical representation of the diffraction model using a point source and a concentric circular baffle.
+---PAGE_BREAK---
+
+The wave equation was expected solvable using integral transformation since this method can cope with the discontinuity at the boundary of the disk. The discontinuity is at radius $r = B$, so radius was planned to be the transformed variable and radius is semi-infinite, which indicates use of the Laplace transformation.
+
+The spherical wave equation uses the independent variables radius $r$ and observation angle $\theta$ assuming rotation symmetry around the z-axis. The spherical wave equation is, according to equation 14:
+
+$$ \frac{\partial^2 p}{\partial r^2} + \frac{2}{r} \frac{\partial p}{\partial r} + \frac{1}{r^2} \left[ \frac{\partial^2 p}{\partial \theta^2} + \cot(\theta) \frac{\partial p}{\partial \theta} \right] + k^2 p = 0, \quad p = p(r, \theta) \exp(i\omega t) $$
+
+Multiplying by $r^2$ the equation becomes:
+
+$$ r^2 \frac{\partial^2 p}{\partial r^2} + 2r \frac{\partial p}{\partial r} + \frac{\partial^2 p}{\partial \theta^2} + \cot(\theta) \frac{\partial p}{\partial \theta} + k^2 r^2 p = 0 $$
+
+The system is singular for $r=0$ so a variable substitution was used to avoid problems during the Laplace transformation. The new variable will be called $\sigma$ and is offsetting the start from the origin to begin at a sphere around the origin with radius $a$, which is small but non-zero. This is effectively a spherical source with known radius and volume velocity and should thus fit well into the model. The substitution becomes:
+
+$$ r = \sigma + a \Rightarrow \sigma = r - a, \quad a > 0 $$
+
+The differential $d\sigma$ is identical to $dr$ and insertion gives the following wave equation to be Laplace transformed.
+
+$$ (\sigma + a)^2 \frac{\partial^2 p}{\partial \sigma^2} + 2(\sigma + a) \frac{\partial p}{\partial \sigma} + \frac{\partial^2 p}{\partial \theta^2} + \cot(\theta) \frac{\partial p}{\partial \theta} + k^2 (\sigma + a)^2 p = 0 \quad 148 $$
+
+Laplace transforming the terms is somewhat tedious and will not be detailed here as (see section 7.4). Assembling the Laplace-transformed terms and re-substituting $\sigma + a$ back into $r$, we finally arrive at:
+
+$$ a^2 \exp(-sa) \left( sp(a, \theta) + \frac{\partial p(a, \theta)}{\partial r} \right) \\ + (s^2 + k^2) \frac{\partial^2 \hat{p}(s, \theta)}{\partial s^2} + 2s \frac{\partial \hat{p}(s, \theta)}{\partial s} + \frac{\partial^2 \hat{p}(s, \theta)}{\partial \theta^2} + \cot(\theta) \frac{\partial \hat{p}(s, \theta)}{\partial \theta} - \hat{p}(s, \theta) = 0 $$
+
+The first line represents the terms due to the spherical excitation source with radius $a$. Assuming that radius is small then $a^2 \to 0$ and the first line can be ignored so the transformed wave equation can be simplified into:
+
+$$ (s^2 + k^2) \frac{\partial^2 \hat{p}(s, \theta)}{\partial s^2} + 2s \frac{\partial \hat{p}(s, \theta)}{\partial s} + \frac{\partial^2 \hat{p}(s, \theta)}{\partial \theta^2} + \cot(\theta) \frac{\partial \hat{p}(s, \theta)}{\partial \theta} - \hat{p}(s, \theta) = 0 \quad 149 $$
+
+The negative sign for the last term was surprising and was expected due to an error but the source of the error was not found.
+---PAGE_BREAK---
+
+### 7.4.1 Comments
+
+Several attempts have been initiated to solve this equation – or similar equations from earlier trials – but they have not been successful. In the first trial was the terms in $\theta$ interpreted as an inhomogeneous Legendre equation (the inhomogeneous part being the Laplace-transformed terms), which was solved but proved to be a dead end; the solution forced the sound pressure to be proportional to $\cos(\theta)$ and this is obviously too limiting to be valuable. It all boils down to the problem of understanding the derivatives in $s$. I am used to Laplace-transformed electrical networks but this equation is quite different from the linear network theory known to me.
+
+The books *Partial Differential Equations* by Nakhle H. Asmar (Pretence Hall, 2nd Ed 2005) and *Green's Functions with Applications* by Dean G. Duffy (Chapmann & Hall, 2001) were carefully studied but the problem remained unsolved although the last book was probably on the right track since Urban et al. solved a similar problem using the Green's function and the Kirchhoff approximation. However, I am not accustomed to Green's function and the more advanced physics and mathematics so I did not dare to follow this path.
+
+Finn Agerkvist suggested consulting the book *Sædvanlige differentialligninger fra fysikken* by Erik Hansen (Polyteknisk forlag, 1976), which solved several minor problems but I did not find a route to a solution to the problem at hand.
+
+Ove Skoggaard offered me two sessions of assistance where the problem was reviewed and my attempts to solve the wave equation were commented. The problem of singularity was addressed and he suggested a route to solve the problem but the Laplace transformed wave equation remained unsolvable to me.
+
+Finn Jacobsen provided me with a reference to Vicente Cutanda, who mailed me the article *On the modelling of narrow gaps using the standard boundary element method* (J. Acoust. Soc. Am., 2001). The article included a very short appendix (20 lines), where the problem of solving the wave equation for a thin circular disk were addressed; it should be expressed in the oblate spheroidal coordinate system to become separable. A solution was shown using spheroidal angular and radial functions; however, his application used a plane circular disk excited by a plane wave, which is not representative of the model used here, and the suggested coordinate system was unknown to me so I decided not to follow the track – I had already used sufficient time on the subject.
+
+Time has not been wasted since I have learned a lot from this study, but the time could have been used improving the remaining two models. It's a pity, but that's how it is.
+
+“Don't worry; be happy” (Bobby McFarrin).
+---PAGE_BREAK---
+
+## 7.5 Laplace transformation
+
+The transformation was developed by Pierre Simon Laplace (France, 1749-1827) for solving differential equations of the initial-value type. It is closely related to the Fourier transform but the definition range of the independent variable is semi-infinite, which is the reason for considering the transformation within this study.
+
+### 7.5.1 One variable
+
+The Laplace transform of function $f(t)$ is defined by the following integral (Asmar, 480). Variable $t$ is commonly associated by time since the transformation was developed for solving transient problems with electrical networks, but it is a general-purpose variable and there are no restrictions on its use for other coordinate systems:
+
+$$L\{f(t)\} = \hat{f}(s) = \int_{0}^{\infty} f(t)e^{-st}dt \qquad 150$$
+
+The function must be exponentially limited within the range $0 \le t < \infty$, in order for the integral to exist. The transform exists for sufficiently large $s$, provided such the function satisfies the following conditions: $f(t) = 0$ for $t < 0$; $f(t)$ is continuous or piece-wise continuous in every interval; $|t'||f(t)| < \infty$ as $t \to 0$ for $n > 1$; and $\exp(-s_0t)|f(t)| < \infty$ as $t \to \infty$ for some number $s_0$.
+
+The unit of the transformed variable $s$ is reciprocal distance (m⁻¹) and represents revolutions per length so the transformed variable represents an “oscillation domain” where $s$ is low for slow variations with $r$ and $s$ is high for fast variations with $r$.
+
+The Laplace transform is linear, so the sum of functions $f(t)$ and $g(t)$ with scaling constants $a$ and $b$ can be transformed term by term:
+
+$$L\{af(t)+bg(t)\} = \hat{a}(s)+\hat{b}(s) \qquad 151$$
+
+Other features of the Laplace transform will be quoted whenever required.
+
+An inverse Laplace transformation exists, but will not be considered within this study.
+
+### 7.5.2 Two variables
+
+The Laplace transformation and is defined for functions of one variable but the present study require transformation of $p(r, \theta)$, which is a function of two variables so the basic definition must be changed to include a second variable. This is done by “renaming” the second variable into a *parameter* instead of *variable*. The parameter remains fixed during the transformation but can be changed at will after the transformation. Laplace transforming a function of two parameters will use the following definition, where $r$ is the transformed variable and $\theta$ is a fixed parameter.
+
+$$L\{f(r,\theta)\} = \hat{f}(s,\theta) = \int_{0}^{\infty} f(r,\theta)e^{-sr}dr \qquad 152$$
+
+The unit is changed by the Laplace transformation so a sound pressure is changed from force per length squared (N/m² = Pa) to force per length (N/m).
+---PAGE_BREAK---
+
+## 7.6 Laplace transformed wave equation
+
+The wave equation to be transformed is (see equation 148):
+
+$$ (\sigma + a)^2 \frac{\partial^2 p}{\partial \sigma^2} + 2(\sigma + a) \frac{\partial p}{\partial \sigma} + \frac{\partial^2 p}{\partial \theta^2} + \cot(\theta) \frac{\partial p}{\partial \theta} + k^2 (\sigma + a)^2 p = 0 $$
+
+The variable has been substituted using $r = \sigma + a$, where $a$ is a non-zero positive real number, in order to avoid a singularity at $r = 0$. The new variable $\sigma$ may equal zero without singularity problems during the Laplace transformation. The derivative is not hanged: $dr = d\sigma$.
+
+Using the linearity of the Laplace transform, the terms can be transformed individually.
+
+### 7.6.1 Laplace transform of $r^2 d^2 p/dr^2$
+
+We start by quoting the definition of the Laplace transform using the new variable.
+
+$$ L\left\{ r^2 \frac{\partial^2 p(r, \theta)}{\partial r^2} \right\} = L\left\{ (\sigma + a)^2 \frac{\partial^2 p(\sigma + a, \theta)}{\partial \sigma^2} \right\} = \int_0^\infty (\sigma + a)^2 \frac{\partial^2 p(\sigma + a, \theta)}{\partial \sigma^2} \exp(-s(\sigma + a))d\sigma $$
+
+The second-order derivative of the exponential will be used to remove the squared term.
+
+$$ \frac{\partial^2}{\partial s^2} \{\exp(-s(\sigma+a))\} = (\sigma+a)^2 \exp(-s(\sigma+a)) $$
+
+Differentiation can be moved outside the integration since the exponential is the only term with $s$-dependency.
+
+$$ L\left\{ r^2 \frac{\partial^2 p(r, \theta)}{\partial r^2} \right\} = \frac{\partial^2}{\partial s^2} \int_0^\infty \frac{\partial^2 p(\sigma + a, \theta)}{\partial \sigma^2} \exp(-s(\sigma + a)) d\sigma $$
+
+Partial integration (see section 7.1.1) will be used to remove the differentiation of the sound pressure and two partial integrations will be required since differentiation is of second order. The first partial integration of the integral gives:
+
+$$ \int_0^\infty \frac{\partial^2 p(\sigma + a, \theta)}{\partial \sigma^2} \exp(-s(\sigma + a)) d\sigma = \left[ \frac{\partial p(\sigma + a, \theta)}{\partial \sigma} \exp(-s(\sigma + a)) \right]_{\sigma=0}^{\infty} \\ -(-s) \int_0^\infty \frac{\partial p(\sigma + a, \theta)}{\partial \sigma} \exp(-s(\sigma + a)) d\sigma $$
+
+The derivative of the sound pressure is bounded for $\sigma$ approaching infinity since the sound pressure decays gradually toward zero for radius approaching infinity, so the only output from the square brackets is the derivative of an initial term.
+
+$$ L\left\{ r^2 \frac{\partial^2 p(r, \theta)}{\partial r^2} \right\} = \frac{\partial^2}{\partial s^2} \left\{ -\frac{\partial p(a, \theta)}{\partial \sigma} \exp(-sa) + s \int_0^\infty \frac{\partial p(\sigma + a, \theta)}{\partial \sigma} \exp(-s(\sigma + a)) d\sigma \right\} $$
+---PAGE_BREAK---
+
+Partial integration will again be used to integrate the derivative of the sound pressure.
+
+$$
+\int_0^\infty \frac{\partial p(\sigma + a, \theta)}{\partial \sigma} \exp(-s\sigma) d\sigma = \left[ p(\sigma + a, \theta) \exp(-s(\sigma + a)) \right]_{\sigma=0}^{\sigma=\infty} \\
+- \int_0^\infty (-s) p(\sigma + a, \theta) \exp(-s(\sigma + a)) d\sigma
+$$
+
+The sound pressure is bounded for σ approaching infinity so the only output from the square brackets is an initial term for σ = 0 and the last term is recognised as the definition of the Laplace transform of the sound pressure when the integration variable is changed from dσ to d(σ + a).
+
+$$
+L\left\{ r^2 \frac{\partial^2 p(r, \theta)}{\partial r^2} \right\} = \frac{\partial^2}{\partial s^2} \left\{ -\frac{\partial p(a, \theta)}{\partial \sigma} \exp(-sa) + s[-p(a, \theta)\exp(-sa) + sL\{p(\sigma + a, \theta)\}] \right\} \\
+= \frac{\partial^2}{\partial s^2} \left\{ -\frac{\partial p(a, \theta)}{\partial \sigma} \exp(-sa) - sp(a, \theta)\exp(-sa) + s^2 \hat{p}(s, \theta) \right\}
+$$
+
+The second-order derivative of the first term is:
+
+$$
+\frac{\partial^2}{\partial s^2} \left\{ \frac{\partial p(a, \theta)}{\partial \sigma} \exp(-sa) \right\} = a^2 \frac{\partial p(a, \theta)}{\partial \sigma} \exp(-sa) = a^2 \frac{\partial p(a, \theta)}{\partial r} \exp(-sa)
+$$
+
+The second-order derivative of the second term is calculated using the rule for differentiation of a product of two functions (see section 7.1.3):
+
+$$
+\begin{align*}
+\frac{\partial^2}{\partial s^2} \{sp(a, \theta)\exp(-sa)\} &= p(a, \theta) \frac{\partial^2}{\partial s^2} \{s\exp(-sa)\} \\
+&= p(a, \theta) \left\{ \frac{\partial^2 s}{\partial s^2} \exp(-sa) + 2 \frac{\partial s}{\partial s} \frac{\partial \exp(-sa)}{\partial s} + s \frac{\partial^2 \exp(-sa)}{\partial s^2} \right\} \\
+&= p(a, \theta) [-2a\exp(-sa) + sa^2 \exp(-sa)]
+\end{align*}
+$$
+
+The second-order derivative of the third term is:
+
+$$
+\begin{align*}
+\frac{\partial^2}{\partial s^2} \{s^2 \hat{p}(s, \theta)\} &= \frac{\partial^2 s^2}{\partial s^2} \hat{p}(s, \theta) + 2 \frac{\partial s^2}{\partial s} \frac{\partial \hat{p}(s, \theta)}{\partial s} + s^2 \frac{\partial^2 \hat{p}(s, \theta)}{\partial s^2} \\
+&= \hat{p}(s, \theta) + 4s \frac{\partial \hat{p}(s, \theta)}{\partial s} + s^2 \frac{\partial^2 \hat{p}(s, \theta)}{\partial s^2}
+\end{align*}
+$$
+
+Assembling:
+
+$$
+\begin{align*}
+-\frac{\partial^2}{\partial s^2} \frac{\partial p(a, \theta)}{\partial \sigma} \exp(-sa) &= -a^2 \frac{\partial p(a, \theta)}{\partial r} \exp(-sa) \\
+-\frac{\partial^2}{\partial s^2} sp(a, \theta) \exp(-sa) &= -p(a, \theta)[-2a\exp(-sa) + sa^2 \exp(-sa)] \\
++\frac{\partial^2}{\partial s^2} s^2 \hat{p}(s, \theta) &= \hat{p}(s, \theta) + 4s \frac{\partial \hat{p}(s, \theta)}{\partial s} + s^2 \frac{\partial^2 \hat{p}(s, \theta)}{\partial s^2}
+\end{align*}
+$$
+---PAGE_BREAK---
+
+Hence, the Laplace transform of the first term of the wave equation:
+
+$$L\left\{r^2 \frac{\partial^2 p(r, \theta)}{\partial r^2}\right\} = a \exp(-sa) \left( [sa-2]p(a, \theta) - a \frac{\partial p(a, \theta)}{\partial r} \right) \\ + \hat{p}(s, \theta) + 4s \frac{\partial \hat{p}(s, \theta)}{\partial s} + s^2 \frac{\partial^2 \hat{p}(s, \theta)}{\partial s^2}$$
+
+Note that the first part of the transform disappears for $a \to 0$ (although $a > 0$ is required).
+
+### 7.6.2 Laplace transform of rdp/dr
+
+We start by quoting the definition of the Laplace transform with the term to Laplace transform and using the new variable.
+
+$$L\left\{r \frac{\partial p(r, \theta)}{\partial r}\right\} = L\left\{(\sigma + a) \frac{\partial p(\sigma + a, \theta)}{\partial \sigma}\right\} = \int_{0}^{\infty} (\sigma + a) \frac{\partial p(\sigma + a, \theta)}{\partial \sigma} \exp(-s(\sigma + a)) d\sigma$$
+
+The derivative of the exponential can be used to eliminate the $\sigma + a$ term:
+
+$$\frac{\partial}{\partial s}\{\exp(-s(\sigma+a))\} = -(\sigma+a)\exp(-s(\sigma+a))$$
+
+Differentiation can be moved outside the integration since the exponential is the only term with $s$-dependency and a negative sign balances the sign from the differentiation:
+
+$$L\left\{r \frac{\partial p(r, \theta)}{\partial r}\right\} = -\frac{\partial}{\partial s} \int_{0}^{\infty} \frac{\partial p(\sigma + a, \theta)}{\partial \sigma} \exp(-s(\sigma + a)) d\sigma$$
+
+Partial integration will be used to remove the differentiation of the sound pressure.
+
+$$\int_0^\infty \frac{\partial p(\sigma + a, \theta)}{\partial \sigma} \exp(-s(\sigma + a)) d\sigma = \left[p(\sigma + a, \theta) \exp(-s(\sigma + a))\right]_{\sigma=0}^{\infty} \\ -(-s) \int_0^\infty p(\sigma + a, \theta) \exp(-s(\sigma + a)) d\sigma$$
+
+The sound pressure is bounded for $\sigma$ approaching infinity so the only output from the square brackets is an initial term for $\sigma = 0$ and the last term is recognised as the definition of the Laplace transform of the sound pressure.
+
+$$L\left\{r \frac{\partial p(r, \theta)}{\partial r}\right\} = -\frac{\partial}{\partial s} \left\{p(a, \theta) \exp(-sa) + s\hat{p}(s, \theta)\right\}$$
+
+The derivative of the first term is:
+
+$$\frac{\partial}{\partial s}\{p(a, \theta)\exp(-sa)\} = -ap(a, \theta)\exp(-sa)$$
+
+The derivative of the second term is:
+
+$$\frac{\partial}{\partial s}\{\hat{s}p(s, \theta)\} = \frac{\partial s}{\partial s}\hat{p}(s, \theta) + s\frac{\partial \hat{p}(s, \theta)}{\partial s} = \hat{p}(s, \theta) + s\frac{\partial \hat{p}(s, \theta)}{\partial s}$$
+---PAGE_BREAK---
+
+Hence, the Laplace transform of the second term of the wave equation.
+
+$$L\left\{r \frac{\partial p(r, \theta)}{\partial r}\right\} = a \exp(-sa)p(a, \theta) - \hat{p}(s, \theta) - s \frac{\partial \hat{p}(s, \theta)}{\partial s} \qquad 154$$
+
+### 7.6.3 Laplace transform of dp/dθ and d²p/dθ²
+
+We start by quoting the definition of the Laplace transform using the new variable.
+
+$$L\left\{\frac{\partial p(r, \theta)}{\partial \theta}\right\} = L\left\{\frac{\partial p(\sigma + a, \theta)}{\partial \theta}\right\} = \int_{0}^{\infty} \frac{\partial p(\sigma + a, \theta)}{\partial \theta} \exp(-s(\sigma + a))d\sigma$$
+
+The differentiation can be moved to the front of the integration since the sound pressure is the only term with $\theta$-dependency.
+
+$$L\left\{\frac{\partial p(r, \theta)}{\partial \theta}\right\} = \frac{\partial}{\partial \theta} \int_{0}^{\infty} p(\sigma + a, \theta) \exp(-s(\sigma + a)) d\sigma$$
+
+A substitution of the integration variable from $d\sigma$ to $d(\sigma + a)$ shows that the integral is the definition of the Laplace transformation; hence, the Laplace transform of the third term of the wave equation.
+
+$$L\left\{\frac{\partial p(r, \theta)}{\partial \theta}\right\} = \frac{\partial}{\partial \theta} \int_{0}^{\infty} p(\sigma + a, \theta) \exp(-s(\sigma + a)) d\sigma$$
+
+Similar arguments leads to the Laplace transform of the second-order derivative with respect to the angle; hence, the Laplace transform of the fourth term of the wave equation.
+
+$$L\left\{\frac{\partial^2 p(r, \theta)}{\partial \theta^2}\right\} = \frac{\partial^2 \hat{p}(s, \theta)}{\partial s^2} \qquad 156$$
+
+This is not surprising since the angle is a fixed variable during the transformation (parameter).
+
+### 7.6.4 Laplace transform of $r^2p$
+
+We start by quoting the definition of the Laplace transform using the new variable.
+
+$$L\{r^2 p(r, \theta)\} = L\{\langle \sigma + a \rangle^2 p(\sigma + a, \theta)\} = \int_0^\infty (\sigma + a)^2 p(\sigma + a, \theta) \exp(-s(\sigma + a)) d\sigma$$
+
+Second-order differentiation of the exponential is used to remove the squared factor.
+
+$$\frac{\partial^2}{\partial s^2} \{\exp(-s(\sigma + a))\} = (\sigma + a)^2 \exp(-s(\sigma + a))$$
+
+The differentiation can be moved to the front of the integration since the exponential is the only term with $s$-dependency.
+
+$$L\{r^2 p(r, \theta)\} = \frac{\partial^2}{\partial s^2} \int_0^\infty p(\sigma + a, \theta) \exp(-s(\sigma + a)) d\sigma$$
+---PAGE_BREAK---
+
+The integral is recognised as the Laplace transform of the sound pressure; hence, the Laplace
+transform of the fifth term of the wave equation.
+
+$$
+L\{r^2 p(r, \theta)\} = \frac{\partial^2 \hat{p}(s, \theta)}{\partial s^2}
+$$
+
+Ørsted • DTU – Acoustical Technology
+
+175
+---PAGE_BREAK---
+
+# 8 References
+
+## 8.1 Books
+
+Leo. L. **Beranek**: *Acoustics*. 1993 Edition, Massachusetts Institute of Technology, 1996.
+
+Dean G. **Duffy**: *Green's Functions with Applications*, Chapman & Hall/CRC, 2001.
+
+O. S. **Heavens** and R. W. **Ditchburn**: *Insight into Optics*. Wiley, 1991.
+
+Finn **Jacobsen**: *Propagation of Sound Waves in Ducts*, Note 31260, Ørsted•DTU.
+
+Franklin F. **Kuo**: *Network Analysis and Synthesis*. Wiley, 1966.
+
+W. Marshall **Leach**: *Introduction to Electroacoustics & Audio Amplifier Design*.
+Kendall/Hunt Publishing Company, 2003.
+
+Brian C. J. **Moore**: *An Introduction to the Psychology of Hearing*, 5th Edition, Elsevier, 2004.
+
+Frank L. Pedrotti and Leno S. **Pedrotti**: *Introduction to Optics*, 2nd Ed, Prentice Hall, 1987.
+
+Lennart Råde & Bertil **Westergren**: *Mathematics Handbook for Science and Engineering*.
+Studentlitteratur, 5th Ed, 2004.
+
+## 8.2 Journal of the Audio Engineering Society
+
+Finn **Agerkvist**: *A Study of Simple Diffraction Models*, 1997.
+
+R. M. Bews & M. J. Hawksford: *Application of the Geometric Theory of Diffraction (GTD) to Diffraction at the Edges of Loudspeaker Baffles*, 1986.
+
+M. **Urban**, et al.; *The Distributed Edge Dipole (DED) Model for Cabinet Diffraction Effects*, 2004.
+
+John **Vanderkooy**: *A Simple Theory of Cabinet Edge Diffraction*, 1991.
+
+J. R. **Wright**: Fundamentals of Diffraction, 1996.
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