Unrecognized function or variable 'quad8'

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Jose Iglesias
Jose Iglesias 2022-2-12
评论： Jose Iglesias 2022-2-16
Greetings,
Currenlty working on a project for a radar class I am taking and I keep getting an error for Unrecognized function or variable 'quad8'. As far as I know quad is supposed to numerciallly evaluate an integral. I tried using quall and intgeral instead but keep getting an error. I am basically troubleshooting this entire code which is lengthy and I can certainly use some advice. The end result is a Power Spectral Density plot and a isodop footprint plot which is describing a radar travelling horizontally with a narrow beam antenna squinted at 45 degrees.I attached pics of what plots should look like. I am including the code here.Thank you in advance for any help you may offer!
clear all;
close;
format long;
global r;
%
%Setup of the parameters
%
u=7.5e+3; %speed of the plane
alpha0=2.5*pi/180; %beam depth in radians
alpha_lim=1.5*alpha0; %limit angle for integration
phi=45*pi/180; %squinted angkle in radiians
theta=30*pi/180; %vertial plotting anglein radians
%Finding the limit in the integration
%path in terms of relative Doppler frequency
%
c = cos(alpha_lim);
s = sin(alpha_lim);
fu = (8*c*c-7)/(-2*sqrt(14)*s+2*sqrt(2)*c)
fl = (8*c*c-7)/(2*sqrt(14)*s+2*sqrt(2)*c)
%
% Computing the power spectral density(psd)
%
fd = linspace(fu,fl,502); % rrelative Doppler frequency
fr = 1./fd; % fr parameter
Num = length(fr);
psd = zeros(1,Num-2); % We exclude the boundary points
% in the psd
footprint_u = zeros(1,Num-2);
footprint_l = zeros(1,Num-2);
footprint_f = zeros(1,Num-2);
for n = 2:Num-1,
r = fr(n);
t1 = sqrt(r*r-1);
t2 = sqrt(7*(r*r-1)-(1-2*sqrt(2)*c*r)^2);
t3 = 1+t1-2*sqrt(2)*c*r;
su = (-sqrt(6)*t1-t2)/t3;
sl = (-sqrt(6)*t1+t2)/t3;
uu = log(su);
ul = log(sl);
footprint_u(n-1) = uu;
footprint_l(n-1) = ul;
footprint_f(n-1) = r;
psd(n-1) = quad8('f_int',ul,uu)/(sqrt(1-1/r/r))^3;
end
psd= psd/max(psd);
fd_plot = fd(2:Num-1);
plot(fd_plot,psd);
%semilogy(fd_plot,psd);
hold on
x_fl=[fd(1) fd(1)];
y_fl=[0 1];
plot(x_fl,y_fl,'k')
%semilogy(x_fl,y_fl,'k')
x_fu=[fd(Num) fd(Num)];
y_fu=[0 1];
plot(x_fu,y_fu,'k')
%semilogy(x_fu,y_fu,'k')
title('Fading Spectrum')
xlabel('f_{D}/f_{D_{0}}')
ylabel('PSD/PSD_{0}')
text(0.281,.5,'lowest f_{d}/f_{d_{0}}=0.292');
text(0.385,.5,'highest f_{d}/f_{d_{0}}=0.414');
%
% generating the Limit Isodops and
% The footprint
%
figure(2)
u_hyp=linspace(0,1,100);
x_hyp_low=cosh(u_hyp)./(sqrt((1/fu)^2-1));
y_hyp_low_sinh(u_hyp);
plot(x_hyp_low,y_hyp_low,'k--')
hold on;
x_hyp_up=cosh(u_hyp)./(sqrt((1/fl)^2-1));
y_hyp_up=sinh(u_hyp);
plot(x_hyp_up,y_hyp_up,'k--')
% The footprint
% Those point belong to thte isodops
% and for those points u=uu or ul
%
u_footprint = zeros(1,2*Num-4);
f_footprint = zeros(1,2*Num-4);
for n = 1:Num-2,
u_footprint1(n)=footprint_u(n);
u_footprint2(n)=footprint_1(n);
f_footprint1(n)=footprint_f(n);
f_footprint2(n)=footprint_f(n);
end
x_footprint1=cosh(u_footprint1)./(sqrt((f_footprint1).^2-1));
y_footprint1=sinh(u_footprint1);
plot(x_footprint1,y_footprint1);
x_footprint2=cosh(u_footprint2)./(sqrt((f_footprint2).^2-1));
y_footprint2=sinh(u_footprint2);
plot(x_footprint2,y_footprint2);
axis equal;
set(gca, 'xlim', [0 0.8], 'ylim', [0 0.8])
title('Footprint')
xlabel('X=x/h')
ylabel('Y=y/h')
grid on
text(0.25, 0.1, 'f_{d}/f_{d}_{0}=0.292')
text(0.5, 0.7, 'f_{d}/f_{d}_{0}=0.414')
%
% This is the integrand for the psd
%
function y = f_int(x)
global r
s = exp(x);
rterm = sqrt(r*r-1);
alpha0 = 2.5*pi/180;
alpha = acos(((1+rterm)*s.*s+2*sqrt(6)*rterm*s+(1-rterm))./ ...
(2*sqrt(2)*r*(s.*s+1)));
y = (cosh(x).^2-1/r/r).*exp(-2*alpha.*alpha/alpha0/alpha0);
end

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采纳的回答

Jose Iglesias
Jose Iglesias 2022-2-13
Thank you tremendously for all your guidance!!
  1 个评论
Voss
Voss 2022-2-13
编辑：Voss 2022-2-13
No problem! How about you do me a favor and unaccept your answer and accept mine? I appreciate it!

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更多回答（1 个）

Voss
Voss 2022-2-12
编辑：Voss 2022-2-12
Ran in:
As mentioned in the answer here (with plenty of reference links in it):
https://www.mathworks.com/matlabcentral/answers/301131-quad8-in-matlab-7#answer_232947
quad8() has been removed and replaced by quadl(), which was then replaced by integral().
Changing quad8 to integral and changing the first argument to a function handle - and fixing a couple of typos - seems to have gotten the code to run:
clear all;
close;
format long;
global r;
%
%Setup of the parameters
%
u=7.5e+3; %speed of the plane
alpha0=2.5*pi/180; %beam depth in radians
alpha_lim=1.5*alpha0; %limit angle for integration
phi=45*pi/180; %squinted angkle in radiians
theta=30*pi/180; %vertial plotting anglein radians
%Finding the limit in the integration
%path in terms of relative Doppler frequency
%
c = cos(alpha_lim);
s = sin(alpha_lim);
fu = (8*c*c-7)/(-2*sqrt(14)*s+2*sqrt(2)*c)
fu =
0.413975431045285
fl = (8*c*c-7)/(2*sqrt(14)*s+2*sqrt(2)*c)
fl =
0.291617380244238
%
% Computing the power spectral density(psd)
%
fd = linspace(fu,fl,502); % rrelative Doppler frequency
fr = 1./fd; % fr parameter
Num = length(fr);
psd = zeros(1,Num-2); % We exclude the boundary points
% in the psd
footprint_u = zeros(1,Num-2);
footprint_l = zeros(1,Num-2);
footprint_f = zeros(1,Num-2);
for n = 2:Num-1,
r = fr(n);
t1 = sqrt(r*r-1);
t2 = sqrt(7*(r*r-1)-(1-2*sqrt(2)*c*r)^2);
t3 = 1+t1-2*sqrt(2)*c*r;
su = (-sqrt(6)*t1-t2)/t3;
sl = (-sqrt(6)*t1+t2)/t3;
uu = log(su);
ul = log(sl);
footprint_u(n-1) = uu;
footprint_l(n-1) = ul;
footprint_f(n-1) = r;
% psd(n-1) = quad8('f_int',ul,uu)/(sqrt(1-1/r/r))^3;
psd(n-1) = integral(@f_int,ul,uu)/(sqrt(1-1/r/r))^3;
end
psd= psd/max(psd);
fd_plot = fd(2:Num-1);
plot(fd_plot,psd);
%semilogy(fd_plot,psd);
hold on
x_fl=[fd(1) fd(1)];
y_fl=[0 1];
plot(x_fl,y_fl,'k')
%semilogy(x_fl,y_fl,'k')
x_fu=[fd(Num) fd(Num)];
y_fu=[0 1];
plot(x_fu,y_fu,'k')
%semilogy(x_fu,y_fu,'k')
title('Fading Spectrum')
xlabel('f_{D}/f_{D_{0}}')
ylabel('PSD/PSD_{0}')
text(0.281,.5,'lowest f_{d}/f_{d_{0}}=0.292');
text(0.385,.5,'highest f_{d}/f_{d_{0}}=0.414');
%
% generating the Limit Isodops and
% The footprint
%
figure(2)
u_hyp=linspace(0,1,100);
x_hyp_low=cosh(u_hyp)./(sqrt((1/fu)^2-1));
% y_hyp_low_sinh(u_hyp);
y_hyp_low=sinh(u_hyp); % possible typo corrected
plot(x_hyp_low,y_hyp_low,'k--')
hold on;
x_hyp_up=cosh(u_hyp)./(sqrt((1/fl)^2-1));
y_hyp_up=sinh(u_hyp);
plot(x_hyp_up,y_hyp_up,'k--')
% The footprint
% Those point belong to thte isodops
% and for those points u=uu or ul
%
u_footprint = zeros(1,2*Num-4);
f_footprint = zeros(1,2*Num-4);
for n = 1:Num-2,
u_footprint1(n)=footprint_u(n);
% u_footprint2(n)=footprint_1(n);
u_footprint2(n)=footprint_l(n); % possible typo corrected
f_footprint1(n)=footprint_f(n);
f_footprint2(n)=footprint_f(n);
end
x_footprint1=cosh(u_footprint1)./(sqrt((f_footprint1).^2-1));
y_footprint1=sinh(u_footprint1);
plot(x_footprint1,y_footprint1);
x_footprint2=cosh(u_footprint2)./(sqrt((f_footprint2).^2-1));
y_footprint2=sinh(u_footprint2);
plot(x_footprint2,y_footprint2);
axis equal;
set(gca, 'xlim', [0 0.8], 'ylim', [0 0.8])
title('Footprint')
xlabel('X=x/h')
ylabel('Y=y/h')
grid on
text(0.25, 0.1, 'f_{d}/f_{d}_{0}=0.292')
text(0.5, 0.7, 'f_{d}/f_{d}_{0}=0.414')
%
% This is the integrand for the psd
%
function y = f_int(x)
global r
s = exp(x);
rterm = sqrt(r*r-1);
alpha0 = 2.5*pi/180;
alpha = acos(((1+rterm)*s.*s+2*sqrt(6)*rterm*s+(1-rterm))./ ...
(2*sqrt(2)*r*(s.*s+1)));
y = (cosh(x).^2-1/r/r).*exp(-2*alpha.*alpha/alpha0/alpha0);
end
  8 个评论
显示 6更早的评论
Voss
Voss 2022-2-13
You could also edit your path so it doesn't include that directory.
Jose Iglesias
Jose Iglesias 2022-2-16
Thank you for all your help. I renamed the file and it works.

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