how can execute the following program/
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%Road profile
t=1:0.1:100;
dt=0.002*sin(2*pi*t)+0.002*sin(7.5*pi*t);
t1=t-3.5;
t2=t-6.5;
t3=t-8.5;
t4=t-11.5;
if((t>=3.5) && (t<5))
zr= -0.0592*t1^3+0.1332*t1^2+dt;
elseif((t>=5) && (t<6.5))
zr=0.0592*t2^3+0.1332*t2^2+dt;
elseif((t>=8.5) && (t<10))
zr=0.0592*t3^3-0.1332*t3^2+dt;
elseif((t>=10) && (t<11.5))
zr=-0.0592-0.1332*t4^2+dt;
else
dt;
end
plot(zr)
when i run it, i have the following error
Operands to the and && operators must be convertible to logical scalar values.
回答（2 个）
Walter Roberson
2012-8-31
0 个投票
This is not surprising in your code. You create t as a vector, so (t>=3.5) will be a vector of logical results, and (t<5) will be a vector of logical results, and then you attempt to use the scalar operator && against those two vectors.
I would suggest that you need to research "logical indexing"
Image Analyst
2012-8-31
I think you're looking to do this:
% Road profile
t = 1:0.1:100;
dt = 0.002*sin(2*pi*t)+0.002*sin(7.5*pi*t);
t1=t-3.5;
t2=t-6.5;
t3=t-8.5;
t4=t-11.5;
% Initialize
zr = dt;
% Assign the various elements
indexes = (t>=3.5) & (t<5)
zr(indexes) = -0.0592*t1(indexes).^3+0.1332*t1(indexes).^2+dt(indexes);
indexes = (t>=5) & (t<6.5)
zr(indexes) = 0.0592*t2(indexes).^3+0.1332*t2(indexes).^2+dt(indexes);
indexes = (t>=8.5) & (t<10)
zr(indexes) = 0.0592*t3(indexes).^3-0.1332*t3(indexes).^2+dt(indexes);
indexes = (t>=10) & (t<11.5)
zr(indexes) = -0.0592-0.1332*t4(indexes).^2+dt(indexes);
% Plot it.
plot(zr);
grid on;
fontSize = 24;
title('Road Profile', 'FontSize', fontSize);
xlabel('t', 'FontSize', fontSize);
ylabel('zr', 'FontSize', fontSize);
% Enlarge figure to full screen.
set(gcf, 'units','normalized','outerposition',[0 0 1 1]);
% Give a name to the title bar.
set(gcf,'name','Demo by ImageAnalyst','numbertitle','off')
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2012-8-31
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