Array indices must be positive integers or logical values.

2021 7 10
2 个回答
回答已采纳
更新时间：2021 7 10
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0 个投票

h = 0.2;
T = 4:h:8
X_s=[-5.87 -4.23 -2.55 -0.89 0.67 2.09 3.31 4.31 5.06 5.55 5.78 5.77 5.52 5.08 4.46 3.72 2.88 2.00 1.10 .23 -0.59]
for X=4:0.2:8 %%%need to take the first and last point off since central difffrencing doesnt work on them
k=[X];
x = round(k/0.2 - 4/0.2 + 1);
V = ((X_s((x)+1))-(X_s((x)-1)))/(2*h);
if X==4
k=[X];
x = round(k/0.2 - 4/0.2 + 1);
V = ((X_s((x)+1))-(X_s((x))))/(2*h);
end
if X==8
k=[X];
x = round(k/0.2 - 4/0.2 + 1);
V = ((X_s((x)))-(X_s((x-1))))/(2*h);
end
Velcoity=V
end
Hey there,
Im now trying to get the array to work with two end points where it can take them and redirect them to equations suited to solving them with the given data. However I find myself getting the error "Array indices must be positive integers or logical values". I've tried reorginizing my if statments to come first and that doesnt seem to work. If i change the start to 4.2 and the end to 7.8 it works like a charm but cant figure out why it wont accept the start as 4 and end as 8 and use the forward and backward diffrencing equations. Any help or knowledge helps and thank you for looking.

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 采纳的回答

Walter Roberson
Walter Roberson 2021-7-10
Ran in:

0 个投票

Ran in:
h = 0.2;
X_s = [-5.87 -4.23 -2.55 -0.89 0.67 2.09 3.31 4.31 5.06 5.55 5.78 5.77 5.52 5.08 4.46 3.72 2.88 2.00 1.10 .23 -0.59]
X_s = 1×21
-5.8700 -4.2300 -2.5500 -0.8900 0.6700 2.0900 3.3100 4.3100 5.0600 5.5500 5.7800 5.7700 5.5200 5.0800 4.4600 3.7200 2.8800 2.0000 1.1000 0.2300 -0.5900
nX = length(X_s);
for x = 1 : nX
if x == 1
V = ((X_s((x)+1))-(X_s((x))))/(2*h);
elseif x == nX
V = ((X_s((x)))-(X_s((x-1))))/(2*h);
else
V = ((X_s((x)+1))-(X_s((x)-1)))/(2*h);
end
Velocity(x) = V;
end
plot(1:nX, X_s, 'k-', 1:nX, Velocity, 'b--')

更多回答（1 个）

G A
G A 2021-7-10

0 个投票

When x=1, then X_s(x-1) = X_s(0), which is not allowed in Matlab
x =
1
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3 个评论
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Mark Prentice
Mark Prentice 2021-7-10
yes, but the if statment is supposed to take the X=4 value and run it through a diffrent equation to avoid a zero value. I cant seem to arrange the whole thing in any way to fix this issue though.
Walter Roberson
Walter Roberson 2021-7-10
Yes, but you have
V = ((X_s((x)+1))-(X_s((x)-1)))/(2*h);
before you test whether X == 4
Mark Prentice
Mark Prentice 2021-7-10
Yes, I've been rearranging the whole system of 3 equations in every and all configurations. Such as putting it last however I'm still getting the error. I'm wondering if theres another variable im missing or a function that makes the 4 value on go into one equation ( if im useing the if functions right). But yes you are correct V = ((X_s((x)+1))-(X_s((x)-1)))/(2*h); should not be first.

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