Function doesn't return a vector from a vector input

2019 11 15
2 个回答
回答已采纳
更新时间：2019 11 15
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function Ffork = Fforkfase3(t)
Feff=35; %Load applied by the driver at the shift lever [N], Maximum value allowed: 180-250 N
Feff_max=90; %Max effort applied by the driver [N]
eff=0.6; %Efficiency of the shift mechanism (normally <70%)
TR=7; %Transmission ratios of the shift mechanism (Normally varies between 7:1 and 12:1)
Ffork0=Feff*eff*TR; %Load on the sleeve [N]
Ffork_max=Feff_max*eff*TR; %Max load on the sleeve [N]
Inc_Ffork=(Ffork_max-Ffork0)/0.2;
if Ffork0+Inc_Ffork*t <= Ffork_max
Ffork=Ffork0+Inc_Ffork*t;
elseif Ffork0+Inc_Ffork*t > Ffork_max
Ffork = Ffork_max;
end
end
Hi,
I have a function with a distinction of cases. The force Ffork should be constant once it reached the maximum value, as you can see in the code.
The input t should be a vector, I called the function in this way:
t=0.08:0.01:0.2727
Ffork = Fforkfase3(t)
I would like to get a vector of Ffork as output. However, I got an error message after running it:
Output argument "Ffork" (and maybe others) not assigned during call to "Fforkfase3".
Could someone tell me my mistake?

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 采纳的回答

Star Strider
Star Strider 2019-11-15

1 个投票

‘Could someone tell me my mistake?’
You need to use the any or all functions in the if statements, depending on what you want to do:
if any(Ffork0+Inc_Ffork*t <= Ffork_max)
Ffork=Ffork0+Inc_Ffork*t;
elseif any(Ffork0+Inc_Ffork*t > Ffork_max)
Ffork = Ffork_max;
end
See the documentation on any (and all) to determine the correct code for your function.

更多回答（1 个）

dpb
dpb 2019-11-15

1 个投票

Yes, you didn't read about logical IF
if Ffork0+Inc_Ffork*t <= Ffork_max
is True iff every element of Ffork0+Inc_Ffork*t is <= Ffork_max and similarly for the eseif clause. The upshot is your code didn't pass either test and so the if clause was never executed.
However, "the MATLAB way" doesn't need the if clause anyways; use the vector nature...
function Ffork = Fforkfase3(t)
Feff=35; %Load applied by the driver at the shift lever [N], Maximum value allowed: 180-250 N
Feff_max=90; %Max effort applied by the driver [N]
eff=0.6; %Efficiency of the shift mechanism (normally <70%)
TR=7; %Transmission ratios of the shift mechanism (Normally varies between 7:1 and 12:1)
Ffork0=Feff*eff*TR; %Load on the sleeve [N]
Ffork_max=Feff_max*eff*TR; %Max load on the sleeve [N]
Inc_Ffork=(Ffork_max-Ffork0)/0.2;
Ffork=min(Ffork0+Inc_Ffork*t,Ffork_max);
end

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