Hello, if you have a moment, I could use your help in resolving this error.

[Danikha Shyken]

Below is the code:

function [q1, q2, q3, q4] = buckBoostControl(inputVoltage, ~, ~)
% Constants for duty cycle and delay values
dutyCycleBoost = [85.71, 81.82, 60.00];
delayBoost = [51.444, 65.448, 144.000];
dutyCycleBuck = 47.37;
delayBuck = 170.532;
% Check input voltage to determine the mode (Boost or Buck)
if inputVoltage == 0.3 || inputVoltage == 0.4 || inputVoltage == 1.2
    % Boost Mode
    dutyCycle = dutyCycleBoost(inputVoltage == [0.3, 0.4, 1.2]);
    delay = delayBoost(inputVoltage == [0.3, 0.4, 1.2]);
    
    q1 = 1;
    q2 = dutyCycle / 100;
    q3 = 0;
    q4 = 1 - q2;
elseif inputVoltage == 2
    % Buck Mode
    dutyCycle = dutyCycleBuck;
    delay = delayBuck;
    
    q1 = dutyCycle / 100;
    q2 = 0;
    q3 = 1 - q1;
    q4 = 1;
else
    error('Input voltage is out of the specified range.');
end
end

Answer 1

[Guillaume]

Hello,

You should help the interpreter to determine the size of your variables. You can define the size in the data properties or in the code with unambiguous assignments. Your code could work like that :

function [q1, q2, q3, q4] = buckBoostControl(inputVoltage, ~, ~)
% Constants for duty cycle and delay values
dutyCycleBoost = [85.71, 81.82, 60.00];
delayBoost = [51.444, 65.448, 144.000];
dutyCycleBuck = 47.37;
delayBuck = 170.532;


% Check input voltage to determine the mode (Boost or Buck)
ind = inputVoltage == [0.3, 0.4, 1.2]; % avoid to repeat 3 times this "inputVoltage == [0.3, 0.4, 1.2]"

if any(ind) % easier 
    % Boost Mode
    dutyCycle = dutyCycleBoost(ind);
    delay = delayBoost(ind);
    
    q1 = 1;
    q2 = dutyCycle(1) / 100; % add (1) to clearly define size 
    q3 = 0;
    q4 = 1 - q2;
elseif inputVoltage == 2
    % Buck Mode
    dutyCycle = dutyCycleBuck;
    delay = delayBuck;
    
    q1 = dutyCycle / 100;
    q2 = 0;
    q3 = 1 - q1;
    q4 = 1;
else
    error('Input voltage is out of the specified range.');
end
end

[Danikha Shyken]

I got it to work. Thank you so much for your help.

[Guillaume]

You are welcome :-)

Answer 2

[dpb]

inputVoltage=0.3;
[q1, q2, q3, q4]=buckBoostControl(inputVoltage)

q1 = 1

q2 = 0.8571

q3 = 0

q4 = 0.1429

function [q1, q2, q3, q4] = buckBoostControl(inputVoltage, ~, ~)
% Constants for duty cycle and delay values
dutyCycleBoost = [85.71, 81.82, 60.00];
delayBoost = [51.444, 65.448, 144.000];
dutyCycleBuck = 47.37;
delayBuck = 170.532;
% Check input voltage to determine the mode (Boost or Buck)
if inputVoltage == 0.3 || inputVoltage == 0.4 || inputVoltage == 1.2
    % Boost Mode
    dutyCycle = dutyCycleBoost(inputVoltage == [0.3, 0.4, 1.2]);
    delay = delayBoost(inputVoltage == [0.3, 0.4, 1.2]);
    
    q1 = 1;
    q2 = dutyCycle / 100;
    q3 = 0;
    q4 = 1 - q2;
elseif inputVoltage == 2
    % Buck Mode
    dutyCycle = dutyCycleBuck;
    delay = delayBuck;
    
    q1 = dutyCycle / 100;
    q2 = 0;
    q3 = 1 - q1;
    q4 = 1;
else
    error('Input voltage is out of the specified range.');
end
end

The function is ok in MATLAB although

if inputVoltage == 0.3 || inputVoltage == 0.4 || inputVoltage == 1.2

is very problematical in doing exact comparisons with floating point values; internal rounding of even the last significant digit away from the value converted from the text representation to internal representation will be sufficient for the comparison to fail when running and the inputVoltage value is not input manually as in the above.

More robust would be

BOOST_V=[0.3, 0.4, 1.2];                    % put constants in variables so can change instead of in code
BUCK_V=2.0; 
if any(ismembertol(inputVoltage,BOOST_V));  % check if any match
  ...

See ismembertol for details and how to set the "nearness" to a match for comparison; the above default will be at about 1E-12 or roughly three digits of precision of a double precision floating point value. How close to the actual values the computation will return will depend upon just how the inputVoltage value is calculated.

As for the syntax error, it appears that Simulink (I presume?) looks at code with a jaundiced eye and if it can't determine the shape of anything internally, it throws up its hands.

Try the following rearrangement of the code which is also just a little more efficient as it doesn't do the lookup indexing twice...

BOOST_V=[0.3, 0.4, 1.2];                    % put constants in variables so can change instead of in code
BUCK_V=2.0; 
ix=ismembertol(inputVoltage,BOOST_V));      % Check input voltage to determine the mode (Boost or Buck)
if any(ix)                                  % found one matched...Boost Mode
    ix=find(ix);                            % convert logical array to index for addressing
    dutyCycle = dutyCycleBoost(ix);
    delay = delayBoost(ix);
    ...

and see if that will satisfy...

[Danikha Shyken]

All good now. I truly appreciate your help.

[Walter Roberson]

The concerns about == with a floating point number are important.

And what should happen if the input voltage is (for example) 0.5 ?

Perhaps you should consider using interp1() .