if elseif statement operands

Hi,

I have a problem with statement if else, it doesn't work. Error:

Operands to the and && operators must be convertible to logical scalar values.

Error in My_Project (line 31) elseif ((f0<=f)&&(f<fl));

%%My_Project

f = [31.5 40 50 63 80 100 125 160 200 250 315 400 500 630 800....
    1000 1250 1600 2000 2500 3150 4000 5000 6300 8000];

f0=220;
fl=2.58;

if (f<f0)
    TL=TLm;
elseif ((f0<=f)&&(f<fl));
    TL=TLm1+TLm2+20.*log10(f.*d)-29;
else
    TL=TLm1+TLm2+6;
end

Thank you,

Dominika

Answer 1

[nl2605]

You have to give a condition for elseif. Something like

elseif ((f0<=f) && (f<f1)== 1) or zero according to your condition.

Answer 2

[Jos (10584)]

f is an array with multiple elements, and f0 is a scalar (single element). The statement " f < f0 " returns an array the same size as f with true and false values for each element of f. You can use the functions ALL or ANY to convert this to a single value, so it can be used in an IF condition.

However, I suspect that you want to create a new array TL that has the same number of elements as f, but some elements of TL are calculated one way and some another way ...

You can try this

% Assuming TLm, TLm1, TLm2, and d are scalars!
TL = repmat(TLm1+TLm2+6, size(f)) % default values
q1 = (f0<=f) & (f<fl) % condition 1
TL(q1) = TLm1 + TLm2 + 20.*log10(f(q1).*d)-29
q2 = f < f0 % condition 2
TL(q2) = TLm

[Dominika]

Thanks for the answer

However, using your code I get TL array with dimentions 1x625 instead of 1x25

[Dominika]

OK, I got it

Maybe it will be usuful for someone

I dealt with the problem using the following code:

%Input
d=0.25;                     % separation of two panels [m]
m1=6000;                    % density of panel 1 [kg/m^3]
m2=8000;                    % density of panel 2 [kg/m^3]
m=m1+m2;
me=2*m1*m2/(m1+m2);         % [kg/m^3]

f = [31.5 40 50 63 80 100 125 160 200 250 315 400 500 630 800....
    1000 1250 1600 2000 2500 3150 4000 5000 6300 8000]; % frequencies third octave band

f0=113./sqrt(me.*d);        % resonance frequency
fl=55/d;
% Mass law transmission loss
TLm=20.*log10(m.*f)-48;     % total construction [dB]
TLm1=20.*log10(m1.*f)-48;   % panel 1 [dB]
TLm2=20.*log10(m2.*f)-48;   % panel 2 [dB]

%Transmission loss calculation
TL = repmat(TLm1+TLm2+6,1); % default values (TLm1+TLm2+6)
q1 = (f0<=f) & (f<fl); % condition 1
TL(q1) = 20.*log10(m1.*f(q1))-48 + 20.*log10(m2.*f(q1))-48 + 20.*log10(f(q1).*d)-29; % (TLm1+TLm2+20log(fd)-29)
q2 = f < f0; % condition 2
TL(q2) = 20.*log10(m.*f(q2))-48; % (TLm)