Find two x values for one y value

I have a function where I need to know values of f for 0.707*max(WI) value. I tried this code, but found only one value:

C=1*10^(-4);
L=3*10^(-1);
R=100
N2=100;
frez=1/(2*pi*sqrt(L*C))                      %f value for Imax(theory of RLC circles)
WI=zeros(N2+1,1);
f=zeros(N2+1,1);
for n=0:N2
    omega=2*pi*n;
    I=1/(sqrt(R^2+(omega*L-(1/(omega*C)))^2));
    WI(n+1)=I;
    fcz=n;
    f(n+1)=fcz;
end
plot(f,WI)
Imax=max(WI)                                  
Igr=0.707*max(WI)                             %value that I searched for
[~, index] = min(abs(WI - 0.707*max(WI)));    %trying to find f values for Igr but only one show up(should be two)
fgr=f(index)

Answer 1

[Star Strider]

C=1*10^(-4);
L=3*10^(-1);
R=100

R = 100

N2=100;
frez=1/(2*pi*sqrt(L*C))                      %f value for Imax(theory of RLC circles)

frez = 29.0576

WI=zeros(N2+1,1);
f=zeros(N2+1,1);
for n=0:N2
    omega=2*pi*n;
    I=1/(sqrt(R^2+(omega*L-(1/(omega*C)))^2));
    WI(n+1)=I;
    fcz=n;
    f(n+1)=fcz;
end

[WImax,idx] = max(WI)                                               % Find Maximum & Index

WImax = 0.0100

idx = 30

idxv = {1:idx; idx:numel(f)}                                        % Create Index Vectors

idxv = 2×1 cell array
    {[                                                                                                                                         1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30]}
    {[30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101]}

for k = 1:2
    xval(k) = interp1(WI(idxv{k}),f(idxv{k}), WImax/sqrt(2))        % Interpolate To Fine X-Values
end

xval = 12.8235

xval = 1×2

12.8235 65.8711

plot(f,WI)
hold on
plot(xval, [1 1]*WImax/sqrt(2), 'rs')                               % Plot Desired Points
hold off

Imax=max(WI)                                  

Imax = 0.0100

Igr=0.707*max(WI)                             %value that I searched for

Igr = 0.0071

[~, index] = min(abs(WI - 0.707*max(WI)));    %trying to find f values for Igr but only one show up(should be two)
fgr=f(index)

fgr = 66

.

[Bartosz Pasek]

Thank You a lot

[Star Strider]

As always, my pleasure!

Answer 2

[Jan]

tmp      = abs(WI - 0.707*max(WI));
minValue = min(tmp);
index    = (tmp == minValue);
fgr      = f(index)

For numerical values it is unlikely that there are no rounding artifacts. Searching for tmp==minValue is critical. Prefer to use a tolerance of a matching size instead.