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maximum value not evaluating

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Hello,
I have tried to evaluate my function for it's maximum, but I keep failing. What am I missing here?
I realize this is a very basic calculus problem, but my function I am determing the max output power for has it's own set of variables that I am optimizing, and I think that is why I cannot use the function "handle"? I am sure my terminology is incorrect.
I get the correct plot, but I just want to spit out the value of the maximum of the plot (ymax value), which in the code I have labeled as P_ave_output(R_l_optimal). Shouldn't this give me the max value of the function?
clear all
syms omega t R_l s A_in Y
%Inputs
A_o = 3;
R_c = 40;
L_c = 0.051;
f=186;
omega_in = 2*pi*f;
A = A_o*sin(omega*t);
M=0.01;
%Theta from part a)
theta = 0.244;
%Spring coefficient (N/m)
k = 13660;
%Damping coefficient (N*s/m)
C_d = 0.07;
%Capacitance Assumption
C=12*10^-6;
%Impedance equations
Z1 = R_c+R_l+L_c*s+1/(s*C);
Z2 = C_d+k/s+M*s;
%From part a) transfer function
%Output voltage V_l
V_l = (theta*(Y*s)*R_l)/Z1;
%Input excitation force Z_in
A_in=(Y*s*Z2+theta*V_l/R_l)/(M);
%output voltage ratio to input force
Transfer_func = simplify(V_l/A_in);
Transfer_func(s) = vpa(simplify(V_l/A_in),5);
Transfer_func(omega) = subs(Transfer_func, {s},{1j*omega});
amplitude_TF = A_o*abs(Transfer_func(omega_in));
angle_TF = angle(Transfer_func(omega_in));
V_l_steady_state = amplitude_TF*(sin(omega_in*t+angle_TF));
P_ave_output = (amplitude_TF/(sqrt(2)))^2/R_l;
P_ave_diff = diff(P_ave_output);
P_ave_diff = simplify(P_ave_diff, 'Steps',500);
P_ave_max = P_ave_diff == 0;
R_l_optimal = vpa(abs((solve(P_ave_max, R_l))),3)
fplot(P_ave_output,[10 200])
xticks([10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200])
%xline(72.3)
ylabel('Average Power (Watts)')
xlabel('Load Resistance R_l (Ohms)')
P_ave_output(R_l_optimal)

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Accepted Answer

Star Strider
Star Strider on 9 Jun 2020
Since ‘P_ave_output’ is not defined as function, MATLAB interprets the current syntax as a subscript.
Try this:
plotMax = vpa(subs(P_ave_output,{R_l},{R_l_optimal}),16)
producing:
plotMax =
0.00003278334191412342
with:
R_l_optimal =
42.5
.

  2 Comments

Shane Palmer
Shane Palmer on 9 Jun 2020
Thank you again. This worked well.
Star Strider
Star Strider on 9 Jun 2020
As always, my pleasure!

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