For loop with linspcace - for a multiple plots?

STP
2019 2 22
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更新时间：2019 3 4
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beta=5;
omega = 2856;
Qo = 10000;
QL = Qo/(1+beta);
omega_half = omega/(2*QL);
U_in1 = 1;
y1 = 0;
alfa = (2*beta)./(1+beta);
t1 = [0 4.2]
dV1dt = @(t,V) ((U_in1*omega*beta)/Qo) - omega_half*V;
[t1 V1] = ode15s(dV1dt, t1, y1)
figure;
plot(t1,V1);
figure
Aout1=V1-1;
plot(t1, Aout1);
figure
P1 = (V1 - U_in1).^2
plot(t1, P1);
.. further more plots
Now If I wish to produce all such plots for different values fo 'beta' and 'Qo'; which has been stated at the top; how can I go about that? I read the for loop documentation but failed to apply it here.
for eg. beta from 2 to 6 like 2.1, 2.2 etc.. and for Qo from 70000 to 120000 with 500 spacing. Thanks in advance to all the volunteers :)

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 采纳的回答

Star Strider
Star Strider 2019-2-22

1 个投票

Use nested loops:
beta= 2 : 0.1 : 6;
omega = 2*pi*2856;
Qo = 7E+4 : 500 : 1.2E+5;
U_in1 = 1;
y1 = 0;
alfa = (2*beta)./(1+beta);
t0 = [0 4.2];
dV1dt = @(t,V,beta,Qo,omega_half) ((U_in1*omega*beta)/Qo) - omega_half*V;
for k1 = 1:2%numel(beta)
for k2 = 1:2%numel(Qo)
QL = Qo(k2)/(1+beta(k1));
omega_half = omega/(2*QL);
[t1 V1] = ode15s(@(t,V)dV1dt(t,V,beta(k1),Qo(k2),omega_half), t0, y1);
figure
plot(t1,V1)
xlabel('t')
ylabel('V_1')
title(sprintf('\\beta = %.1f Q_o = %6d', beta(k1),Qo(k2)))
figure
Aout1=V1-1;
plot(t1, Aout1)
xlabel('t')
ylabel('A_{out}')
title(sprintf('\\beta = %.1f Q_o = %6d', beta(k1),Qo(k2)))
figure
P1 = (V1 - U_in1).^2;
plot(t1, P1)
xlabel('t')
ylabel('P_1')
title(sprintf('\\beta = %.1f Q_o = %6d', beta(k1),Qo(k2)))
end
end
Note that this will produce 12423 figures! It will probably be better to combine all of them into one figure for each loop iteration using the subplot function, to reduce that to 4141 figures. .

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Jan
Jan 2019-2-22
Even 4141 figures will exhaust the resources of the computer massively.
STP
STP 2019-2-22
编辑：STP 2019-2-22
Yeah, Thanks! it is exhausting for the resources. But I could reduce the span of the parameters and then use subplot!. If I want to add more parts on the same plot for eg:
omega = 2*pi*2856;
Qo = 100000;
beta=5
QL = Qo./(1+beta);
alfa = (2*beta)./(1+beta);
omega_half = omega/(2*QL);
U_in1 = 1;
y1 = 0;
t1 = [0 4.2]
dV1dt = @(t,V) ((U_in1*omega*beta)/Qo) - omega_half*V;
[t1 V1] = ode15s(dV1dt, t1, y1)
% hold on ;
t2 = [4.2 5]
U_in2 = -1;
y2= V1(end);
dV2dt = @(t,V) ((U_in2*omega*beta)/Qo) - omega_half*V;
[t2, V2] = ode15s(dV2dt, t2, y2)
t3 = [5 8]
U_in3 = 0;
y3= V2(end);
dV3dt = @(t,V) ((U_in3*omega*beta)/Qo) - omega_half*V;
[t3, V3] = ode15s(dV3dt, t3, y3)
V = [V1; V2; V3];
t = [t1; t2; t3];
plot(t,V);
How is this achieved in the same way? I tried but its just overlapping on one time period when it should be in 3 different timeperiods.
Star Strider
Star Strider 2019-2-22
To use subplot and your variable vectors, do this:
N = 5;
beta= linspace(2, 6, N);
omega = 2*pi*2856;
Qo = linspace(7E+4, 1.2E+5, N);
U_in1 = 1;
y1 = 0;
alfa = (2*beta)./(1+beta);
t0 = [0 4.2];
dV1dt = @(t,V,beta,Qo,omega_half) ((U_in1*omega*beta)/Qo) - omega_half*V;
for k1 = 1:numel(beta)
for k2 = 1:numel(Qo)
QL = Qo(k2)/(1+beta(k1));
omega_half = omega/(2*QL);
[t1 V1] = ode15s(@(t,V)dV1dt(t,V,beta(k1),Qo(k2),omega_half), t0, y1);
figure
subplot(3,1,1)
plot(t1,V1)
ylabel('V_1')
title(sprintf('\\beta = %.1f Q_o = %6d', beta(k1),Qo(k2)))
subplot(3,1,2)
Aout1=V1-1;
plot(t1, Aout1)
ylabel('A_{out}')
title(sprintf('\\beta = %.1f Q_o = %6d', beta(k1),Qo(k2)))
subplot(3,1,3)
P1 = (V1 - U_in1).^2;
plot(t1, P1)
xlabel('t')
ylabel('P_1')
title(sprintf('\\beta = %.1f Q_o = %6d', beta(k1),Qo(k2)))
end
end
This produces figures, each with 3 subplot axes.
Experiment to get the result you want.
STP
STP 2019-2-25
编辑：STP 2019-2-25
Hi Star Strider. Thankyou :)
Star Strider
Star Strider 2019-2-25
My pleasure!
If my Answer helped you solve your problem, please Accept it!
STP
STP 2019-2-27
编辑：STP 2019-2-27
Hi, Its a little similar- so I am commenting here instead of creating a new question. For the loop - if I wish to plot for all values in a single plot- which is usually achieved by 'hold on' instead of using figures or subplots- so as to see how the curve changes for diff values in one plot. For the same thing how can index or like a small descripion be applied on the fig so as to know what parameter is giving which output as with hold on; it comes all together.
Star Strider
Star Strider 2019-2-27
I am not certain what you want, so I will hazard a guess.
I have been experimenting, and came up with this idea:
N = 10;
beta= linspace(2, 6, N);
omega = 2*pi*2856;
Qo = linspace(7E+4, 1.2E+5, N);
U_in1 = 1;
y1 = 0;
alfa = (2*beta)./(1+beta);
% t0 = [0 4.2];
t0 = linspace(0, 4.2, 50);
dV1dt = @(t,V,beta,Qo,omega_half) ((U_in1*omega*beta)/Qo) - omega_half*V;
for k1 = 1:numel(beta)
for k2 = 1:numel(Qo)
QL = Qo(k2)/(1+beta(k1));
omega_half = omega/(2*QL);
[t1 V1(k1, k2,:)] = ode15s(@(t,V)dV1dt(t,V,beta(k1),Qo(k2),omega_half), t0, y1);
end
end
figure
surfc(t1, beta, squeeze(mean(V1,2)))
xlabel('t')
ylabel('\beta')
figure
surfc(t1, Qo, squeeze(mean(V1,1)))
xlabel('t')
ylabel('Q_o')
NQo = 1;
figure
surfc(t1, beta, squeeze(V1(:,NQo,:)))
xlabel('t')
ylabel('\beta')
title(sprintf('Qo = %3.1E', Qo(NQo)))
Nbeta = 10;
figure
surfc(t1, Qo, squeeze(V1(Nbeta,:,:)))
xlabel('t')
ylabel('Q_o')
title(sprintf('\\beta = %.2f', beta(Nbeta)))
It reduces the 3D ‘V1’ matrix to 2D by taking the mean the other parameter (first two figures), or simply chooses one value of the other parameter (last two figures). I use surfc here to also draw sontours beneath the plot. That gives a bit more information.
Experiment to get the result you want.
STP
STP 2019-3-4
Hi,
I have a question on similar terms- so I am putting it here. As you showed me how to vary Q and beta's both in a loop. I pt Q as one value and I vary beta and plot. Now if I wish to plot P vs beta; I will have to store the value of P at 4.2 for all beta's in a variable, and then plot it with the beta's on x axis - so as to see for what beta am I getting max P. How to achieve that? Again it needs to be part of the loop. Many thanks!
Star Strider
Star Strider 2019-3-4
I am lost. What is ‘P’? It does not appear anywhere.
Since I have no idea what you are now doing, I suggest that you experiment with the code I posted to get the result you want.
STP
STP 2019-3-4
P1 = (V1 - U_in1).^2
which is plot from 0 to 4.2 so if i want to plot all the values of P1 for different betas at 4.2 vs the beta.
for
beta-4-6
P1= ..
plot(t1,P1) // here the value of P1 at 4.4 (end of t1))= stored in a variable)
end
plot(all beta's used, variable which has all P1 values at 4.2 stored) <-- objective
Hopefully the above helps; nevertheless Thanks for your reply!:)
Star Strider
Star Strider 2019-3-4
As always, my pleasure!
Try this:
N = 5;
beta= linspace(2, 6, N);
omega = 2*pi*2856;
Qo = linspace(7E+4, 1.2E+5, N);
U_in1 = 1;
y1 = 0;
alfa = (2*beta)./(1+beta);
t0 = [0 4.2];
dV1dt = @(t,V,beta,Qo,omega_half) ((U_in1*omega*beta)/Qo) - omega_half*V;
for k1 = 1:numel(beta)
for k2 = 1:numel(Qo)
QL = Qo(k2)/(1+beta(k1));
omega_half = omega/(2*QL);
[t1 V1] = ode15s(@(t,V)dV1dt(t,V,beta(k1),Qo(k2),omega_half), t0, y1);
figure
subplot(3,1,1)
plot(t1,V1)
ylabel('V_1')
title(sprintf('\\beta = %.1f Q_o = %6d', beta(k1),Qo(k2)))
subplot(3,1,2)
Aout1=V1-1;
plot(t1, Aout1)
ylabel('A_{out}')
title(sprintf('\\beta = %.1f Q_o = %6d', beta(k1),Qo(k2)))
subplot(3,1,3)
P1 = (V1 - U_in1).^2;
plot(t1, P1)
xlabel('t')
ylabel('P_1')
title(sprintf('\\beta = %.1f Q_o = %6d', beta(k1),Qo(k2)))
P42(k1,k2,:) = P1(end);
end
end
figure
plot(beta, P42)
grid
xlabel('\beta')
ylabel('\itP\rm')
lgnd = sprintfc('Q_o = %.2E', Qo);
legend(lgnd, 'Location','NW')
I believe the last plot is the one you want. The new ‘P42’ matrix stores the values of ‘P1’ at the last value of each integration vector (where ‘t1=4.2’). The rows are ‘beta’, and the columns are ‘Qo’.
Make appropriate changes otherwise.

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