why this function is not plotting?

a =5;

T = 2E3;

Z = linspace(0,0.1,0.01);

U = (1+2.*Z)./(2.*a.*T);

plot(Z,U)

Answer 1

[Star Strider]

Look t the linspace result —

a =5;
T = 2E3;
Z = linspace(0,0.1,0.01)

Z =

  1×0 empty double row vector

U = (1+2.*Z)./(2.*a.*T);  
plot(Z,U)


a =5;
T = 2E3;
Z = linspace(0,0.1,150)                                 % 'linspace' Now Produces A Vector

Z = 1×1000

0 0.0001 0.0002 0.0003 0.0004 0.0005 0.0006 0.0007 0.0008 0.0009 0.0010 0.0011 0.0012 0.0013 0.0014 0.0015 0.0016 0.0017 0.0018 0.0019 0.0020 0.0021 0.0022 0.0023 0.0024 0.0025 0.0026 0.0027 0.0028 0.0029

U = (1+2.*Z)./(2.*a.*T);  
plot(Z,U)

.

[Steven Lord]

I suspect @SAHIL SAHOO expected this line of code:

Z = linspace(0,0.1,0.01)

Z =

  1×0 empty double row vector

to go from 0 to 0.1 in steps of 0.01. If that's the intention, linspace is not the right function to use. The colon, : operator is.

y = 0:0.01:0.1

y = 1×11

0 0.0100 0.0200 0.0300 0.0400 0.0500 0.0600 0.0700 0.0800 0.0900 0.1000

@Star Strider I suspect you changed the third input in your call to linspace to 150 after you evaluated the code but before you published, as that call shouldn't have created a 1000 element vector. I don't know if you want to re-evaluate your answer to remove that discrepancy.

[SAHIL SAHOO]

%thanks for the help.

%any idea how to shaded between the two curves ?

clc
clear all
a =5;
T = 2E3;
Z = linspace(0,0.1,150)                             

Z = 1×150

0 0.0007 0.0013 0.0020 0.0027 0.0034 0.0040 0.0047 0.0054 0.0060 0.0067 0.0074 0.0081 0.0087 0.0094 0.0101 0.0107 0.0114 0.0121 0.0128 0.0134 0.0141 0.0148 0.0154 0.0161 0.0168 0.0174 0.0181 0.0188 0.0195

U = (1+2.*Z)./(2.*a.*T);
V = (a.*Z)./(1+2.*Z);

plot(Z,log10(U))
hold on
plot(Z,log10(V))
hold off
grid on

========================================

[Star Strider]

Shading between them is straightforward —

% clc
% clear all
a =5;
T = 2E3;
Z = linspace(0,0.1,150);
U = (1+2.*Z)./(2.*a.*T);
V = (a.*Z)./(1+2.*Z);
Lv = Z>0;
UL10 = log10(U(Lv));
VL10 = log10(V(Lv));

figure
patch([Z(Lv) flip(Z(Lv))], [UL10 flip(VL10)], 'g', 'FaceAlpha',0.25)
hold on
plot(Z,log10(U))
plot(Z,log10(V))
hold off
grid on

The use of ‘Lv’ here is to avoid using values of ‘Z’ equal to 0 becausse the log of 0 in any base is -Inf, and the patch function will not work with non-finite values.

.