produce a plot of V for characteristic impedance values ranging from Z0 = 0.01; 0.1; 1; 10; 100; 1000; 10000 and for values of ZL = 1x10^-3 to 1x10^6. Use a step size for ZL of 1. Plot your results on a Semi-logarithmic plot with the horizontal axis as the log scale. You should have 1 plot with 7 curves on it. where V = (ZL - Z0 / ZL + Z0)
I wrote this code:
ZL = 10^-3:1:10^6;
Z0 = [0.01, 0.1, 1, 10, 100, 1000, 10000];
V = (ZL - Z0)/(ZL+Z0);
I got this: matrix dimensions must agree
I used .- and .+
it said wrong operator

댓글 수: 1

hint:
(1:2).' + (3:5)
hint: .+ and .- do not exist as operators, but ./ does.

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답변 (1개)

Image Analyst
Image Analyst 2020년 4월 12일

0 개 추천

Use a for loop:
Here's a start.
ZL = 10^-3:1:10^6
Z0 = [0.01, 0.1, 1, 10, 100, 1000, 10000]
for k = 1 : length(Z0)
thisZ0 = Z0(k);
V = (ZL - thisZ0) ./ etc.
plot(.....etc.
hold on
end
See if you can finish it.

댓글 수: 2

NASA
NASA 2020년 4월 12일
it worked thanks, can i do the scaling inside the loop?
Image Analyst
Image Analyst 2020년 4월 12일
편집: Image Analyst 2020년 4월 12일
Sure. Or use semilogx().

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2020년 4월 12일

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2020년 4월 12일

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