how i can found the region of absolute stability ?
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
sadeem alqarni
am 7 Jul. 2018
Kommentiert: sadeem alqarni
am 8 Jul. 2018
n=0:30:180
h=((768 *cos((1/2)*n) + 96* cos(n) - 864) ./ (40*cos (1/2*n)+ 2*cos(n)+ 102))
figure
plot(n,h, 'b*-', 'LineWidth', 2)
grid on;
xlabel('n', 'FontSize', 20);
ylabel('h', 'FontSize', 20);
Akzeptierte Antwort
Image Analyst
am 7 Jul. 2018
Because you're using / (array division) instead of ./ (element by element division). Try this:
n=0:30:180
h=((768 *cos((1/2)*n) + 96* cos(n) - 864) ./ (40*cos (1/2*n)+ 2*cos(n)+ 102))
figure
plot(n,h, 'b*-', 'LineWidth', 2)
grid on;
xlabel('n', 'FontSize', 20);
ylabel('h', 'FontSize', 20);
% Do it again with more resolution.
n = linspace(min(n), max(n), 1000);
h=((768 *cos((1/2)*n) + 96* cos(n) - 864) ./ (40*cos (1/2*n)+ 2*cos(n)+ 102))
hold on;
plot(n,h, 'r-', 'LineWidth', 2)
grid on;
xlabel('n', 'FontSize', 20);
ylabel('h', 'FontSize', 20);

1 Kommentar
Weitere Antworten (2)
sadeem alqarni
am 8 Jul. 2018
2 Kommentare
Image Analyst
am 8 Jul. 2018
How is this an "Answer" to your original question? Anyway, as you can see from my plot, it depends on how you define stable. The values are constantly changing so they're not stable, however the overall pattern seems pretty stable - it seems to oscillate pretty much the same way over the time period plotted.
Siehe auch
Kategorien
Mehr zu 2-D and 3-D Plots finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
