how to plot x(t) = 0.9^k*cos(pi*k/5)*u(k) with simulink

9 次查看(过去 30 天)
I tried to plot x(t) = 0.9^k*cos(pi*k/5)*u(k) with simulink.
I used constant, sine wave, Pulse generator and product block in order to express 0.9, cos(pi/5), and u(t). But I couldn't figure out how to express 0.9^k.
Please help...

采纳的回答

Paulo Silva
Paulo Silva 2011-6-25
Something is wrong with that equation, can you see the error?
here's the MATLAB code, I leave the simulink for you to do, it's easy after knowing the basics.
k=1:100; %samples
u(k)=ones(1,numel(k)); %the input is a discrete step
x(k) = 0.9.^k.*cos(pi*k/5).*u(k);
plot(k,x(k))
hold on
plot(k,u(k),'r')
axis([0 numel(k) -2 2])
legend('x(k)=0.9^k*cos(pi*k/5)*u(k)','u(k)')
I couldn't wait so I just tested it on simulink, never done something similar before so I improvised, the results are similar to those of the script.
First thing you should do is change the configuration parameters.
Start time -> 1 Stop time -> 100 Solver Type -> Fixed step Sample time -> 1
Now the blocks
Clock -> Fcn -> Scope
In the Fcn put the expression -> 0.9^u*cos(pi*u/5)
Do the simulation
PS: Don't blame me if there's something wrong, try it and check if all is correct.
  3 个评论
Samuel
Samuel 2011-6-26
It was typo. It is x(k).
I plot this eq. with code below
anyway, I will try again..Thanks.
m = 10;
k = -m:m;
x = (0.8).^k.*cos((pi*k)./5).*(k>=0);
subplot(2,1,1);
stem(k,x); grid on;
axis([-m,m,-1,1]);
set(gca,'XTick',-m:m);
set(gca,'XTicklabel',-m:m);
set(gca,'YTick',-1:1);
set(gca,'YTicklabel',-1:1);
title('1i) x(k)=(0.8)^k*cos(pi*k/5)*u(k)');
xlabel('k');
ylabel('x(k)');

请先登录,再进行评论。

更多回答(1 个)

Samuel
Samuel 2011-6-26
Thank you very much Paulo.
I successed ploting this equation based on your idea. I used ram function instead of clock as input of FCN and producted pulse with output of FCN because it is discrete squence signal.
Thanks again..!

类别

Help CenterFile Exchange 中查找有关 Sources 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by