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

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...

 채택된 답변

Paulo Silva
Paulo Silva 2011년 6월 25일

0 개 추천

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.

댓글 수: 3

Samuel
Samuel 2011년 6월 25일
Is the equation wrong????
I plot this equation with matlab. This is discrete time signals.
x(t) should be x(k) or something similar
Samuel
Samuel 2011년 6월 26일
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)');

댓글을 달려면 로그인하십시오.

추가 답변 (1개)

Samuel
Samuel 2011년 6월 26일

0 개 추천

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..!

카테고리

태그

질문:

2011년 6월 25일

Community Treasure Hunt

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

Start Hunting!

Translated by