Get access to the MATLAB software package. Read the MATLAB appendices M1.1 M1.3

Question

Get access to the MATLAB software package. Read the MATLAB appendices M1.1 M1.3 at the end of Chapter 1 of your text. Perform the following tasks: Plot the following functions: sin(2t), sin(2(t - 1/2)) and sin(2(t + 1/3)) over the interval [0,5] on a single graph. Plot the following functions: u(t), u(t - 1), u(t) - u(t - 1) and u(2t) - u(2t - 1), each over the interval [0,2]. Remark: Always present your simulations results in a clear, concise and professional manner. This means: Provide the source code. Title all plots and label all axes on all plots. If more than one plot is on a single graph, supply a legend. Explain your results. Is it what you expect If not, why not An unintelligible presentation will simply not be graded at all and result in zero credit.

Explanation / Answer

t = 0:0.1:5; % [0,5] interval

y1 = sin(2*t);
y2 = sin(2*(t-1/2));
y3 = sin(2*(t+1/3));

figure
plot(t,y1,'r',t,y2,'b',t,y3,'g')
title('Graph of Sine functions')
legend('sin(2t)','sin(2(t-1/2))','sin(2(t+1/3)','Location','northoutside')
xlabel('0 < t < 5') % x-axis label
ylabel('Sine values') % y-axis label

x = 0:0.1:2; % [0,2] interval

y4 = heaviside(x); % u(t)
y5 = heaviside(x-1); % u(t-1);
y6 = heaviside(x) - heaviside(x-1); %u(t) - u(t-1)
y7 = heaviside(2*x) - heaviside(2*x -1); %u(2t) - u(2t-1)

figure   
plot(x,y4)
title('Graph of u(t)')
xlabel('0 < t < 2') % x-axis label
ylabel('u(t)') % y-axis label

figure   
plot(x,y5)
title('Graph of u(t-1)')
xlabel('0 < t < 2') % x-axis label
ylabel('u(t-1)') % y-axis label

figure   
plot(x,y6)
title('Graph of u(t) - u(t-1)')
xlabel('0 < t < 2') % x-axis label
ylabel('u(t) - u(t-1)') % y-axis label

figure   
plot(x,y7)
title('Graph of u(2t) - u(2t-1)')
xlabel('0 < t < 2') % x-axis label
ylabel('u(2t) - u(2t-1)') % y-axis label

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.