Write a MATLAB program to sketch the following discrete-time signals in the time range of –10 ≤ n ≤ 10. Please label all the graph axes clearly. If the sequence is complex, plot the magnitude and angle separately. i) x(n) = u(n) – u(n – 3)

n=-10:10;
u(1:21)=ones(1,21); %creates a unit step sequence for u(n)
                   %how to write a delayed version of sequence i.e for u(n-3) ?

Answer 1

[Aliya Patel]

t = (-10:1:10)';  %%Can change the interval time by replacing 1 with 0.1
step1 = t>=0;  %% For u[n]
step2 = t>=3;  %%For u[n-3]
x = step1-step2;
plot(t,x)    %%scatter can be used instead of plot
xlabel('time');
ylabel('amplitude');
title('x(n)=u(n)-u(n-3)');

[Satadru Mukherjee]

This is wrong , your code is not taking care of the time index :-/

[DGM]

Indexing looks fine, but representing this using plot() is visually misleading. I suspect that motivation is why Aliya suggested scatter(), though I think stem() would probably be better.

Answer 2

[Satadru Mukherjee]

n1=-10:10;
x=(n1>=0);
n2=n1+3;
y=x;
u=min(min(n1),min(n2));
t=max(max(n1),max(n2));
r=u:1:t;
z1=[];
temp=1;
for i=1:length(r)
    if(r(i)<min(n1) || r(i)>max(n1))
      z1=[z1 0];
    else
        z1=[z1 x(temp)];
        temp=temp+1;
    end
end
z2=[];
temp=1;
for i=1:length(r)
    if(r(i)<min(n2) || r(i)>max(n2))
      z2=[z2 0];
    else
        z2=[z2 y(temp)];
        temp=temp+1;
    end
end
z=z1-z2;
subplot(3,1,1);
stem(r,z1);
xlabel('Time sample');
ylabel('Amplitude');
title('First signal');
subplot(3,1,2);
stem(r,z2);
xlabel('Time sample');
ylabel('Amplitude');
title('Second signal');
subplot(3,1,3);
stem(r,z);
xlabel('Time sample');
ylabel('Amplitude');
title('Signal after Subtraction');

[Marvin Patel]

what does z1= [z1 x(temp)] do and why is it required

[Walter Roberson]

It appends x(temp) to the end of z1. They wrote the code using that technique because people tend to learn the [] concatenation before they gain the experience to initialize an array to full size and write at a particular in the array.

[Bhargav Jyoti Saikia]

Thank you very much. This program gave a clear idea as to how to implement these functions. Took little time to understand but after little debugging got the idea. There might be easier way or easier to understand simple method to achieve the same functionality.

Answer 3

[Roman Mia]

n=-10:10;
u=[zeros(1,13), ones(1, 8)];
stem(n,u);

Answer 4

[Jyothi]

import numpy as np import matplotlib.pyplot as plt

1. Define the range of n n = np.arange(0, 11)

1. Calculate the exponential signal signal = np.exp(-0.1 * n)

1. Plot the signal plt.stem(n, signal, use_line_collection=True) plt.xlabel('n') plt.ylabel('e^(-0.1n)') plt.title('Discrete Exponential Signal e^(-0.1n)') plt.grid(True) plt.show()

[Walter Roberson]

This appears to be python code, but MATLAB was being asked about rather than Python.

The python code does not appear to answer the question that was asked.