MATLAB Answers

How to graph Asin(pi*f*t) with a for loop?

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Brian James
Brian James on 12 Apr 2017
Edited: Vic on 21 Jan 2018
In this problem you will compute and then plot three sinusoids.
% The form of each sinusoid will be u=Asin(2*pi*f*t).
%
% a) Initialize t to run from 0 to 1 in 0.01 increments
clear all;
clc;
t=[0:.01:1];
% b) Initialize A to [1/3 2/3 1] and set f = 1
A=[1/3 2/3 1];
f=1;
% c) Initialize u using the zeros command with the length of A and the
% length of t as arguments
u=zeros(length(A),length(t))
% d) Initialize a character array to ['k';'r';'g'] - this array will control
% the line color during plotting
lincol=['k';'r';'g'];
% e) Open a figure window
figure(1),
% f) Use a "for" loop to compute the rows of "y" based on the values of "a"
% Also inside the "for" loop, plot each row of "u" vs "t" with
% linewidth 1.5 and colors as specified in the character arrray that
% you initialized above (a=-0.2, black; a=0, red; a=0.1, green).

Answers (2)

Vandana Ravichandran
Vandana Ravichandran on 14 Apr 2017
You can use the "sin" function in MATLAB to compute u=Asin(2*pi*f*t). As an example, one iteration of the for loop can be computed and plotted in the following way:
A = 1/3;
t = 0:0.01:1;
f = 1;
result = A * sin(2*pi*f*t);
plot(t, result,'Color','k')
You can extend the above in order to use a for-loop

Vic
Vic on 21 Jan 2018
Edited: Vic on 21 Jan 2018
I do not think this plots the signal properly but this is my attempt. Someone may edit it.
A = 5; %magnitude which decays with time
freqSamplingRate = 8000; %needs to be less than half of the input frequency, f.
tt = -2:(1/freqSamplingRate):2;
a = 0.34; %calculated decaying factor
f = 1000; %frequency(Hz)
for r = 1:2 %loop cycles
xx = A*exp(-a*tt).*cos(2*pi*f*tt); %the signal which is outputted
plot(tt,xx)%plotting the output signal
xlabel('time(s)'), ylabel('Amplitude (units)'); %naming axis
soundsc(xx); %sounding the output signal
pause(3); %pause between each loop
end %ending of loop

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