# sine wave plot

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aaa on 24 Apr 2012
Commented: Walter Roberson about 9 hours ago
Hi,
I am having some trouble plotting a sine wave and i'm not sure where i am going wrong.
i have
t = [0:0.1:2*pi]
a = sin(t);
plot(t,a)
this works by itself, but i want to be able to change the frequency. When i run the same code but make the change
a = sin(2*pi*60*t)
the code returns something bad. What am i doing wrong? How can i generate a sin wave with different frequencies?
Walter Roberson about 1 hour ago
payal soni
please start a new question for that, and show an image of the equation. You will a probably also need to talk more about what G is.

Rick Rosson on 24 Apr 2012
%%Time specifications:
Fs = 8000; % samples per second
dt = 1/Fs; % seconds per sample
StopTime = 0.25; % seconds
t = (0:dt:StopTime-dt)'; % seconds
%%Sine wave:
Fc = 60; % hertz
x = cos(2*pi*Fc*t);
% Plot the signal versus time:
figure;
plot(t,x);
xlabel('time (in seconds)');
title('Signal versus Time');
zoom xon;
HTH.
Rick
MD RAJIBUL HOSSAIN RUBEL on 28 May 2021
There are two numbers given as -5+2i 3-2i using Plot and Stem function show the output of this equation

Mike Mki on 29 Nov 2016
Dear Mr. Rick, Is it possible to create knit structure in Matlab as follows: Robert on 28 Nov 2017
aaa,
What goes wrong: by multiplying time vector t by 2*pi*60 your discrete step size becomes 0.1*2*pi*60=37.6991. But you need at least two samples per cycle (2*pi) to depict your sine wave. Otherwise you'll get an alias frequency, and in you special case the alias frequency is infinity as you produce a whole multiple of 2*pi as step size, thus your plot never gets its arse off (roundabout) zero.
Using Rick's code you'll be granted enough samples per period.
Best regs
Robert

Junyoung Ahn on 16 Jun 2020
clear;
clc;
close;
f=60; %frequency [Hz]
t=(0:1/(f*100):1);
a=1; %amplitude [V]
phi=0; %phase
y=a*sin(2*pi*f*t+phi);
plot(t,y)
xlabel('time(s)')
ylabel('amplitude(V)')

shampa das on 26 Dec 2020
Edited: Walter Roberson on 31 Jan 2021
clc; t=0:0.01:1; f=1; x=sin(2*pi*f*t); figure(1); plot(t,x); fs1=2*f; n=-1:0.1:1; y1=sin(2*pi*n*f/fs1); figure(2); stem(n,y1); fs2=1.2*f; n=-1:0.1:1; y2=sin(2*pi*n*f/fs2); figure(3); stem(n,y2); fs3=3*f; n=-1:0.1:1; y3=sin(2*pi*n*f/fs3); figure(4); stem(n,y3); figure (5); subplot(2,2,1); plot(t,x); subplot(2,2,2); plot(n,y1); subplot(2,2,3); plot(n,y2); subplot(2,2,4); plot(n,y3); soumyendu banerjee on 1 Nov 2019
%% if Fs= the frequency u want,
x = -pi:0.01:pi;
y=sin(Fs.*x);
plot(y)

wilfred nwakpu on 1 Feb 2020
%%Time specifications:
Fs = 8000; % samples per second
dt = 1/Fs; % seconds per sample
StopTime = 0.25; % seconds
t = (0:dt:StopTime-dt)'; % seconds
%%Sine wave:
Fc = 60; % hertz
x = cos(2*pi*Fc*t);
% Plot the signal versus time:
figure;
plot(t,x);
xlabel('time (in seconds)');
title('Signal versus Time');
zoom xon;

sevde busra bayrak on 24 Aug 2020
sampling_rate = 250;
time = 0:1/sampling_rate:2;
freq = 2;
%general formula : Amplitude*sin(2*pi*freq*time)
figure(1),clf
signal = sin(2*pi*time*freq);
plot(time,signal)
xlabel('time')
title('Sine Wave')

Faizan Arshad on 24 Apr 2021
how to plot 3 sign wave with phase difference?

Mehrab Pretum on 22 Jun 2021
Generate an analog signal using the following equation ,
Signal = 2*sin(2*pi*20*t)+0.4*cos(2*pi*100*t)+0.1*sin(2*pi*500*t)+0.05*randn(size(t));
. Show the signal in time and frequency domain and calculate the capacity using Shannon capacity formula .
• Show the quantize signal considering 6 equally distributed levels and provide image for one cycle of the original signal and quantized signal ( using subplot ) .
Walter Roberson on 24 Jun 2021
I am not clear on how this will help aaa achieve the sine wave plot they were looking for 9 years ago??

First Last on 28 Jun 2021
t = [0:0.1:2*pi] a = sin(t); plot(t,a)
if true
% code
end