# sine wave plot

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aaa on 24 Apr 2012
Commented: DGM on 5 May 2023 at 17:51
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?
Gokul Krishna N on 13 Oct 2021
Just been reading the comments in this question. Hats off to you, sir @Walter Roberson

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
Walter Roberson on 21 Nov 2022
I do not find any questions posted by you ?

Mike Mki on 29 Nov 2016
Dear Mr. Rick, Is it possible to create knit structure in Matlab as follows:
DGM on 27 Nov 2022

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)')
##### 2 CommentsShow 1 older commentHide 1 older comment
Walter Roberson on 8 Jan 2023
I think the intent was to give 100 samples per cycle.

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

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')

Ana Maria on 15 Mar 2023
Implement a function to generate a column vector containing a sine wave, sin(2πf(t)t), with a growing frequency, f(t) from f(0) = f1 to f(T) = f2. The inputs of the function are the duration, T in seconds, the frequencies, f1 and f2, in Hz and the sampling rate, fs, in samples per second x = chirpT one(T, f1, f2, fs)
##### 2 CommentsShow 1 older commentHide 1 older comment
DGM on 15 Mar 2023
Edited: DGM on 16 Mar 2023
You're copying and pasting an assignment text. This is not an answer, so it doesn't belong here as an answer. I'm compelled to keep things where they belong and remove them when they don't.
This is ultimately your task to perform. The information already present on this page is largely sufficient to complete it. I'm sure with enough effort, you can find even more specific examples elsewhere on the forum.
If you want to ask a question, please open a new question using the 'ask' button at the top of the page. If and when you do, ask an actual question, but also be prepared to prove that you've exhausted what due diligence provides.
EDIT:
To prove the point, I'm just going to grab the answer directly above and make one simple change. Other than changing the specific parameters (a matter of choice), the only real change is that instead of being a scalar, freq is a vector generated from two specified values.
% these are parameters
samplerate = 500;
duration = 2;
flim = [0 8];
% both these vectors have the same size
time = 0:1/samplerate:duration; % time is a linear vector
freq = linspace(flim(1),flim(2),numel(time)); % freq is a linear vector
Generating a vector of uniformly-spaced values is very basic MATLAB stuff. The remaining change necessary to make freq work as a vector is also basic (literally one single character), but I have to leave something for you to do.

Avinash Mishra on 5 May 2023 at 9:04
1. Write a MATLAB script to plot a graph with the following specifications.
2. Consider two vectors t and y.
3. t starts with 0, ends at 15, and contains 400 elements.
y = exp(-0.5*t). *Cos (2*pi. *t).
• Plot y against t. Use a blue (b) dashed (--) line for the plot.
Mark the minimum value m of the vector y by adding a point to the plot. This point should be a red asterisk marker and must be added after the blue line.
DGM on 5 May 2023 at 17:51
Why did you paste your homework as an answer? This is not an answer to the question, nor is it a question.

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