# sine wave plot

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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?

##### 6 Comments

Walter Roberson
on 10 Aug 2021

In order to solve that, you need some hardware to do analog to digital conversion between your 3V source and MATLAB.

3V is too large for audio work, so you are not going to be able to use microphone inputs to do this. You are going to need hardware such as a National Instruments ADC or at least an arduino (you might need to put in a resistor to lower the voltage range.)

The software programming needed on the MATLAB end depends a lot on which analog to digital convertor you use.

The appropriate analog to digital convertor to use is going to depend in part on what sampling frequency you need to use; you did not define that, so we cannot make any hardware recommendations yet.

Gokul Krishna N
on 13 Oct 2021

Just been reading the comments in this question. Hats off to you, sir @Walter Roberson

### Accepted Answer

Rick Rosson
on 24 Apr 2012

Please try:

%%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

##### 2 Comments

Nauman Hafeez
on 28 Dec 2018

How to calculate Fs for a particular frequency signal?

I am generating a stimulating signal using matlab for my impedance meter and it gives me different results on different Fs.

### More Answers (9)

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 Comments

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

##### 0 Comments

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);

##### 0 Comments

soumyendu banerjee
on 1 Nov 2019

%% if Fs= the frequency u want,

x = -pi:0.01:pi;

y=sin(Fs.*x);

plot(y)

##### 0 Comments

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;

##### 0 Comments

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')

##### 0 Comments

Ana Maria
on 15 Mar 2023

##### 2 Comments

Walter Roberson
on 15 Mar 2023

Could you explain how the process you set out will solve the original question posted in 2012 ?

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.

Adewole
on 17 Aug 2023

##### 2 Comments

Rik
on 17 Aug 2023

How does this answer the question? You can find guidelines for posting homework on this forum here. If you have trouble with Matlab basics you may consider doing the Onramp tutorial (which is provided for free by Mathworks). If your main issue is with understanding the underlying concept, you may consider re-reading the material you teacher provided and ask them for further clarification.

This answer will be deleted in 24h unless you respond.

DGM
on 18 Aug 2023

I was getting ready do delete this when I realized something. If we trust that this is a rather literal duplication of the assignment text, then this is a pretty terrible assignment. Given that x is a coarse discrete set of points, what exactly is this trivial ratio (i.e. the solution) supposed to teach the student other than some vague precaution regarding aliasing? Given that the calculation of the solution is so utterly simplistic, I find it hard to believe that any lesson so complicated is intended.

On the other hand, is x really supposed to be a discrete set of 11 integers, or are we supposed to be considering all values on the closed interval [0 10]? That would actually make a decent problem both in terms of math fundamentals and using MATLAB, but it's contrary to the text as given. Poster's mistake? Instructor's mistake?

Either way, the answer is exactly

2*(pi - 2*asin(4/5))/10

or maybe it's exactly

3/11

They're not equivalent, so which is which? Which is right? Is either one right? If not, why?

I should also point out that even if you copied and pasted the right one and submitted it, any decent TA would still give you zero points, since you'd have no work to show, and chances are you might think that my final question was worth ignoring.

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