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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?

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

Jessica
on 26 Nov 2020

Att: Sameeksha Singh

Do not ask questions within other persons questions.

Here is how you do it:

Top right there is a "Ask the Community" button.

Click and copy/paste your question there.

Make sure that you have as much code or specific facts with, not so much text as possible.

MUHAMMAD BASIM
on 14 Dec 2020

Walter Roberson
on 14 Dec 2020

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)')

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')

shampa das
on 26 Dec 2020 at 20:04

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);

Radwa Ahmed
on 5 Jan 2021 at 3:38

9- Consider modulating a carrier cos(10t) with the following signals:

(a) š„1 (š”) = š ā|š”| . Use MATLAB to find the Fourier transform of š¦1 (š”) = š„1 (š”) cos(10š”) and plot y1(t) and its magnitude spectra. (b) š„2 (š”) = 0.2[š(š” + 5) ā 2š(š”) + š(š” + 5)], where r(t) is the ramp signal. Use MATLAB to š„2 (š”) ššš š¦1 (š”) = š„2 (š”) cos(10š”) and compute and plot the magnitude of their Fourier transforms.

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