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sine wave plot

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aaa
aaa on 24 Apr 2012
Answered: sevde busra bayrak on 24 Aug 2020
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?

  3 Comments

Ahmed Grera
Ahmed Grera on 3 Sep 2017
How many cycles do you need in drawing?
Govinda Nahak
Govinda Nahak on 2 Oct 2017
in sine function in MATLAB it is always sin(wt). here frequency w is in radian/sec not f (in HZ) so w will give you the no.of the cycle.
suppose w=1 it is one cycle and so on
if you want to use the sin(2*pi*60*t) you can use the sind(2*pi*9.545*t). why i use the 9.545 bcz we should convert the f to w in the time interval of 2*pi.
Jorge Ignacio Cisneros Saldana
How can we make it 3 phase system? for 50 Hz and 230 V peak to peak

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Accepted Answer

Rick Rosson
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

Rajasekaran
Rajasekaran on 14 Mar 2013
Thanks for your reply & detailed answer.
Nauman Hafeez
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.

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More Answers (6)

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

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Robert
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

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soumyendu banerjee
soumyendu banerjee on 1 Nov 2019
%% if Fs= the frequency u want,
x = -pi:0.01:pi;
y=sin(Fs.*x);
plot(y)

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wilfred nwakpu
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;

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Junyoung Ahn
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)')

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sevde busra bayrak
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')

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