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 (7)
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)
1 Comment
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
Ranjita
on 30 Sep 2024
clc
clear all
fs = 10000;
T=1/fs
f1 = 100;
f2= 50;
L= 10000;
t = (0:L-1)*T;
x1 =sin(2*pi*f1*t)+4*cos(2*pi*f2*t)
figure
subplot(2,2,1)
plot(t,x1)
axis([0 0.1 -1 6]);
title('SS Function');
xlabel('time');
ylabel('magnitude');
%frequency domain conversion and plotting
Y_x1=fftshift(fft(x1));
subplot(2,1,2)
plot (-(fs/2-fs/L)-1:(fs/L):(fs/2-fs/L),abs(Y_x1))
axis([-700 700 0 max(abs(Y_x1))+10000]);
title('Magnitude spectrum of S1 Function');
xlabel('Frequency(Hz)');
ylabel('magnitude');
sgtitle('Frequency Domain Representation of S1 Function');
0 Comments
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