sine wave plot

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aaa
aaa 2012-4-24
回答： Steven Lord 2025-9-23
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 个评论
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Walter Roberson
Walter Roberson 2021-8-10
@Arif Raihan
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
Gokul Krishna N 2021-10-13
Just been reading the comments in this question. Hats off to you, sir @Walter Roberson

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采纳的回答

Rick Rosson
Rick Rosson 2012-4-24
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
  3 个评论
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Nauman Hafeez
Nauman Hafeez 2018-12-28
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.
alex
alex 2025-9-23
class you are mate, bang on

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更多回答（8 个）

Mike Mki
Mike Mki 2016-11-29
Dear Mr. Rick, Is it possible to create knit structure in Matlab as follows:
Junyoung Ahn
Junyoung Ahn 2020-6-16
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 个评论
DARSHAN
DARSHAN 2023-1-8
why should we multiply f with 100?
Walter Roberson
Walter Roberson 2023-1-8
I think the intent was to give 100 samples per cycle.

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Robert
Robert 2017-11-28
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
shampa das 2020-12-26
编辑：Walter Roberson 2021-1-31
Ran in:
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
soumyendu banerjee 2019-11-1
%% if Fs= the frequency u want,
x = -pi:0.01:pi;
y=sin(Fs.*x);
plot(y)
sevde busra bayrak
sevde busra bayrak 2020-8-24
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')
Ranjita
Ranjita 2024-9-30
Ran in:
clc
clear all
fs = 10000;
T=1/fs
T = 1.0000e-04
f1 = 100;
f2= 50;
L= 10000;
t = (0:L-1)*T;
x1 =sin(2*pi*f1*t)+4*cos(2*pi*f2*t)
x1 = 1×10000
4.0000 4.0608 4.1174 4.1696 4.2171 4.2598 4.2973 4.3294 4.3561 4.3770 4.3920 4.4009 4.4037 4.4000 4.3898 4.3730 4.3496 4.3193 4.2821 4.2381 4.1871 4.1292 4.0643 3.9926 3.9139 3.8284 3.7362 3.6374 3.5320 3.4202
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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');
Steven Lord
Steven Lord 2025-9-23
Ran in:
If you're using release R2018b or later, rather than computing sin(pi*something), I recommend using the sinpi function (and there is a corresponding cospi function.)
x = 0:0.25:2
x = 1×9
0 0.2500 0.5000 0.7500 1.0000 1.2500 1.5000 1.7500 2.0000
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s1 = sin(x*pi)
s1 = 1×9
0 0.7071 1.0000 0.7071 0.0000 -0.7071 -1.0000 -0.7071 -0.0000
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s2 = sinpi(x)
s2 = 1×9
0 0.7071 1.0000 0.7071 0 -0.7071 -1.0000 -0.7071 0
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Note that elements 5 and 9 of s1 and s2 are visually different. In s1 they are very close to, but not exactly equal to, 0. In s2 since we're taking the sine of exact multiples of pi (x(5) is exactly 1 and x(9) is exactly 2) we get actual 0 values.
format longg
[s1([5 9]); s2([5 9])]
ans = 2×2
1.0e+00 * 1.22464679914735e-16 -2.44929359829471e-16 0 0
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And in this particular example from the original question:
t = [0:0.1:2*pi];
inner = 2*60*t
inner = 1×63
0 12 24 36 48 60 72 84 96 108 120 132 144 156 168 180 192 204 216 228 240 252 264 276 288 300 312 324 336 348
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When we compare their values with the rounded version of those values using a very tight tolerance, we see that the values of inner are all very, very close to integer values. [isapprox was introduced in release R2024b.]
all(isapprox(inner, round(inner), 'verytight'))
ans = logical
1
That means that if we use sinpi all the values should be very close to 0.
a = sinpi(inner)
a = 1×63
1.0e+00 * 0 0 0 2.23223583872552e-14 0 0 4.46447167745105e-14 4.46447167745105e-14 0 0 0 0 8.9289433549021e-14 0 8.9289433549021e-14 0 0 8.9289433549021e-14 0 8.9289433549021e-14 0 0 0 1.78578867098042e-13 1.78578867098042e-13 0 0 0 1.78578867098042e-13 1.78578867098042e-13
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maximumDifferenceFromZero = max(a, [], ComparisonMethod="abs")
maximumDifferenceFromZero =
3.57157734196084e-13
I'd say that's effectively 0 for most purposes.

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