sine wave plot

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?

[Ahmed Grera]

How many cycles do you need in drawing?

[Govinda Nahak]

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

Answer 1

[Rick Rosson]

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

[Rajasekaran]

Thanks for your reply & detailed answer.

[Nauman Hafeez]

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.

Answer 2

[Mike Mki]

Dear Mr. Rick, Is it possible to create knit structure in Matlab as follows:

Answer 3

[Robert]

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

Answer 4

[soumyendu banerjee]

%% if Fs= the frequency u want,
x = -pi:0.01:pi;
y=sin(Fs.*x);
plot(y)

Answer 5

[wilfred nwakpu]

   %%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;

Answer 6

[Junyoung Ahn]

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)')

Answer 7

[sevde busra bayrak]

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')