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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?

Gokul Krishna N
about 17 hours ago

Just been reading the comments in this question. Hats off to you, sir @Walter Roberson

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

MD RAJIBUL HOSSAIN RUBEL
on 28 May 2021

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)')

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);

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')

Ayesha Noor
on 5 Aug 2021

help me solving this numerical.

Consider x(t) = 1.5sin(πt)[u(t) − u(t − 1)] and h(t) = 1.5[u(t) − u(t − 1.5)] − u(t − 2) + u(t − 2.5). Compute the output y(t) = x(t) ∗ h(t) using MATLAB.

Mehrab Pretum
on 22 Jun 2021

Generate an analog signal using the following equation ,

Signal = 2*sin(2*pi*20*t)+0.4*cos(2*pi*100*t)+0.1*sin(2*pi*500*t)+0.05*randn(size(t));

. Show the signal in time and frequency domain and calculate the capacity using Shannon capacity formula .

• Show the quantize signal considering 6 equally distributed levels and provide image for one cycle of the original signal and quantized signal ( using subplot ) .

Walter Roberson
on 24 Jun 2021

First Last
on 28 Jun 2021

Edited: Walter Roberson
on 5 Aug 2021

t = [0:0.1:2*pi]

a = sin(t);

plot(t,a)

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