Plotting discontinuous discrete signal with shift and fold

10 visualizaciones (últimos 30 días)
Miza
Miza el 23 de Feb. de 2011
Respondida: Deepak el 28 de Feb. de 2024
Hello !
I have a given question below
x[n]= {1+n/3 -3<=n<=-1,
1 0<=n<=3,
0 other wise}
I have to plot this signal and I have developed the following code for that
clc
n=-10:1:10;
for i=1:length(n)
if n(i)>=-3 && n(i)<=-1
x(i)=1+(n(i)/3);
elseif n(i)>=1 && n(i)<=3
x(i)=1;
else
x(i)=0;
end
stem(x)
end
axis([-20 20 0 2])
but the resulting waveform is plotted for positive values of n, Kindly help me to plot the above sequence,

Respuestas (4)

Gnaneswar Nadh satapathi
Gnaneswar Nadh satapathi el 6 de Nov. de 2013
clc clear all close all n=-10:1:10; for i=1:length(n) if n(i)>=-3 && n(i)<=-1 x(i)=1+(n(i)/3); elseif n(i)>=1 && n(i)<=3 x(i)=1; else x(i)=0; end
end stem(n,x)

Sk Group
Sk Group el 8 de Feb. de 2021
MATLAB CODE:
function [n1,x1] = sigshift(x,n,k)
n = 1:10;
k = 2;
x = exp(n);
if k>0
disp(‘Positive’);
n1 = n(1):n(end)+k;
x1 = [zeros(1,k) x];
else
disp(‘Negative’);
n1 = n(1)+k:n(end);
x1 = [x zeros(1,abs(k))]; % abs is for absolute value of (k) because quantity can never be (-ve) negative %
end
MATLAB CODE:
function [y n1] = sigfold(x,n)
n1 = -fliplr(n)
y = fliplr(x)
end

Kunal
Kunal el 27 de Jul. de 2023
discrete time signal folding code

Deepak
Deepak el 28 de Feb. de 2024
how to folda given signal

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