Clear Filters
Clear Filters

How to make a nonperiodic signal periodic?

23 views (last 30 days)
I have the below code. If you run it, you'll see a graph that runs from t = 0 to t = 7. What I'd like to do is make this signal continuous, as in it runs forever from t -∞, +∞. Of course I looked it up and I see many users use stem or syms functions to make signals but they never have signals with this many piece wise components.
Thanks so much!
clc
close all
%
t=linspace(-1,8,2356);
v_t=-2*t.*fun_unit_step_gen(t,0);
v_t= v_t+2*(t-1).*fun_unit_step_gen(t,1);
v_t= v_t+3*fun_unit_step_gen(t,2);
v_t= v_t+(t-2).*fun_unit_step_gen(t,2);
v_t= v_t-5*(t-3).*fun_unit_step_gen(t,3);
v_t= v_t+3*(t-4).*fun_unit_step_gen(t,4);
v_t= v_t+5*(t-5).*fun_unit_step_gen(t,5);
v_t= v_t-4*(t-6).*fun_unit_step_gen(t,6);
v_t= v_t-fun_unit_step_gen(t,7);
%
plot(t,v_t,'r','LineWidth',2);
axis([-1 9 -4 4])
grid
title('Function f1(t)')
hold on
px=[0,.001];py=[-5,5];
plot(px,py,'-.k','LineWidth',1)
py=[0,.001];px=[0,9];
plot(px,py,'-.k','LineWidth',1)
hold off
  3 Comments
William Rose
William Rose on 19 Sep 2023
Since you did not provide the function fun_unit_step_gen(), I made a guess. Is my guess correct?
t=linspace(-1,8,2356);
v_t=-2*t.*(t>=0);
v_t= v_t+2*(t-1).*(t>=1);
v_t= v_t+3*(t>=2);
v_t= v_t+(t-2).*(t>=2);
v_t= v_t-5*(t-3).*(t>=3);
v_t= v_t+3*(t-4).*(t>=4);
v_t= v_t+5*(t-5).*(t>=5);
v_t= v_t-4*(t-6).*(t>=6);
v_t= v_t-(t>=7);
%
plot(t,v_t,'r','LineWidth',2);
axis([-1 9 -4 4])
grid
title('Function f1(t)')
hold on
px=[0,.001];py=[-5,5];
plot(px,py,'-.k','LineWidth',1)
py=[0,.001];px=[0,9];
plot(px,py,'-.k','LineWidth',1)
hold off
The funciton as plotted will be discontinuous at t=0, 7, 14,..., if you repeat it at those intervals. Is that OK?
William Rose
William Rose on 19 Sep 2023
If you want dot-dash lines along the axes, you can do it more simply with xline() and yline():
plot(-1:8,-3+6*rand(1,10),'-r');
xline(0,'-.k'); yline(0,'-.k'); grid on

Sign in to comment.

Accepted Answer

Fabio Freschi
Fabio Freschi on 19 Sep 2023
Edited: Fabio Freschi on 19 Sep 2023
I had the same idea of @William Rose, here is a possible implementation
clear variables, close all
% anonymous function
v_t = @(t)-2*t.*(t>=0)+...
2*(t-1).*(t>=1)+...
3*(t>=2)+...
(t-2).*(t>=2)+...
-5*(t-3).*(t>=3)+...
3*(t-4).*(t>=4)+...
5*(t-5).*(t>=5)+...
-4*(t-6).*(t>=6)+...
-(t>=7);
% anonymous function with period T using mod
T = 8;
v_t_per = @(t,T)v_t(mod(t,T));
% time axis
t=linspace(-8,16,2356*3);
% plot
figure, hold on, grid on
plot(t,v_t(t),'r','LineWidth',2);
plot(t,v_t_per(t,T),'b:','LineWidth',2);
  11 Comments
Paul
Paul on 20 Sep 2023
Edited: Paul on 21 Sep 2023
There is also the Discrete Time Fourier Transform (DTFT) that can be applied to either finite duration or infinite duration signals (periodic or not), and the DTFT is a function of continuous frequency with a "continuous spectrum" in the sense that you're interpreting that phrase, i.e., delta-F -> 0, which is different than what I assumed the OP meant by that phrase. Further clarification needed from the OP as to what the goal of the problem actually is.
If you want to analyze continuous-time signals that can be expressed in closed form, as is the case with v_t, then fourier is a good place to start. If you want to analyze discrete-time signals, then fft and/or freqz are useful for finite duration signals, and fft is useful for periodic signals. As always one has to pick the right tool for the job and interpret the outputs correctly based on the problem statement.

Sign in to comment.

More Answers (2)

Image Analyst
Image Analyst on 19 Sep 2023
How about making it for one chunk, and then using repmat to make copies of it? You can't go from t → -∞, +∞ but you can go for some finite number of elements (indexes).

William Rose
William Rose on 19 Sep 2023
If your goal is for the signal to repeat so that x(7...14)=x(0...7), and x(-7...0)=x(0...7), then I recommend using modulo division by 7 of the time argument.

Products


Release

R2022b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!