# How to make a nonperiodic signal periodic?

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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
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
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

### Accepted Answer

Fabio Freschi
on 19 Sep 2023

Edited: Fabio Freschi
on 19 Sep 2023

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
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.

### More Answers (2)

Image Analyst
on 19 Sep 2023

##### 0 Comments

William Rose
on 19 Sep 2023

##### 0 Comments

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