Generating a pulse signal

Hİ guys, I basically have this code and want to generate below figure. But I can't understand how it works. I tried the help function for the impulse but still couldn't understand it.

I only have hard time generating the pulse signal.

f=3; %frequency [Hz]
t0=(0:0.0009:1);
a=1;    %amplitude [V]
phi=0;  %phase
y=a*sin(2*pi*f*t0+phi)-2;
f=5; %frequency [Hz]
t1=(2:0.0009:3);
a=1;    %amplitude [V]
phi=0;  %phase
z=a*cos(2*pi*f*t1+phi)+2;
t = (1:0.01:2);
impulse = t==1;
impulse1 = t==1.1;
impulse2 = t==1.2;
impulse3 = t==1.3;
impulse4 = t==1.4;
impulse5 = t==1.5;
impulse6 = t==1.6;
impulse7 = t==1.7;
impulse8 = t==1.8;
impulse9 = t==1.9;
impulse10 = t==2;
x=double(impulse).*1; 
x1=double(impulse1).*1;
x2=double(impulse2).*1;
x3=double(impulse3).*1;
x4=double(impulse4).*1;
x5=double(impulse5).*1;
x6=double(impulse6).*1;
x7=double(impulse7).*1;
x8=double(impulse8).*1;
x9=double(impulse9).*1;
x10=double(impulse10).*1;
d=linspace(0,3,3335);
c=[y x x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 z];
subplot(4,1,1)
plot(d,c,'black','LineWidth',1.5)

[Can Kayan]

above code works but i cant understand how it does. That's the part I need help with

Answer 1

[Star Strider]

The posted code is an extremely cumbersome way of coding an extremely straightforward problem.

The ‘fcn’ function here should make it more understandable —

t = linspace(0, 3, 1800);
f=3; %frequency [Hz]
a=1;    %amplitude [V]
phi=0;  %phase
% y=a*sin(2*pi*f*t0+phi)-2
fcn = @(t,f,a,phi) (a*sin(2*pi*f*t+phi)-2) .* ((t>=0) & (t<1)) + (rem(t,0.1)<=1E-2) .* ((t>=1) & (t<2)) + (a*sin(2*pi*f*t+phi)+2) .* ((t>=2) & (t<=3));
%                              ↑ ← FIRST 'sin' CALL                       ↑ ← PULSE TRAIN                             ↑ ← SECOND 'sin' CALL

figure
plot(t, fcn(t,f,a,phi))
grid

.

[Can Kayan]

Thanks a lot for the quick answer !!

[Star Strider]

My pleasure!

If my Answer helped you solve your problem, please Accept it!

.

[Jon]

Using the anonymous function is a nice clean way of getting the plot and a vast improvement over the original approach. One detail. The OP's original figure does not match the code below it. The figure shows the final portion as a sin, but the code gives it as a cos. @Star Strider's answer matches the OP's figure, but not the code. Anyhow Star Strider's answer can easily be changed to use a cos for the second one. You just have to figure out what you actually want.

[Jon]

Also in the OP's post the frequency for the two sinusoids is different, but only one frequency is used in the anonymous function. You would just need to modify the anonymous function a little if you actually need two different frequencies

[Star Strider]

Corrected ‘fcn’ and plot —

t = linspace(0, 3, 3335);
f=[3 5]; %frequency [Hz]
a=1;     %amplitude [V]
phi=0;   %phase
fcn = @(t,f,a,phi) (a*sin(2*pi*f(1)*t+phi)-2) .* ((t>=0) & (t<1)) + (rem(t,0.1)<=1E-2) .* ((t>=1) & (t<2)) + (a*cos(2*pi*f(2)*t+phi)+2) .* ((t>=2) & (t<=3));
%                              ↑ ← 'sin' CALL                               ↑ ← PULSE TRAIN                              ↑ ← 'cos' CALL

figure
plot(t, fcn(t,f,a,phi))
grid

NOTE — The original ‘z’ is coded as cos, however plotted as sin. This cannot be accounted for by ‘phi’ that is 0 in both function calls. The first value of ‘z’ in the original code is 3 and not 2 as plotted.

.

Answer 2

[Jon]

I like @Star Strider's approach, but as I had already coded up the approach below before I saw that, I thought I would provide this just to show an alternative approach.

% define first sinusoidal segment
tStart = 0;
tEnd = 1; 
f=3; %frequency [Hz]
a=1;    %amplitude [V]
phi=0;  %phase
t1 = tStart:0.0009:tEnd;
x1=a*sin(2*pi*f*t1+phi)-2;


% define impulses
tStart = 1;
tEnd = 2;
t2 = tStart:0.001:tEnd;
idx = 1:100:1000; % indices in t2 where impulses occur
impulseMag = 1; % magnitude of impulses
% assign impulses at corresponding times where they occur
x2 = zeros(size(t2)); % initialize with no impulses
x2(idx) = impulseMag;  % assign impulses

% define second sinusoidal segment
tStart = 2;
tEnd = 3;
f=5; %frequency [Hz]
a=1;    %amplitude [V]
phi=0;  %phase
t3 = tStart:0.0009:tEnd
x3=a*cos(2*pi*f*t3+phi)+2;

% make overall signal
t = [t1 t2 t3];
x = [x1 x2 x3];


% plot results
figure
subplot(4,1,1)
plot(t,x,'black','LineWidth',1.5)