Equation for a Triangle wave shape [closed]

Question

So I have this function, and is it writter analytically. I cannot understand how do we get 4Um/T and 2UM- 4Um/T or -4Um + 4Um/T. What I mean is that I have tried formula like this when I take u(t) axis and divide it by t? such as UM/(T/4) but I do not know whether this approach is correct and if not then how do I generally solve this?

Answer 1

It boils down to finding the slopes \$m\$ and time shifts \$T_s\$ based on the equation of a straight line voltage ramp:$$u=m(t-T_s)\tag{1}$$ which can be manipulated to :$$u=mt+b\text{, where }T_s=\frac{-b}{m} $$

There are only two slopes: \$m\$ and \$-m\$.

There are three time shifts \$T_s\$ equaling one of: 0, \$T/4\$, and \$3T/4\$

The rising slope is easily calculated from the point-slope form of a straight line:$$m=\frac{y_2-y_1}{x_2-x_1}$$

The values now can be substituted into equation 1 to produce the results shown.

Answer 2

Simple line equation:

$$ y = m \ x + n $$

where \$m\$ is the slope, and \$n\$ is the point that the line crosses y-axis.

For your case \$y\$ is \$u(t)\$ and \$x\$ is \$t\$. For the waveform given, in a period we see three line segments: one between {0, T/4}, another between {T/4, 3T/4} and the final one between {3T/4, T}.

Write the line equations for all of these segments and that's it.

I'm going to give you the first segment:

$$ u(t) = m \ t + n \\ m=\frac{du(t)}{dt} = \frac{U_m-0}{T/4-0}=4\ \frac{U_m}{T} \\ n = 0 \\ \therefore u(t)=\Big(4\ \frac{U_m}{T}\Big)t $$

And I'm sure you can take it from here.

Answer 3

I first define a triangle function, which can be done in at least two ways. In the first half-period I generate a positive triangle, in the following half-period I add a negative triangle. for the subsequent periods I move the two previous triangles by one period and in general by k periods T. The pulse train is obtained simply summing the two triangles for each period. You can shift the pulse train along the time axis simply adding the desired displacement δ to the variable t of the function f(t+δ,τ).

Answer 4

With regard to finding the Fourier transform, the easiest way is to see that a triangle wave is the integral of a square wave which is the integral of an alternating pulse train. The alternating pulse train is a repetition of a single alternating pulse. Repetition transforms into sampling (a line spectrum, representable alternatively as a Fourier series), a single alternating pulse has a sine as its transform, and integration transforms into division by \$2\pi jf\$.

Getting every factor right requires a bit of diligence, but most of the work of juggling with piecewise linear function definitions can be conveniently avoided in that manner.

Answer 5

The actual not-understood thing was splitting the time to intervals and covering each interval with a different equation of a line. The accepted answer shows the division and the different line equations perfectly.

Then you have got some guidance how to find the line equations by yourself.

One of the answers presents "how to write the formula of the triangle as a series of ramp pulses which are written with the step function" The answer assumes indirectly that you may well want to forget the original problem and be satisfied by a new formula for the triangle if the formula is straightforward so that a single formula covers all time values.

Finally there's a guess you do not actually miss at all the formula of the triangle pulse, but the Fourier series of it and there's given a hint how to calculate it from the Fourier series of the square wave.

I add my variation of the "forget the whole problem, use this single formula for all time values" My formula is:

u(t)=(2Um/Pi)arcsin(sin(2Pi*t/T))

Single formula is enough because the periodic function sin returns all time values to the range -T/4....T/4 i.e. forces to use the ramp which goes through the origin. The formula is tested in function plotter website desmos with Um=1 and T=2: