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I was going through the solutions of a past paper and came across this question where it is asked to determine the laplace transform of a transient waveform. I'm a bit confused with how it balances the waves and getting 10000 slope. Can anybody shed some light on it?

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The equation obtained is as follows

$$i(t) = (5000 slope ramp at t=0) -(10000 slope ramp at t=2)+ (5000 slope ramp at t=4)($$ $$ I(s) = \frac{5000}{s^2}(1-2e^{-0.002s}+e^{-0.004s})$$

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In the interval [0, 0.002], the signal has a slope of \$100/0.002= 50000\$. Then in [0.002, 0.004] its \$-50000 \$ and in [0.004, \$\infty\$] it is \$0\$.

Now let us try to represent the original signal as a sum of ramp (shifted ramp) signals. ie.,

$$ y = y_1 + y_2+y_3$$

Interval 1, [0, 0.002]: The slope is 50000 so a ramp with slope 5000 is required. So $$y_1 = 50000t$$

Interval 2, [0.002, 0.004]: The slope here is -50000. So another ramp (\$y_2\$) starting at \$t=0.002\$ should be added here to make the slope = \$-50000\$.

$$y = 50000t+y_2$$

Taking the derivative on both sides

$$\frac{dy}{dt} = 50000 +\frac{dy_2}{dt} =-50000 $$ $$\frac{dy_2}{dt} =-100000$$

So the slope of \$y_2 = -100000\$ and

$$y_2 = -100000(t-0.002)$$

Interval 3, [0.004, \$\infty\$]: following the similar analysis,

$$y_3 = 50000(t-0.004)$$

Then,

$$y = 50000t -100000(t-0.002) + 50000(t-0.004)$$

Taking Laplace transform, $$Y(s) = \frac{50000}{s^2}(1-2e^{-0.002s}+e^{-0.004s})\tag1$$

EDIT: Calculating Laplace transform obtained in (1)

If \$F(s)\$ is the laplace transform of \$f(t)\$, then by property of Laplace transform, $$f(t)\Leftrightarrow F(s) $$ $$f(t-t_0)\Leftrightarrow e^{-t_0s}F(s) $$

We know that: $$t \Leftrightarrow \frac{1}{s^2}$$ Using the property mentioned above, $$(t-0.002) \Leftrightarrow \frac{1}{s^2}\times e^{-0.002s}$$ $$(t-0.004) \Leftrightarrow \frac{1}{s^2}\times e^{-0.004s}$$

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  • \$\begingroup\$ Can u please explain a bit about how you converted the y(t) to Y(s)? I can understand how the first term (tranformation of 50,000t) was obtained but I can't understand the next two terms. \$\endgroup\$ – S.Dan Mar 5 '15 at 12:17
  • \$\begingroup\$ I was not aware of a time shifting property before. Got it (y) \$\endgroup\$ – S.Dan Mar 5 '15 at 13:58
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The easiest way is to define \$f(t)\$ as a piecewise function

$$f(t)=\begin{cases}\frac{100}{0.002}\cdot t,&0\le t\le 2\cdot 10^{-3}\\ 100-\frac{100}{0.002}\cdot (t-2),&2\cdot 10^{-3}<t\le 4\cdot 10^{-3}\\ 0,&\text{otherwise}\end{cases}\tag{1}$$

Then with the definition of the Laplace transform

$$F(s)=\int_{-\infty}^{\infty}f(t)e^{-st}dt$$

you just split the integral and use the top expression in (1) in the interval \$[0,2]\$, and the bottom expression in the interval \$[2,4]\$.

Another way to solve such problems is to compute the Laplace transform of the derivative of \$f(t)\$, which is a piecewise constant function. The derivative of \$f(t)\$ can be written using the step function \$u(t)\$:

$$f'(t)=k\cdot[u(t)-2u(t-2)+u(t-4)]\tag{2}$$

where \$k=100/0.002\$, and where I've used units of milliseconds. From (2) you can immediately write down the result by noting that the Laplace transform of \$u(t-t_0)\$ is \$e^{-st_0}/s\$. As soon as you have the Laplace transform of \$f'(t)\$, you just need to divide by \$s\$ to obtain the Laplace transform of \$f(t)\$.

And, by the way, \$100/0.002=50000\$, and not \$5000\$.

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  • \$\begingroup\$ I definitely wouldn't call your first method the easiest way - it's the most obvious way, but it's pretty messy and unnecessary. \$\endgroup\$ – Greg d'Eon Mar 4 '15 at 13:52
  • \$\begingroup\$ @Kynit: It's easier than that cancelling ramp business in the OP. I meant easy in terms of defining \$f(t)\$. The easiest solution to the problem is the one with the step functions. \$\endgroup\$ – Matt L. Mar 4 '15 at 13:55
  • \$\begingroup\$ The "cancelling ramp business" is exactly the same as your step functions. You just wrote the derivatives instead of the functions. \$\endgroup\$ – Greg d'Eon Mar 4 '15 at 13:59
  • \$\begingroup\$ @Kynit: I know that, but it's exactly the derivatives that make it clearer and less error prone. But it's up to everybody to choose their favorite method. \$\endgroup\$ – Matt L. Mar 4 '15 at 14:05
  • \$\begingroup\$ Meh. The way I learned it, time shifts are simple and seem like they require one less step. To each their own. Take a +1 for your trouble. \$\endgroup\$ – Greg d'Eon Mar 4 '15 at 14:08

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