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RLC Circuit

I was trying to solve this problem, in which we have

$q(t)=\begin{cases} 0,\ t\lt 0 \\ \sin(t),\ t\ge0 \end{cases}$

$\therefore\ q(t)=u(t-0)\cdot\sin(t)=u(t)\cdot\sin(t)$

$u(t)$ is the Heaviside function applied in $t=0$.

Our goal is to determine the tension given by the source.

Using Kirchhoff's Laws, I get:

$ V=L\cdot\dfrac{d^2q}{dt^2}+R\cdot\dfrac{dq}{dt}+\dfrac{q}{C} $

$ V=\dfrac{d^2q}{dt^2}+3\dfrac{dq}{dt}+5q $

Given that we know the equation for $q(t)$, I calculated $q'(t)$ and $q''(t)$ and substituted them in my equation for $V$, which results in:

$ V=u(t)\cdot(4\sin(t)+3\cos(t))+\delta(t)\cdot(3\sin(t)+2\cos(t))+\delta'(t)\cdot\sin(t) $

And then I got stuck. What do I do with $ \delta(t)\ \mbox{and}\ \delta'(t) $? What are their values? I know that $ \delta(t) $ is zero for every value of $t$ except for $t=0$, but I have no clue how $ \delta'(t) $ works.

Can you help me?

The answer is $ V=u(t)\cdot(3\cos(t)+4\sin(t))+2\delta(t) $, according to my professor, but how do I get there from what I got from my calculations?

Vinícius Lara
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2 Answers2

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For $t>0$ this is just $4\sin t + 3\cos t$. $\delta$ is not a function, and so neither are it's derivatives. In reality, the function $q$ is differentiable away from $0$ only, so $V$ will only be defined away from $0$ as well. Thus, you could say that $V$ is a function from $\mathbb{R}-\{0\}$ to $\mathbb{R}$ defined as $$ V(t) = \begin{cases} 0 & t < 0 \\ 4\sin t + 3\cos t & t>0 \end{cases} $$

From a physical point of view, this may be unsatisfying. However, this particular $q$ is not possible in a physical sense either, thus the discrepancy.

The solution provided by the professor is reached by considering $\delta$ and it's derivatives as functions which are $0$ for all $t\neq 0$. In this light, $\delta^{(k)}(t)\sin(t) = \delta^{(k)}(t)\sin(0)=0$ and $\delta^{(k)}(t)\cos(t)=\delta^{(k)}(t)\cos(0)=\delta^{(k)}(t)$.

Alex Jones
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  • I get your thoughts, but how can I kind of "ignore" $\delta(t)$ and $\delta'(t)$? Is $\delta'(t)=0$ (i updated the question)? Thank you! – Vinícius Lara Jun 24 '18 at 23:52
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    There's nothing to ignore here. You're working under the misconception that $\frac{d}{dt} u(t) = \delta (t)$, which is not true in the usual sense. Instead, $\frac{d}{dt} u(t) = 0$ when $t\neq 0$, and is not defined for $t=0$. Since $\delta$ and its derivatives aren't functions, there is no sense in which $\delta^{(k)}(t)$ "equals" anything for a particular $t$. – Alex Jones Jun 25 '18 at 00:00
  • It's not true that $\delta^{(k)}(t) \sin t = \delta^{(k)}(t) \sin 0$. – Maxim Jun 25 '18 at 11:54
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Informally, you can think about $\delta(t)$ and $u(t)$ as things that make sense only under the integral sign. If you have $\delta(0)$ or $u(0)$ in your answer, that means something went wrong, because an integral shouldn't depend on how you define $u(t)$ at one point.

You can write $$\int_{-\infty}^\infty \{\delta(t) f(t)\} \phi(t) dt = f(0)\phi(0) = \int_{-\infty}^\infty \{\delta(t) f(0)\} \phi(t) dt.$$ This means $f(t)\delta(t) = f(0)\delta(t)$, because the results of integration will be the same for any $\phi$.

Assuming that $\phi$ is zero at $\pm\infty$ and applying integration by parts, you can write $$\int_{-\infty}^\infty \delta'(t) \phi(t) dt = -\int_{-\infty}^\infty \delta(t) \phi'(t) dt = -\phi'(0),$$ and, from this, derive $f(t) \delta'(t)$ = $f(0) \delta'(t) - f'(0) \delta(t)$. In particular, $\delta'(t) \sin t = -\delta(t)$. This has a rigorous meaning in terms of distributions.

But the rule for $\delta'(t)$ isn't actually necessary, because $$(u(t) \sin t)' = \delta(t) \sin t + u(t) \cos t = u(t) \cos t, \\ (u(t) \sin t)'' = (u(t) \cos t)' = \delta(t) \cos t - u(t) \sin t = \delta(t) - u(t) \sin t, \\ (u(t) \sin t)'' + (3 u(t) \sin t)' + 5 u(t) \sin t = \\ u(t)(3 \cos t + 4 \sin t) + \delta(t).$$ This differs from your answer by $-\delta(t)$.

Maxim
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  • I really do not know how your result is not the same as my teacher (who did through Mapple) both look correct to me. – Vinícius Lara Jun 25 '18 at 20:57
  • We don't know how the result you're referring to was obtained. Do you have the Maple code? – Maxim Jun 25 '18 at 21:09
  • Ohh, sorry. My teacher copied the Maple response wrong (i don't know how). But my end question, why $q''(t) = -u(t)sin(t) + cos(t)\delta(t) + sin(t) \delta'(t) + \delta(t)cos(t) = 2\delta(t) - u(t)sin(t)$ I used the ${\displaystyle \delta ^{(k)}[\varphi ]=(-1)^{k}\varphi ^{(k)}(0).}$ but it did not help me. – Vinícius Lara Jun 25 '18 at 21:29
  • This is correct: $q''(t) = -u(t) \sin t + {\delta(t) \cos t + \delta'(t) \sin t} + \delta(t) \cos t$. The two terms in curly braces is what you get if you differentiate $\delta(t) \sin t$ (which, as we know, is zero). But the second equality is not correct, because you're replacing $\delta'(t) \sin t$ with zero. The correct formula for $\delta'(t) \sin(t)$ is given in my answer. – Maxim Jun 25 '18 at 21:48