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The response of a linear, time-invariant system to a unit step is \$s(t)=(1-e^{\frac{-t}{RC}})u(t)\$, where u(t) is the unit step. What is the impulse response of the system?

(A)\$e^{-t/RC} \$

(B)\$e^{-2t/RC} \$

(C)\$\frac{1}{RC}e^{-t/RC}u(t)\$

(D)\$\delta (t) \$

Doubt:

The derivative of the unit step response generates the unit impulse response.

Therefore we have to find the derivative of the expression: \$s(t)=(1-e^{\frac{-t}{RC}})u(t)\$ w.r.t \$t\$.

DOUBT 1) The derivation of the the first term in the expression \$s(t)=(1-e^{\frac{-t}{RC}})u(t)\$, should yield $\delta (t)$. But there is no option which contains $\delta (t)$ in combination with other terms.

DOUBT 2) How to find the derivative of the term: \$-e^{-t/RC}u(t)\$. I am confused about it because \$u(t)\$ is appended to it. Should it be like:

\$-e^{-t/RC}\$ exists only for +ve values of time. Therefore, derivative of \$-e^{-t/RC}u(t)\$ is \$\frac{1}{RC}e^{-t/RC}u(t)\$.

Therefore, the answer should be:

\$\delta(t)+\frac{1}{RC}e^{-t/RC}u(t)\$

But there is no such option.

In short, my question is:

How to find the derivative of:

$$s(t)=(1-e^{\frac{-t}{RC}})u(t)$$ w.r.t \$t\$ ?

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    \$\begingroup\$ Hint: Use the factoring and arithmetic properties for derivatives.... \$(1-e^{t/RC})u(t) = u(t)-u(t)e^{t/RC}\$. Now use the chain rule for that exponent. \$\endgroup\$ – KingDuken Apr 9 '18 at 16:51
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    \$\begingroup\$ @KingDuken Is the answer: $\delta(t)+\frac{1}{RC}e^{-t/RC}u(t)$ \$\endgroup\$ – Anwesa Roy Apr 9 '18 at 16:54
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    \$\begingroup\$ @ThePhoton The graph of $s(t)=(1-e^{\frac{-t}{RC}})u(t)$ starts from 0, increases gradually and then tends towards 1 but never reaches 1. \$\endgroup\$ – Anwesa Roy Apr 9 '18 at 17:13
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    \$\begingroup\$ @ThePhoton Sir, probably it does. There is a discontinuity at $t=0$ for the signal $s(t)=(1-e^{\frac{-t}{RC}})u(t)$, because left hand limit $\ne$ right hand limit. So, there should probably be an impulse at $t=0$. And, for the term, $u(t)-e^{\frac{-t}{RC}}u(t)$ ,at instances $t>0$, the diffrence between $t^+$ and $t^-$ is high towards the starting end and gradually decreases and after that tends to zero but never reaches 0. \$\endgroup\$ – Anwesa Roy Apr 9 '18 at 17:34
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    \$\begingroup\$ @AnwesaRoy Congratulations, you made your first MathJax equation on this site. \$hghg\$. \$\endgroup\$ – Harry Svensson Apr 9 '18 at 17:50
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Forewords: I like Meenie Leis's answer. There's just some minor formatting problems, but the fruit is there for the taking. But if you are presenting peanuts to someone who is allergic to peanuts, then you have to try apples, or some other fruit. So I acknowledge Meenie's answer as a good one.

Here is my attempt at giving apples.


So we want the impulse response of the following function which, is the result of a step function.

\$s(t)=(1−e^{\frac{-t}{RC}})u(t)\$

In other words, we want to transform our \$s(t)\$ to \$S(s)\$, multiply by \$s\$, and then transform back to \$s(t)\$.


According to the laplace table from here, the following important transforms can be made:

\$ \begin{array}{l l l} e^{-at} & \mathcal{L} & \frac{a}{s+a}\\ \text{u(t) & 1} & \mathcal{L} & \frac{1}{s}\\ \end{array} \$

We also know the following things:

\$ \begin{array}{l l l} a(t)*b(t) & \mathcal{L} & A(s)B(s)\\ a(t)b(t) & \mathcal{L} & A(s)*B(s)\\ \end{array} \$

where \$*\$ means convolution.


So right off the bat we have a multiplication in time domain which results in a convolution in Laplace domain. But if we look on the left hand side, it says that \$\text{u(t) & 1}\$ give the same result. In other words, we can replace the \$u(t)\$ with a 1. Nice.

Note: this is only possible to do with double sided Laplace transforms, so you can't always do this trick.

So let's Laplace transform the following equation instead:

\$1−e^{\frac{-t}{RC}}~~~~ \mathcal{L}~~~~ \frac{1}{s}-\frac{1}{s+\frac{1}{RC}}\$

And let's multiply by \$s\$

\$ \begin{align} s\bigg(\frac{1}{s}-\frac{1}{s+\frac{1}{RC}}\bigg) &= 1-\frac{s}{s+\frac{1}{RC}}\\\\ &= \frac{s+\frac{1}{RC}}{s+\frac{1}{RC}}-\frac{s}{s+\frac{1}{RC}}\\\\ &= \frac{s+\frac{1}{RC}-s}{s+\frac{1}{RC}}\\\\ &= \frac{\frac{1}{RC}}{s+\frac{1}{RC}}\\\\ &= \frac{1}{RC}\frac{1}{s+\frac{1}{RC}}\\ \end{align} \$

Let's mark out very important things.

\$\color{red}{\frac{1}{RC}}\color{green}{\frac{1}{s+\frac{1}{RC}}}\$

The first part, in red, is just a scalar. So we only have to transform the green part.

\$ \color{red}{\frac{1}{RC}}\color{green}{\frac{1}{s+\frac{1}{RC}}} ~~~~ \mathcal{L}^{-1}~~ \color{red}{\frac{1}{RC}}\color{green}{e^{\frac{-t}{RC}}}\$

The original function had a heaviside function attached to it. This means that the derivative is 0 for t < 0. So the derivative will be 0 for t < 0 and whatever it is for t>=0. So let's simply attach the heaviside function due to the original equation.

So going to the laplace domain, multiplying by \$s\$, and then going back to the time domain gives you the following answer: \$\frac{1}{RC}e^{\frac{-t}{RC}}u(t)\$

That answer matches with option C.

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Use Product Rule.

$$\frac{d}{dt}(1-e^{\frac{-t}{RC}})u(t)=\frac{d}{dt}(1-e^{\frac{-t}{RC}}).u(t)+\frac{d}{dt}u(t).(1-e^{\frac{-t}{RC}})$$ $$=(0+\frac{1}{RC}e^{\frac{-t}{RC}}).u(t)+\delta(t).(1-e^{\frac{-t}{RC}})$$ \$\delta(t)\$ is defined only at t = 0. And its value = \$\infty\$ at t = 0. Therefore , $$= \frac{1}{RC}e^{\frac{-t}{RC}}.u(t)+\infty (1-e^0)$$ $$= \frac{1}{RC}e^{\frac{-t}{RC}}.u(t)+\infty .0 $$ $$= \frac{1}{RC}e^{\frac{-t}{RC}}.u(t)$$

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – clabacchio Apr 12 '18 at 7:33

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