Unit impulse / step response of a 1st order differential equation

Question

You are given the equation $10v'(t) + 0.6 v(t) = f(t)$

$v(t)$ is the velocity of the object

Determine the unit impulse response AND the unit step response.

How would i approach this question? do i set v(0 -) = 1 and solve or..?

Answer 1

We are given:

$$10v'(t) + 0.6 v(t) = f(t)$$

Since the impulse input is the derivative of the unit step input, the impulse response of a linear system is the derivative of its unit step response. Hence, in order to get the impulse response, first obtain the unit step response, and then differentiate the unit step response with respect to time. This procedure is easier.

As a first step, we set $f(t) = u(t)$, the Heaviside unit step function, and solve:

$$10 v'(t) + \dfrac{6}{10} v(t) = u(t), ~~v(0-) = 0$$

This results in:

$$v_{unit}(t) = \dfrac{5}{3} e^{-\frac{3 t}{50}} \left(e^{\frac{3 t}{50}}-1\right) u(t)$$

A plot of the unit response is:

Next, we want to find the impulse (Dirac delta function) response by taking the derivative of $v_{unit}(t)$. This results in:

$$v_{impulse}(t) = \begin{cases}\dfrac{1}{10}e^{-\frac{3t}{50}}, & t > 0, \\ \infty, & t = 0.\end{cases}$$

A plot of the impulse response is:

Answer 2

We have to solve the equation in two particular cases, $$a v^\prime + b v = \delta(t)$$ $$a v^\prime + b v = \theta(t)$$ where $a = 10$, $b = 0.6$, $\delta(t)$ is the Dirac delta function, and $\theta(t)$ is the Heaviside step function. Note that ${d\over dt}\theta(t) = \delta(t)$.

Let's take the second case first. We seek a continuous solution $v(t)$ that vanishes for $t \leq 0$. For $t > 0$, the equation says $$a v^\prime + b v = 1\,.$$ This is a familiar equation with solution of the form $$v(t) = Ae^{\alpha t} + B\,.$$ Substituting this ansatz, we find $$A(a\alpha + b)e^{\alpha t} + bB = 1\,.$$ This must hold for all $t > 0$, which requires $\alpha = -{b\over a}$ and $B = {1\over b}$. Our solution so far for $t>0$ is thus $$v(t) = Ae^{-{b\over a}t} + {1\over b}\,.$$ Continuity of $v(t)$ at $t=0$ requires $v(0^+)=v(0^-) = 0$, or $A + {1\over b} = 0$. This determines $A = -{1\over b}$ and the desired solution is finally $$v(t) = {1\over b}(1-e^{-{b\over a}t})\,.$$ Physically, this like the following situation: A particle in a resistive medium is initially at rest. Then at $t=0$, a constant force is switched on. The particle eventually reaches terminal velocity.

Now for the second problem, $$av^\prime + bv = \delta(t)\,.$$ As noted by Amzoti, we can easily find the solution to this problem by differentiating the solution to the previous problem. For if a function $u(t)$ satisfies $a u^\prime + bu = \theta$, then differentiating yields $a(u^\prime)^\prime + b(u^\prime) = \delta$, that is, $u^\prime$ satisfies the impulse equation.

But I will solve the impulse equation from scratch anyway in case helpful in this or similar problems.

We seek a function $v(t)$, vanishing for $t < 0$, discontinuous at $t=0$, and satisfying $av^\prime + bv = \delta(t)$. For $t > 0$, the equation reads $$av^\prime + bv = 0\,,$$ which is easily solved to obtain $$v(t) = Ae^{-{b\over a}t}\,.$$ Clearly the magnitude of $A$ will set the magnitude of the discontinuity in $v$ at $t=0$. How big should the discontinuity be? To find out, integrate both sides of the differential equation on a small neighborhood of $t=0$: $$\int_{-\epsilon}^\epsilon{av^\prime(t)\,dt} + \int_{-\epsilon}^\epsilon{bv(t)\,dt} = \int_{-\epsilon}^\epsilon{\delta(t)\,dt}\,.$$ Performing two of the integrations, we have $$a(v(\epsilon) - v(-\epsilon)) + \int_{-\epsilon}^\epsilon{bv(t)\,dt} = 1\,.$$ Take the limit as $\epsilon \rightarrow 0$. The remaining integral vanishes, because the integral of $v$ is continuous. So we have $$a\Delta v|_{t=0} = 1$$ and the discontinuity at $t=0$ must be $\Delta v = {1\over a}$. But using the solution $Ae^{-{b\over a}t}$ above (and remembering $v=0$ for $t<0$), this gives $A = {1\over a}$. Thus our solution for the impulsive case is $$v(t) = {1\over a}\,e^{-{b\over a}t}\,.$$ You can verify that this is indeed the derivative of the previous solution.

Physically, this is like the following situation: A particle in a resistive medium is initially at rest. At $t=0$, it is given a sharp whack. It gains an impulsive kick, but with no continuing force applied, the particle slows down because of the resistive medium.

Since the question was tagged with the Laplace transform, I'll show one other method for the impulsive case. Taking the Laplace transform of both sides of $av^\prime + bv = \delta$, we have $$a s F(s) - b F(s) = 1$$ so that $$F(s) = {1\over as + b} = {1\over a} {1\over s - -({b\over a})}\,.$$ A table of Laplace transforms now gives the solution as $v(t) = {1\over a}e^{-{b\over a}t}$.