Given unit impulse response, obtain differential equation

  • #1
zenterix
524
74
Homework Statement
Find the linear time-invariant differential operator ##p(D)## having unit impulse response ##2u(t)##.
Relevant Equations
Note that ##u(t)## is the Heaviside unit step function.
I initially solved this problem in quite a roundabout way by thinking about a mass-spring-dashpot system modeled by

$$m\ddot{x}+b\dot{x}+kx=f(t)$$

Since the response is constant at ##x=0## and has a jump at ##t=0## to ##x=1##, my reasoning was that there can be no spring otherwise there would be a force at ##x=1## that would change ##x##.

Thus, ##k=0##.

A delta function ##\delta(t)## as input to an ##n##-th order differential equation generates a jump in the ##(n-1)##-th derivative function. This indicates that the equation we are looking for should be first order.

Besides, from the response we see there is no acceleration.

At this point then, we have

$$b\dot{x}=\delta(t)$$

which has constant solution

$$x(t)=\frac{1}{b}=2\implies b=\frac{1}{2}$$

That is, the differential equation ##\frac{1}{2}\dot{x}=\delta{t}## has unit impulse response ##2u(t)##.

This differential equation is ##p(D)x(t)=\delta(t)## where ##p(D)=\frac{1}{2}D##.

Then, later I found an easier way.

Since we have the response ##x(t)=2u(t)## all we have to do is differentiate

$$\dot{x}(t)=2\delta(t)$$

$$\frac{1}{2}\dot{x}(t)=\delta(t)$$

which is a differential equation.

My question is about solving this problem in the following way

This function has a jump in value, so the operator must be of first order.

##(aD+bI)(2u)=2a\delta(t)+2bu(t)##, so ##b=0## and ##a=\frac{1}{2}##.

Thus ##p(D)=\frac{1}{2}D##.

Why is it possible to infer ##b=0## and ##a=\frac{1}{2}## here?

PS This question is based on this problem set (with solutions). The snippet above is the solution presented by the problem set.
 
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  • #2
The object is to find [itex]a[/itex] and [itex]b[/itex], given [itex]\delta(t) = 2a\delta(t) + 2bu(t)[/itex].

Clearly [itex]a = \frac12[/itex] and [itex]b = 0[/itex] is a solution to this, but is it the only one? Yes, as we can easily show:

If [itex]\delta(t) = 2a\delta(t) + 2bu(t)[/itex] then for any [itex]f[/itex] we have [tex]\begin{split}
\int_{-1}^1 f(t)\delta(t)\,dt &= \int_{-1}^{1} (2a\delta(t) + 2bu(t))f(t)\,dt \\
f(0) &= 2af(0) + 2b\int_0^1 f(t)\,dt. \end{split}[/tex] Now this holds for every [itex]f[/itex], so in particular it holds for [itex]f(t) = \cos(2\pi t)[/itex], for which [itex]f(0) = 1[/itex] and [itex]\int_0^1 f(t)\,dt = 0[/itex]. Then [itex]1 = 2a[/itex] requires [itex]a = \frac12[/itex]. It must also hold for [itex]f(t) = t[/itex], yielding [itex]0 = b[/itex].
 
  • #3
Ok, so we start with ##x(t)=2u(t)##.

We try first order operator ##p(D)=(aD+bI)##

$$p(D)x(t)=(aD+bI)(2u(t))=2a\delta(t)+2bu(t)$$

and at this point we equate this to ##\delta(t)##. Why ##\delta(t)##?

What is the exact reason?

It seems that all this work is unnecessary in any case. Isn't it best to just differentiate ##x(t)##?
 
  • #4
Are you not trying to find [itex]p(D)[/itex] such that [itex]p(D)(2u(t)) = \delta(t)[/itex]?
 
  • #5
pasmith said:
Are you not trying to find [itex]p(D)[/itex] such that [itex]p(D)(2u(t)) = \delta(t)[/itex]?
The problem asks us to find ##p(D)## such that ##p(D)x(t)=\delta(t)## has solution ##2u(t)##.

You are right.

I overlooked the fact that the problem says find the LTI differential operator ##p(D)## having ##2u(t)## as unit impulse response.

Therefore, yes, we want to find ##p(D)## in ##p(D)x(t)=\delta (t)##.

I was thinking of the problem as find a differential equation ##p(D)x(t)=f(t)## such that ##x(t)=2u(t)## is the solution.

In the latter case, differentiating ##x(t)=2u(t)## leads to ##\dot{x}=2\delta(t)##.

We can think about the system in two ways:

1) ##p(D)=D## and ##f(t)=2\delta (t)##.

2) ##p(D)=\frac{D}{2}## and ##f(t)=\delta (t)##

In case 2) we have unit impulse.

In case 1), we have double the impulse.

Both cases are the same thing. I was just thinking about the fact that ##p(D)## represents a system and ##f(t)## the signal.

Even though the equation is essentially the same in both cases, it seems we can think about it in two different ways.
 
Last edited:
  • #6
The problem has been answered, but as for methodology: Transforms are your friend here!

For example, you have:
$$
P(D)[2u(t)] = \delta(t)
$$
Take the Fourier transform
$$
2 P(ik) \left[ \frac{1}{ik} + \pi \delta(k) \right] = 1
\qquad \Leftrightarrow \qquad
P(ik) = \frac{ik}{2}[1 - 2\pi P(0) \delta(k)] = \frac{ik}{2}.
$$
It's like magic! Do the substitution ##\xi = ik## to obtain ##P(\xi) = \xi/2## and therefore ##P(D) = D/2##.

This will work well also for more complicated impulse responses.
 
Last edited:
  • #7
How about if ##x(t)=u(t)t##?

Since ##\dot{x}(t)=u(t)## we have a jump discontinuity in the derivative.

We expect ##p(D)## to be second order so that a delta function causes such a jump.

$$p(D)=aD^2+bD+c\tag{1}$$

We're looking for the coefficients such that

$$p(D)x(t)=a\ddot{x}+b\dot{x}+cx=\delta(t)\tag{2}$$

is true for our given ##x(t)=u(t)t##.

For ##t>0## this means

$$a\ddot{x}+b\dot{x}+cx=0\tag{3}$$
$$\dot{x}(0^+)=\frac{1}{a}\tag{4}$$
$$x(0^+)=0\tag{5}$$

Subbing in ##\dot{x}(t)=u(t)##,

$$a\delta(t)+bu(t)+ctu(t)=0\tag{5}$$

$$b+ct=0$$

which must be true for all ##t## and thus ##b=c=0##.

From (4)

$$\dot{x}(0^+)=1=\frac{1}{a} \implies a=1$$

Thus, ##p(D)=D^2##.
 
  • #8
Then the fourier transform is
$$
P(ik)\left[\frac{1}{i^2k^2} - i\pi \delta’(k)\right] = 1
$$
so
$$
P(D) = D^2
$$
 

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