- #1

zenterix

- 524

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- 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.

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.

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

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 wayThis function has a jump invalue, 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.