how to solve a differential equation with impulse input?

Differential equation such as a simple first order non-homogeous equation:

y'(t)+2*y(t) = x(t),

now x(t)=delta(t), since I want to find the impulse response of this system 
to the impulse input.

I did not find this in Oppeheim's Signal & Systems book... that's very 
strange. I am very confused by his "initial rest condition" arguement.

Can anybody show some rules about solving diff equation with impulse input 
in general? And point me to some resources?

Thanks a lot!

> Differential equation such as a simple first order non-homogeous equation:
>
> y'(t)+2*y(t) = x(t),

are you sure this is the right equation ?
asuming you were sure ...

> now x(t)=delta(t), since I want to find the impulse response of this 
> system to the impulse input.
>
> I did not find this in Oppeheim's Signal & Systems book... that's very 
> strange. I am very confused by his "initial rest condition" arguement.
>
> Can anybody show some rules about solving diff equation with impulse input 
> in general? And point me to some resources?

for solving diff equation like this i would recommend laplace-transformation
the solution for y'(t) + 2*y(t) = 0 is y(t) = exp(-2*t)
there is only a DC-component in the frequency response
so the responce will be that y after the impulse is constant, but different 
of y before
to make things visual you can calculate the frequency respons by using the 
fourier-tranformation
Y(w)/X(w) = (2 + jw)/(2 - w^2)
a pole on -2*j, making the system stable

in mechanical terms it's a massless object connected with a spring and a 
damper
the initial rest condition implies that the kinetic energy and potential 
energy of the spring are initially zero 

On Sun, 14 Aug 2005, kiki wrote:

> Differential equation such as a simple first order non-homogeous equation:
>
> y'(t)+2*y(t) = x(t),
>
You want to find a particular solution to the equation
which depends upon x(t).  Call that f(t).

Now (the rest condition?)
	y' + 2y = 0
has the solution
	e^-2t

Thus the solutions are
	c.e^-2t + f(t)

> now x(t)=delta(t), since I want to find the impulse response of this system
> to the impulse input.
>
Yuck,equationwithoutspaces.

> I did not find this in Oppeheim's Signal & Systems book... that's very
> strange. I am very confused by his "initial rest condition" arguement.
>
> Can anybody show some rules about solving diff equation with impulse input
> in general? And point me to some resources?
>

William Elliot wrote:

   ...

> Yuck,equationwithoutspaces.

There's often good reason for that in a newsgroup. It keeps the equation 
from being broken across lines no matter what quoting does to it.

   ...

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

jan hauben wrote:
>>Differential equation such as a simple first order non-homogeous equation:
>>
>>y'(t)+2*y(t) = x(t),
> 
> 
> are you sure this is the right equation ?
> asuming you were sure ...
> 
> 
>>now x(t)=delta(t), since I want to find the impulse response of this 
>>system to the impulse input.
>>
>>I did not find this in Oppeheim's Signal & Systems book... that's very 
>>strange. I am very confused by his "initial rest condition" arguement.
>>
>>Can anybody show some rules about solving diff equation with impulse input 
>>in general? And point me to some resources?
> 
> 
> for solving diff equation like this i would recommend laplace-transformation
> the solution for y'(t) + 2*y(t) = 0 is y(t) = exp(-2*t)
> there is only a DC-component in the frequency response
> so the responce will be that y after the impulse is constant, but different 
> of y before
> to make things visual you can calculate the frequency respons by using the 
> fourier-tranformation
> Y(w)/X(w) = (2 + jw)/(2 - w^2)
> a pole on -2*j, making the system stable
> 
> in mechanical terms it's a massless object connected with a spring and a 
> damper
> the initial rest condition implies that the kinetic energy and potential 
> energy of the spring are initially zero 

I often find that my intuition is better served by solving for the step 
response, then differentiating.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

Jerry Avins wrote:

> William Elliot wrote:
> 
>    ...
> 
>> Yuck,equationwithoutspaces.
> 
> There's often good reason for that in a newsgroup. It keeps the
> equation from being broken across lines no matter what quoting
> does to it. 

Except if there's a minus sign in the formula. Better to make it 
readable, then castigate fourth-level respondents who don't rectify 
messy quotes ;)


Martin

-- 
Quidquid latine dictum sit, altum viditur.

Jerry Avins wrote:
> 
> jan hauben wrote:
> >>Differential equation such as a simple first order non-homogeous equation:
> >>
> >>y'(t)+2*y(t) = x(t),
> >
> >>now x(t)=delta(t), since I want to find the impulse response of this
> >>system to the impulse input.
> >>

y'(t)+2*y(t) = x(t) => (multyplying with integrating factor)

exp(2*t)*y'(t)+2*exp(2*t)*y(t) = exp(2*t)*x(t) =>

( exp(2*t)*y(t) )' = exp(2*t)*x(t) =>

exp(2*t)*y(t) = C + int_{-oo}^{t} exp(2*s)*x(s) ds =>

y(t) = exp(-2*t)*C + exp(-2*t)*int_{-oo}^{t} exp(2*s)*x(s) ds

for x(t) = delta(t) it follows that

y(t) = exp(-2*t)*C + exp(-2*t)*H(t)

with H(t) the Heaviside function (H(t)=1 for t>0 and 0 for t<0; dH(t)/dt
= dirac(t)).

Wilbert

On Mon, 15 Aug 2005, Jerry Avins wrote:

> William Elliot wrote:
>
> > Yuck,equationwithoutspaces.
>
> There's often good reason for that in a newsgroup. It keeps the equation
> from being broken across lines no matter what quoting does to it.
>
That is not a good reason because most equations aren't that long
and an equation which is that long and all the moreindeedofspaces
canbebrokenintotwolines which can be done at a major +, for example.

kiki wrote:
> Differential equation such as a simple first order non-homogeous equation:
> 
> y'(t)+2*y(t) = x(t),
> 
> now x(t)=delta(t), since I want to find the impulse response of this system 
> to the impulse input.
> 
> I did not find this in Oppeheim's Signal & Systems book... that's very 
> strange. I am very confused by his "initial rest condition" arguement.
> 
> Can anybody show some rules about solving diff equation with impulse input 
> in general? And point me to some resources?
> 
> Thanks a lot!
> 
> 
> 
Playing fast and loose not only with the dirac delta functional but with 
the conversion from differential to integral equations gets you:

y'(t) = -2*y(t) + x(t)

implies that

           t
         /
y(t) =  | (-2*y(t) + x(t)) dt + y(0-).
         /
          0-

Where 0- is infinitesimally lower than zero and 0+ is infinitesimally 
higher.  Now solve the integral from 0- to 0+.  You can argue that y(t) 
is finite, so:

y(0+) = 1 + y(0-).

Now finish solving as if it were a regular old diff-eq.

-- 

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Tim Wescott wrote:
> kiki wrote:
>> Differential equation such as a simple first order non-homogeous
>> equation: y'(t)+2*y(t) = x(t),
>>
>> now x(t)=delta(t), since I want to find the impulse response of this
>> system to the impulse input.
>>
>> I did not find this in Oppeheim's Signal & Systems book... that's
>> very strange. I am very confused by his "initial rest condition"
>> arguement. Can anybody show some rules about solving diff equation with 
>> impulse
>> input in general? And point me to some resources?
>>
>> Thanks a lot!
>>
>>
>>
> Playing fast and loose not only with the dirac delta functional but
> with the conversion from differential to integral equations gets you:
>
> y'(t) = -2*y(t) + x(t)
>
> implies that
>
>           t
>         /
> y(t) =  | (-2*y(t) + x(t)) dt + y(0-).
>         /
>          0-
>
> Where 0- is infinitesimally lower than zero and 0+ is infinitesimally
> higher.  Now solve the integral from 0- to 0+.  You can argue that
> y(t) is finite, so:
>
> y(0+) = 1 + y(0-).
>
> Now finish solving as if it were a regular old diff-eq.

Correct.  Tim has the right way to look at the problem.  To expand on this 
approach a bit:
The original problem is
y'+2y=delta(t), y(0)=0.   (1)
The highest order derivative on the LHS must be a delta, so y is a step 
function at t=0.
Integrate the ODE from t=0- to t=0+ to get
y(0+)+0=1.
Thus, the original system, Eq. 1, is equivalent to the system
y'+2y=0, y(0)=1,  (2)
whose solution the original poster can presumably obtain.
y(t) is discontinuous at t=0.