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I am new in the field of systems and signals, and I have a rather basic for the majority of the group, question:

Can we find the impulse response function of homogeneous ODE, instead of its zero-input response?

for example, we have the following 2nd order homogeneous ODE:

$a_{2}\ddot{x}+a_{1}\dot{x}+a_{0}{x} = 0$

, where the output is the $x(t)$ given the initial conditions

I understand that if it were: $a_{2}\ddot{x}+a_{1}\dot{x}+a_{0}{x} = f(t)$

the $f(t)$ would be its input, $x(t)$ its output given the input, and we could find the impulse response by replacing the $f(t)$ with $\delta (t)$.

Now that the input is zero, how can we find what the output would look like with respect to any input?

Is this:

$a_{2}\ddot{x}+a_{1}\dot{x}+a_{0}{x} = \delta (t)$

even allowed, for an initially homogeneous equation?

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  • $\begingroup$ Alex, formally speaking a homogeneous ODE with initial conditions will not constitude an LTI system and won't admit an impulse response: a response which can be used through convolution to compute outputs for arbitrary admitted inputs... $\endgroup$
    – Fat32
    Oct 18 '19 at 20:47
  • $\begingroup$ I don't understand this question. The response of a system described by an ODE with constant parameters is the sum of the zero-input (natural) response and the zero-state (forced) response (which assumes zero initial conditions, i.e., zero initial state). $\endgroup$ Oct 19 '19 at 6:50
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The answer to your question

Can we find the impulse response function of homogeneous ODE, instead of its zero-input response?

is "no", because only a system with an input and an output can have an impulse response, a homogeneous ODE doesn't have an impulse response.

The impulse response of a system, possibly described by an ODE, is the zero-state response to an input signal $x(t)=\delta(t)$.

Of course, a system can also have a zero-input response, which is obtained by solving the corresponding homogeneous ODE with the appropriate initial conditions, but this response is not directly related to the system's impulse response.

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Basically, you can solve this kind of EDO using two different techniques (which also have c different meaning).

  1. Classical Method - you will need the full EDO, containing input and outputs with initial conditions applied over the whole solution y(t), where y(t) = yhomogeneous(t) + yforced(t). The first term, you solve the homogeneous EDO considering NO CONDITIONS. Then, following the pattern of input you exactly find the forced contribution. Then, you add both contributions and apply the initial conditions... In this technique you totally lose the feeling about the system itself.

  2. zero state response and zero input response. In this case, the whole solution is written as y(t) = y0(t) (zero input response) + x(t)h(t) (zero state response, as convolution of both signals. The text cutof the asterics). To find y0(t) you do not consider the input x(t), and apply condition at t= 0^-. It happens because it is completely independent of the input. The system responds just because of the storage energy. The second term, zero state response, it is the convolution of input and impulse response of the system. So, you need to find the impulse response of EDO. It is done considering the whole EDO , e.g., D4y -D3y +2D2y -2Dy +y = Dx -3x . First you have to solve the homogeneous equation and apply some special conditions generalized... y=Dy=D2y=0 and D3y=1. Lets call it yn(t).... Then you apply the input operator over this yn(t). It means, D (yn(t)) -3yn(t)... After that, you finally have h(t)... So, you do the convolution of h(t) and x(t) and find the zero state response y

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I'm not sure if this answers the question, but in my view, the impulse response and natural response are related. i.e. for $Tx'(t) + x(t) = Ku(t)$: natural response is: $x_n(t) = x(0)e^{-t/T}$ and unit impulse response is: $x_i(t) = Ke^{-t/T}$ so both decay exponentially with time constant $T$ and $x_n(t) = x_i(t)$ if $K = x(0)$

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