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The solution to an inhomogeneous differential equation can be split up into homogeneous solution and a particular solution (forced response).

Another way to split up the solution to an inhomogeneous differential equation is in a zero-input response and a zero-state response.

The zero-input response is the system's response to its own internal initial conditions - no input signal is applied. The zero-input response is the homogeneous solution to the system's differential equation, using the initial conditions at $t=0^-$.

The zero-state response is the system's response to only the input signal - all initial conditions set to $0$. The zero-state response is found by convolving the system's impulse response $h(t)$ with the input signal $x(t)$.

This excerpt is from Lathi's signal processing and linear systems:

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Lathi claims that for a linear system, one can show that the decomposition property holds.

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Question:

How can I show that the decomposition property for a non-linear system, for example $\dot{y}(t) + y(t) = x(t) + 1$, does not hold?

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1 Answer 1

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Assume that $y_0(t)$ is the zero-input response. Then $y_0(t)$ must satisfy

$$\dot{y}_0(t)+y_0(t)=1\tag{1}$$

because $x(t)=0$.

Now let $y_1(t)$ be the zero-state response to an input $x(t)$, satisfying

$$\dot{y}_1(t)+y_1(t)=x(t)+1\tag{2}$$

If the decomposition property holds, the function $y_2(t)=y_0(t)+y_1(t)$ must be the response to $x(t)$ with possibly non-zero initial conditions. I.e., $y_2(t)$ should satisfy

$$\dot{y}_2(t)+y_2(t)=x(t)+1\tag{3}$$

However, adding Equations $(2)$ and $(3)$ gives

$$\dot{y}_2(t)+y_2(t)=x(t)+2\tag{3}$$

which shows that for the given system the decomposition property doesn't hold.

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  • $\begingroup$ Thanks for the answer Matt, but I'm not entirely convinced yet. How does the last equation show that the decomposition property does not hold? Adding equation 1 and 2 yields equation 3. $\endgroup$
    – Carl
    Feb 6, 2023 at 12:35
  • $\begingroup$ @Carl: The combined response $y_2(t)$ should satisfy the given input/output equation of the system if the decomposition property were to hold, but Eq. (3) shows that it doesn't. $\endgroup$
    – Matt L.
    Feb 6, 2023 at 12:40
  • $\begingroup$ Oh I think I get it now. So the system response $y(t)$ should satisfy the original differential equation $\dot{y}(t) + y(t) = x(t) + 1$. But when we solve the differential equation by splitting the response into the zero-input response $y_0(t)$ and zero-state response $y_1(t)$, then equation 3 shows that this method actually yields the solution to $\dot{y}(t) + y(t) = x(t) + 2$ which is not the original differential equation. Hence, in this case, splitting the system response into zero-state and zero-input response yields the wrong total response, and decomposition does not hold. Right? $\endgroup$
    – Carl
    Feb 6, 2023 at 12:53
  • $\begingroup$ @Carl: Yes, that's what I meant. $\endgroup$
    – Matt L.
    Feb 6, 2023 at 12:55

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