# Show that decomposition does not hold for non-linear system

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

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?

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.

• 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.
– Carl
Feb 6 at 12:35
• @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. Feb 6 at 12:40
• 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?
– Carl
Feb 6 at 12:53
• @Carl: Yes, that's what I meant. Feb 6 at 12:55