# Zero-Input/Zero-State Response vs. Homogenous/Particular Solution

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1. Sep 9, 2015

### tomizzo

I have a question regarding the solutions to linear-ordinary differential equations. I had originally learned that the solutions to such differential equations consist of a homogenous solution and particular solution. The homogenous response is due to initial conditions while the particular response is due to the forcing function.

However, I've recently heard the terminology of Zero-Input/Zero-State response. More specifically, the summation of these two responses gives the solution to the linear ODE. I assume that the zero-input response is similar to the homogenous response, but I'm not sure about the zero-state response. How exactly is the zero-input/zero-state response response different from the idea of homogenous/particular solutions?

I've attempted to search for an answer to this question, but have had no luck.

Any help?

Thanks,

2. Sep 9, 2015

### RUber

Zero input sounds like they mean the forcing function is zero...i.e. the homogeneous solution.
Zero state is a term I have not heard, but is likely what drives your particular solution.

3. Sep 9, 2015

### WWGD

Note that homogeneous refers to a really smart homosexual (homo genius) ;).

4. Sep 10, 2015

### Strum

5. Sep 10, 2015

### jasonRF

In linear systems analysis, at least the way electrical engineers do it, the approach using zero input at zero state responses is often useful. The idea is that with real systems you typically know the initial conditions (voltage across a resistor ...) before the input is applied. Usually we denote that time as $t=0^-$. Then we apply the input, which can be something as simple as closing a switch connected to a battery. The zero input response uses these intitial conditions with no input, and the zeros state response sets all initial conditions to zero then applies the input.

If you want to apply the traditional homogeneous+complementary solution approach there is a complication, which I will try to describe here. The time right after the input is applied is often called $t=0^+$; note that in general the "initial conditions" at $t=0^+$ are NOT the same as those at $t=0^-$. IHowver, if you want to apply the traditional approach you need to derive the initial conditions at $t=0^+$. For example, if we have a circuit with all inductors, resistors and capacitors then we know that the current through an inductor must be continuous and the voltage across a capacitor must be continuous - this lets us derive what the new initial condition is (the voltage across our resistor again, which does not have to be continuous) at $t=0^+$. Once we do that, we can find the homogeneous and particular solutions and we are done.

Note that the zero input response includes terms that you would find in the homogeneous solution, but they will have different coeffients. This is because the zero state response will have have the particular solution plus terms that are found in the homogeneous solution. If you do everything right, both ways give the same answer.

Note that this is why electrical engineers usually define laplace transforms to be
$F(s) = \int_{0^-}^\infty dt \, e^{-s t} f(t)$. Note the $0^-$ as the lower limit of integration. This lets us use the initial conditions at $t=0^-$ that we know without having to derive those at $t=0^+$.

I hope that helps

jason

6. Sep 10, 2015

### jasonRF

Extra piece of information. If you are solving a simple problem (for which the conditions at $t=0^+$ are easy to derive) for a single input, then the traditional approach is probably easier and faster. Even better, the Laplace transform, as I wrote it above, is probably by far the easiest way to go (it is what I would usually do).

When solving for many different inputs, or for some kinds of theoretical analysis the zero state and zero input may be preferred.

jason