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Zero-input and zero-state response is a simple method to solve for circuits that would otherwise require horrible differential equations to solve. By observing how parts of a circuit act at different times, you can determine the results as if you had gone through the whole differential equation mess.

Whenever you solve for the differential equations for ugly circuits with capacitors and inductors in them, you will undoubtedly end up with terms of the Ae(-t/T) form, where t represents time and A and T are constants. Since that's such a common pattern, it's only natural to take a look at why this occurs so often. The number e to some power usually refers to some kind of buildup or decay, and that is exactly what's happening here.

The whole idea behind zero-input and zero-state response is that you find how all of the elements act at their zero-input and zero-state times, and take their characteristics at those points in time. There are two possible things to determine using this technique for any given element, voltage and current. If you wanted to solve for an element's voltage as a function of time, for instance, you could do so. In order to do this, you need to be able to determine the element's initial voltage, final voltage, and the Thévenin equivalent resistance of the whole circuit. Here's the basic equation for voltage according to zero-input and zero-state response:

V(t) = VI(e-t/T) + VF(1 - e-t/T) 

And the response for current is very similar:

I(t) = II(e-t/T) + IF(1 - e-t/T) 

That capital T in the equation should be the Greek letter tau. Tau stands for the time constant of a circuit. In a circuit with just resistors and capacitors, the time constant is simply the Thévenin equivalent resistance, RTH, times the equivalent capacitance, CEQ. Let's carry out an example, finding the voltage of a capacitor, VC, as a function of time for a simple RC circuit.

    |                   |
    |                   |
   /+\                  |
  /   \                 |        
 |  V  |V Volt        -----        
  \   / voltage              C
   \-/  source        -----
    |                   |
    |                   |

Using Thévenin's Theorem, you can see that the Thévenin equivalent resistance is just R. The equivalent capacitance is just C. That determines our time constant, T = RC.

Next we must determine the intitial and final voltage across the capacitor. There is an easy trick for this: if capacitors are not charged and a current is suddenly introduced to them, they act as short circuits. When they have their maximum charge stored on them, they act as open circuits. Thinking about the circuits in this manner will allow you to quickly determine VI and VF. As for inductors, they act in the opposite manner: as open circuits when they are discharged, and as short circuits when they charge up fully.

We assume that there is no charge on the capacitor to begin with and that the voltage source steps up to V at time t=0. So, at time t=0, the voltage across the capacitor is zero. As time goes on, charge will build up on the capacitor until it lets no current by. At this point, the capacitor acts as an open circuit. That would break our current loop and current would stop flowing. Thus, the capacitor must have the same voltage drop as the source, otherwise current would start flowing from one end to the other. We can just take the equation from above and fill it in:

V(t) = V(1 - e-t/RC) 

That's much easier than using Kirchhoff's Laws and having to bang through some differential equations. One more nice thing about zero-input and zero-state response is that you deal with the initial conditions up front: you determine VI and VF right away, instead of having to worry about constants of integration. The equation in itself is not complicated at all, and one term will usually disappear because one of the initial or final terms will usually be zero.

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