Zero Input Response

Posted by Jeff Schuler on Sep 25, 2009 in Uncategorized |

Characterizing the zero-input response

The total response of a given system can be expressed as the sum of two components: the zero-input component and the zero-state component:

Total response = zero-input response + zero-state response.

Our system: Q(D)y_{1}(t) = P(D)f_{1}(t)

Where y_{1}(t) represents our input and f_{1}(t) represents our output.

The zero-input response, which we will be solving for here, is the system response when the input f(t) = 0 so that is the result of the internal system conditions.  It is independent of the external input f(t).  Therefore, in order to solve a differential equation that represents the zero-input response component of a system, we will need to have the initial conditions of the system, or solve for them.  Keep this in mind while working your problem.

A zero-input response example problem

Okay, let’s go ahead and solve a problem shall we?

zero-input-responseOur first step is to find a differential equation that will represent the system for t \ge 0 .  This is because we only care about the behavior after t = 0 for a zero-input response problem.  We can notice that when our switch opens, this is a natural response.  In other words, we do not switch into a circuit that has a current or voltage source in it, just a capacitor, inductor and two resistors.  In order to solve our ZIR problem, we also need initial conditions do we not?  We will worry about those after we get our differential equation.

The problem tells us that we need to find V_{c}(t) , so lets look at this circuit after the switch opens and decide what kind of equation we need to write.  Can you see that once the switch opens we have a capacitor, an inductor, and two resistors in series?  If not, you need to study up on parallel and series circuits.  After we combine these two resistors, we have three elements in series, a series RLC circuit.

The sum of voltages around this loop should equal zero by KVL

V_{c} + V_{L} + V_{R1} + V_{R2} = 0

We should know the following relationships to help us with our substitutions:

V_{L} = L\frac{di}{dt} ;   V_{c}(t) = \frac{1}{C}\int_{t_{0}}^{t}i(t)dt + v(0) ;   I_{c} = C\frac{dV}{dt} I_{L}(t) = \frac{1}{L}\int_{0}^{t}V(t)dt + i(0)

Because we are solving for V_{c} , we will leave this term alone and replace as many of the others as we can.  Note that all four of the elements are in series and therefore share the same current (i).

V_{c} + L\frac{di}{dt} + (R_{1} + R_{2})*i

We need this equation in terms of constants and V_{c} so we need to perform substitutions on all terms that involve the current (i).  From the relationship I_{c} = C\frac{dV}{dt} we note that \frac{di}{dt} = C\frac{dV^{2}}{dt^{2}} and can perform the substitutions that we need:

V_{c}(t) + LC\frac{dV^{2}}{dt^{2}} + (R_{1} + R_{2}) C \frac{dV_{c}}{dt} = 0

We divide all terms by the amount in front of our highest order term which in this case is LC.

\frac{dV^{2}}{dt^{2}} + \frac{(R_{1} + R_{2})}{L}\frac{dV_{c}}{dt} + \frac{1}{LC}V_{c}(t) = 0

and our differential equation becomes: D^{2} + 3D + 5

By using the quadratic equation, we can form our characteristic equation (\lambda + \frac{3}{2} + j\frac{\sqrt{11}}{2})(\lambda + \frac{3}{2} - j\frac{\sqrt{11}}{2}) = 0

Roots: -\frac{3}{2} \pm j\frac{\sqrt{11}}{2}

Now we have some work to do, and a little bit of thinking.  Using the roots that we just derived, we can now write a general equation for this system, remembering that our roots are complex:

V_{c}(t) = C_{1} e^{\frac{3}{2} t} cos(\frac{\sqrt{11}}{2}t) + C_{2} e^{\frac{3}{2} t} sin(\frac{\sqrt{11}}{2}t)

Evaluating this at t = 0

V_{c}(0) = C_{1} We know that V_{c}(0) is 2V because after time has elapsed prior to the switch opening the capacitor has become an open and assumes all available potential across it.  When the switch opens, voltage can’t change instantaneously across a capacitor so it initially remains at the same potential as it was prior to the switch opening.

We should also note that the current i_{c}(0) is -1 Amp.  Why is that?  The initial voltage across the inductor (which is a short prior to switch opening) is 2 V, leaving all of this potential for the 2 ohm resistor (remember that the cap is an open and no current flows in that branch).  Therefore there is a 1 Amp current flowing in the clockwise direction when the switch opens.  CURRENT CAN CHANGE INSTANTANEOUSLY IN A CAPACITOR.  Therefore, immediately after the switch opens, the 1 Amp current flows through the capacitor in the direction opposite it’s polarity, resulting in -1 Amp.

We will need to evaluate the derivative of this function in order to get a second equation (need 2 equations to solve for 2 unknowns C_{1} and C_{2} ).  We now know that C_{1} = 2 V .

\frac{dV_{c}(0)}{dt} = \frac{3}{2}C_{1}e^{\frac{3}{2} t}cos(\frac{\sqrt{11}}{2}t) + \frac{\sqrt{11}}{2}C_{2}e^{\frac{3}{2} t}cos(\frac{\sqrt{11}}{2}t)

(I have removed the SINE terms because at 0 they evaluate to 0, in this case.)

\frac{dV_{c}}{dt} = \frac{I_{c}(0)}{C} = -10

Therefore,

-10 = 3 + \frac{\sqrt{11}}{2} C_{2} C_{2} = \frac{-26}{\sqrt{11}}

Now that we have both of our constants, we can insert them into our equation which represents the zero-input response of our system where y_{0}(t) = V_{c}(t) .

V_{c}(t) = 2 e^{\frac{3}{2} t} cos(\frac{\sqrt{11}}{2}t) + \frac{-26}{\sqrt{11}} e^{\frac{3}{2} t} sin(\frac{\sqrt{11}}{2}t) t\ge 0

Conclusion

This example was a little complex for teaching a new concept, but I believe it covers a wide variety of important concepts and steps that you will likely see.  If you can do this whole problem, not only do you understand how to evaluate the ZIR of a system, but you also understand characterisic roots and modes for complex roots, series RLC circuit analysis, initial conditions, basic inductor and capacitor physics concepts, as well as frequency and time domain analysis for such circuits.  I hope this helped!

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