Signals and Systems/Time Domain Analysis/System Response

Example 1
Consider the driven RLC circuit below. Determine the current in the circuit, y(t), given an input of $$x(t)=8e^{-2t}$$. The initial current is 0A and the initial voltage across the capacitor is 2V.



Setting Up the Differential Equation
Recall that the voltages across inductors, resistors and capacitors are related to the current as follows:
 * $$v_L=L \frac{di}{dt}$$
 * $$v_R=iR \,$$
 * $$v_C=\frac{1}{C} \int i \, dt$$

By Kirchhoff's Voltage Law, we obtain the loop equation, replacing i with y(t):


 * $$x(t)=L \frac{dy(t)}{dt}+Ry(t)+\frac{1}{C} \int y(t) \,dt$$

Differentiating with respect to t gives us:


 * $$\dot x(t)=L \ddot y(t)+R \dot y(t)+\frac{1}{C} y(t)$$

Substituting in the values gives us our differential equation describing the system:


 * $$\dot x(t)= \ddot y(t)+ 4 \dot y(t)+ 3 y(t)$$

We want to put this into the following form:


 * $$Q(D)y(t)=P(D)x(t)\,$$
 * $$ \left( D^2+4D+3 \right) y(t) = D x(t)\,$$

Now we have our equation, we can begin to find the responses.

Zero-input Response
Zero-input response (ZIR) is the behaviour of the circuit when the input is null. Thus the differential equation to solve is:


 * $$\left( D^2 + 4D + 3 \right) y(t) = 0 $$

What we expect to see here is the capacitor discharging its initial charge through the resistor and inductor, subject to the initial conditions. The ratio of the component values will set the damping. Let's look at this first to get an idea of the behaviour of the circuit.

The equation can also be written as:


 * $$ \left( D^2 + 2\zeta\omega_0 D + \omega_0^2 \right) y(t) = 0 $$,

where &zeta; is the damping ratio, and &omega;0 is the resonant frequency, found in the DE as the constant term in the polynomial:
 * $$\omega_0=\sqrt{3} \, \mbox{rad/s}$$

We therefore can solve for &zeta;:
 * $$2 \zeta \omega_0 = 4\,$$
 * $$\zeta = \frac{4}{2 \sqrt{3}} \approx 1.15$$

This system is therefore over-damped, so we expect to see the current rising and decaying to zero without oscillation.

Initial Conditions
The initial conditions also need to be factored in before we can compute the ZIR. We know from the problem that the initial current is zero. However, the current's derivative is not explicitly given. However, the voltage across the capacitor is, so we can go back the loop voltage equation:


 * $$x(t)=v_L+v_R+v_C\,$$

We know that the input is zero, and that there is 2V across the capacitor. Also, as the current is zero, the resistor also has zero current across it. Thus:


 * $$0=v_L+0+2\,$$
 * $$v_L=-2 \,\mbox{V}$$

The voltage across the inductor is linked to the derivative of the current:


 * $$v_L=L \dot y_0(t)$$
 * $$\dot y_0(t)= -2 $$

We now have enough initial conditions:


 * $$y_0(t)=0; \quad \dot y_0(t)=-2$$

Solving the DE
The characteristic equation is:


 * $$\lambda^2 + 4 \lambda + 3 = 0\,$$
 * $$(\lambda+1)(\lambda+3)=0\,$$

We have our characteristic roots:


 * $$\lambda_1=-1; \quad \lambda_2=-3$$

The general solution is:


 * $$y_0(t)=c_1 e^{-t} + c_2 e^{-3t}\,$$

Applying initial conditions:


 * $$    y_0(0)=  0 =   c_1 + c_2\,$$
 * $$\dot y_0(0)= -2 = - c_1 - 3 c_2\,$$

Solving these simultaneous equations gives us:


 * $$c_1=-1; \quad c_2=1$$

The zero-input response is now:


 * $$y_0(t)=-e^{-t}+e^{-3t}\,$$

Unit-impulse Response
To find the unit input response, h(t), we consider the same differential equation as for the zero-input case above, but we consider the circuit with all values (current and voltage) at zero. To do this, we specify the following initial conditions, to find a sum of the characteristic modes of the system, yn(t).


 * $$y_0(t)=0; \quad \dot y_0(t)=1\,$$

The general solution is the same as before:

The general solution is:


 * $$y_n(t)=c_1 e^{-t} + c_2 e^{-3t}\,$$

However, solving for the coefficients yields a different specific solution:
 * $$c_1=\frac1 2; \quad c_2=-\frac1 2$$


 * $$y_n(t)=\frac1 2 e^{-t} - \frac1 2 e^{-3t}$$

Now, recall that the original DE describing the circuit had a differential operator, P(D) on the right-hand side, acting on the input. Because we are now applying an input, we must take this into account. This is done by applying it to yn(t). In this case, it is a single derivative. The result of this is the unit-impulse response.


 * $$h(t)=P(D)y_n(t)\,$$
 * $$h(t)=(D)y_n(t)\,$$


 * $$h(t)= -\frac1 2 e^{-t} + \frac3 2 e^{-3t}$$

Note that this is non-zero at t=0, which appears to contradict our conditions. However, as the unit impulse lasts for an infinitesimal period, the circuit must suddenly "jump" to a different total energy instantly, or else it couldn't ever be excited by an impulse. In real life, a finite-duration signal will cause the circuit to ramp up rather than jump.

Zero-state Response
The zero-state response, ys(t) is the response of the initially relaxed circuit to the input x(t). This is found by the convolution of the unit-impulse response and the input:


 * $$y_s(t)=h(t) * x(t)\,$$

By the definition of convolution:


 * $$y_s(t)=\int_0^t h(\tau) x(t-\tau) \, d\tau$$
 * $$y_s(t)=\int_0^t \left( -\frac1 2 e^{-\tau} + \frac3 2 e^{-3\tau} \right) \left( 8 e^{-2(t-\tau)} \right) \, d\tau$$

Multiplying through:


 * $$y_s(t)=\int_0^t \left( -4 e^{-\tau} e^{-2(t-\tau)}+ 12 e^{-3\tau} e^{-2(t-\tau)} \right) \, d\tau$$
 * $$y_s(t)=\int_0^t \left( -4 e^{-\tau} e^{2\tau}e^{-2 t)}+ 12 e^{-3\tau} e^{2\tau}e^{-2 t)} \right) \, d\tau$$

Because the integral is with respect to &tau;, we can bring out the t term:


 * $$y_s(t)=e^{-2 t}\int_0^t \left( -4 e^{-\tau} e^{2\tau}+ 12 e^{-3\tau} e^{2\tau} \right) \, d\tau$$


 * $$y_s(t)=e^{-2 t}\int_0^t \left( -4 e^{\tau} + 12 e^{-\tau} \right) \, d\tau$$


 * $$y_s(t)=e^{-2 t} \left[ -4 e^{\tau} - 12 e^{-\tau} \right]_0^t$$


 * $$y_s(t)=e^{-2 t} \left( -4 e^{t} - 12 e^{-t} + 4 + 12 \right)$$

Finally, multiplying out and combining exponentials gives us the zero-state response:


 * $$y_s(t)=-4 e^{-t} + 16 e^{-2t} - 12 e^{-3t}\,$$

Note that the current is now zero at t=0 again. For any realistic signal, this will happen, due to the inductor disallowing steps in current.

Total Response
The total response is given by the sum of the zero-input and zero-state components:


 * $$y_t(t)=y_{0}(t)+h(t) * x(t)\, $$
 * $$y_t(t)=-e^{-t}+e^{-3t}-4 e^{-t} + 16 e^{-2t} - 12 e^{-3t}\,$$
 * $$y_t(t)=-5 e^{-t} + 16 e^{-2t} - 11 e^{-3t}\,$$

We can plot the zero-input, zero-state, unit-impulse and total responses together: