The Zero-Input Response

Posted on 2024-04-28 Edited on 2024-10-28 In Electrical Engineering Views:

Sources:

B. P. Lathi & Roger Green. (2021). Chapter 2: Time-Domain Analysis of Continuous-Time Systems. Signal Processing and Linear Systems (2nd ed., pp. 151-162). Oxford University Press.

The zero-input response

The zero-input response, denoted as $y_0(t)$, is the solution of $Q(D) y(t)=P(D) x(t)$ in the past post when the input $x(t)=0$ so that \[ Q(D) y_0(t)=0 \] or \[ \begin{align} & (D^N+a_1 D^{N-1}+\cdots+a_N ) y_0(t) \label{eq_2_3} \\ & =0 \nonumber \end{align} \]

A solution to this equation can be obtained systematically. However, we will take a shortcut by using heuristic reasoning.

Observation:

$\eqref{eq_2_3}$ shows that a linear combination of $y_0(t)$ and its $N$ successive derivatives is zero, not at some values of $t$, but for all $t$. Such a result is possible if and only if $y_0(t)$ and all its $N$ successive derivatives are of the same form. Otherwise, their sum can never add to zero for all values of $t$.

A heuristic solution

We know that an exponential function $e^{\lambda t}$ has this property. So, let us assume that \[ \textcolor{brown} { y_0(t)=c e^{\lambda t} } \] is a solution to $\eqref{eq_2_3}$ . Then \[ \begin{aligned} D y_0(t) & =\frac{d y_0(t)}{d t}=c \lambda e^{\lambda t} \\ D^2 y_0(t) & =\frac{d^2 y_0(t)}{d t^2}=c \lambda^2 e^{\lambda t} \\ & \vdots \\ D^N y_0(t) & =\frac{d^N y_0(t)}{d t^N}=c \lambda^N e^{\lambda t} \end{aligned} \]

Substituting these results in $\eqref{eq_2_3}$, we obtain \[ c\left(\lambda^N+a_1 \lambda^{N-1}+\cdots+a_{N-1} \lambda+a_N\right) e^{\lambda t}=0 \]

For a nontrivial solution of this equation, \[ \begin{align} & \lambda^N+a_1 \lambda^{N-1}+\cdots+a_{N-1} \lambda+a_N \label{eq_2_4} \\ & =0 \nonumber \end{align} \]

This result means that $c e^{\lambda t}$ is indeed a solution of $\eqref{eq_2_3}$, provided $\lambda$ satisfies $\eqref{eq_2_4}$. Note that the polynomial in $\eqref{eq_2_4}$ is identical to the polynomial $Q(D)$ in $\eqref{eq_2_3}$, with $\lambda$ replacing $D$.

Therefore, $\eqref{eq_2_4}$ can be expressed as \[ Q(\lambda)=0 \]

Expressing $Q(\lambda)$ in factorized form, we obtain \[ Q(\lambda)=\left(\lambda-\lambda_1\right)\left(\lambda-\lambda_2\right) \cdots\left(\lambda-\lambda_N\right)=0 \]

Clearly, $\lambda$ has $N$ solutions: $\lambda_1, \lambda_2, \ldots, \lambda_N$, assuming that all $\lambda_i$ are distinct (The non-distinct case is analyzed later), each $\lambda_i$ is a solution.

Consequently, $\eqref{eq_2_3}$ has $N$ possible solutions: \[ c_1 e^{\lambda_1 t}, c_2 e^{\lambda_2 t}, \ldots, c_N e^{\lambda_N t} \]

, with $c_1, c_2, \ldots, c_N$ as arbitrary constants.

We can readily show that the zero-input response is a linear combination of the characteristic modes of the system so that \[ \begin{align} & \textcolor{blue} {y_0(t)} \nonumber \\ \textcolor{blue} {=} & \textcolor{blue} {c_1 e^{\lambda_1 t}+c_2 e^{\lambda_2 t}+\cdots+c_N e^{\lambda_N t}} \label{eq_2_6} \end{align} \] where $c_1, c_2, \ldots, c_N$ are arbitrary constants determined by $N$ constraints (the auxiliary conditions) on the solution.

Terminologies:

The polynomial $Q(\lambda)$ is the characteristic polynomial of the system. It has nothing to do with the input since its derived from the zero-input case, i.e., $Q(D) y_0(t)=0$.
The equation \[ Q(\lambda)=0 \] is called the characteristic equation of the system.
The roots of the characteristic equation are $\lambda_1, \lambda_2$, $\ldots, \lambda_N$. So they are called the characteristic roots of the system. The terms characteristic values, eigenvalues, and natural frequencies are also used for characteristic roots.
The exponentials $e^{\lambda_i t}(i=1,2, \ldots, n)$ in the zero-input response are the characteristic modes (also known as natural modes or simply as modes) of the system.

Repeated roots

The solution of $\eqref{eq_2_3}$ as given in $\eqref{eq_2_6}$ assumes that the $N$ characteristic roots $\lambda_1, \lambda_2, \ldots$, $\lambda_N$ are distinct. If there are repeated roots (same root occurring more than once), the form of the solution is modified slightly. By direct substitution, we can show that the solution of the equation \[ (D-\lambda)^2 y_0(t)=0 \] is given by \[ y_0(t)=\left(c_1+c_2 t\right) e^{\lambda t} \]

In this case, the root $\lambda$ repeats twice. Observe that the characteristic modes in this case are $e^\lambda$ $t e^{\lambda t}$.

Continuing this pattern, we can show that for the differential equation \[ \textcolor{red} { (D-\lambda)^r y_0(t)=0 } \] the characteristic modes are \[ \textcolor{red} {e^{\lambda t}, t e^{\lambda t}, t^2 e^{\lambda t}, \ldots, t^{r-1} e^{\lambda t}} \] , and that the solution is \[ \textcolor{red} { y_0(t)=\left(c_1+c_2 t+\cdots+c_r t^{r-1}\right) e^{\lambda t} } \]

Consequently, for a system with the characteristic polynomial \[ \textcolor{green} { Q(\lambda)=\left(\lambda-\lambda_1\right)^r\left(\lambda-\lambda_{r+1}\right) \cdots\left(\lambda-\lambda_N\right) } \] the characteristic modes are \[ \textcolor{green} { e^{\lambda_1 t}, t e^{\lambda_1 t}, \ldots, t^{r-1} e^{\lambda_1 t}, e^{\lambda_{r+1} t}, \ldots, e^{\lambda_N t}} \]

and the solution is \[ \textcolor{green} { y_0(t)=\left(c_1+c_2 t+\cdots+c_r t^{r-1}\right) e^{\lambda_1 t}+c_{r+1} e^{\lambda_{r+1} t}+\cdots+c_N e^{\lambda_N t} } \]

Complex roots

The procedure for handling complex roots is the same as that for real roots.

For a real system, complex roots must occur in pairs of conjugates if the coefficients of the characteristic polynomial $Q(\lambda)$ are to be real (Consider a complex number x, if a real number z is the multiplication of x and another complex number y, then y must be the conjugate of x).

Therefore, if $\alpha+j \beta$ is a characteristic root, $\alpha-j \beta$ must also be a characteristic root. The zero-input response corresponding to this pair of complex conjugate roots is \[ \begin{equation} y_0(t)=c_1 e^{(\alpha+j \beta) t}+c_2 e^{(\alpha-j \beta) t} \label{eq_2_7} \end{equation} \]

For a real system, the response $y_0(t)$ must also be real. This is possible only if $c_1$ and $c_2$ are conjugates. Let \[ c_1=\frac{c}{2} e^{j \theta} \quad \text { and } \quad c_2=\frac{c}{2} e^{-j \theta} \]

This yields \[ \begin{align} & y_0(t) \nonumber \\ & =\frac{c}{2} e^{j \theta} e^{(\alpha+j \beta) t}+\frac{c}{2} e^{-j \theta} e^{(\alpha-j \beta) t} \nonumber \\ & =\frac{c}{2} e^{\alpha t}\left[e^{j(\beta t+\theta)}+e^{-j(\beta t+\theta)}\right] \nonumber \\ & =c e^{\alpha t} \cos (\beta t+\theta) \label{eq_2_8} \end{align} \] Therefore, the zero-input response corresponding to complex conjugate roots $\alpha \pm j \beta$ can be expressed in a complex form $ $ or a real form $\eqref{eq_2_8}$.

Examples

Find $y_0(t)$, the zero-input response of the response for an LTIC system described by (a) the simple-root system $\left(D^2+3 D+2\right) y(t)=D x(t)$ with initial conditions $y_0(0)=0$ and $\dot{y}_0(0)=-5$. (b) the repeated-root system $\left(D^2+6 D+9\right) y(t)=(3 D+5) x(t)$ with initial conditions $y_0(0)=3$ and $\dot{y}_0(0)=-7$. (c) the complex-root system $\left(D^2+4 D+40\right) y(t)=(D+2) x(t)$ with initial conditions $y_0(0)=2$ and $\dot{y}_0(0)=16.78$.

Q1

Note that $y_0(t)$, being the zero-input response $(x(t)=0)$, is the solution of $\left(D^2+3 D+\right.$ 2) $y_0(t)=0$. The characteristic polynomial of the system is $\lambda^2+3 \lambda+2$. The characteristic equation of the system is therefore $\lambda^2+3 \lambda+2=(\lambda+1)(\lambda+2)=0$. The characteristic roots of the system are $\lambda_1=-1$ and $\lambda_2=-2$, and the characteristic modes of the system are $e^{-t}$ and $e^{-2 t}$. Consequently, the zero-input response is \[ y_0(t)=c_1 e^{-t}+c_2 e^{-2 t} \]

Differentiating this expression, we obtain \[ \dot{y}_0(t)=-c_1 e^{-t}-2 c_2 e^{-2 t} \]

To determine the constants $c_1$ and $c_2$, we set $t=0$ in the equations for $y_0(t)$ and $\dot{y}_0(t)$ and substitute the initial conditions $y_0(0)=0$ and $\dot{y}_0(0)=-5$, yielding \[ \begin{aligned} 0 & =c_1+c_2 \\ -5 & =-c_1-2 c_2 \end{aligned} \]

Solving these two simultaneous equations in two unknowns for $c_1$ and $c_2$ yields \[ c_1=-5 \text { and } c_2=5 \]

Therefore, \[ y_0(t)=-5 e^{-t}+5 e^{-2 t} \]

This is the zero-input response of $y(t)$. Because $y_0(t)$ is present at $t=0^{-}$, we are justified in assuming that it exists for $t \geq 0$.

($y_0(t)$ may be present even before $t=0^{-}$. However, we can be sure of its presence only from $t=0^{-}$ onward.)

Q2

The characteristic polynomial is $\lambda^2+6 \lambda+9=(\lambda+3)^2$, and its characteristic roots are $\lambda_1=-3, \lambda_2=-3$ (repeated roots). Consequently, the characteristic modes of the system are $e^{-3 t}$ and $t e^{-3 t}$. The zero-input response, being a linear combination of the characteristic modes, is given by \[ y_0(t)=\left(c_1+c_2 t\right) e^{-3 t} \] We can find the arbitrary constants $c_1$ and $c_2$ from the initial conditions $y_0(0)=3$ and $\dot{y}_0(0)=-7$ following the procedure in part (a). The reader can show that $c_1=3$ and $c_2=2$. Hence, \[ y_0(t)=(3+2 t) e^{-3 t} \quad t \geq 0 \] ## Q3

The characteristic polynomial is $\lambda^2+4 \lambda+40=(\lambda+2-j 6)(\lambda+2+j 6)$. The characteristic roots are $-2 \pm j 6 .^{+}$The solution can be written either in the complex form $ $ or in the real form $ $. The complex form is $y_0(t)=c_1 e^{\lambda_1 t}+c_2 e^{\lambda_2 t}$, where $\lambda_1=-2+j 6$ and $\lambda_2=-2-j 6$. Since $\alpha=-2$ and $\beta=6$, the real-form solution is [see $\eqref{eq_2_8}$] \[ y_0(t)=c e^{-2 t} \cos (6 t+\theta) \]

Differentiating this expression, we obtain \[ \dot{y}_0(t)=-2 c e^{-2 t} \cos (6 t+\theta)-6 c e^{-2 t} \sin (6 t+\theta) \]

To determine the constants $c$ and $\theta$, we set $t=0$ in the equations for $y_0(t)$ and $\dot{y}_0(t)$ and substitute the initial conditions $y_0(0)=2$ and $\dot{y}_0(0)=16.78$, yielding \[ \begin{aligned} 2 & =c \cos \theta \\ 16.78 & =-2 c \cos \theta-6 c \sin \theta \end{aligned} \]

Solution of these two simultaneous equations in two unknowns $c \cos \theta$ and $c \sin \theta$ yields \[ c \cos \theta=2 \text { and } c \sin \theta=-3.463 \]

Squaring and then adding these two equations yield \[ c^2=(2)^2+(-3.464)^2=16 \Longrightarrow c=4 \]

Next, dividing $c \sin \theta=-3.463$ by $c \cos \theta=2$ yields \[ \tan \theta=\frac{-3.463}{2} \] and \[ \theta=\tan ^{-1}\left(\frac{-3.463}{2}\right)=-\frac{\pi}{3} \]

Therefore, \[ y_0(t)=4 e^{-2 t} \cos \left(6 t-\frac{\pi}{3}\right) \]

Practical Initial Conditions and the Meaning of $0^{-}$and $0^{+}$

In Q.1, the initial conditions $y_0(0)$ and $\dot{y}_0(0)$ were supplied. In practical problems, we must derive such conditions from the physical situation. For instance, in an RLC circuit, we may be given the conditions (initial capacitor voltages, initial inductor currents, etc.).

From this information, we need to derive $y_0(0), \dot{y}_0(0), \ldots$ for the desired variable as demonstrated in the next example.

In much of our discussion, the input is assumed to start at $t=0$, unless otherwise mentioned. Hence, $t=0$ is the reference point.

The conditions immediately before $t=0$ (just before the input is applied) are the conditions at $t=0^{-}$, and those immediately after $t=0$ (just after the input is applied) are
the conditions at $t = 0^{+}$.

In practice, we are likely to know the initial conditions at $t=0^{-}$rather than at $t=0^{+}$. The two sets of conditions are generally different, although in some cases they may be identical.

The total response $y(t)$ consists of two components: the zero-input response $y_0(t)$ [response due to the initial conditions alone with $x(t)=0$ and the zero-state response resulting from the input alone with all initial conditions zero. At $t=0^{-}$, the total response $y(t)$ consists solely of the zero-input response $y_0(t)$ because the input has not started yet.

Hence, the initial conditions on $y(t)$ are identical to those of $y_0(t)$. Thus, $y\left(0^{-}\right)=y_0\left(0^{-}\right), \dot{y}\left(0^{-}\right)=\dot{y}_0\left(0^{-}\right)$, and so on.

Moreover, since $y_0(t)$ is the response due to initial conditions alone and does not depend on the input $x(t)$, hence application of the input at $t=0$ does not affect $y_0(t)$. This means the initial conditions on $y_0(t)$ at $t=0^{-}$and $0^{+}$are identical; that is, $y_0\left(0^{-}\right), \dot{y}_0\left(0^{-}\right), \ldots$ are identical to $y_0\left(0^{+}\right), \dot{y}_0\left(0^{+}\right), \ldots$, respectively.

It is clear that for $y_0(t)$, there is no distinction between the initial conditions at $t=0^{-}$, 0 , and $0^{+}$. They are all the same.

But this is not the case with the total response $y(t)$, which consists of both the zero-input and zero-state responses. Thus, in general, $y\left(0^{-}\right) \neq y\left(0^{+}\right), \dot{y}\left(0^{-}\right) \neq \dot{y}\left(0^{+}\right)$, and so on.

Independence of the zero-input and zero-state reponses

The zero-input component is computed without using the input $x(t)$.

The zero-state response is computed from the knowledge of the input $x(t)$ alone; the initial conditions are assumed to be zero (system in zero state).

The two components of the system response (the zero-input and zero-state responses) are independent of each other.