1
\$\begingroup\$

enter image description here

This shows that a system transfer function assumes zero initial conditions.

By the definition of the natural response - I know that it is due to initial conditions of the system. enter image description here

So, how can the author here say that the natural response is due to the poles of the system transfer function when he previously stated that the transfer function has zero initial conditions?

Can someone please clarify to me? I think I am confusing natural, forced, zero-state, zero-input, etc.

Book is Control systems engineering by Nise.

\$\endgroup\$
2
  • 3
    \$\begingroup\$ I'd rather not repeat what's already a very good answer on dsp.ee. \$\endgroup\$ Mar 11, 2022 at 12:55
  • \$\begingroup\$ I am not sure how the author defines natural response, but maybe the author is referring to what is often also called the transient response. Namely, under the assumption of zero initial conditions one can still split the response into a transient part and steady-state part. \$\endgroup\$
    – fibonatic
    Feb 22, 2023 at 11:59

3 Answers 3

2
\$\begingroup\$

I believe that the concepts presented below could help to get your question answered. The total response of a system can be classified in different ways, depending on behavior you want to put in evidence (always as a sum of two components):

Zero Input Response (ZIR) + Zero State Response (ZSR)

  • Zero Input Response: Is the response of the system to the initial conditions, with the input set to zero.

  • Zero State Response: Is the response of the system to the input, with initial conditions set to zero. The transfer function definition involves this type of response.

$$ H(s)=\frac{Y(s)}{X(s)}=\frac{\mathcal{L}{\text{{Zero State Response}}}}{\mathcal{L}{\text{{Input}}}} $$

Natural Response + Forced Response

  • Natural Response: Is the response involving only the system's natural modes (also called characteristic modes, poles, eigenvalues, natural frequencies or roots of the characteristic polynomial). it is worth noting that the Zero State Response may contain natural modes. The natural response is also called Homogeneous or Complementar Solution.
  • Forced Response: Is the response which does not involve the natural modes of the system. It is also called Particular Solution.

Transient Response + Steady-state Response

  • Transient Response: It is the part of the response which vanishes with time.
  • Steady-state Response: It is the part of the response which remains when the transient response vanishes.

In order to consolidate these concepts, I'm going to show two examples:

========== EXAMPLE 1 ==========

Consider the differential equation below which describes a linear time invariant system:

$$ \frac{d^2y(t)}{dt^2}+12\frac{dy(t)}{dt}+32y(t)=32x(t) $$ with \$x(t)=u(t)\$, i.e., the unit step input.

The associated transfer function is:

$$ H(s)= \frac{32}{s^2+12s+32}=\frac{32}{(s+4)(s+8)} $$

With zero initial conditions and \$ X(s)=\frac{1}{s} \$:

$$ Y(s)=\frac{32}{s(s+4)(s+8)} \tag{1} $$

In time domain:

$$ y(t)=\left( 1-2e^{-4t}+e^{-8t} \right )u(t) $$

Zero State Response: \$ \left( 1-2e^{-4t}+e^{-8t} \right )u(t) \$

Zero Input Response: Not applicable.

Natural Response: \$ \left( -2e^{-4t}+e^{-8t} \right )u(t) \$

Forced Response: \$ 1u(t) \$

Transient Response: \$ \left( -2e^{-4t}+e^{-8t} \right )u(t) \$

Steady-state Response: \$ 1u(t) \$

The multiplication by unit step function is due to use of unilateral Laplace Transforms.

Now, we're going to use nonzero initial conditions, for example \$y(0_-)=0\$ and \$y'(0_-)=1\$:

$$ \frac{Y(s)}{X(s)}= \frac{32}{s^2+12s+32} $$

So,

$$ s^2Y(s)-sy(0_-)-y'(0_-) +12\left [ sY(s)-y(0_-) \right ]+32Y(s) = 32X(s) $$ $$ Y(s) = \frac{\frac{32}{s}+1}{s^2+12s+32} $$ $$ Y(s) = \frac{32}{s(s+4)(s+8)} + \frac{1}{(s+4)(s+8)} \tag{2} $$

Compare equations (1) and (2). The second has an additional term due to nonzero initial conditions. In other words, the information contained in transfer function ends up participating on the total response.

In time domain:

$$ y(t)=\left( 1-\frac{7}{4}e^{-4t}+\frac{3}{4}e^{-8t} \right )u(t) $$

Zero State Response: \$ \left( 1-2e^{-4t}+e^{-8t} \right )u(t) \$

Zero Input Response: \$ \left( \frac{1}{4}e^{-4t}-\frac{1}{4}e^{-8t} \right )u(t) \$

Natural Response: \$ \left( -\frac{7}{4}e^{-4t}+\frac{3}{4}e^{-8t} \right )u(t) \$

Forced Response: \$ 1u(t) \$

Transient Response: \$ \left( -\frac{7}{4}e^{-4t}+\frac{3}{4}e^{-8t} \right )u(t) \$

Steady-state Response: \$ 1u(t) \$

========== EXAMPLE 2 ==========

Consider the following transfer function:

$$ H(s) = \frac{s+2}{s^2+4} $$

and, an input as:

$$ X(s) = \frac{1}{(s+1)} $$

With zero initial conditions:

$$ Y(s) = \frac{s+2}{(s^2+4)(s+1)} $$

$$ Y(s) = \frac{-0.1 -0.3j}{s-2j} + \frac{-0.1+0.3j}{s+2j} + \frac{0.2}{s+1} $$

In time domain:

$$ y(t) = \left [ (-0.1-0.3j)e^{2jt} +(-0.1+0.3j)e^{-2jt} + 0.2e^{-t} \right ]u(t) $$

Finally:

$$ y(t) = \left [ \sqrt{0.4}\cos(2t-108.43^o)+0.2e^{-t} \right ]u(t) $$

Zero State Response: \$ \left [ \sqrt{0.4}\cos(2t-108.43^o)+0.2e^{-t} \right ]u(t) \$

Zero Input Response: Not applicable.

Natural Response: \$ \left [ \sqrt{0.4}\cos(2t-108.43^o) \right ]u(t) \$

Forced Response: \$ 0.2e^{-t}u(t) \$

Transient Response: \$ 0.2e^{-t}u(t) \$

Steady-state Response: \$ \left [ \sqrt{0.4}\cos(2t-108.43^o) \right ]u(t) \$

Note that, unlike Example 1, here the Natural Response does not vanish (oscillatory behavior due to a pair of complex conjugate poles). In other hand, the forced response now vanishes.

\$\endgroup\$
0
\$\begingroup\$

I agree with you that this is not clear as I have the same problem in the definition of natural response. It is defined as the system response to the initial conditions while transfer function is found under zero initial conditions. I read the following definitions of the natural response in a textbook as "In general, the shapes of the components of the natural response are determined by the locations of the poles of the transfer function." and "Natural response is the system response to an initial condition or suddenly applied input signal".

This makes more sense to me than other definitions.

\$\endgroup\$
0
\$\begingroup\$

Classical approach for solving linear differential equations

Let \$y(t)\$ be the output of a linear system, \$f(t)\$ be the input, and let \$D\$ be a differential operator.

Consider a linear system with a set of initial conditions just before the input is applied, \$D^i y(0^-), i = 0,1,2,...\$ and a set of initial conditions just after the input is applied, \$D^i y(0^+), i=0,1,2,... \$. Obviously, the input is applied at \$t=0\$ so it is of the form \$f(t) = \phi(t)u(t)\$ where \$u(t) = \begin{cases} 1 \ \ \ \text{for} \ \ \ t>0 \\ 0 \ \ \ \text{for} \ \ \ t<0\end{cases}\$.

Let \$y_\text{total}(t)\$ be the total response of the system satisfying the initial conditions \$D^i y(0^+)\$, that is

$$Q(D)y_\text{total}(t) = P(D)f(t), \ \ \ D^i y_\text{total}(0^+) = D^i y(0^+). $$

Claim: The total response can be found as the sum of the natural response \$y_n(t)\$ and forced response \$y_\phi(t)\$. That is,

$$\begin{align*} Q(D)y_n(t) &= 0 \\ Q(D)y_\phi(t) &= P(D)f(t), \ \ \ \text{and} \\ D^i[y_n + y_\phi](0^+) &= D^i y(0^+). \end{align*} $$

Modern approach for solving linear differential equations

Consider the same system as before. Now, let \$y_\text{total}(t)\$ be the total response of the system satisfying \$D^i y(0^-)\$, that is

$$Q(D)y_\text{total}(t) = P(D)f(t), \ \ \ D^i y_\text{total}(0^-) = D^i y(0^-). $$

Claim: The total response can be found as the sum of the zero-input response \$y_{zi}(t)\$ and zero-state response \$y_{zs}(t)\$. That is,

$$\begin{align*} Q(D)y_{zi}(t) &= 0, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ D^i y_{zi}(0^-) = D^i y(0^-) \\ Q(D)y_{zs}(t) &= P(D)f(t), \ \ \ D^i y_{zs}(0^-) = 0 \end{align*} $$

Classical method vs modern method

In the modern approach, the initial conditions at time \$t=0^-\$ are used instead of at \$t=0^+\$ to find the total response. This can be done, because the zero-input components can be isolated from the zero-state components.

In the classical approach, the initial conditions at time \$t=0^+\$ are used, because the zero-input components cannot be isolated from the zero-state components. This is due to the fact that the coefficients in the natural response are determined by applying the inital conditions to \$[y_n(t) + y_\phi(t)]\$.

Example

Let's say we have a system \$(D^2+3D+2)y(t) = Df(t) \$ with initial conditions \$y(0^-) = y(0^+) = 0 \ \ \ \text{and} \ \ \ \dot{y}(0^-) = \dot{y}(0^+) = 5\$ and where \$f(t) = 10e^{-3t}u(t)\$. It can be shown that the modern approach yields the solution

$$ y(t) = \underbrace{(-5e^{-t}+5e^{-2t})}_\text{zero-input response} + \underbrace{(-5e^{-t} + 20e^{-2t} - 15e^{-3t})u(t)}_\text{zero-state response}, \ \ \ t \geq 0^- $$

To find the natural and forced response, we group together all the components stemming from the natural modes:

$$ y(t) = \underbrace{(-10e^{-t}+25e^{-2t})}_\text{natural response} + \underbrace{15e^{-3t}}_\text{forced response}, \ \ \ t \geq 0^+$$

Because the natural response consists of components coming from both the zero-input response and zero-state response the initial conditions at \$t=0^+\$ must be used.

Your question

This shows that a system transfer function assumes zero initial conditions.

True. The transfer function is the Laplace transform of the system's impulse response, which is a zero-state response (i.e. zero initial conditions).

By the definition of the natural response - I know that it is due to initial conditions of the system.

False. The natural response consists of both zero-input components (stemming from initial conditions) and zero-state components (stemming from the input signal).

So, how can the author here say that the natural response is due to the poles of the system transfer function when he previously stated that the transfer function has zero initial conditions?

The initial conditions is implicitly set to \$0\$ when only investigating the transfer function, but the natural response consists of both zero-state components and zero-input components. Therefore, even with zero initial conditions an input signal will still produce a natural response, along with a forced response. The form of the natural response is determined by the poles of the transfer function.

Discussion

Why even partition the initial conditions into one set for \$t=0^-\$ and one set for \$t=0^+\$? Why not define the initial conditions for \$t=0\$ and only consider this case?

This is because an input signal asserted at \$t=0\$, for example, \$f(t) = \sin(t) u(t) \$, is undefined at \$t=0\$ since \$u(0)\$ is undefined. The assertion of an input signal to a system may instantaneously change the initial conditions present at \$t=0^-\$. But since the input signal is effectively undefined at \$t=0\$ the possible change to the initial conditions at \$t=0\$ becomes unpredictable. To avoid this problem of determinism the initial conditions are considered either at \$t=0^-\$ or \$t=0^+\$.

Moreover, an impulse \$\delta(t)\$ in the input causes an instantaneous jump in the initial condition of second highest order. For example, if a system of order \$2\$ is excited with an impulse then \$\dot{y}(0^+) - \dot{y}(0^-) = 1\$. But what happens to the system and initial conditions at \$t=0\$ is ill-defined.

Source for example

Signal Processing and Linear Systems - by B.P.Lathi

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.