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I am new to DSP and was going through different responses of a system subjected to an input. My understanding of zero input response is: it is the response/output of the system when the input signal is set to zero. In other words if a system is described by a linear constant coefficient difference equation the zero input response is the homogeneous solution.

However if the $\mathcal Z$-transform of the input is a rational function $X(z)=N(z)/Q(z)$ and that of the LTI system function is $H(z)=B(z)/A(z)$ and the system is initially relaxed, then $Y(z)= H(z)X(z) = N(z)B(z)/A(z)Q(z)$. Assuming distinct zeros(real only) and poles(real only) of $X(z)$ and $H(z)$ then

$$Y(z) = \sum_{k=1}^N \frac{A_k}{1-p_kz^{-1}} + \sum_{k=1}^L \frac{Q_k}{1-q_kz^{-1}}$$

which gives

$$y(n) = \sum_{k=1}^N A_k(p_k)^{n}u(n) + \sum_{k=1}^L Q_k(q_k)^{n}u(n)$$

where $p_k$ and $q_k$ are the poles of system $H(z)$ and input signal $X(z)$ respectively and $u(n)$ is the unit step function. Now the first term is referred to as the natural response of the system $H(z)$. It's very confusing to grasp the difference between zero input and natural response.

Edit: The reference of the question is to book DSP : Principles , Algorithms and Applications by John Proakis and D Manolakis pdf of the book is here Page no 203 and 204. The two paragraphs after formula 3.6.4 explains the difference between zero input response and natural response

Thank you Peter and Matt for your answers and comments.

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First it's important to realize that many authors use the terms zero-input response and natural response as synonyms. This convention is used in the corresponding wikipedia article, and for instance also in this book. Even Proakis and Manolakis are not entirely clear about it. In the book you quoted you can find the following sentence on page 97:

[...] the output of the system with zero input is called the zero-input response or natural response.

This suggests that the two terms can be used interchangeably. Further down the page, we find the following sentence:

Thus the zero-input response is a characteristic of the system itself, and it is also known as the natural or free response of the svstem.

Again, this strongly suggests that the authors believe that both terms are equivalent.

However, on the pages you mentioned they appear to make a difference between the two. And the difference is as follows. The zero-input response is the response which is caused by non-zero initial conditions. It only depends on the system properties and on the values of the initial conditions. The zero-input response becomes zero if the initial conditions are zero.

The natural response is the part of the total response the shape of which is only determined by the poles of the system, and which doesn't depend on the poles of the (transform of the) input signal. The natural response does depend on the input signal in terms of constants but its form is entirely determined by the system's poles. Unlike the zero-input response, the natural response does not vanish for zero initial conditions.

The total response of the system can be written as the following two sums:

  1. zero-input response + zero-state response
  2. natural response + forced response

The zero-state response is the response for zero initial conditions, and the forced response is the part of the response the form of which is determined by the form of the input signal.

I hope this becomes clear in the following example. Let's investigate the following system:

$$y[n]+ay[n-1]=b^nu[n],\qquad y[-1]=c\tag{1}$$

where $u[n]$ is the unit step sequence. The total response can be computed using $\mathcal{Z}$-transform techniques:

$$y[n]=\left[\frac{1}{a+b}b^{n+1}+\left(c-\frac{1}{a+b}\right)(-a)^{n+1}\right]u[n]\tag{2}$$

The zero-input response is the part of the total response that is determined by the initial condition and that does not depend on $b$:

$$y_{ZI}[n]=c(-a)^{n+1}u[n]\tag{3}$$

Obviously, $y_{ZI}[n]=0$ for $c=y[-1]=0$, i.e., for zero initial condition.

The natural response is the part of the total response the shape of which is determined by the system's pole:

$$y_N[n]=\left(c-\frac{1}{a+b}\right)(-a)^{n+1}u[n]\tag{4}$$

Note that it depends on the initial conditions as well as on the input signal (via the constant $b$).

Also note that it is the shape of the zero-state response that depends on the poles of the system as well as on the poles of the input signal transform. All other responses mentioned here only depend on one of the two sets of poles. The shapes of of the zero-input response and of the natural response depend only on the system's poles, whereas the shape of the forced response is determined by the poles of the input signal. The expression for $y[n]$ quoted in your question from Proakis and Manolakis is the zero-state response (because the system is initially relaxed), and the first sum is the forced response, and the second sum is the natural response. Since the zero-input response is zero in this case, the sum of natural response and forced response (i.e., the total response) equals the zero-state response

In mathematical terms, the natural response is the homogeneous solution of the difference equation, where the constants are determined such that the sum of the particular solution (the forced response) and the homogeneous solution satisfy the given initial condition. Clearly, the zero-input response is also a solution to the homogeneous equation, but the difference with the natural response is that the zero-input response alone satisfies the initial conditions, because it is combined with the zero-state response, which assumes zero initial conditions. On the other hand, the natural response alone does not satisfy the initial conditions. The initial conditions are satisfied only by combining the natural response with the particular solution of the difference equation (the latter being the forced response).

As mentioned above, we can write the total solution as

$$y[n]=y_{ZI}[n]+y_{ZS}[n]$$

(zero-input response plus zero-state response)

and as

$$y[n]=y_N[n]+y_F[n]$$

(natural response plus forced response). For the given example, we have

$$y_{ZI}[-1]=y[-1]$$

i.e., it is $y_{ZI}[n]$ that takes care of the initial condition. That's also why $y_{ZI}[n]=0$ if the initial condition is zero. $y_{ZI}[n]$ must satisfy the homogeneous equation

$$y_{ZI}[n]+ay_{ZI}[n-1]=0,\qquad y_{ZI}[-1]=y[-1]$$

So if $y[-1]=0$, $y_{ZI}[n]=0$ for all $n$. The natural response also satisfies the homogeneous equation, but not with the initial condition $y_N[-1]=y[-1]$. What is satisfied is $y_N[-1]+y_F[-1]=y[-1]$. This is why the natural response is generally non-zero, even for zero initial conditions. And the natural response is the homogeneous solution which we need to combine with the particular solution (forced response) found in the standard way. We usually have no direct means to find the specific particular solution which, when combined with the special homogeneous solution represented by the zero-input response, will give the complete solution of the difference equation. For this we need another homogeneous solution, and this is the natural response.

Again using the above example will hopefully clarify this. For an exponential forcing signal, the standard (and most straight-forward) way to obtain a particular solution is choosing a scaled version of the forcing function:

$$y_p[n]=Ab^n\tag{A1}$$

(for the sake of simplicity I leave out the unit step $u[n]$, assuming that we consider $n\ge 0$, unless we talk about the initial condition). The constant $A$ is determined by plugging $(A1)$ into the difference equation:

$$Ab^n+aAb^{n-1}=b^n$$

giving $A=\frac{b}{a+b}$. The general form of the homogeneous solution is

$$y_h[n]=B(-a)^n\tag{A2}$$

Of course $y_h[n]=0$ (i.e., $B=0$) is one specific solution, but that's not the one we're looking for. We need to determine the constant $B$ in such a way that the sum of the particular and the homogeneous solution satisfies the initial condition:

$$y[-1]=y_p[-1]+y_h[-1]=\frac{A}{b}-\frac{B}{a}$$

From this equation we get

$$B=\frac{a}{a+b}-ay[-1]$$

which shows that the homogeneous solution we need is non-zero if $y[-1]=0$. $y_p[n]$ and $y_h[n]$ found in this way are identical to the forced response and the natural response, respectively, as shown in $(4)$ and - implicitly - in $(2)$.

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OK, now that I've had a chance to read it a bit (not just on mobile!), this is what I think is happening.

We have a linear, constant coefficient difference equation:

$$ y[n] + \sum_{k=1}^N \alpha_k y[n-k] = \sum_{m=0}^M \beta_m x[n-m] $$

and we wish to find the $y$ for a given $x$.

In general, the solution $Y$ will comprise two components: $$ y = y_p + y_h $$ where $y_p$ is the particular solution and $y_h$ is the homogenous solution.

The particular solution is obtained by setting the system input to $x$. The homogenous solution is obtained by setting the input of the system to 0 (zero) and selecting an arbitrary initial condition for the system state.

What Proakis and Manolakis assume is that the initial system state is all zero (just above equation 3.6.2, and as you've highlighted in your question).

So that's the difference between the zero-input (and zero-state) solution and the zero-input (and arbitrary state or homogenous) solution: what you select the initial conditions to be. The homogeneous solution requires arbitrary initial conditions, otherwise every homogeneous solution would be $y_h = 0$.

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  • $\begingroup$ If your last sentence is meant to imply that the solution of a LCCDE is completely characterized by its particular solution if the initial conditions are zero (because then, you say, $y_h=0$), then I would say it's wrong. I think you refer to the fact that a homogeneous difference equation needs a non-zero initial condition to "get started", but that's a different thing. In general, the solution is $y_p+y_h$ with $y_h\neq 0$, even if initial conditions are zero. $\endgroup$ – Matt L. Mar 29 '16 at 8:54
  • $\begingroup$ @MattL. Can you give me an example where the initial conditions and the forcing function are zero and for which $y_h = 0$ is NOT a valid solution? I take your point, I just can't think of how to find a non-trivial homogeneous solution without assuming arbitrary initial conditions --- which obviously includes zero. $\endgroup$ – Peter K. Mar 29 '16 at 11:26
  • $\begingroup$ If you have a homogeneous difference equation with zero initial conditions (and with $y_p=0$), then you're right. But my point is this: if you want a solution to a difference equation with forcing function, and you decompose it as $y=y_p+y_h$ then it's not correct to say that $y_h=0$ if the initial conditions are zero. Take the example from my answer. The homogeneous solution is given by Eq. (4), and it is not zero even if $y[-1]=c=0$. The reason is that the initial condition is satisfied by $y_h$ combined with $y_p$, even if those conditions are zero. So $y_h[-1]\neq 0$, even if $y[-1]=0$ $\endgroup$ – Matt L. Mar 29 '16 at 12:00
  • $\begingroup$ @MattL. My point is that $y[n] + ay[n-1] = 0$ has a perfectly valid "homogeneous solution" $y_h[n] = 0, \forall n$ if you assume zero initial conditions. You have to assume non-zero initial conditions to get find what you've written (the true homogeneous solution). $\endgroup$ – Peter K. Mar 29 '16 at 12:06
  • $\begingroup$ Of course I agree with your example; this type of example is what I referred to in the first sentence of my previous comment (the case where $y_p=0$). But you also have a homogeneous solution if the forcing function is not zero. In these cases $y_h\neq 0$, even for zero initial conditions. And I thought that the last sentence in your answer claims the opposite, at least that's the way I understand it. So in sum, you generally have $y_h\neq 0$ even with zero initial conditions (if $y_p\neq 0$). $\endgroup$ – Matt L. Mar 29 '16 at 12:12

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