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Suppose we have some system given by

$$\begin{aligned} \dot{x}(t) &= Ax(t) +Bu(t) \\ y(t) &= Cx(t)+Du(t) \end{aligned}$$

where $x(t)$ are the state variables, $y(t)$ is the output and $u(t)$ is the input. All matrices are constant. The same question applies to the discrete case

$$\begin{aligned} x[n+1] &= Ax[n] +Bu[n] \\ y[n] &= Cx[n]+Du[n] \end{aligned}$$

It is known that a system with non-zero initial conditions cannot be LTI. However, if $x(0)\neq0$, I don't see why the system above would not be LTI as it is expressed. As far as I know, if a system is expressed that way, it has to be linear and, as matrices do not depend on $t$, it should be time-invariant too.

So we have a system that has to be LTI as it is expressed in state space with constant matrices, but it can't be LTI because it has $x(0)\neq0$.

I can't see the mistake in the reasoning that leads me to this absurd contradiction. Can someone point it out?

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  • $\begingroup$ Hi Tendero: I've been dealing with a problem like that ( much simpler actually ) and the problem with initial conditions is that they make it difficult to write the step response in a general way ( because the initial conditions can change each time ) since it is dependent upon initial conditions. I'm not sure if this is related to your question but it might be. Again, I come from econometrics so very different world. Definitely hope someone can explain this. $\endgroup$ – mark leeds Mar 14 '18 at 17:41
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    $\begingroup$ I really think you and your reference are being misled by semantics. I will look up some references but LTI and Linearity are structural items not dependent upon particular predecessors (i.e. history); which in "linear" systems are "initial conditions". $\endgroup$ – rrogers Mar 20 '18 at 21:28
  • $\begingroup$ @rrogers You can check Oppenheim's Signals and Systems book. In it's section named "CAUSAL LTI SYSTEMS DESCRIBED BY DIFFERENTIAL AND DIFFERENCE EQUATIONS" (it's 2.4 in my edition), the topic is approached. Please enlighten me if I'm misunderstanding something, but the author clearly states that a system with non-initial rest conditions is not LTI. $\endgroup$ – Tendero Mar 23 '18 at 20:40
  • $\begingroup$ Far be it from me to correct your book (even Oppenheim's); but consider whether the "system" includes the values of variables or the hardware acted on. Say an RC network, would you call for a new theory if the voltage on the cap was non-zero; no it is called a transient response and considered by the same equation. Look at the Laplace transform of am LTI differential equation; it has "initial" conditions corresponding to the derivatives. I would say the "system" doesn't depend upon when measurements are taken or the inputs. $\endgroup$ – rrogers Mar 25 '18 at 19:14
  • $\begingroup$ Please glance at: web.mit.edu/2.14/www/Handouts/StateSpaceResponse.pdf, vtechworks.lib.vt.edu/bitstream/handle/10919/78864/… . They both incorporate initial conditions into their explanations of LTI without even considering changing the equations. Another viewpoint (really the same): given the indefinite integral you have to establish a start and an end when evaluated. The start is the direct analog of initial condition. BTW: most initial conditions can be emulated by the delta function and it's derivatives. $\endgroup$ – rrogers Mar 25 '18 at 19:20
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I am only an undergraduate student so perhaps my answer is a bit naive, but according to Oppenheim it is not just nonzero initial conditions that cause an linear constant coefficient differential/difference equation to be non-LTI. A differential/difference equation with fixed zero initial conditions cannot be LTI either. For a linear constant coefficient differential/difference equation describe a causal, LTI system, the initial conditions have to satisfy the condition of initial rest: that is, the output does not become nonzero until the input becomes nonzero.

With regards to your question (the state space representation), notice that the input to the system is $u(t)$ and the output is $y(t)$. The "zero input/zero output" property of linear systems only applies to $$y(t) = C x(t) + Du(t)$$ if $x(t) = 0$, if we only consider $u(t)$ to be the input to the system, but appears to me that the notion of linearity can be extended to state-space representations to account for the state vector $x(t)$. In any case, the initial conditions that Oppenheim refers to when talking about differential equations (conditions on the output $y(t)$ and its derivatives) are not the same as the initial conditions you referenced in your question (conditions on the state vector $x(t)$). Again, I don't know if I'm correct, and I've always been confused by this myself, but perhaps this could help.

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  • $\begingroup$ I think this could be the right answer, as I have forgotten that Oppenheim refers to initial conditions for $y(t)$, and the states $x(t)$ in state-space representation are not the output. I'm not really sure I have understood this yet, but I'm accepting the answer because I really think you hit the nail on the head. $\endgroup$ – Tendero Apr 15 '18 at 16:51
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If you look at Chapter 5 of:

https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-011-introduction-to-communication-control-and-signal-processing-spring-2010/readings/MIT6_011S10_notes.pdf

which is entitled "Properties of LTI State Space Models" , equation 5.33 doesn't seem to have a problem with initial conditions, or any other book (I stand corrected, there is one book) that I'm aware of. Unless Oppenheim was touched with insanity, I'm inclined to accept his characterization that initial conditions don't disqualify a LTI system as "not linear" by his use of the term "zero input linear".

At the beginning of the notes, (and in Oppenheim and Shaefer 3rd edition) a LTI system is given as:

$$ y[n]=\sum_{k=-\infty}^{\infty} x[k] h[ n -k] $$ which doesn't require $h[n]$ to be causal or stable. $x[n]$ doesn't have to satisfy $x[n]=0 \;\text{for} \; n <0$ .

There is emphasis in the text that one needs to consider the entire history of $x[n]$, not just for $n \ge 0$.

let $$x[n] = \hat{x}[n]+\tilde{x}[n] $$ where $$ \hat{x}[n] = \begin{cases} x[n]\; \text{for}\; n<0 \; \text{and}\\ 0 \; \text{for}\; n\ge 0 \end{cases}$$

and $$ \tilde{x}[n] = \begin{cases}0\; \text{for} \; n<0 \; \text{and}\; \\ x[n] \; \text{for}\; n\ge 0 \end{cases}$$ by linearity.

$$ y[n]=\sum_{k=-\infty}^{-1} \hat{x}[k] h[ n -k] + \sum_{k=0}^{\infty} \tilde{x}[k] h[ n -k] $$ if $y[n]$ is causal, $$ y[n]=\underbrace{\sum_{k=-\infty}^{-1} \hat{x}[k] h[ n -k]}_{\text{zero input linear}} + \underbrace{\sum_{k=0}^{n} \tilde{x}[k] h[ n -k]}_{\text{zero state linear}}, \quad n \ge 0 $$

The essential point is that initial conditions account for prior input. Where $n=0$ is referenced for $x[n]$ is arbitrary, which is another manifestation of time invariance. Initial conditions are not some arbitrary value vexing the system. if $x[n] = 0 $ for $n <0 $ the initial conditions are zero.

Let's try something else. Let $z[n]=\tilde{x}[n+1]$ (advance by one sample) and with $\tilde{x}[n]$, the system was LTI without controversy. But now, $$ y[n]=\underbrace{ z[-1] h[n]}_{\text{zero input linear}} + \underbrace{\sum_{k=0}^{n} z[k] h[ n -k]}_{\text{zero state linear}}, \quad n \ge 0 $$ and now we have an initial condition. A forward shift of 1 sample would make an LTI system nonlinear?

The logical fallacy at the root of question is to use the definition of zero state linearity and applying it to the zero input case.

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  • $\begingroup$ It's in his own book, Signals and Systems, that Oppenheim states that a system described by a differential/difference equation with non-initial rest conditions is not LTI. If you can, check the section "CAUSAL LTI SYSTEMS DESCRIBED BY DIFFERENTIAL AND DIFFERENCE EQUATIONS" (it's 2.4 in my edition). Does that line up with your answer? I'm really confused here. $\endgroup$ – Tendero Mar 23 '18 at 20:20
  • $\begingroup$ These notes are his and came out in 2001. I grew up with Lathi. and he explicitly states that a system has to be BOTH zero state and zero input linear. note the term causal $\endgroup$ – Stanley Pawlukiewicz Mar 23 '18 at 20:23
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    $\begingroup$ Think of this example. I have some charged capacitor with voltage $V_0$. If I connect a voltage source $v_1(t)$ to it, the voltage on the capacitor will change with time, but it will start at $V_0$. If I connect a voltage source $v_2(t)$, the same will happen. But, if I connect a voltage source $v_3(t)=v_1(t)+v_2(t)$, the voltage at the capacitor (which would be the output here) will not be $2V_0$ at $t=0$. Instead, it will continue to be $V_0$. Thus the superposition principle doesn't work here, and therefore the system is not linear. At least, that's my reasoning. Do you see any mistake? $\endgroup$ – Tendero Mar 23 '18 at 20:25
  • $\begingroup$ Think about his example. What idiot randomly shoves charge capacitors into a system. (must have missed ESD training)The only realistic situation is the charged capacitor represents an accumulated history. $\endgroup$ – Stanley Pawlukiewicz Mar 23 '18 at 20:29
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    $\begingroup$ Chill, there is no need for profanity. We are talking about ideal concepts here, this is purely mathematical and we wouldn't even be talking about linearity if we focus on rigorous real-life cases, as nature is inherently non-linear. I really think that my example shows the problem presented in the OP in a pretty simple way. And by the way I wouldn't just discredit the "DSP's bible" just because other notes are more recent. I don't think this answer is helpful, at least as it is right now. $\endgroup$ – Tendero Mar 23 '18 at 20:35

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