Impulse Response to LTI

I am new to DSP, and I am self-studying using mostly Proakis. I have a question. There are some examples in the text where you will be given the impulse response of an LTI system, and then asked to solve something/prove something, so forth. My confusion involves the following: How do I know that there exists an LTI system with the given impulse response? For instance, if a problem is posed like one of the examples in Chapter 2, "given a relaxed LTI system with impulse response:

$$h[n] = a^n u[n]$$

where $-1 < a < 1$," (and $u[n]$ is the discrete-time unit step function) how do I know that this impulse response actually comes from an LTI system? Sorry if this question is simple. I'm probably not seeing something.

There is a general question and a specific question.

To answer the specific question, you can guess at an LTI and check if it satisfies your spec:

$$h[n] = a^n u[n] \qquad \text{where} \ |a| < 1$$

Here is my guess:

$$y[n] = a \cdot y[n-1] + x[n] \qquad \text{where} \ y[-1]=0$$

you can set $x[n] = \delta[n]$ and you will find that $y[n] = h[n]$ for all integer $n$ and you are done with the specific question.

To see if a conceptual discrete-time LTI can be made to satisfy a general impulse response spec, $h[n]$, there are a few requisite properties to satisfy:

1. Is the system causal? This means the output $y[n]$ does not react to future samples of the input $x[n]$. And if it's both LTI and causal, that means the impulse response does not react to the input impulse before that input impulse takes on non-zero values. And that means $$h[n] = 0 \qquad \forall \ n<0$$ You can see that this is satisfied by your example spec.

2. Is the LTI system FIR or IIR? If it's FIR, that means that the non-zero length of the impulse response, $h[n]$ is finite in length (let's call that FIR length "$M$") and that means $$h[n] = 0 \qquad \forall \ n<0, \ n \ge M$$ Now it turns out that your example is not FIR (which means it is IIR, and Infinite Impulse Response LTI system), but if it were FIR, then your general system can always be realized (or "actualized") as $$y[n]=\sum\limits_{m=0}^{M-1} h[m] \, x[n-m]$$ That can always be done for an FIR. It might not be the most efficient way to make an FIR for extremely large $M$, but it is a method that, if you can afford it, will always work.

3. If it's an IIR (which means the condition in paragraph 2 above is not satisfied), then you have some restrictions about what can be realized and what cannot. And you have some other conditions to nail down. One is "BIBO stability" which means, for any bounded input $x[n]$, the output $y[n]$ is also bounded. Now, I don't know all of the implications, but that does mean that systems that blow up, like $$h[n] = a^n u[n] \qquad \text{where} \ +1 < |a|$$ can be realized (same input/output equation as in 2), but they will blow up. I don't think you want that. BIBO stability also means that impulse responses like $$h[n] = \tfrac{1}{n+1} u[n]$$ will also blow up because $\tfrac{1}{n+1}$ does not converge fast enough. If, say, $x[n]=u[n]$ (which is bounded), the output $y[n]$ will blow up.

4. So it turns out that BIBO stable LTI will need to either be FIR (they're always stable from an input/output POV) or, if the LTI is IIR, then the impulse response must decay exponentially. That means, for large $n$, $$\Big|h[n] \Big| \le A \ a^n \ u[n] \qquad \text{for some} \ |a| < 1$$ If that's true, you can conceptually make an IIR filter that approximates any $h[n]$ that satisfies the causality condition (paragraph 1) and the BIBO stability conditions (paragraph 3). The way we learn to do that is by taking a DSP course after learning LTI system theory.

• Sir, but I am still confused about the validity of the statement that you can determine whether a system is LTI or not just from its impulse response. I think no, because I have a counterexample: $y[n] = x[5-n]$ has an impulse response of $\delta[5-n] = \delta[n-5]$ which corresponds to an LTI system, $y[n] = x[n-5]$, after convolution with $x[n]$. So in general, it is impossible to tell if the system is LTI by just looking at its impulse response. Is my approach sensible? Is there a deeper reason why? Thank you! – Pranshu Malik Apr 14 '19 at 16:43
• @PranshuMalik, determining that a system is LTI requires more information than just the impulse response. You need the system description (what is the output $y[n]$ as a function of the input $x[i]\qquad \forall i \le n$). But, once you have determined that the system is LTI, then the impulse response $h[n]$ is sufficient to completely describe what the output $y[n]$ is, as a function of any input $x[n]$. – robert bristow-johnson Apr 15 '19 at 18:50
• Makes sense yes. Thank you! – Pranshu Malik Apr 15 '19 at 18:52

Let's consider only 1D case: All LTI systems have an impulse response of the form $h[n]$ , a unique 1D sequence, by their definition. And furthermore, only LTI systems do have impulse responses of the form $h[n]$ and not of the form $h_k[n]$ that can describe non-LTI systems. Hence from these two you can conclude that given an impulse response of the form $h[n]$, a 1D unique sequence, then there is always an LTI system that produces it. However note that the associated LTI system may exist only on the paper, as a theoretical model, rather than an actually realizable physical system or a finitely computable algorithm.

This last statement is due to the fact that not all LTI systems do have an LCCDE (linear constant coefficient difference equation) representation. For example an ideal lowpass filter, being an LTI system, has the impulse response of the form $h[n] = \frac{\sin(\omega_c n)}{\pi n}$ which is non-causal and has infinite length. That system however does not have an LCCDE representation; it can not be exactly realized in a computational form, will only be approximated.

Therefore; all and only LTI stytems do have impulse responses of teh form $h[n]$ as 1D unique seeunces and not of the form $h_k[n]$, hence given an $h[n]$ there is always an LTI system on the paper. However only certain LTI system can be realized exactly from their LCCDE representations and they are those LTI systems with rational transfer function property.

• Just a nitpick. I think all systems have impulse responses (i.e. you can put an impulse at the input of any system and see what shows up at the output), but LTI are the only ones that can be fully characterized by them. – Tendero Jan 26 '18 at 14:28
• Reading more creafully reveals that i said impulse response of the form $h[n]$ ;-) So I hope it's clear now? It's not just an impulse response $h_k[n] = \mathcal{T}\{ \delta[n-k]\}$ , which as you've indicated any system would have, but the one $h[n]$ for which $h_k[n] = h[n-k]$ holds... – Fat32 Jan 26 '18 at 14:38
• Time-variant systems might have several impulse responses – Laurent Duval Jan 26 '18 at 14:42
• I was just adding to the previous comment "I think all systems have impulse responses" – Laurent Duval Jan 26 '18 at 14:52
• @Fat32 I think the disagreement is with your statement that "only LTI systems do have impulse responses", which is inaccurate, because all systems have impulse responses. I would change that sentence to "only LTI systems are fully characterized by their impulse responses". – MBaz Jan 26 '18 at 15:47

[EDIT 2018/04/28 : discussion on the specific case below] On the generic quedtion. Being LTI for a system is a fact or an hypothesis: in the first case, this is knowledge, in the second, this hypothesis helps you draw conclusions on that system.

Once this hypothesis/fact is either verified (see comments later on that) or only assumed, you know that, for the whole class of LTI systems, each LTI system's behavior can be completely described (only) by its response to an impulse, also called the "impulse response". In other words: if a system is LTI, then in theory there exists an impulse response than relates inputs and outputs.

Symmetrically, given an impulse response defined by coefficient at lag $n$, you can assume that the system is LTI, even if other systems can provide responses to an impulse.

Nota: for theoretical systems (like in exercices), it is easy to constrain the LTI property. For real systems, it is unlikely that one can test exhaustively linearity and time-invariance, but a few tests and knowledge can provide you with a good guess that the system is, at least, approximately LTI.

In your specific example, you are given a function which is LTI by definition: a function of $n$ which is linear, and time invariant.

• Thanks for the response. The issue of symmetry in your answer is what I am after here. Going one way, I understand that given an LTI system, the impulse response of this system will be h[n], a function of n alone, not of n and k, but going in the other direction, given an arbitrary function of n alone, how can it be assumed that this arbitrary function is the impulse response of some LTI system? For instance, if someone gives me a function h[n] = a^n*u[n]+n^2*sin[2*n], how do I know that there exists an LTI system that produces this specific impulse function? – mrmingus Jan 26 '18 at 19:31
• To elaborate a little more, is it possible for a system that is either nonlinear or time-varying to produce an impulse function that is a function of n alone? – mrmingus Jan 26 '18 at 19:37
• Given your example, you can check whether $h$ is both linear AND time-invariant. Many exercices belong to this vein. In this case, the system is not LTI. – Laurent Duval Jan 27 '18 at 0:34
• //, each LTI system's behavior can be completely described (only) by its response to an impulse, also called the "impulse response".// ---- that's not entirely true. the input/output relationship of an LTI system is described fully by its impulse response. but if the system is not completely observable and you have pole-zero cancellation, (and let's say the pole and zero that cancel are outside of the unit circle), your system behavior will not be fully described by the impulse response. – robert bristow-johnson Mar 28 '18 at 20:08
• That was the reason for my use of the word "behavior", if I don't get its meaning wrong (not my natural language) – Laurent Duval Mar 28 '18 at 20:11