# Where can I read up on zero state and zero input response for Discrete Time LTIs?

There seems to be something I'm just not understanding about this topic. Every question I work out that asks for zero state or input I get wrong.

Hence I think I am just not understanding the definition properly, could someone guide me to a preferably beginner friendly text that covers this topic in detail?

One of the Examples:

In 6b) I honestly have no idea how to tackle the question, but here's what I did:

I tried convoluting h[n] with u[n]. My reasoning is, that this way I can obtain y[n]. Then through some magic, I can get the zero state.

Here's how I derived y[n]:

• Could you edit your question with a concrete example, and perhaps your working of the example? We want to help, but we don't know what point you're sticking on. – TimWescott Nov 11 '19 at 20:38
• @TimWescott very well I'll do that, but since I was getting stuck on all of them I figured I must've misunderstood the entire concept – Novicegrammer Nov 11 '19 at 20:44

Yes it's a little confusing at first. And it only deepens if you also take an LTI system perspective into account, which imposes its own constraints into the corresponding difference equation structure.

Consider the following difference equation structure for a causal system:

$$\sum_{k=0}^{N} a_k y[n-k] = \sum_{k=0}^{M} b_k x[n-k]$$

with auxiliary conditions of $$y[n]=0$$ for $$n<0$$; known as initial rest.

The Zero State response of this system is equivalent to computing the output $$y[n]$$ for a given input $$x[n]$$ with initial rest assumption; i.e., setting all past values of $$y[n]$$ (aka the states of the filter) to zero before the signal is applied. This output $$y[n]$$ will equivalently be obtained by the well known convolution sum

$$y[n] = x[n] \star h[n] = \sum_{k} x[k]h[n-k]$$

where $$h[n]$$ is the impulse response of the associated LTI system described by the above difference equation under initial rest conditions.

The Zero Input response of an LTI system must be identically zero; which is evident by the definition of an LTI system : it must posses initial rest hence zero initial conditions and if you also set $$x[n]=0$$ identically, then the output must also be zero identically. So an LTI system can not have zero input response.

What happens in most of those questions is that, there is an LCCDE with arbitrary initial conditions; hence it won't strictly represent an LTI system, but it's output can be superposed into two components: one coming from input alone (assuming zero initial conditions) denoted as zero-state response, and the other coming from non-zero initial states alone, assuming no input but a response due only to the non-zero states of the recursive system.

• Thanks for the answer. However it is still unclear to me how the leap from the "definition of zero state response" to "find this using the convolution sum" happens. Could you please clarify? – Novicegrammer Nov 11 '19 at 21:21
• No there is no leap: to find the zero-state response $y[n]$, you formally solve the LCC difference equation for the given input $x[n]$ with all initial conditions set to zero. Also, alternatively, this zero-state output $y[n]$ can be calculated (without solving the difference equation) according to the LTI input-output convolution relationship $y[n] = x[n] \star h[n]$. – Fat32 Nov 11 '19 at 21:42
• @MattL. is that you to downvote ? something wrong in the answer ? – Fat32 Nov 11 '19 at 23:54
• I believe I understand this now, the convolution sum is derived under the assumption of a relaxed LTI, which coincides with the definition of a zero state response. So if I have a relaxed LTI, the zero state response is the convolution sum, by definition. – Novicegrammer Nov 12 '19 at 7:48
• Your understanding is ok but the derivation of the convolution sum is actually based on computing the output of an arbitrary system $y[n] = T\{x[n]\}$, by representing the input as $x[n] = \sum_{k} x[k] \delta[n-k]$ and using the Linearity and Time-Invariance properties of the system which yields the convolution sum as: $$y[n] = T\{x[n]\} = T\{\sum_k x[k] \delta [n-k] \}$$ , $$y[n] = \sum_k x[k] T\{\delta[n-k] \}$$, $$y[n] = \sum_k x[k] h[n-k]$$ This convolution sum is incidentally equivalent to th zero-state solution of an LCCDE associated with that $T\{\cdot\}$ system. – Fat32 Nov 12 '19 at 11:35