What's the impulse response of this first order system?

Question

I think my textbook may be wrong, I really can't get this. I'll ask to my teacher, but in the meanwhile I'd like to discuss this with you.

A first order system:

$$ y(k+1) = a(k)y(k) + g(k) $$

Given $y(k_0)$ and a sequence $g(k)$ defined from $k\ge k_0$, its solution is well-known:

$$\begin{align} y(k_0 + 1) &= a(k_0)y(k_0)+g(k_0)\\ y(k_0 + 2) &= a(k_0+1)y(k_0+1)+g(k_0+1)\\ &= a(k_0+1)a(k_0)y(k_0)+a(k_0+1)g(k_0) + g(k_0+1)\\ y(k_0+3) &= a(k_0+2)y(k_0+2) + g(k_0+2)\\ &= a(k_0+2)a(k_0+1)a(k_0)y(k_0) + a(k_0+2)a(k_0+1)g(k_0) + a(k_0+2)g(k_0+1) +\cdots\\ \end{align}\\ \vdots\\ y(k) = \left(\prod_{i=k_0}^{k-1}a(k_0 + i)\right)y(k_0) + \sum_{\tau = k_0}^{k-1}\left(\prod_{i = \tau + 1}^{k-k_0-1}a(k_0+i)\right)g(k_0 + \tau) $$

Defining a function $\phi(k,h)$ as:

$$ \phi(k, h) = \begin{cases} \begin{align} a(k-1)\cdots a(h) &h < k\\ 1 &h = k \end{align} \end{cases} $$

Solution can be written as:

$$ y(k) = \phi(k, k_0)y(k_0) + \sum_{\tau=k_0}^{k-1}\phi(k, \tau + 1)g(\tau) $$

So I cite my textbook (translation):

What happens to $y_f$when the initial condition is zero?

$$ y_f(k) = \sum_{\tau=k_0}^{k-1}\phi(k, \tau + 1)g(\tau) $$

When the input is the unit impulse centered in $\tau$, the sum reduces to the only addend $\phi(k, \tau +1)g(\tau)$: this is the response to the unit impulse centered in $\tau$.

Basically it says that if $g(\cdot) = \delta(\cdot - \tau)$, where $\delta$ is the unit impulse while $\delta(\cdot - \tau)$ is the same unit impulse delayed by $\tau$ (that is centered in $\tau$), then the sum reduces to $\phi(k, \tau +1)g(\tau)$.

I can't get the whole point of impulse response in the first place. And I can't see how the sum reduces itself to the only added above. Is that wrong?

Answer 1

The idea of impulse response is quiet simple. It shows you how you system will react on a short (one tack) impulse. Those equations usually describe actual physical models, you I'll try to use mechanical analogy. Imagine that you have a "simple pendulum" (an idealization of a "real pendulum"), which is motionless. Then you hit it: ideally you give it some energy in a single moment of time. So you unbalance the system, which begins its' movement dissipating the received energy. The way how (trajectory, amplitude or phase characteristics, elapsing time etc. it depends on what you are analyzing) the system behaves (loses its' energy) after such an impact is the impulse response. Now about the second part of the question. You are using the so-called sifting property of the Kronecker delta-function $$\sum_{i=-\infty}^\infty a_i \delta_{ij} =a_j$$ It is so, because the delta-function is not zero only in one moment of time. In you case $g(\cdot) = \delta(\cdot - \tau)$ it is $\tau$. So the sum will be reduced to only one term.

About the Kronecker delta.
By definition: $$\delta_{ij} = \left\{\begin{matrix}0, & \mbox{if } i \ne j \\1, & \mbox{if } i=j, \end{matrix}\right.$$ Or if you use Dirac delta function: $$\delta(x) = \begin{cases} +\infty, & x = 0 \\ 0, & x \ne 0 \end{cases}$$ But in practice Dirac delta is harder to interpret because of the $\infty$. So, if you go with Kronecker delta, then what you have under the summation sign will be zero (because of the definition of delta) when the first index of the Kronecker delta doesn't match the second, and if it is so the sum reduces only to one term. Let's say that you have $g(\tau,n) = \delta(\tau-n)$, then: $$ \sum_{\tau=k_0}^{k-1}\phi(k, \tau + 1)g(\tau,n)=\sum_{\tau=k_0}^{k-1}\phi(k, \tau + 1)\delta(\tau-n)=\begin{cases} 0, & \tau\ne n \\ \phi(k, n+1), & \tau =n \end{cases}$$