# How can output sequence be equal to the sum of copies of the impulse response, scaled and time-shifted signals?

I am sorry, this is a very basic question. But i am having hard time understanding is how it is possible.

I know impulse response is the output of the system when impulse sequence is given as input with initial conditions set to 0.

Scaling is to increase the amplitude of the signal, i.e. if i multiply the input by 2, output will also be multiplied by 2.

Time shifted signal is if I delay the input then the output is also delayed by the same factor.

Now can some one please illustrate this with an example how any sequence can be decomposed into sum of copies of the impulse response, scaled and time-shifted signals?

Thanks a lot in advance.

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You seem to be considering a linear time-invariant system although you have not said so explicitly. Are you wanting to know how you can say any given sequence, e.g. one that was obtained on the basis of coin flips, can be expressed as a sum of scaled and time-shifted impulse responses? That is, given an arbitrary sequence, find the input to the system that will produce the given sequence as the output? If so, search for information about deconvolution –  Dilip Sarwate Sep 28 '12 at 11:42
For more information on why the impulse response is important, see the answer to this question. –  Jason R Sep 28 '12 at 13:17
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## 2 Answers

One interpretation of your question could be as follows:

Given that a system has the following two properties:

1. the scaling or homogeneity property that if the response to input $x(t)$ is output $y(t)$, then for any choice of $\alpha$, the system response to scaled input $\alpha\cdot x(t)$ is scaled output $\alpha\cdot y(t)$,

2. the time-invariance property that for all choices of $\tau$, the response to time-delayed input $x(t-\tau)$ is time-delayed output $y(t-\tau)$,

then why does the system have the additivity or superposition property that the response to input $x_1(t)+x_2(t)$ is $y_1(t) + y_2(t)$ where the system response to $x_i(t)$ is $y_i(t)$, $i = 1,2~$ ???? $~~~~~~~~~~~$ More generally, why is the system response to input $\alpha\cdot x_1(t-\tau_1) + \beta\cdot x_2(t-\tau_2)$ given by $\alpha\cdot y_1(t-\tau_1) + \beta\cdot y_2(t-\tau_2)~$?

The answer is that a system with properties 1 and 2 does not necessarily have the additivity or superposition property. If the superposition property also holds, then the system is called a linear time-invariant system. But this is an additional assumption that you need to make (or prove).

Commonly, homogeneity and additivity are combined together into the linearity property that says that the response to input $\alpha\cdot x_1(t)+\beta\cdot x_2(t)$ (that is, a linear combination of inputs $x_1(t)$ and $x_2(t)$) is $\alpha\cdot y_1(t) + \beta\cdot y_2(t)$ (that is, the same linear combination of outputs $y_1(t)$ and $y_2(t)$).

A couple of points that should be tucked away into the back of one's mind:

• A system can be linear without being time-invariant (e.g. a modulator $x(t) \to x(t)\cos(\omega t)$, or time-invariant without being linear (e.g. a square-law circuit $x(t) \to [x(t)]^2$

• A additive system which produces output $y(t) + y(t) = 2y(t)$ in response to input $x(t) + x(t) = 2x(t)$ and so seems to have the scaling property does not in fact have the scaling property. Persuade yourself that this is true by attempting to prove that the response to $0.5x(t)$ is $0.5y(t)$. In short, scaling and additivity are two different properties and a system that enjoys one of them does not necessarily enjoy the other.

A second interpretation of your question could be as follows:

For a linear time-invariant system, the output is supposed to be the sum of scaled and time-delayed versions of the impulse response, but I don't see how this is so. For example, the standard convolution result (for discrete-time systems) says $$y[n] = \sum_m x[m]h[n-m]$$ where $h[\cdot]$ is the impulse (or unit) response of the system. But this seems to be completely backwards since the impulse response is running backwards in time (as in $-m$ in the argument of $h$ in the above formula compared to $x[m]$ in which time is running forwards.

This is indeed a legitimate concern, but actually the convolution formula is very successful in concealing the result that the output is the sum of scaled and time-delayed versions of the impulse response. What's going on is as follows.

We break down the input signal $x$ into a sum of scaled unit pulse signals. The system response to the unit pulse signal $\cdots, ~0, ~0, ~1, ~0, ~0, \cdots$ is the impulse response or pulse response $$h[0], ~h[1], \cdots, ~h[n], \cdots$$ and so by the scaling property the single input value $x[0]$, or, if you prefer $$x[0](\cdots, ~0, ~0, ~1, ~0,~ 0, \cdots) = \cdots ~0, ~0, ~x[0], ~0, ~0, \cdots$$ creates a response $$x[0]h[0], ~~x[0]h[1], \cdots, ~~x[0]h[n], \cdots$$

Similarly, the single input value $x[1]$ or creates $$x[1](\cdots, ~0, ~0, ~0, ~1,~ 0, \cdots) = \cdots ~0, ~0, ~0, ~x[1], ~0, \cdots$$ creates a response $$0, x[1]h[0], ~~x[1]h[1], \cdots, ~~x[1]h[n-1], x[1]h[n] \cdots$$ Notice the delay in the response to $x[1]$. We can continue further in this vein, but it is best to switch to a more tabular form and show the various outputs aligned properly in time. We have $$\begin{array}{l|l|l|l|l|l|l|l} \text{time} \to & 0 &1 &2 & \cdots & n & n+1 & \cdots \\ \hline x[0] & x[0]h[0] &x[0]h[1] &x[0]h[2] & \cdots &x[0]h[n] & x[0]h[n+1] & \cdots\\ \hline x[1] & 0 & x[1]h[0] &x[1]h[1] & \cdots &x[1]h[n-1] & x[1]h[n] & \cdots\\ \hline x[2] & 0 & 0 &x[2]h[0] & \cdots &x[2]h[n-2] & x[2]h[n-1] & \cdots\\ \hline \vdots & \vdots & \vdots & \vdots & \ddots & \\ \hline x[m] & 0 &0 & 0 & \cdots & x[m]h[n-m] & x[m]h[n-m+1] & \cdots \\ \hline \vdots & \vdots & \vdots & \vdots & \ddots \end{array}$$ The rows in the above array are precisely the scaled and delayed versions of the impulse response that add up to the response $y$ to input signal $x$. But if you ask a more specific question such as

What is the output at time $n$?

then you can get the answer by summing the $n$-th column to get $$y[n] = x[0]h[n] + x[1]h[n-1] + x[2]h[n-2] + \cdots + x[m]h[n-m] + \cdots = \sum_{m=0}^{\infty} x[m]h[n-m],$$ the beloved convolution formula that befuddles generations of students because the impulse response seems to be running backwards in time.

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If we consider a discrete signal, say

x[n] = {1,5,3} , having three impulses at n = 0, 1 and 2 with amplitude 1, 5 and 3.

now, we can write

x[n] = 1*$\delta[n]$ + 5*$\delta[n-1]$ + 5*$\delta[n-2]$

or we generalize it,

x[n] = $\sum^{\infty}_{-\infty} x[k]\delta(n-k)$

For Linear time invariant system we know that,

for a given input, x[n] = $x[m]\delta[n-m]$, a system response as h[n], output $y_m[n]$ = $x[m]h[n-m]$

Therefore, using commutative property,

y[n] = $\sum^{\infty}_{-\infty} y_k[n]$ = $\sum^{\infty}_{-\infty} x[k]h[n-k]$

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