4.7 Properties of the digital convolution

Commutativity

By a change of variable $n-k=k^{,}$ , or $k=n-k^{,}$ in the formula for convolution

$$y(n)=\sum _{k=-\infty }^{\infty }x(k)h(n-k)=\sum _{k^{,}=-\infty }^{\infty }x(n-k^{,})h(k^{,})$$

and by replacing the temporary variable k’ by k, we get

$$y(n)=\sum _{k=-\infty }^{\infty }x(n-k)h(k)=\sum _{k=-\infty }^{\infty }h(k)x(n-k)=h(k)\ast x(k-n)$$

That is the order of convolution is reversed. Thus we have two formulae of convolution:

and

In practive we usually let the longer sequence stay fixed, and shift the shorter one.

The commutative characteristic of convolution means that we can swap the input signal with the impulse response of a system without affecting the output. This idea is depicted in [link] .

Associativity

It can be shown that

[link] shows the system meaning of the associativity, where two systems in series (in cascade) can be replaced by only one whose impulse response is the convolution of the two individual impulse responses.

Solution

First $\mid a\mid$ and $\mid b\mid$ should be smaller than 1 to ensure the convergence of the sequences. Notice that both impulse responses are causal. The overall impulse response is

$$h(n)=h_1(n)\ast h_2(n)=\sum _{k=-\infty }^{\infty }{h}_1(k)h_2(n-k)$$

The actual limits of summation are $k=0$ and $k=n$ (see [link] later), hence

$$\begin{array}{}h(n)=\sum _{k=0}^n{a}^k{b}^{k-n}u(k)u(k-n)\\ \begin{array}{ccc}& & \end{array}=b_n\sum _{k=0}^n(\frac{a}{b}{)}^k\end{array}$$

Using the formula of finite geometric series.

here $x=a/b$ , we get

$$h(n)=b^n\frac{1-(\frac{a}{b}{)}^{n+1}}{1-\frac{a}{b}}=\frac{b^{n+1}-a^{n+1}}{b-a}$$

Distributivity

It can be shown

The system meaning is illustrated in [link] where two systems connected in parallel can be replaced by one whose impulse response is the sum of the two ones.

Impulse response for causal system and signal

Since impulse response is a characterization (among other characterizations) of systems. As such, the causality of a system would be reflected on its impulse response. From the convolution Figure the output at instant $n_0$ is:

$$y(n_0)=\sum _{k=0}^{\infty }h(k)x(n_0-k)+\sum _{k=-\infty }^{-1}h(k)x(n_0-k)$$

In order the output signal $y(n_0)$ does not depend on future $(n>n_0)$ values of input signal $x(n)$ , the second term of above equation should be zero, i.e. $h(k)=0$ for . As k is a dummy variable, we conclude

Thus, the causality of a system implies that its impulse response is zero and vice versa. The output at time $n_0$ is now the first term of the equation

$$y(n_0)=\sum _{k=0}^{\infty }h(k)x(n_0-k)$$

For any time n,

Had the convolution $x(n)\ast h(n)$ been used, the result would be

In above, only the causality of the system is considered. Now, the imput signal is also causal, the result is

And equivalently

Notice that the summation limits for both cases are the same and the upper limit increases with n, also that the output signal $y(n)$ at time n depends only the convolution summation up to n and independent of future values after n. So we can write the upper limit as $\infty$ without effecting the result.

Solution

Notice that both $x(n)$ and $h(n)$ are causal and of infinite duration. The given condition ensures the convergence of $h(n)$ . We choose to evaluate $h(n)\ast x(n)$ , using [link]

$$\begin{array}{}y(n)=h(n)\ast x(n)=\sum _{k=-\infty }^{\infty }h(k)x(n-k)=\sum _{k=0}^n{a}^{k}u(k)u(n-k)\\ \begin{array}{cccc}\begin{array}{cccc}& & & \end{array}& & & \end{array}\begin{array}{cccc}& & & \end{array}\begin{array}{cc}& \end{array}=\sum _{k=0}^n{a}^k\end{array}$$

So, the result is.

$$\begin{array}{}y(0)=1\\ y(1)=1+a\\ y(2)=1+a+a^2\\ \text{.}\text{.}\text{.}\\ y(n)=1+a+a^2+\text{.}\text{.}\text{.}+a^n\end{array}$$

The output signal $y(n)$ does not go to $\infty$ but grows asymptotically to the finite value of $\frac{1}{1-a}$ (see Example ).

System identification

In DSP sometimes we need to determine a system, assumed LTI (or LSI), when we know the input signal and output signal, e.g. by setting up an experiment. This problem is called system identification . Specifically, we must determine the impulse response of the system, and then the signal difference equation if necessary.

Adaptive filters using FIR filters are often used to identify unknown DSP systems. In control theory, also, sytem identification is a familiar problem.

For causal systems the output is given by the reduced convolution ( [link] ) which is repeated here

$$y(n)={\displaystyle \sum _{k=0}^{n}h(k)x(n-k),n\ge 0}$$

At $n=0$

$$y(0)=h(0)x(0)$$

giving

$$h(0)=\frac{y(0)}{x(0)}$$

provided $x(0)\ne 0$ . At $n\ge 1$ , We can write

$$y(n)=h(n)x(0)+{\displaystyle \sum _{k=0}^{n-1}h(k)x(n-k)}$$

giving

$$h(n)=\frac{y(n)-{\displaystyle \sum _{k=0}^{n-1}h(k)x(n-k)}}{x(0)},n\ge 1$$

Solution

Note that both input and output signals are causal. We proceed the evaluation of the impulse response as follows.

$$h(0)=\frac{y(0)}{x(0)}=0$$

$$h(1)=\frac{y(1)-h(0)x(1)}{x(0)}=\frac{1-0(2)}{1}=1$$

$$h(2)=\frac{y(2)-h(0)x(2)-h(1)x(1)}{x(0)}=\frac{4-0(-1)-1(2)}{1}=2$$

$$h(3)=\frac{y(3)-h(0)x(3)-h(1)x(2)-h(2)x(1)}{x(0)}=\frac{3-0(-2)-1(-1)-2(2)}{1}=0$$

Continuing we will see that for . Thus the system impulse response is

$$h(n)=[0,1,2]$$

The system is causal as expected.

Another method is to transform the problem to the z-domain (chapter 4).