Output signal as convolution of impulse response and step signal

Question

Hi All: This question is kind of an add on question to the gorgeous answer by Juancho provided at this link:

How does this "simple filter" work?

The answer explains how the limit of exponential smoothing given an impulse response $x_0$, (so that new $x_i$'s are not arriving. only $x_0$ arrives) can be computed by convolving the impulse with a step signal.

The difference between the framework at that link and my framework is that actual series is not driven by an impulse response. By this I mean that the same process, occurs over and over because a "new" $x_i$ is arriving during each period so one could think of there being an impulse response as described but occurring in each period. So, there is a step signal convolved with the first impulse response. Then, in the next period, another step signal is convolved with a second impulse, etc, etc. In other words, now there is $x_1$, $x_2$, $x_3$, ..... $x_n$, where each $x_i$ gets operated on in the same manner as described by Juancho. Assuming that my setup is clear, my question is: is there a simple recursion for this non-impulse response case. Does my case have a name ? It seems like there should be a way to use recursion but I can't figure it out. Thank you very much for any references, answers, books, whatever.

Answer 1

The answer explains how the limit of exponential smoothing given an impulse response $x_0$, (so that new $x_i$'s are not arriving. only $x_0$ arrives) can be computed by convolving the impulse with a step signal.

Actually, Juancho's answer explains how the steady state limit of the exponential smoothing filter, given the filter's impulse response and a step input $x_0 u[n]$ (where $u[n]$ is the step function), can be computed by convolving the filter's impulse response with the input signal.

In your case the input signal is not a step function so the output of the filter would not be an exponential as was the case in the linked post. Regardless, the output of the filter can still be computed with the recursion formula:

$$ y[n]=(1−a)x[n]+ay[n−1] $$ or alternatively by computing a very long convolution between the infinite filter impulse response (truncated to some large length $N$ for practical purposes) and your input sequence.

As an illustration, the following Matlab script computes the output of the filter with a step input x1:

N=50;
M=4;
a = 0.8;

h = (1-a)*a.^[0:N];

x1=[zeros(1,M) 5*ones(1,N)];
y1a=conv(x1, h);             % compute output with convolution
y1b=filter(1-a, [1 -a], x1); % compute output with recursion
figure(1);
hold off; stem(x1);
hold on;  stem(y1a,'rs');
hold on;  stem(y1b,'k*');

Similarly, the same operation can be performed on a more arbitrary input sequence x2:

k = 0.5;
x2=[zeros(1,M) 5*ones(1,N)] + k*randn(1,N+M);
y2a=conv(x2, h);
y2b=filter(1-a, [1 -a], x2);
figure(2);
hold off; stem(x2);
hold on;  stem(y2a,'rs');
hold on;  stem(y2b,'k*');

Answer 2

It is a normal IIR filter, and a very simple one at that.

Everything that can be described with a linear time invariant difference equation is an IIR filter and can be designed and analyzed using well established mathematical methods.

You calculate the output of the filter for ANY given input (impulse, step, the Dow Jones, Wagner's Nibelungen, etc.) with the equation $$y[n] =\sum_{i=0}^{N}b_i\cdot x[n-i]-\sum_{i=1}^{N}a_i\cdot y[n-i]$$

In your case you have $b_0=0.2$ and $a_1 = -0.8$ with all the other a and b coefficients being zero, so the equation simplifies to $$y[n] = 0.2 \cdot x[n] + 0.8 \cdot y[n-1] $$

That is indeed a simple first order low pass filter with a -3dB point at a normalized frequency of 0.0355 or thereabouts. It's not a very good low pass filter since it doesn't go to zero at the Nyquist frequency. A somewhat better choice would be

$$y[n] = 0.1 \cdot x[n] + 0.1 \cdot x[n-1] + 0.8 \cdot y[n-1] $$

There are plenty of smoothing algorithms that are not IIR filters but your specific example is just a simple first order low pass.

EDIT: based on the question in the comment The equation above describes the operation for ALL input signals, not just an impulse. Let's consider the input signal x = [1.0 1.1 1.05]; So we have x[0] = 1.0, x[1] = 1.1, x[2] = 1.05, and x[3...] = 0; You would calculate the output as follows

y[0] = 0.2*x[0] + 0.8*y[-1] = 0.2; // since y[-1] = 0;
y[1] = 0.2*x[1] + 0.8*y[0]  = 0.38;
y[2] = 0.2*x[2] + 0.8*y[1]  = 0.514;
y[3] =            0.8*y[2]  = 0.4112;     // since x[3] = 0;
y[4] =            0.8*y[3]  = 0.32896; 
etc.