# How to apply discrete filters to a signal

Let's say I have a signal like x[k] = [-20 -50 -30 50 30 -60 60 -60 60 10 5 10 5 5], and I want to apply a lowpass and a highpass filter to this signal (separately). For example the impusle response of the filters are as follows:

Lowpass: h[k] = [-1 2 6 2 -1], k = -1,0,...,3

Highpass: g[k] = [-1 2 -1]; k = -1, 0, 1

How can I calculate the first four signal values after applying these filters?

Another way to simply get your result for this kind of a problem (where $$h[n]$$ is very short) is to use the following method :

Let your output of the discrete convolution sum be $$y[n]$$ : $$y[n] = x[n] \star h[n]$$

Then by expanding $$h[n]$$ into impulses, the convolution will be distributed over addition ( using the highpass filter {-1,2,-1}; k = -1,0,1; to demonstrate ) :

$$y[n] = x[n] \star \{ -\delta[n+1] + 2 \delta[n] - \delta[n-1] \}$$

$$y[n] = - x[n+1] + 2 x[n] -x[n-1]$$

By simple argumentation of convolution nonzero ranges, it can be seen that $$y[n]$$ starts at the index $$n=-1$$ hence the first sample of the output is:

$$y[-1] = -x + 2 x[-1] - x[-2] = -x = 20$$

Note that $$x[n]=0$$ for $$n<0$$; The second sample of $$y[n]$$ will be $$y$$ which is:

$$y = -x + 2 x - x[-1] = 50 + -40 = 10$$

and so on... You can apply the procedure for your other filter and other output samples

The discrete convolution of a signal $$x[k]$$ with an impulse response $$h[k]$$, where $$k$$ is the discrete time index, is computed as

$$y[k] = x[k] \ast h[k] = \sum_{n=-\infty}^{\infty} x[n] \cdot h[k-n].$$

for a given sample index $$k_0$$, e.g. $$k_0 = 0$$, you can compute the first sample of your exercise as

$$y_\text{hp}[k_0] = \sum_{n=-1}^1 x[n] \cdot g[k_0-n] \\ = x[-1]\cdot g + x \cdot g + x \cdot g[-1],$$ which will result in

$$y_\text{hp}[n_0] = 0 \cdot -1 + -20 \cdot 2 + -50 \cdot -1 = 10.$$

For the rest of the question, you could compute the rest by hand this way. Another quick way is for example documented in these slides, a graphical approach is shown in this video. A Tutorial also featuring continous signals can be found here.

• Please check your second equation, right now it doesn't make any sense. – Matt L. Aug 1 '19 at 9:27
• You are right, thank you for pointing it out! – Jonas Schwarz Aug 1 '19 at 15:51