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:
$$ y[-1] = -x[0] + 2 x[-1] - x[-2] = -x[0] = 20 $$
Note that $x[n]=0$ for $n<0$; The second sample of $y[n]$ will be $y[0]$ which is:
$$ y[0] = -x[1] + 2 x[0] - x[-1] = 50 + -40 = 10 $$
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
$$ y_\text{hp}[k_0] = \sum_{n=-1}^1 x[n] \cdot g[k_0-n] \\ = x[-1]\cdot g[1] + x[0] \cdot g[0] + x[1] \cdot g[-1],$$ 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.
