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I would like to ask, how should I proceed when calculating convolution of LTI systems with the following parameters:

x[n] = 3δ[n+1] + δ[n−1] + 4δ[n−2] − 2δ[n−4]
h[n] = −δ[n + 1] + δ[n] − δ[n − 1]

I do know the result, but I am unfamiliar with the procedure. I would be very much grateful, if someone could elaborate on that. The result should be:

y[n] = {−3, 3, −4, −3, 3, −2, −2, 2}
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    You know $y[n]=x[n]\ast h[n]=\sum_{k=-\infty}^{\infty}x[k]h[n-k]$? – 2017-01-20
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    I'm not really clear on what's the "k", would you mind showing how would you get the first element of "y[n]"? Thank you. – 2017-01-20

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As a response to your question, let me explain the equation, which is discrete convolution: \begin{equation} y[n]=x[n]\ast h[n] \quad = \sum_{k=-\infty}^{\infty}x[k]h[n-k] \end{equation} This equation comes from the fact that we are working with LTI systems but maybe a simple example clarifies more.

Call $y[n]$ the output, $x[n]$ the input and $h[n]$ the impulse response(maybe better known to you as a transfer function). Say our input sequence is $x[n]=\{x[0]=1,x[1]=2\}$ and $h[n]=\{h[0]=2,h[1]=3\}$. Then we are interested in for example $y[1]$ which depends on all previous inputs. The way to look at this is the following, at $n=0$ we apply the input and see $y[0]=x[0]h[0]=2$, now what happens at $n=1$? Again, we apply an input, which results in $y[1]=x[1]h[0]$, why $h[0]$? Because we just applied the input, $h[n]$ tells us what the effect of an input is over time. So should we include the effects of our first input? Yes! Then see that the output becomes \begin{align} y[0]&=x[0]h[0]\\ y[1]&=x[0]h[1]+x[1]h[0]\\ y[2]&=x[1]h[1]\\ y[n>2]&=0 \end{align} For more involved signals you don't want to manually think about each time step, you can use convolution! Intuitively think about flipping $h[n]$ about the vertical axis, i.e. $h[k]\to h[-k]$ and then shift it along the horizontal axis in the direction of your time steps, i.e. $h[-k]\to h[n-k]$. Then your output is proportional to the area of overlap between $x$ and $h$. This is exactly what the equation above does, the $k$'s are really just dummy variables. Try it out with simple examples and see that it works!

As a sidenote, you have a lot of impulses, see that their overlap is particularly easy.

If you have no clue I would take a look at the lectures/book by Alan Oppenheim.