How to sketch/plot this discrete time signal?

$(n+5)u[n+5]+2nu[n]+(n-10)u[n-10]$ is the question.

The $u[n]$ is messing me up. If I substitute $n$ as $0$, what would be $u[5]$ or $u[-10]$?

• sorry, I should have used (). Sep 7, 2017 at 6:52
• Nothing bad happened :) thanks for clarifying :) Sep 7, 2017 at 6:54
• so, $u$ is the unit step function, right? Sep 7, 2017 at 6:55
• yep it is and thanks for editing, I had no clue! Sep 7, 2017 at 6:56
• then, what I'd do now is simply draw the three parts of your sum (the one with $u[n+5]$, the one with $u[n]$ and the one with $u[n-10]$) separately in three graphs right below each other, then add them up. Hint: what is the lowest $n$ for which any of these three change, what is the highest? Sep 7, 2017 at 6:58

Begin with defining the base signal as $x[n] = n ~u[n]$ and recognize the output as a linear compositions (superpositions) of the base signal as: $$y[n] = x[n+5] + 2 x[n] + x[n-10]$$

Now what you have to answer is the result of translations and scalings applied on the basic signal $x[n]$. This ability is a simple yet an indispensibly necessary skill for a clear understanding and successful applications of all subsequent signal processing theory. Therefore you must train yourself enough for this task.

Incidentally, the signal $x[n+5]$ is an advanced (shift left) version of $x[n]$ whereas $x[n-10]$ is a delayed (shift right) version.

Graphically the result is obtained by summing the three components drawn on the on the same plot. A more detailed approach would use the piecewise definitions of the components to define the sum. It would proceed as follows:

Let $x[n]$be defined as: $$x[n] = n ~\text{u}[n] = \begin{cases} n , \text{ for } n \geq 0 \\ 0 , \text{ for } n < 0 \\ \end{cases}$$

Then you can define the three superposed signals as:

\begin{align} x_1[n] = x[n+5] = (n+5) ~u[n+5] &= \begin{cases} n+5, &\text{ for } &n+5\geq 0 \\ 0, &\text{ for } &n+5< 0 \\ \end{cases} \\ x_2[n] = 2x[n] = 2n ~u[n] &= \begin{cases} 2n,~~~~ &\text{ for } &2n \geq 0 \\ 0,~~~ &\text{ for } &2n < 0 \\ \end{cases} \\ x_3[n] = x[n-10] = (n-10) ~u[n-10] &= \begin{cases} n-10, &\text{for} &n-10 \geq 0 \\0, &\text{for} &n-10<0 \\ \end{cases} \\ \end{align}

Modifiy the intervals: \begin{align} x_1[n] = x[n+5] = (n+5) ~u[n+5] &= \begin{cases} n+5, &\text{ for } &n \geq -5 \\ 0, &\text{ for } &n<-5 \\ \end{cases} \\ x_2[n] = 2x[n] = 2n ~u[n] &= \begin{cases} 2n,~~~~ &\text{ for } &n \geq 0 \\ 0,~~~ &\text{ for } &n<0 \\ \end{cases} \\ x_3[n] = x[n-10] = (n-10) ~u[n-10] &= \begin{cases} n-10, &\text{for} &n \geq 10 \\0, &\text{for} &n<10 \\ \end{cases} \\ \end{align}

Therefore $y[n] = x_1[n] + x_2[n] + x_3[n]$ is: $$y[n] = \begin{cases} 0 , &\text{ for } &-\infty<n<-5 \\ n+5 , &\text{ for } &-5 \leq n <0 \\ 3n+5 , &\text{ for } &0 \leq n <10 \\ 4n-5 , &\text{ for } &10 \leq n <\infty \\ \end{cases}$$

I don't know did you by plotting also meant a MATLAB plot but it's a good way for checking your results if you're unsure. For discrete signals you can use the stem plot:

n = [-12:1:12];
f =(n+5).*stepfun(n,-5)+2*n.*stepfun(n,0)+(n-10).*stepfun(n,10);
stem(n,f,'marker','o','color','r')
grid on;


From the plot you can see that e.g. for n=10 the y-axis shows 35 which you will get by inserting n=10 into y[n]=3n+5 provided by Fat32 which confirms his results.