- #1
ongxom
- 26
- 0
Homework Statement
Plot this signal
y(t)=3u(t+3)-r(t+2)+2r(t)-2u(t-2)-r(t-3)-2u(t-4)
The Attempt at a Solution
I tried to draw each single function, but do not know the way to plot for the whole signal like the bottom graph
Wolfram alpha would be good for checking your work, but you should first try to do the graphs manually.Both of graph is plotted using wolframalpha, for the u-functions :
y(t)=3u(t+3)-2u(t-2)-2u(t-4)
Why would you think that? The function -2u(t-4) "activates" at t=4.The 3u(t+3) function is correct, but for -2u(t-4), i think it should have a line starts from (2;-2)
(4,-1) is correct.but the graph only shows a line starts from (4;-1)
That would be from the third time interval (I3), below:and where is the line for -2u(t-2). Where does the line start from (2;1) to (4;1) come from ?
You don't need any special tools to solve these kinds of problems. A pencil and some graph paper will do the job.I tried to plot all thing above in a graph, but does not know how to combine them into one.
Not really. (-∞;0) to (-1;0) to (-1;-1) to (0;-1) and (0;1) to (1;1) to (1;0) to (∞;0)The correct graph have lines start from (-1;-1) to (0;-1) and (0;1) to (1;1).
2u(t) is an additive term in the composite graph. You won't "see it" in the graph as you would if it stood alone. It is responsible for the 2-unit jump from (0;-1) to (0;1).I am stucking at see the way 2u(t) represent in the graph.
True.-∞<t<-2 : No signal, 0 for amplitude.
False. r(t) is the unit ramp function, with amplitude 0 for t<0 and a "ramping" output for t >= 0 (output starts at zero and increases linearly with a slope of 1).-2<t<0 : -r(t+2) is active at t=-2, -1 for amplitude.
"amplitude" is wrong. You are thinking about the behavior of the u-function.0<t<3 : -r(t+2) and 2r(t) are both active, 2-1 for amplitude
Again, the "constant amplitude" concept only applies to u-functions.3<t<+∞ : -r(t+2) and 2r(t) and -r(t-3) are all active, 2-1-1 for amplitude.
You started out this post with a collection of horizontal and diagonal lines. Those diagonal lines are the the result of the r-functions. By the way, the red dash-dot-dot-dashed line is not shown correctly (it should start at t=3).I have no idea what the graph would look like, the u-functions only have lines paralled with t-axis.