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## 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

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- Thread starter ongxom
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- #1

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Plot this signal

y(t)=3u(t+3)-r(t+2)+2r(t)-2u(t-2)-r(t-3)-2u(t-4)

I tried to draw each single function, but do not know the way to plot for the whole signal like the bottom graph

- #2

lewando

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Only for u function

Only for r function

What things I have to do next to get the correct graph ?

Only for r function

What things I have to do next to get the correct graph ?

- #4

lewando

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Both of graph is plotted using wolframalpha, for the u-functions :

y(t)=3u(t+3)-2u(t-2)-2u(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) but the graph only shows a line starts from (4;-1), and where is the line for -2u(t-2). Where does the line start from (2;1) to (4;1) come from ?

- #6

lewando

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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 ?

Realize that y(t)=3u(t+3)-2u(t-2)-2u(t-4) is composed of 4 time intervals:

I1: -∞ < t < -3 [nothing is happening because no u-function is "active" --for amplitude, think 0]

I2: -3 <= t < 2 [3u(t+3) is the only "active" --for amplitude, think 3]

I3:

I4: 4 <= t < ∞, [3u(t+3) AND -2u(t-2) AND -2u(t-4) are all "active" --for amplitude, think 3-2-2 ]

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-2 if t-4>0 and 0 if t-4<0

so it should start from (4;-2), am I right ?

- #8

lewando

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Yes. Technically its: -2 if t-4 >**=** 0 and 0 if t-4 < 0

- #9

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Another example x(t)=2u(t)-u(t-1)-u(t+1)

For 2u(t) = 2 if t>0 and 0 if t<0, a line will start from (0;2)

For -1u(t-1) = -1 if t>1 and 0 if t<1, line starts from (+1;-1)

For -1u(t+1) = -1 if t>-1 and 0 if t<-1. line starts from (-1;-1)

I tried to plot all thing above in a graph, but does not know how to combine them into one. The correct graph have lines start from (-1;-1) to (0;-1) and (0;1) to (1;1).

I am stucking at see the way 2u(t) represent in the graph.

For 2u(t) = 2 if t>0 and 0 if t<0, a line will start from (0;2)

For -1u(t-1) = -1 if t>1 and 0 if t<1, line starts from (+1;-1)

For -1u(t+1) = -1 if t>-1 and 0 if t<-1. line starts from (-1;-1)

I tried to plot all thing above in a graph, but does not know how to combine them into one. The correct graph have lines start from (-1;-1) to (0;-1) and (0;1) to (1;1).

I am stucking at see the way 2u(t) represent in the graph.

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- #10

lewando

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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.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.

Before jumping all over this and other examples, did you understand what I wrote in post #6? If not, what is not clear? I am trying to understand what you don't understand about the OP so please focus on post #6 for now.

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- #12

lewando

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y

You know, maybe you are still trying to do too much all at once. Why not just take two of the function terms and add them together:

y

What would that look like?

- #13

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Let me do it :

-∞<t<-2 : No signal, 0 for amplitude.

-2<t<0 : -r(t+2) is active at t=-2, -1 for amplitude.

0<t<3 : -r(t+2) and 2r(t) are both active, 2-1 for amplitude

3<t<+∞ : -r(t+2) and 2r(t) and -r(t-3) are all active, 2-1-1 for amplitude.

I have no idea what the graph would look like, the u-functions only have lines paralled with t-axis.

-∞<t<-2 : No signal, 0 for amplitude.

-2<t<0 : -r(t+2) is active at t=-2, -1 for amplitude.

0<t<3 : -r(t+2) and 2r(t) are both active, 2-1 for amplitude

3<t<+∞ : -r(t+2) and 2r(t) and -r(t-3) are all active, 2-1-1 for amplitude.

I have no idea what the graph would look like, the u-functions only have lines paralled with t-axis.

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- #14

lewando

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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.

-r(t+2) starts ramping at t=-2 with a slope of -1.

"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.

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