Draw heaviside and ramp function

In summary: NO! STOP! Not another word about the graph until we are done with this discussion of the time intervals. Let's finish that first. If you don't understand something, ASK!OK, I will not answer the graph question, but you should know where i am stuck at. I think I am wrong at drawing the ramps.Ok, so you have the four time intervals, right? (I'm not going to list them again. If you don't know what I'm talking about, go back and re-read my posts.)For the first two:-∞<t<-2 : No signal, 0 for amplitude. OK. So what is yramp-first-2-terms(t) during this interval?-2<t
  • #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
PKCxBl7.png
 
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  • #2
Try breaking the problem down into 2 parts--a graph resulting from only the u-components, and another from only the r-components. Post what you get.
 
  • #3
Only for u function
image.gif


Only for r function
image.gif


What things I have to do next to get the correct graph ?
 
  • #4
Your u-function looks right. Your r-function has a problem with the rightmost segment. Fix that and then just add them together. It may be easier to start with the r-function and then add the offsets from the u-function. You may even be able to do it visually if your x-axes are the same scale.
 
  • #5
Let me ask some question first :
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
Both of graph is plotted using wolframalpha, for the u-functions :
y(t)=3u(t+3)-2u(t-2)-2u(t-4)
Wolfram alpha would be good for checking your work, but you should first try to do the graphs manually.


The 3u(t+3) function is correct, but for -2u(t-4), i think it should have a line starts from (2;-2)
Why would you think that? The function -2u(t-4) "activates" at t=4.


but the graph only shows a line starts from (4;-1)
(4,-1) is correct.


and where is the line for -2u(t-2). Where does the line start from (2;1) to (4;1) come from ?
That would be from the third time interval (I3), below:

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: 2 <= t < 4, [3u(t+3) AND -2u(t-2) are both "active" --for amplitude, think 3-2]
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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  • #7
In the definition of unit step function, the function-2u(t-4) should be
-2 if t-4>0 and 0 if t-4<0

so it should start from (4;-2), am I right ?
 
  • #8
Yes. Technically its: -2 if t-4 >= 0 and 0 if t-4 < 0
 
  • #9
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.
 
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  • #10
I tried to plot all thing above in a graph, but does not know how to combine them into one.
You don't need any special tools to solve these kinds of problems. A pencil and some graph paper will do the job.


The correct graph have lines start from (-1;-1) to (0;-1) and (0;1) to (1;1).
Not really. (-∞;0) to (-1;0) to (-1;-1) to (0;-1) and (0;1) to (1;1) to (1;0) to (∞;0)


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


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.
 
  • #11
Thank you, I got the #6. Now come to the ramp function, I have tried to separate time intervals but it seems there is nothing like the amplitude for heaviside, it doesn't work.
 
  • #12
What doesn't work? The time intervals for the combined ramp functions can be obtained from inspection:

yramp(t) = -r(t+2) + 2r(t+0) -r(t-3)

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:

yramp-first-2-terms(t) = -r(t+2) + 2r(t+0)

What would that look like?
 
  • #13
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.
 
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  • #14
So jumping ahead (against my advice :frown:), you are now working on yramp(t) = -r(t+2) + 2r(t+0) -r(t-3).

-∞<t<-2 : No signal, 0 for amplitude.
True.
-2<t<0 : -r(t+2) is active at t=-2, -1 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).

-r(t+2) starts ramping at t=-2 with a slope of -1.
0<t<3 : -r(t+2) and 2r(t) are both active, 2-1 for amplitude
"amplitude" is wrong. You are thinking about the behavior of the u-function.
3<t<+∞ : -r(t+2) and 2r(t) and -r(t-3) are all active, 2-1-1 for amplitude.
Again, the "constant amplitude" concept only applies to u-functions.
I have no idea what the graph would look like, the u-functions only have lines paralled with t-axis.
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).
 

1. What is the definition of the Heaviside function?

The Heaviside function, also known as the unit step function, is a mathematical function that is defined as 0 for negative inputs and 1 for positive inputs. It can be written as H(x) = 0 for x < 0 and H(x) = 1 for x ≥ 0.

2. How is the Heaviside function used in engineering and physics?

The Heaviside function is commonly used in engineering and physics to model discontinuous or sudden changes in a system. It is also used to represent a switch between two different states, such as on and off.

3. What is the mathematical expression for the ramp function?

The ramp function, also known as the unit ramp function, is defined as r(x) = 0 for x < 0 and r(x) = x for x ≥ 0. It can also be written as r(x) = xH(x), where H(x) is the Heaviside function.

4. How is the ramp function used in calculus?

The ramp function is commonly used in calculus to model linearly increasing functions. It is also used in the Laplace transform to solve differential equations.

5. Can the Heaviside and ramp functions be graphed?

Yes, both the Heaviside and ramp functions can be graphed. The graph of the Heaviside function is a step function with a vertical jump at x = 0, while the graph of the ramp function is a straight line with a slope of 1 starting at x = 0.

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