signals and systems Unit Step Function a question about

[circuit975]

I am having a hard time solving the following question about signals and systems, and I need to plot its graph:

(2t/5) *[u(t)- u(t-6)]

Tools offer a solution, but I want to understand the logic behind how it is drawn. in other words, he expressed it with a unit digit function in this expression, but we need logic, how do we draw it?

If anyone knows and can help, it will be greatly appreciated.

 

[danadak]

Unit Step Function

www.desmos.com

https://ocw.mit.edu/courses/es-1803-differential-equations-spring-2024/mites_1803_s24_topic20.pdf

Regards, Dana

 

[circuit975]

Unit Step Function

www.desmos.com

https://ocw.mit.edu/courses/es-1803-differential-equations-spring-2024/mites_1803_s24_topic20.pdf

Regards, Dana

I've studied too many articles like this, but I didn't understand. can you explain this

 

[Tony Stewart]

Can you plot each part, then solve the product?

 

[circuit975]

Can you plot each part, then solve the product?

actually, I think it's something like that. but I don't know if it's right or wrong?

 

[Tony Stewart]

Plug me in the numbers and see for yourself

 

[danadak]

So if you expand the expression you get 2tu(t)/5 - 2tu(t-6)/5

So first term is a ramp of 2t/5 starting at the unit step (which enables
the ramp when = 1, eg for all t > 0) which occurs at t = 0.

Then you have another ramp starting at......

Then you sum (perform subtraction as second term is - going ramp)

 

[MrAl]

I am having a hard time solving the following question about signals and systems, and I need to plot its graph:

(2t/5) *[u(t)- u(t-6)]

View attachment 149392

Tools offer a solution, but I want to understand the logic behind how it is drawn. in other words, he expressed it with a unit digit function in this expression, but we need logic, how do we draw it?

If anyone knows and can help, it will be greatly appreciated.

Hi,

My guess is that you do not understand what the unit step function is, or how it works with notation like this. It might seem strange at first because we don't usually see this stuff until we get into circuits that need it which can be long after we started learning about circuit theory.

Let's say we have a function f(x) and we define that function like this:
f(x) for x<0 equals 0,
f(x) for x>=0 equals 1.

Now what happens if x=-7?
Since when x<0 the function is equal to 0, that means f(-7)=0.

What happens if x=8?
Since when x>=0 the function is equal to 1, that means f(8)=1.

That should be easy to follow.

Now what about the function u(t). u(t) is just like that:
u(t) for t<0 equals 0,
u(t) for t>=0 equals 1.

But sometimes we see other things inside the parentheses like t-7 rather than just one value.
What we do then is also very simple, we perform the subtraction first.
u(-7) as we know equals 0, but what about u(t-7).
Well, all we have to do is perform the subtraction first t-7 and see what we get. Now if t=0 then we see:
u(0-7) which of course equals u(-7) and we know that u(-7) equals 0. On the other hand, if t=9 then we see:
u(9-7) which of course equals u(2) and we know that since 2 is greater than zero u(2) must equal 1.

You might do a couple of those by hand before proceeding further here.
u(t-3) for t running from 0 to 10 would have output 0 from t=0 to 3 and have output 1 from t=3 to 10.
Note that the output for t=0 to 3 is not perfectly correct it is actually from t=0 to 2.999999999 because at t=3 it changes to 1. We have to run t for all t not just integers, so right up until we reach 3 the output is 0 but as soon as we reach 3 it goes to an output of 1. I am using mostly integers here but the time has to run over all fractions also like t=2.1, t=2.2, t=2.3, etc., and even t=2.001, t=2.002, etc. When we plot this we have to realize that there is some small step involved but in theory the step is infinitesimally small.

Now what about when we see two of them in the same expression like [u(t)-u(t-3)].
We have to do both of those to get this. First we can look at each one individually:
u(t) and
u(t-3)
and treat them like separate problems. When we do this we see that:
u(t) must be 1 for all t from 0 and higher, and
u(t-3) must be 1 for all t from 3 and higher, and 0 for t from t=0 to 2.
We can plot both of those separately.
Next, wee see that together inside the brackets they are subtracted:
u(t)-u(t-3)
and so we have to subtract them over t from 0 and up.
Since u(t) is 1 for all time from 0 and up, and u(t-3) is 0 from t=0 to 2, from t=0 to 2 we end up with output
[1-0]=1 for t running from 0 to 2.
Once we get to t=3 however, the second term kicks in too. That is 1 for t=3 and up. That means now we have:
[1-1]=0 for t running from t=3 and up.
These two solutions [1-0]=1 and [1-1]=0 can be plotted, and here it would look like a pulse from t=0 to 3 and at t=3 it would go to zero.

We also have a multiplier here too though. It is 2*t/5, but let's say it is just 2*t for now. I think you know this is a ramp, but because we have those two terms inside the brackets we actually have two ramps. The first ramp is:
2*t*[1-0]=2*t
and the second is:
2*t*[0-0]=0
This means for t running from 0 to 3 we see a ramp that goes up as 2*t which means it ramps from 0 to 6, and after that we see 0 for t from 3 and up because 2*t*0=0.

Now the actual problem has 2*t/5, but you should be able to figure out how that changes the plot of the ramp.

If this still does not make too much sense, try plotting each term individually first. First do u(t) that's the most important, then do u(t-6). If you like, you can use u(T) instead and then calculate T=t-6 first, and that is because we do define u(t) but then we make t=t-6 which is a little strange at first, so it might seem clearer if we use T=t-6, then plot u(T).

After you do a few of these it gets kind of easy. You can then start to understand it when you see more terms inside like this:
[u(t)-u(t-1)+u(t-7)-u(t-8)]
The simple way to view this is that +u(T) turns 'on' a pulse equal to 1, and -u(T) turns 'off' the pulse which equals 0. So it's just turning on and off pulses that's about it. That's because a pair like
u(t)-u(t-k)
turns a pulse on then turns it off at t=k.