# Current Signals at Switched times problem

1. Jan 15, 2017

### JoeMarsh2017

1. The problem statement, all variables and given/known data

2. Relevant equations
i(t)=5u(t)+ 3u(t^2)

3. The attempt at a solution

2. Jan 15, 2017

### Staff: Mentor

The question is somewhat ambiguous. Does your textbook provide answers to any similar questions, because that might help figure out what they want. Their answer to question 2.7 might be enlightening.

Your sketch looks more puzzling than the question.

3. Jan 15, 2017

### JoeMarsh2017

My instructor is actually making his own text book for the course, so we dont have a text book to reference to this semester. He is cutting problems out of another book and posting them in a PDF for us to work on...no clue what the book is...

4. Jan 15, 2017

### Staff: Mentor

5. Jan 15, 2017

### JoeMarsh2017

Yes, we are working on Unit Step functions, ramps, and combining signals right now. I am stuck on another part because I am still learning how to use MATLAB.

i(t)=u(5t)+u(3t^2)

I am still confused but is this where it becomes a "power function" like A^2/n! ?
Joe

6. Jan 15, 2017

### Staff: Mentor

The examples in the article all show that the expression inside the unit step's parentheses is a simple expression such as (t) or (t–4), but no multiples of t, and no powers of t.

So, for example, a ramp starting at t=0 could be described as t • U(t)
and a steeper ramp as, e.g., 10t • U(t)

If the ramp signal is described as just 10t then at times before t=0, i.e., negative time, the ramp function would have a non-zero value. But when we don't want it to be a ramp for negative time we multiply it by the Heaviside step U(t) to indicate the ramp starts at t=0, and before that it is everywhere zero.

Were we to want a signal to not appear until t=4, the unit step needed would be U(t–4).

From wikipedia:
The Heaviside step function, or the unit step function, ... is a discontinuous function whose value is zero for negative argument and one for positive argument.