Step Function and Transformations of a Constant

[fasih178]

I can't seem to understand the logic behind these functions. Since H(x) is a constant, I feel like it would not change at all for the transformations in part a, c, and d, since it is perfectly horizontal.

First, in (a) if we shift this function by 10 units to the right, we only change the domain right? That is, we change only from where y=1 begins (X≥10 now).

In (b) do we shift right by 1/2? Shouldn't we be shifting left by 1 cuz of the +1? Also, since the graph is merely two constants, we can't again shift it horizontally, so we'll be changing the domain again as in (a)?

C is all clear.

For (d), I've heard the solution is a square wave? When you see something like H(x) = 1 transformed to H(sin(x)), what do you do algebraically? Do you input any function into another or something? If you can, please show this on a page; but an explanation here would also do just fine. Alternatively, you can also redirect me to a video or page where all this is explained.

 

[pka]

I can't seem to understand the logic behind these functions. Since H(x) is a constant, I feel like it would not change at all for the transformations in part a, c, and d, since it is perfectly horizontal.

For (d), I've heard the solution is a square wave? When you see something like H(x) = 1 transformed to H(sin(x)), what do you do algebraically? Do you input any function into another or something? If you can, please show this on a page; but an explanation here would also do just fine. Alternatively, you can also redirect me to a video or page where all this is explained.

\(H(x)=\begin{cases}1 &\: x\ge 0\\0 &\: x< 0\end{cases}\) HERE is a plot of the \(\sin(x)\)
Note that the \(\sin(x)\ge 0\text{ if } x\in[2k\pi,(2k+1)\pi],k\in\mathbb{Z}\) and is zero otherwise.

 

[fasih178]

I'm sorry to sound so dumb, but this makes no sense to me. What does H(1-2x) and H(sin(x)) even mean if H(x) = 1 and 0 i.e., has no placeholder for x?—it's just a constant. Also, is there any way to solve this algebraically rather than graphically?

 

[Steven G]

In H(x), what goes between the two parenthesis is called the argument. By definition of H, if the argument is 0 or greater, then H will be 1. If the argument is negative then H is 0.

Now draw the sin(x) graph. Whenever it is 0 or positive then H(sin(x)) will be 1. Whenever sin(x) is negative then H(sin(x)) will be 0.

Lets start from x=0. sin(x) is 0 or positive when x is in [0, pi], then it is negative when x is in (pi, 2pi), then it is 0 or positive when x is in [2pi, 3pi], then it is negative (3pi, 4pi),.... The same idea when x is negative.

So H(sin(x)) = 1 if x is in [0, pi], [2pi, 3pi], [4pi, 5pi],....,[-2pi, -pi], [-4pi, -3pi],....
H(sin(x)) = 0 if x is in (pi, 2pi), (3pi, 4pi), (5pi, 6pi),..., (-pi, 0), (-3pi, -2pi),...

H(1-2x)
If 1-2x>0, then H(1-2x) = 1
Now 1-2x> 0 when x< 1/2
So H(1-2x) = 1 if x < 1/2 and H(1-2x) = 0 if x> 1/2

This problems makes you think carefully about what H(x) means. I like this problem very much!