Unit Step Functions

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Hi!

My question is about multiplying unit step functions. How are they multiplied?

Example:
y[n] = u[n] * u[n-3]. My textbook's ("Signals and Systems") answer:

y[n] = { 0 for n < 3; n-2 for n >=3 }
 
Disclaimer: It's been awhile, since I used step functions, so read my comments carefully because I might not know what I'm talking about.

How are they multiplied?

u[n] * u[n-3]

This multiplication?

I'm not sure what the engineers might be doing (i.e., signal systems), but I'm thinking that the expressions u[n] and u[n - 3] are each symbolic numbers (function notation) that represent output values of step-function u, when the inputs are n and n-3, respectively.

In other words, the product that defines function y is already simplified. I don't think that we can multiply that expression.

If you can tell us what the actual definition is for function u, then perhaps we could multiply u[n]*u[n-3] using their definitions, and simplify this product in terms of n.

To determine the piecewise definition for function y that you posted as the answer, I'm thinking that you must have more information about function u.

Are you familiar with the Heavyside function? That's the basic unit step function.

H[n] = {0, n < 0; 1, n >= 0}

So, I'm thinking that your u[n] is similarly defined, except that the jump discontinuity is shifted three units to the right (i.e., at n = 3) and further that u[n] is defined in terms of n, when n is three or more. But, this is kinda a guess.

u[n] = {0, n < 3; [not sure what goes here], n}

This is why y[n] is defined as zero, when n<3, yes? In other words, u[n] is zero whenever the input n is to the left of the jump discontinuity at n = 3. And, whenever n is to the left of 3, the input n-3 is clearly also to the left of the jump discontinuity; hence, u[n-3] is also zero when n<3. That makes y[n] = 0*0, when n is to the left of the jump discontinuity.

The part that I cannot figure out is why y[n] starts growing to the right of the jump discontinuity (it's growing because y[n] is defined in terms of n when n>=3; that is, as n gets bigger, so does y[n]). I just can't remember a unit step function that has a slope other than zero to the right of the jump discontinuity.

Hopefully, though, my comments show that, at least for n<3, we can analyze the expression u[n]*u[n-3] without doing any multiplication. Maybe, you can now reason why y[n] is n-2, for n>=3.

Do you have more information about the step function u[n]?

Again, if I'm way off base in this post, feel free to ignore it.

Cheers.
 
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Yes sir, the Heaviside function is what I was referring to. I can reason why y[n] = 0 for n < 3. As you pointed out, u[n-3] = 0 for all n < 3, so it follows that y[n] = u[n] * 0 = 0 for all n < 3. But my intuition also tells me y[n] = 1 for n >=3, since u[n] and u[n-3] both equal 1 for all n > 3. Hopefully, someone can explain why the answer is (n-2) instead of one.

MS
 
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Yes sir, the Heaviside function is what I was referring to. I can reason why y[n] = 0 for n < 3. As you pointed out, u[n-3] = 0 for all n < 3, so it follows that y[n] = u[n] * 0 = 0 for all n < 3. But my intuition also tells me y[n] = 1 for n >=3, since u[n] and u[n-3] both equal 1 for all n > 3. Hopefully, someone can explain why the answer is (n-2) instead of one.
There is either a misprint in the textbook or you have left some factor out of the post.
If \(\displaystyle \left( {\forall x} \right)\left[ {u(x) \in \left\{ {0,1} \right\}} \right]\) then \(\displaystyle \left( {\forall x} \right)\left[ {u(x)\cdot u(x-3) \in \left\{ {0,1} \right\}} \right]\).

Please review both the text and the post for errors.
 
There is either a misprint in the textbook or you have left some factor out of the post.
If \(\displaystyle \left( {\forall x} \right)\left[ {u(x) \in \left\{ {0,1} \right\}} \right]\) then \(\displaystyle \left( {\forall x} \right)\left[ {u(x)\cdot u(x-3) \in \left\{ {0,1} \right\}} \right]\).

Please review both the text and the post for errors.
The problem as stated in the book:
"Evaluate the following discrete-time convolution sums:
(a) y[n] = u[n]*u[n-3]
(b) y[n] = (1/2)^n *u[n-2]*u[n]
....
Answers:
(a) y[n] = { 0 for n<3, n-2 for n>=3 }
(b) y[n] = { 0 for n<2, 1/2 - (1/2)^n for n>=2 }
..."
 
Yes, that is very different and I just finished working out part a and b using your link's explanation. Thanks Mr. Khan.

MS
 
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