# Another Signals/Systems Questions

Discussion in 'Math' started by crazyengineer, Jul 8, 2011.

1. ### crazyengineer Thread Starter Member

Dec 29, 2010
156
2
Sorry, I have more questions when I was studying signals and systems

1) Why is the z transform of this
$$\frac{z}{(z-2)^{2}}$
$

equal

$$n2^{n-1}u[n]$

$

I'm confuse about where they got the n-1?

2) What's the importance of
$$h_{k}[n-k]$$ for linaer time invariant systems?

2. ### steveb Senior Member

Jul 3, 2008
2,433
469
For the first one ...

As a minor correction for the record, as you know, it's actually the inverse z-transform, not the z-transform.

The reason is just that that is what the calculation gives. You may be getting confused by the factor of "a" that's in front of the usual transform given in tables, as follows.

$n a^n \Leftrightarrow{{az}\over{(z-a)^2}}$

From this you can see that you need to multiply by 2 to use the table to get the inverse transform, and then you have to take your final answer and divide by 2, which gives you the 2^(n-1) rather than the 2^n.

3. ### steveb Senior Member

Jul 3, 2008
2,433
469
This one has me confused. I'm not sure what h_k is refering to. Usually "h" is referring to the impuse response of a system which is very relevant to LTI systems, since the response of any input signal can be calculated if the impulse response is known. However, I dont' know why the subscript k is used in your example. The k seems to be a time delay in the impulse response, but I don't know why the notation uses "k" for both subscript and time delay. Usually, it's either one or the other, and using both seems confusing to me.

Can you provide the context for the question. Where did you see this notation and what is it referring to?

4. ### crazyengineer Thread Starter Member

Dec 29, 2010
156
2
Sorry for the slow response!

1) Is the reason why it should be n-1 because when n=0, 2^(n-1) will be 1/2?

2) In my signal and systems book they call said
" let $$h_{k}[n]]$$ denote the response of the linear system to the shifted unit impulse $$\delta[n-k]$$"

Normally I would scan a page from the book, but my scanner is out

5. ### steveb Senior Member

Jul 3, 2008
2,433
469
For number 1, I don't think that is the real reason, although it is related to the real reason. The real reason is that the mathematical transform gives that answer as a result. I guess I should ask why you think it should be 2^n rather than 2^(n-1). I took a guess that the reason you thought it should be the former, rather than the latter, was because you used a transform table and saw the answer as 2^n. However, this answer is for a different function (factor of "a" in front, as I showed above).

For number 2, yes $h_{k}[n]$ denotes the response of the linear system to the shifted unit impulse $\delta[n-k]$. But, that's not what you originally wrote. You wrote $h_{k}[n-k]$ which is nonsensical in this context. A better way to say it is that $h_{k}[n]$ is the same as $h[n-k]$ where $h[n]$ is the impulse response function (i.e. the response of the linear system to the non-shifted unit impulse $\delta[n]$).

So, the answer to your original question for number 2 is as follows. Both $h[n]$ and $h_k[n]$ are important to LTI systems because they can be used to determine the output response of a system to most any input signal. This is done through the convolution integral, which I'm sure is explained in your text book. (Actually, I used a previous edition of your textbook back in college 1985. This book is outstanding!)

The basic idea is that if you know how a system responds to a non-shifted impulse function, then you can consider any input signal to be a time-series of added and scaled time-shifted impulse functions. The output then becomes a time series of added and scaled time-shifted impulse response functions $h_k[n]$ or $h[n-k]$.

Last edited: Jul 10, 2011
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6. ### steveb Senior Member

Jul 3, 2008
2,433
469
I think my above answers are perhaps to vague. Let's work it through step by step, starting with the usual transform found in a typical table.

$n a^n u[n] \Leftrightarrow{{az}\over{(z-a)^2}}$

Let's put in your value of 2 in place of "a", as follows.

$n 2^n u[n]\Leftrightarrow{{2z}\over{(z-2)^2}}$

Now, lets factor out the 2 so that it matches your example exactly. This is done by dividing both sides by 2. Even thought this is not an equation, this operation is allowed. If you don't see this, go back to the transform integral and you'll see that constants can go inside or outside the integral.

$({{1}\over{2}})n 2^n u[n] \Leftrightarrow{{z}\over{(z-2)^2}}$

Now let's merge the factor of 1/2=2^(-1) into the 2^n

$n 2^{(n-1)} u[n]\Leftrightarrow{{z}\over{(z-2)^2}}$

And, there you have you final answer.

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