-2
$\begingroup$

I compute the output of a LTI system, can someone tell me if my answer is right..? and help me with my others questions?

The impulse response is: $h(n) = \left(\frac{1}{2}\right)^nu(n)$ , entry is $x(n)=u(n)-u(n-1)$ in which $u(n)$ is unit sequence.

(1) We know that the outpout of this LTI system is $y(n)=x(n)*h(n)$

(2) If replace we take $y(n)=(u(n)-u(n-1))*h(n)=u(n)h(n)-u(n-1)h(n)$

(3) $u(n)*h(n)=h(n)$ and $u(n-1)*h(n)=h(n-1)$

As a result: $y(n)=h(n)-h(n-1) = \left(\frac{1}{2}\right)^nu(n) - \left(\frac{1}{2}\right)^{n-1}u(n-1)$

My Questions:

  1. First of all is this solution right?
  2. How we know that the equations (3) stand?
  3. Always in these systems in the entry is the unit sequence?
$\endgroup$
  • $\begingroup$ In writing $$y(t)=(u(t)-u(t-1))*h(t)=u(t)h(t)-u(t-1)h(t)$$ you are using $*$ to mean multiplication (as it does in most computer programming languages, Excel spreadsheets, etc.) whereas here $*$ is supposed to denote convolution. I always write $\star$ for convolution whenever LaTEx is available just to avoid confusion with $*$ as in multiplication and $\,^*$ as complex conjugation. $\endgroup$ – Dilip Sarwate Jan 25 '15 at 15:20
  • $\begingroup$ @DilipSarwate is the solution right? $\endgroup$ – Ewan Terry Jan 25 '15 at 15:40
  • $\begingroup$ @Matt L. Hi I saw that you know this field and you have answered similar questions. Could you please help me? $\endgroup$ – Ewan Terry Jan 25 '15 at 15:44
  • $\begingroup$ @EwanTerry: The problem is that your question is a mess. It is not clear if you talk about continuous time or discrete time (you use $t$ as a time variable, but you talk about 'unit sequence'); you confuse convolution with multiplication; you confuse the step function $u(t)$ or $u[n]$ with the impulse $\delta(t)$ or $\delta[n]$. If you manage to clear up these misunderstandings by editing your question appropriately then people here will be able and willing to help you. $\endgroup$ – Matt L. Jan 25 '15 at 16:11
  • $\begingroup$ @MattL. I talk about discrete time. I change it. I am sorry I am not Latex user $\endgroup$ – Ewan Terry Jan 25 '15 at 16:28
1
$\begingroup$

There are many misconceptions in your question and in your proposed solution. I won't solve the problem for you, but I'll explain the problems in your solution and I'll give you a hint how to find the correct answer.

First of all, I suppose that $u[n]$ is the unit step function, and consequently

$$(h*u)[n]\neq h[n]$$

The confusion about multiplication and convolution has already been pointed out by Dilip Sarwate in his comment.

Now for the correct way to solve this problem:

If $a[n]$ is the system's step response, i.e. $a[n]=(h*u)[n]$, then, due to linearity and time-invariance, we can write the response $y[n]$ to the given input $x[n]=u[n]-u[n-1]$ as

$$y[n]=a[n]-a[n-1]\tag{1}$$

So the only thing you need to know is the system's step response $a[n]$, which is obtained by convolving $h[n]$ with $u[n]$. If you write down the convolution sum then you should find that in this case

$$a[n]=u[n]\cdot \sum_{k=0}^{n}h[k]\tag{2}$$

With the given impulse response $h[n]$, evaluating (2) is simple because it is a geometric series. As soon as you have $a[n]$, the output is directly obtained from (1).

A much simpler route is to realize that the input $x[n]=u[n]-u[n-1]$ is equal to a unit impulse $x[n]=\delta[n]$. This means that the output $y[n]$ is simply given by the impulse response: $y[n]=h[n]$. Note that if you do things right, then this solution and the general solution given by (1) are identical.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.