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?
  • $\begingroup$ 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. $\endgroup$ Commented Jan 25, 2015 at 15:20
  • $\begingroup$ @DilipSarwate is the solution right? $\endgroup$
    – Ewan Terry
    Commented Jan 25, 2015 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
    Commented Jan 25, 2015 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.
    Commented Jan 25, 2015 at 16:11
  • $\begingroup$ @MattL. I talk about discrete time. I change it. I am sorry I am not Latex user $\endgroup$
    – Ewan Terry
    Commented Jan 25, 2015 at 16:28

1 Answer 1


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


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.


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