I've been trying to solve a Discrete LTI System problem where the overall impulse response of the following System, whose input and output sequences are, respectively, $x(n)$ and $y(n)$, has to be found:

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This is what I've tried:

Calling the output sequence that comes out of the adder $e(n)$, I know that $y(n) = e(n) \star h(n)$. I also know that $e(n) = x(n) - y(n)$, which leads to $y(n) = x(n) \star h(n) - y(n) \star h(n)$. Thats the point where I got stuck. I don't know how to proceed in order to isolate $y(n)$ in the last expression. I would like to isolate it in order to make $x(n) = \delta(n)$ and evaluate the overall Impulse Response.

Also, I am not supposed to use any transform techniques as the book I am following has not yet arrived at those subjects. If possible, don't show the final calculations, only what I should do in order to separate $y(n)$. I would like to figure the rest out by myself.

Thanks in advance.


2 Answers 2


Take the z-transform of your equation to get $y(z) = x(z) h(z) - y(z) h(z)$ which you can simplify to $$ y(z) \left( 1 + h(z) \right) = x(z) h(z) $$ and therefore get the transfer function $$ H(z) := \frac{y(z)}{x(z)} = \frac{h(z)}{1+h(z)} $$

Now plug in the z-transform $h(z)$ of your response $h(n)$ and perform an inverse z-transform to get the total impulse response $H(n)$.

An approach that is more basic and does not require z-transforms works for your particular example by solving the recursion equation explicitly. The inner system with the response $h(n) = u(n) \left( \frac{1}{2}\right)^n$ can be written with the recursive difference equation $y'(n) = x'(n) + \frac{1}{2} y'(n-1)$ where $x'$ and $y'$ are the output and input of the subsystem. Then we can write the full system in terms of the following equations:

$$ y(n) = y'(n)$$ $$ x'(n) = x(n) - y(n)$$ $$ y'(n) = x'(n) + \frac{1}{2} y'(n-1) $$

Eliminating $x'$ and $y'$ results in

$$ y(n) = x(n) - y(n) + \frac{1}{2} y(n-1) $$

which simplifies to


This system has the impulse response $$H(n) = \frac{1}{2} u(n) \left(\frac{1}{4}\right)^n$$

  • $\begingroup$ I believe that he wrote about usage of transforms - it is not possible. $\endgroup$
    – jojeck
    Commented Mar 16, 2014 at 0:13
  • $\begingroup$ I should not use a Transform for solving this problem, but thanks anyway! $\endgroup$
    – Thiago
    Commented Mar 16, 2014 at 1:48


Make a list of the value of $u[n]$ as a function of $n$, e.g.

$$\begin{array}{l|c|c|c|c|c|c|l} \hline n \to &-2 &-1 &0 &1 &2 &\cdots \\ \hline u[n] &0 &0 &1 &1 &1 & \cdots\\ \hline \end{array}$$

Now add a new row marked $h[n]$ whose values I will leave you to fill in. Then read this answer to figure out what $y[n]$ is.

  • $\begingroup$ Sorry, but I didn't get it. I added a row for $h(n)$, which is a "causal" geometric series of ratio 0.5 but I don't know how to continue. $\endgroup$
    – Thiago
    Commented Mar 16, 2014 at 2:10

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