I am trying to design a discrete-time inverse system to eliminate an undesired echo in a data transmission problem.

The echo caused by the transmission channel is represented as attenuation by a factor of 0.9 and a delay corresponding to one time unit of the input sequence.

The received signal $y[n]$ can be expressed as follow;

$$y[n] = x[n] + 0.9x[n-1]$$

So basically, I'm trying to determine the unit impulse response of a causal inverse system to discover $x[n]$ from $y[n]$.

Conceptually, I'm thinking I need to connect a system $h_2[n]$ in series with the system $h_1[n]$ such that $h_1[n]*h_2[n] = \delta[n]$.

Can anyone offer some advice on how best to go about this ? I've been searching online but can't find a concrete example to use as a starting point.

  • 1
    $\begingroup$ this doesn't sound like nonlinear distortion, so I would remove the word "distortion" from it $\endgroup$
    – endolith
    Mar 29 '16 at 18:34
  • $\begingroup$ Your correct, it is linear. I've removed references to distortion. Thanks $\endgroup$
    – MarkMark
    Mar 29 '16 at 18:42
  • 2
    $\begingroup$ your inverse system is an IIR type with impulse response of $(-0.9)^n u[n]$ you can implement this system with a recursive computation where $y_i[n] = -0.9 y_i[n-1] + x[n]$ $\endgroup$
    – Fat32
    Mar 29 '16 at 21:02
  • 1
    $\begingroup$ Please do not write [SOLVED] in the title of any questions. If you have a solution, write it in the answer box and give it the check mark. That is the way it works on all StackExchange sites. $\endgroup$
    – Peter K.
    Mar 30 '16 at 11:37

Correct answer is in the comment already but for completion: Your transfer function is

$$H(z) = b_0 + b_1 \cdot z^{-1}, \mbox{ with } b_0 = 1, b_1 = .9$$ which makes the inverse $$H^{-1}(z) = \frac{1}{b_0 + b_1 \cdot z^{-1}} $$ which corresponds to the difference equation $$y(n) + 0.9 \cdot y(n-1) = x(n) $$ or $$y(n) = x(n) -0.9 \cdot y(n-1) $$


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.