The discrete-time system of figure below a step response: $$s[n] = \left[ 5 - 4 \left( \frac{4}{5} \right)^n \right] u[n]$$

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Find the impulse response of the system and verify it from the system difference equation. sr i want ans of this question.with explanation

  • $\begingroup$ Difference of gaussians? Where are the gaussians? $\endgroup$ Oct 1 '17 at 16:33

Considering an LTI system, its impulse response $h[n]$ and the step response $s[n]$ are related by $$h[n] = s[n]-s[n-1]$$ and $$s[n] = u[n] \star h[n] = \sum_{k=-\infty}^{n} h[k] = \sum_{k=0}^{\infty} h[n-k] $$

Given your definition of $s[n]$ therefore its $h[n]$ is $$h[n] = \left[5 - 4 \left( \frac{4}{5} \right)^n \right] u[n] -\left[5 - 4 \left( \frac{4}{5} \right)^{n-1} \right] u[n-1]$$

After simplifications it would become: $$h[n] = 5\delta[n] + 5 \left( \frac{4}{5} \right)^{n} u[n-1] - 4 \left( \frac{4}{5} \right)^{n} u[n]$$ further simplifies to

$$ \boxed{ h[n] = 5 \left( \frac{4}{5} \right)^{n} u[n] - 4 \left( \frac{4}{5} \right)^{n} u[n] = \left( \frac{4}{5} \right)^{n} u[n] }$$

Looking at the signal flow graph and using Z-transform variables you may achieve the same result by considering:

$$Y(z) = \frac{4}{5} Y(z) z^{-1} + X(z)$$ $$Y(z)( 1 - \frac{4}{5} z^{-1}) = X(z)$$ $$ H(z) = \frac{Y(z)}{X(z)} = \frac{1}{1 - \frac{4}{5} z^{-1}} $$

From which the inverse Z-transform (with the causal ROC) yields: $$ \boxed{ h[n] = \left( \frac{4}{5} \right)^n u[n] }$$ which is the same result obtained from the time-domain approach.

  • $\begingroup$ Hi. I understood the first solution but how do you get the first equation for $Y(z)$ at the beginning of the second solution ? I know this material is in books and I have many but still I find new examples tricky. Thanks. Oh, for other novices to EE, another great, albeit extremely intro book that is great is one by Steiglitz.titled "Introduction to Discrete Systems". $\endgroup$
    – mark leeds
    Oct 2 '17 at 5:01
  • $\begingroup$ @markleeds Hi. Have a look at Feedback system analysis. Any fundamental introduction would suffice. $\endgroup$
    – Fat32
    Oct 2 '17 at 9:04
  • 1
    $\begingroup$ Found some useful things so it makes sense. thanks. $\endgroup$
    – mark leeds
    Oct 2 '17 at 17:36

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