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The problem:

Consider a discrete-time LTI system. If the output signal is: $$ y[n]=5 \left( \frac{1}{5} \right) ^n u[n] -2^{-n} u[n] $$ , then the input signal will be: $$ x[n]=\left( \frac{4}{5} \right) ^n u[n] $$ Find the transfer function and the unit impulse response.

We first find the z-transforms: $$ Y(z)= 5\left( \frac{5z}{5z-1} \right) -\frac{2z}{2z-1} $$ $$ \Rightarrow Y(z)= \frac{z(40z-23) }{(5z-1)(2z-1)} $$ $$ X(z)= \frac{\frac{5z}{4}}{\frac{5z}{4}-1}= \frac{5z}{5z-4} $$ $$ H(z)=\frac{Y(z)}{X(z)}= \frac{(5z-4)(40z-23)}{5(5z-1)(2z-1)}=4-\frac{15}{5z-1}+\frac{3}{5(2z-1)} $$

Now, WolframAlpha says that: $$ 4-\frac{15}{5z-1}+\frac{3}{5(2z-1)}=\frac{92}{5}+ \sum_{n=0}^{\infty} \left( \frac{3\cdot 2^{-n}}{5}- 3\cdot 5^{1-n} \right) z^{-n} $$

If you see any mistakes, please do let me know. Now, I need to find the unit impulse response $h[n] $. The inverse $z-$transform of the infinite sum is trivial (I think), but I do not know how can I get rid of $\frac{92}{5}$ .

Is there any other way to find $h[n] $ ?

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    $\begingroup$ Why not simply invert $H(z)$ directly? You know how to invert 1 and $\frac{1}{z-u}$ I guess? $\endgroup$ – Paul Jul 13 '19 at 8:31
  • $\begingroup$ @Paul , thanks for the comment. I was using a table that does not include $\frac{1}{z-u}$ . I found another table that is more comprehensive: control.dii.unisi.it/sdc/altro/TabellaTrasformataZ.pdf . However, the first answer below, from LutzL, has been the most insightful part for me regarding this question. $\endgroup$ – evaristegd Jul 13 '19 at 19:50
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As the system is LTI, the response to $$ \tilde x[n]=\left(\frac45\right)^nu[n-1]=\frac45\left(\frac45\right)^{n-1}u[n-1] $$ is $$ \tilde y[n]=\frac45\left(5\left(\frac15\right)^{n-1}u[n-1]-2^{1-n}u[n-1]\right) $$ so that the answer to the unit pulse $x-\tilde x$ is $$ h[n]=y[n]-\tilde y[n]=5\left(\frac15\right)^n(u[n]-4u[n-1])-2^{-n}\left(u[n]-\frac85u[n-1]\right) $$ Now use that $u[n]=u[n-1]+\delta[n]$ to simplify to $$ h[n]=\underbrace{\left(20-\frac85\right)}_{=\frac{92}5}δ[n]- 15\left(\frac15\right)^n u[n]+\frac35\cdot 2^{-n} u[n] $$ which is exactly what you got, so it is highly probable that also your intermediate steps were correct.

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