(i) $ h[n]=\left(\dfrac{1}{3}\right)^nu[n]+\left(-\dfrac{1}{3}\right)^nu[-n-1]$

here's how i tried to solve :

since we know, step response = running sum of impulse response i.e, $$s[n]=\sum_{k=-\infty}^{n} h[k] $$ we can write

$$ s[n]=\sum_{k=-\infty}^{n}\left(\dfrac{1}{3}\right)^k\left[u[k]+(-1)^ku[-k-1]\right]$$ $$s[n]=\sum_{k=-\infty}^{-1}\left(-\dfrac{1}{3}\right)^k +\sum_{k=0}^{n}\left(\dfrac{1}{3}\right)^k$$ $$s[n]=(-\infty .......+3^4-3^3+3^2-3)+\dfrac{3}{2}.\left(1-\dfrac{1}{3}^{n+1}\right)$$ $$s[n]=\left(-3+3^2-3^3+3^4......-\infty\right)+\left[\dfrac{3}{2}-\dfrac{1}{2}\left(\dfrac{1}{3}\right)^n\right]$$and since sum within parethesis is diverging therefore, $$s[n]=\infty$$ but, answer in the book is given as $$s[n]=\left(2^{n+1}-1\right)u[n]$$

above s[n] denotes step response

please tell me where i'm doing wrong(or if i'm doing conceptual mistake) ,and provide solution if possible ....thank you.

  • $\begingroup$ The given step response does not match the impulse response. Are you sure that the answer belongs to that problem? If so, then the book is wrong. What book is it? $\endgroup$
    – Matt L.
    Commented Jan 21, 2018 at 9:54
  • $\begingroup$ yes answer belongs to the same problem .......the book is indian authored P.Ramesh Babu and R.Anandnatrajan page number 3.65 .... $\endgroup$
    – user33321
    Commented Jan 21, 2018 at 9:58
  • $\begingroup$ so,what should be the right answer $\endgroup$
    – user33321
    Commented Jan 21, 2018 at 9:58
  • $\begingroup$ no it is given with positive power..there is a flaw in the book....i do agree with you that power must be negative for the second term (in that case G.P will give finite sum of -1/4) $\endgroup$
    – user33321
    Commented Jan 21, 2018 at 10:40

1 Answer 1


The given impulse response is unbounded for $n\rightarrow -\infty$, and, consequently, the corresponding step response is also unbounded for all $n$, as you've shown. The step response given in the answer corresponds to a totally different impulse response:


which is also unbounded for $n\rightarrow\infty$, but since the corresponding system is causal, the step response assumes finite values for finite $n$.


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