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My discrete-time signals textbook (Oppenheim) jumps through the derivation of steady-state and transient LTI system response to a complex exponential input- $$y[n]=y_{SS}[n]+y_t[n]$$ It says, given input \$x[n]=e^{j\omega n}u[n]\$ and impulse response \$h[n]\$, the system response can be found using convolution as: $$ y[n] = \left\{\begin{aligned} &0 &&: n < 0\\ &e^{j\omega n}\sum_{k=0}^{n}h[k]e^{-j\omega k} &&: n\ge 0 \end{aligned} \right.$$

But where did the \$n\$ as the upper sum limit come from? When I try to do this myself, I end up with: $$ y[n]=x[n]\ast h[n]=\sum_{k=-\infty}^{\infty}h[k]x[n-k]\\ =\sum_{k=-\infty}^{\infty}h[k]e^{j\omega(n-k)}u[n-k]\\ =\sum_{k=0}^{\infty}h[k]e^{j\omega n}e^{-j\omega k}\\ =e^{j\omega n}\sum_{k=0}^{\infty}h[k]e^{-j\omega k} $$

Where did I go wrong?

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    \$\begingroup\$ 1. What is u[n-k] when k > n? 2. Can you assume h[n] is causal? \$\endgroup\$ – The Photon Feb 10 '17 at 23:30
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    \$\begingroup\$ 1. Ah, whoops, all terms in sum with k > n are 0, and 2. yes h[n] is causal. It makes sense now conceptually, just weird that I can't get the math to end up that way too \$\endgroup\$ – Ben Granger Feb 10 '17 at 23:38
  • \$\begingroup\$ switched convolution order and worked it out. Thanks! \$\endgroup\$ – Ben Granger Feb 10 '17 at 23:48
  • \$\begingroup\$ Post the solution as an answer \$\endgroup\$ – Voltage Spike Feb 11 '17 at 8:12
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Given: the system is LTI and causal. $$\Rightarrow y[n]=h[n]\ast x[n]=\sum_{k=-\infty}^{\infty}h[k]e^{j\omega (n-k)}u[n-k]\\~\\ \because \textrm{ causality implies }h[k]=0\ \forall k<0\\ \because u[n-k]=0\ \forall k>n\\~\\ \therefore \sum_{k=-\infty}^{\infty}h[k]e^{j\omega (n-k)}u[n-k]=e^{j\omega n}\sum_{k=0}^{n}h[k]e^{-j\omega k} $$ Combining that with the fact that the output is causal- $$ y[n] = \left\{\begin{aligned} &0 &&: n < 0\\ &e^{j\omega n}\sum_{k=0}^{n}h[k]e^{-j\omega k} &&: n\ge 0 \end{aligned} \right. $$

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