I've been wrestling with this question for way too long. I appreciate any insight you can offer. Thanks.

If the $9$-point sample sequence $x(n)$ represents a time domain signal sampled at $48\textrm{ kHz}$, what frequency does the eighth term $X[7]$ in its DFT correspond to?


If you have a signal of 9 samples, namely $ {\left\{ {x}_{n} \right\}}_{n = 0}^{8} $ its DFT is given by $ {\left\{ {X}_{k} \right\}}_{k = 0}^{8} $

Now if the sampling rate is $ {F}_{s} $ the bin frequency interval is given by $ \frac{{F}_{s}}{N} $.

Now, you have 9 samples, 1 is the DC ($ k = 0 $) and the rest spans $ \left[ \frac{-{F}_{s}}{2}, \frac{{F}_{s}}{2} \right] $.

You can look on the first 5 which are $ \left\{ 0, \frac{{F}_{s}}{9}, \frac{2 {F}_{s}}{9}, \frac{3 {F}_{s}}{9}, \frac{4 {F}_{s}}{9} \right\} $.

Now, the 7 bin would have to be symmetric and hence would be $ \frac{-3 {F}_{s}}{9} $.

  • $\begingroup$ Dan, please mark the question as answered. Thank You. $\endgroup$
    – Royi
    Aug 9 '15 at 16:43

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