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I am working through previous exam questions for a class I'm currently taking in signal theory, and one problem, although it looks easy, stumps me.

PROBLEM

Two real $N$-periodic time seires $f_n$ and $g_n$ are given, where $n = 0, 1, 2, ..., N-1$.

In addition, the complex time series $h_n$ is given as

$$h_n = f_n + i g_n$$

and the discrete Fourier transform

$$H_m = F_m + i G_m$$,

where $m = 0,1,. . ., N-1$.

Find the discrete Fourier transofrm $F_m$ and $G_m$ to the time series $f_n$ and $g_n$ expressed through $H_m$

ATTEMPTED SOLUTION

As I said, I'm really stumped here. Other than writing:

$$F_m = H_m - i G_m$$

and

$$G_m = i (F_m - H_m)$$

I don't see what else to do. Is this all there is to it? It seems way too easy, so I'm pretty sure I'm not doing this correctly.

If anyone can give me any help, I would greatly appreciate it!

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    $\begingroup$ hint: look at the symmetry properties of the FT for purely real and purely imaginary time signals. The problem youa re supposed to solve is to get G from H without knowing F (and vice versa). So your solutions are not good enough. $\endgroup$ – Hilmar May 8 '14 at 20:20
  • $\begingroup$ Yes, thank you very much. I believe I figured it out. I will post my own answer below. $\endgroup$ – Kristian May 9 '14 at 13:04
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Note that even though

$$H_m=F_m+iG_m$$

$F_m$ and $G_m$ are NOT the real and imaginary parts of $H_m$, respectively. In general they are both complex sequences. Only $f_n$ and $g_n$ are real-valued, so in

$$h_n=f_n+ig_n$$

$f_n$ and $g_n$ really denote the real and imaginary parts of $h_n$. Now you need to know that for a real-valued sequence $x_n$ of length $N$ its DFT satisfies

$$X_k=X^*_{N-k}\tag{1}$$

Similarly, the DFT of a purely imaginary sequence $x_n$ satisfies

$$X_k=-X^*_{N-k}\tag{2}$$

From (1) and (2) it follows that the DFT of the real part of a complex sequence $h_n$ corresponds to the even part of its DFT, i.e.

$$F_m=\frac{1}{2}[H_m+H^*_{N-m}]$$

and the DFT of its imaginary part corresponds to the odd part of $H_m$ (divided by $j$):

$$G_m=\frac{1}{2j}[H_m-H^*_{N-m}]$$

Note that both $F_m$ and $G_m$ satisfy (1) because both correspond to real-valued sequences $f_n$ and $g_n$.

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Proposed answer to my own question:

We have that:

$$F_m = \frac{1}{2}(H_m + H^*_m)$$

and

$$G_m = \frac{1}{2i}(H_m - H^*_m)$$

where $H^*_m$ represents the complex conjugate of $H_m$.

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  • $\begingroup$ I'm afraid this is not correct, please see my answer. $\endgroup$ – Matt L. May 9 '14 at 18:09

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