# Reconstruction of time domain data using half of the frequency domain

Started by July 2, 2014
```Hello all,

I'm an RF engineer trying to deal with some data which was taken by
sampling in the frequency domain and I don't have any expedience dealing
with this type of data&hellip;  I've run into a bit of a rut.

I have a set of radar data sampled in the frequency domain (an I and Q pair
is collected at specified range-delay positions).  This was done in order
to overcome several hardware limitations of getting high resolution data
via time domain sampling.  However, I'm faced with the following dilemma:
I only have I and Q data for positive frequencies in 2GHz of bandwidth in
the K band, and I need the full spectrum (including negative frequencies)
in order to get an accurate result from the inverse Fourier transform.
these samples are all I have to play with.

Since my time domain signal should be all real, I've assumed that the
spectrum should be conjugate symmetric ( F(-w) = F*(w)) and replicated,
flipped, and scaled the I and Q arrays appropriately.  This seems to work,
as I get the time(range) domain results that I was anticipating.  However,
'Seems to work' and 'is mathematically valid' are two very different
things.  Is what I'm doing mathematically valid?  I know that sampling in
the frequency domain is used in other applications; does anyone know of any
'standard' ways of overcoming only having half of the full complex
spectrum?

Thanks!

_____________________________
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```
```Hi,

what you're doing sounds correct. A real-valued signal has a conjugate
(Hermitian) spectrum.

The frequency domain coefficients scale "exp(i omega t)" terms.
For a real-valued signal it is known that "exp(i omega t)" and "exp(-i
omega t") come in pairs, according to the exponential definition of cosine
and sine. It's easy to show if I state that the imaginary part of the
time-domain function is zero, there is no other way,.because terms with
differing omega are independent of each other.

Where you need to pay attention is 0 Hz(at least theoretically - the last
time I checked, DC-radar wasn't invented yet)

_____________________________
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```
```On Wed, 02 Jul 2014 19:14:48 -0500, "Shan9492" <100758@dsprelated>
wrote:

>Hello all,
>
[Snipped by Lyons]
>
>Since my time domain signal should be all real, I've assumed that the
>spectrum should be conjugate symmetric ( F(-w) = F*(w)) and replicated,
>flipped, and scaled the I and Q arrays appropriately.  This seems to work,
>as I get the time(range) domain results that I was anticipating.  However,
>'Seems to work' and 'is mathematically valid' are two very different
>things.  Is what I'm doing mathematically valid?  I know that sampling in
>the frequency domain is used in other applications; does anyone know of any
>'standard' ways of overcoming only having half of the full complex
>spectrum?
>
>
>Thanks!

Hi,
If your freq-domain samples are obtained by
performing a discrete Fourier transform (DFT),
which includes the fast Fourier transform (FFT),
on real-valued time samples, then the freq-domain
symmetry you cited will always be true.

For example, if you're performing an 8-point DFT
on eight x(n) time samples, your eight
complex-valued freq-domain samples will be
(the '*' symbol means conjugate):

X(0) [real-only]
X(1) = X*(7)
X(2) = X*(6)
X(3) = X*(5)
X(4) [real only]
X(5) = X*(3)
X(6) = X*(2)
X(7) = X*(1)

So the two real values, X(0) & X(4), plus the
three complex values, X(1), X(2), & X(3),
contain *ALL* the freq-domain information.

Let's look at a simple 8-point DFT of the
x(n) samples of x(n) = 1,2,0,0,0,0,0,0,
and show that X*(7) = X(1).

Writing the standard DFT summation equation
for X(m),'m' is the freq-domain index
(0 -to- 7).

Because there are only two non-zero x(n)
samples, the DFT summation becomes:

X(m) = x(0)*exp(-j2p0m/8)
+ x(1)*exp(-j2p1m/8)

= 1 + 2exp(-j2pm/8).     (1)

('p' = pi.) The 1st term in Eq. (1) is the
n = 0 term, and the 2nd term in Eq. (1) is
the n = 1 term.

So let's show that X(1) = X*(7).  From
Eq. (1) with m=1, X(1) is:

X(1) = 1 + 2exp(-j2p/8)
= 1 + cos(-2p/8) -jsin(-2p/8)

And from Eq. (1) with m=7, X(7) is:

X(7) = 1 + 2exp(-j2p7/8)
= 1 + cos(-2p7/8) -jsin(-2p7/8)

Because cos(-2p7/8) = cos(-2p/8), we can
rewrite X(7) as:

X(7) = 1 + cos(-2p/8) -jsin(-2p7/8)

The conjugate of X(7) is:

X*(7) = 1 + cos(-2p/8) + jsin(-2p7/8).

Because sin(-2p7/8) = -sin(-2p/8), we can
rewrite X*(7) as:

X*(7) = 1 + cos(-2p/8) -jsin(-2p/8)

= X(1).

Which is what we set out to show.

[-Rick-]

```
```Thank you both very much!!

>On Wed, 02 Jul 2014 19:14:48 -0500, "Shan9492" <100758@dsprelated>
>wrote:
>
>>Hello all,
>>
>   [Snipped by Lyons]
>>
>>Since my time domain signal should be all real, I've assumed that the
>>spectrum should be conjugate symmetric ( F(-w) = F*(w)) and replicated,
>>flipped, and scaled the I and Q arrays appropriately.  This seems to
work,
>>as I get the time(range) domain results that I was anticipating.
However,
>>'Seems to work' and 'is mathematically valid' are two very different
>>things.  Is what I'm doing mathematically valid?  I know that sampling
in
>>the frequency domain is used in other applications; does anyone know of
any
>>'standard' ways of overcoming only having half of the full complex
>>spectrum?
>>
>>
>>Thanks!
>
>Hi,
>  If your freq-domain samples are obtained by
>performing a discrete Fourier transform (DFT),
>which includes the fast Fourier transform (FFT),
>on real-valued time samples, then the freq-domain
>symmetry you cited will always be true.
>
>For example, if you're performing an 8-point DFT
>on eight x(n) time samples, your eight
>complex-valued freq-domain samples will be
>(the '*' symbol means conjugate):
>
>  X(0) [real-only]
>  X(1) = X*(7)
>  X(2) = X*(6)
>  X(3) = X*(5)
>  X(4) [real only]
>  X(5) = X*(3)
>  X(6) = X*(2)
>  X(7) = X*(1)
>
>So the two real values, X(0) & X(4), plus the
>three complex values, X(1), X(2), & X(3),
>contain *ALL* the freq-domain information.
>
>Let's look at a simple 8-point DFT of the
>x(n) samples of x(n) = 1,2,0,0,0,0,0,0,
>and show that X*(7) = X(1).
>
>Writing the standard DFT summation equation
>for X(m),'m' is the freq-domain index
>(0 -to- 7).
>
>Because there are only two non-zero x(n)
>samples, the DFT summation becomes:
>
>  X(m) = x(0)*exp(-j2p0m/8)
>             + x(1)*exp(-j2p1m/8)
>
>       = 1 + 2exp(-j2pm/8).     (1)
>
>('p' = pi.) The 1st term in Eq. (1) is the
>n = 0 term, and the 2nd term in Eq. (1) is
>the n = 1 term.
>
>So let's show that X(1) = X*(7).  From
>Eq. (1) with m=1, X(1) is:
>
>  X(1) = 1 + 2exp(-j2p/8)
>     = 1 + cos(-2p/8) -jsin(-2p/8)
>
>And from Eq. (1) with m=7, X(7) is:
>
>
>  X(7) = 1 + 2exp(-j2p7/8)
>    = 1 + cos(-2p7/8) -jsin(-2p7/8)
>
>Because cos(-2p7/8) = cos(-2p/8), we can
>rewrite X(7) as:
>
>  X(7) = 1 + cos(-2p/8) -jsin(-2p7/8)
>
>The conjugate of X(7) is:
>
>  X*(7) = 1 + cos(-2p/8) + jsin(-2p7/8).
>
>Because sin(-2p7/8) = -sin(-2p/8), we can
>rewrite X*(7) as:
>
>  X*(7) = 1 + cos(-2p/8) -jsin(-2p/8)
>
>        = X(1).
>
>Which is what we set out to show.
>
>[-Rick-]
>
>

_____________________________
Posted through www.DSPRelated.com
```