Frequency domain Sampling & Reconstruction of Analog Signals
We know that continuous-time finite energy signals have continuous spectra. Here we will
consider sampling of such signals periodically and reconstruction of signals from samples of
their spectra.
Consider an analog signal x(t) with a spectrum X(F). Then suppose we obtain samples of
X(F) every F hertz as shown below in Fig. 1.
X(F)
X(kF)
F
kF
F
Fig .1
Then the question is given these samples X(kF) where -∞<k<∞ can the signal x(t) or
spectrum X(F) be recovered?
Note: This problem is mathematically the dual to that of sampling a continuous time
signal in the time-domain.
If we sample the spectrum every F then
X kF  

 x(t )e
 j 2kFt
dt

If we define the reciprocal of F as below:
Ts 
1
F
above expression can be written as:
X kF  

 x(t )e
 j 2kt / Ts
dt

We can then subdivide the integration range of the integral into an infinite number of intervals
of width Ts and change the variable of integration to translate each integral into fundamental
 Ts
T
t  s .
range
2
2
X kF  
 
  j 2kt / Ts
dt
  x(t  nTs ) e


Ts / 2  n  
Ts / 2
(3)
The signal inside the brackets is periodic with fundamental period 1/F. Let us denote it as
xp(t).

x p t  
 xt  nT 
s
n  
Since this signal is periodic we can expand it into a Fourier Series representation.
x p t  

 ck e j 2kt / Ts 
k  

c e
k  
j 2kFt
k
hence
1
ck 
Ts
Ts
2
 x t e
p
 j 2kFt
(4)
dt
 Ts
2
Comparing equation (4) to (3)
1
ck  X (kF )
Ts
(5)
ck  FX kF 
We can conclude from equation (5) that :
The samples of the spectrum X(F) correspond to the Fourier coefficients of a periodic signal
1
xp(t) with period Ts 
.
F
xp(t)
Ts >2
t
-Ts
-

0
Ts
xp(t)
Ts <2
t
-Ts
0
Ts
If the analog signal x(t) ıs time limited to  seconds and Ts is selected such that Ts > 2 there
wıll be no aliasing and the signal spectrum X(F) can be perfectly reconstructed from the
samples X(kF) by using the interpolation formula:
X F  
sin  / F F  kF 

 X (kF )  / F F  kF 
k  
What we have just seen is important for conceptual purposes.
Frequency sampling of continuous time signals is encountered in practical frequency analysis
applications .
However in most applications the analog signal is converted to discrete time signal and the
frequency sampling is performed on the spectrum of the discrete time signal.
Frequency Domain Sampling & Reconstruction of Discrete Time Signals
An aperiodic finite energy signal has continuous spectra.
For an aperiodic signal x[n] the spectrum is :
X w 

 xne
 jwn
(1)
n  
Suppose we sample X[w] periodically in frequency at a sampling of w radians between
successive samples. We know that DTFT is periodic with 2, therefore only samples in the
fundamental frequency range will be necessary. For convenience we take N equidistant
2
samples in the interval 0<=w<2. The spacing between samples will be w 
as shown
N
below in Fig.
X[w]
w
0

2
Fig. 1
Let us first consider selection of N, or the number of samples in the frequency domain.
If we evaluate equation (1) at w 
2k
N

 2k 
X

xne  j 2kn / N


 N  n  
k  0,1,2,......., ( N  1)
We can divide the summation in (1) into infinite number of summations where each sum
contains N terms.
1
N 1
2 N 1
 2k 
 j 2kn / N
 j 2kn / N




X

.......

x
n
e

x
n
e

xne j 2kn / N




N


n N
n 0
n N

 lN  N 1
  xne
 j 2kn / N
l  n lN
If we then change the index in the summation from n to n-lN and interchange the order of
summations we get:
  j 2kn / N
 2k  N 1  
 N      xn  lN e
n  0 l  

for k  0,1,2,......, ( N  1)
(3)
denote the quantity inside the bracket as xp[n]. This is the signal that is a repeating version of
x[n] every N samples. Since it is a periodic signal it can be represented by the Fourier Series.
N 1
x p n   ck e j 2kn / N
n  0,1,2,........, ( N  1)
k 0
with FS coefficients:
ck 
1
N
N 1
 x ne
 j 2kn / N
p
n0
k  0,1,2,......., ( N  1)
(4)
Comparing the expressions in equations (4) and (3) we conclude the following:
ck 
1  2
X
N  N

k

k  0,1,......., ( N  1)
Therefore it is possible to write the expression xp[n] as below:
x p n 
1
N
N 1
 2  j 2kn / N
k e

 X  N
k 0
n  0,1,....., ( N  1)
(7)
The above formula shows the reconstruction of the periodic signal xp[n] from the samples of
the spectrum X[w]. But it does not say if X[w] or x[n] can be recovered from the samples.
Let us have a look at that :
Since xp[n] is the periodic extension of x[n] it is clear that x[n] can be recovered from xp[n] if
there is no aliasing in the time domain. That is if x[n] is time-limited to less than the period N
of xp[n].This is depicted in Fig. 3 below:
x[n]
n
0
L
xp[n]
N>=L
No aliasing
n
0
L
N
xp[n]
N<L
aliasing
n
0
N
Fig . 3
Hence we conclude :
The spectrum of an aperiodic discrete-time signal with finite duration L can be exactly
2k
recovered from its samples at frequencies wk 
if N >= L.
N
We compute xp[n] for n=0,1,.....,N-1 using equation (7)
Then X[w] can be computed using equation (1).