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DFT (Frequency Domain Sampling)
The Fourier series describes periodic signals by discrete spectra, where as the DTFT describes discrete signals by periodic spectra. These results are a consequence of the fact that sampling on domain induces periodic extension in the other. As a result, signals that are both discrete and periodic in one domain are also periodic and discrete in the other. This is the basis for the formulation of the DFT.
Consider aperiodic discrete time signal x (n) with FT X(w) =
Since X (w) is periodic with period 2π , sample X(w) periodically with N equidistance samples with spacing
K = 0, 1, 2…..N1
The summation can be subdivided into an infinite no. of summations, where each sum
contains
Put n = nlN
We know that xp(n) = n= 0 to N1
k=0 to N1
Therefore k=0 to N1
DFT  x_{p} (n) = n = 0 to N1
This provides the reconstruction of periodic signal x_{p}(n) from the samples of spectrum
X(w).
The spectrum of aperiodic discrete time signal with finite duration L<N, can be exactly
recovered from its samples at frequency Wk=
Prove: x(n) = xp (n) 0 ≤ n ≤ N1
Using IDFT
If we define p(w) =
Therefore: X (w) =
At w =2πk/N P (0) =1
And P (w 2πk/N)= 0 for all other values
Ex: x(n) = a^{n} u(n) 0<a<1
The spectrum of this signal is sampled at frequency W_{k=} k=0, 1…..N1, determine reconstructed spectra for a = 0.8 and N = 5 & 50.
X (w) =
X (wk) = k=0, 1, 2… N1
0≤n≤N1
Aliasing effects are negligible for N=50
If we define aliased finite duration sequence x(n)
xˆ(n)= x_{p}(n) 0≤n≤N1
= 0 otherwise
∴Although XÆ (w) ≠ X (w), the samples at Wk= are identical.
Ex: &
Apply IDFT
using Taylor series expansion
= 0 except r = n+mN
The result is not equal to x (n), although it approaches x (m) as N becomes ∞ .
Ex: x (n) = {0, 1, 2, 3} find X (k) =?
= 2+2j
X (2) = 2
X (3) = 22j
DFT as a linear transformation
k = 0 to N1
n = 0, 1…N1
Let x_{N} = X_{N} =
The N point DFT may be expressed in matrix form as
Ex: x (n) = {0, 1, 2, 3}
IDFT
Q.
Find y (n) = x (n) h (n) using frequency domain. Since y (n) is periodic with period 2.
Find 2point DFT of each sequence.
X (0) = 1.5 H (0) = 1.5
X (1) = 0.5 H (1) = 0.5
Y (K) = X (K) H (K)
Y (0) = 2.25 Y (1) = 0.25
Using IDFT y (0) = 1; y (1) = 1.25
= 1 * 0.5 + 0.5 * 1 = 1
= 1 * 1 + 0.5 * 0.5 = 1.25
1 * 0.5 + 0.5 * 1 = 1
~y (n) ={1, 1.25, 1, 1.25…..}
Q. Find Linear Convolution of same problem using DFT
Sol. The linear convolution will produce a 3sample sequence. To avoid time aliasing we convert the 2sample input sequence into 3 sample sequence by padding with zero.
For 3 point DFT
X (0) = 1.5 H (0) = 1.5
X (1) = 1+0.5 H (1) = 0.5+
X (2) = 1+0.5 e H (2) = 0.5+ e
Y (K) = H (K) X (K)
Y (0) = 2.25
Y (1) = 0.5 + 1.25 + 0.5 e
Y (2) = 0.5 + 1.25+ 0.5 e
Compute IDFT
y(0) = 0.5
y(1) =1.25
y(2) =0.5
y(n) = { 0.5, 1.25, 0.5} Ans
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