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…..N-1
The summation can be subdivided into an infinite no. of summations, where each sum
contains
Put n = n-lN
We know that xp(n) = n= 0 to N-1
k=0 to N-1
Therefore k=0 to N-1
DFT ------------ xp (n) = n = 0 to N-1
This provides the reconstruction of periodic signal xp(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 ≤ N-1
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) = an u(n) 0<a<1
The spectrum of this signal is sampled at frequency Wk= k=0, 1…..N-1, determine reconstructed spectra for a = 0.8 and N = 5 & 50.
X (w) =
X (wk) = k=0, 1, 2… N-1
0≤n≤N-1
Aliasing effects are negligible for N=50
If we define aliased finite duration sequence x(n)
xˆ(n)= xp(n) 0≤n≤N-1
= 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) = -2-2j
DFT as a linear transformation
k = 0 to N-1
n = 0, 1…N-1
Let xN = XN =
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 2-point 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 3-sample sequence. To avoid time aliasing we convert the 2-sample 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
1. What is the purpose of frequency domain sampling in digital signal processing? |
2. How does frequency domain sampling work? |
3. What is the relationship between time domain and frequency domain in digital signal processing? |
4. How does frequency domain sampling affect signal analysis and processing? |
5. What are the advantages of frequency domain sampling over time domain sampling? |
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