DFT and DTFT Interpolation
Delve into the relationship between Discrete Fourier Transform (DFT) and Discrete-Time Fourier Transform (DTFT) through interpolation. Explore how to derive DTFT from DFT samples, the synthesis and analysis formulas, the importance of sequence length in relation to N-point DFT, sampling strategies, and more.
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DTFT from DFT samples by interpolation 1
N-point DFT (X[k]) of a given length-N sequence, x[n], is the frequency values of its DTFT,? ???,evaluated at ? uniformly spaced frequency points. ? = ??=2?? 0 ? (? 1) ? Given the N-point DFT, ie. X[k], its DTFT can be uniquely determined from X[k] ? 1 ? ???= ?[?]? ??? ?=0 ? 1 ? 1 1 ? ?=0 ?? ? ??? = ?[?]?? ?=0 ? 1 ? 1 =1 ? ? ? 2?? ? ? ? ?=0 ?[?] ?=0 S 2
?1 1 ? ? ?? 2?? =1 ? ?=0 ?[?] 2??? 1 ? ? ? sin?? 2?? ? 1 =1 ? ? ? 2?? ? 1 2 2 ? ?=0 ? ? ? sin?? 2?? 2? Interpolation function for getting ? ??? from X[k] 3
DTFT Sampling The DTFT is ? ???= ? ? = ?= ?[?]? ??? ? 1? ? ? ?2? ??? The DFT analysis formula is: ? ? = ?=0 If x[n] is a L-point signal, i.e. it is non-zero only in the 0 ? ? 1 range the DTFT can be written as: ? 1 ?[?]? ??? ? ? = ?=0 If we compare these two formulas we see that: for the N-point DFT values to be samples of the DTFT the length L of the sequence x[n] must be less than N (i.e. L <= N) X k = X w |?=2?? 4 ?
If we are given X(w) and we wish to find x[n] we can use the synthesis formula: ? 1 ?(?)?????? ? ? = 2? ? However performing the integral may be inconvenient. The following is an easier approach: 1) Sample DTFT X(w) to get DFT values X[k] for k= 0, , (N-1) 2) Take inverse DFT of X[k] to get back x[n] Does this always work ? No. For this to work L <=N. 5
Example: x[n] 1 n 0 2 DTFT 3 1 ? ???=1 ? ?4? ??? 2? ??? ?/2 1 ? ??= ? ?3? ? ???= 2 ?=0 DTF 3 1 ? ?2? 4??= 1 + ? ?? 2?+ ? ???+ ? ?3? 2? ? ? = ?=0 = 4 k=0 0 k = 1,2,3 6
Sampling DTFT ? ? = ? ??2? = ? ?3?? ? = 0 ? = 1,2,3 ??? ?? ??? ??/4= 4 4? 0 4 7
Sampling in frequency: Sample ? ??? at ??=2?? ? 0 ? (? 1) ? ??? ??? = ? ???2? ? ? ??? ? ??? ? - ? ??? ??? = ? ? ? ? = x[n] ? ? ?? 0 2? 4? ? ?= ? = ? ? ?? ?= 8
??? = ? ? ? Original signal x[n] is replicated in time at integer multiples of M x[n] n xs[n] n -M 0 M 9
Sampling in frequency => replication in time Sampling in time => replication in frequency ???? ? ??? and ??? = ? ??? If ? ? ? = 0, , ? 1 ?=2?? ? ???? ??? where ??? = ?= then ??? ? ? ?? To recover x[n] from ??? ,x[n] must be time limited to <= M values Otherwise we have aliasing. 10
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 below: x[n] n 0 L xp[n] N>=L No aliasing n 0 L N xp[n] N<L aliasing n 0 N 11
Example : ? ? = ??? ? where ? < 1 1 X w = 1 ?? ?? Here the time domain signal is npot time-limited (i.e. L = ) Now suppose we take N samples of this DTFT i.e ? ?? = ? ? ?=2? 2? ?? 12
for a = 0.7 ? ? = ??? ? 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 -5 0 0 5 10 15 20 25 30 1.4 1.2 1 0.8 0.6 N=6 samples 0.4 0.2 -10 0 -5 0 5 10 15 20 x[n] san not be recovered from replicated time-domain signal since L > N Note for n >20 x[n] starts to vanish so if N=20 or more is chosen then x[n] can be recovered from the time-domain signal 13
