Relationship Among CTFS, CTFT, DTFT, and DFT

3.5 Relationship Among CTFS, CTFT, DTFT, and DFT

3.5.1 Relationship Between CTFS and DFT/DFS

To investigate the relationship between CTFS and DFT/DFS, suppose we have a continuous-time periodic signal ˜x P (t ) with period P and its discrete-time version

˜x N [n] = ˜x P (nT ), which is obtained by sampling ˜x P (t ) at the rate of N times per period P. Since ˜x P (t ) is a continuous-time periodic signal with period P, we can use Eq. (2.1.5a) to write its CTFS representation as

1 ∞ ˜x P (t ) =

X k e j 2π kt/N T P

0 =2π/P=2π/N T NT

k=−∞

(3.5.1) j Substituting t = nT into this equation and using the fact that e 2π kn/N is unique

only for n mod N yields

X k e j 2π kn/N

(since e j 2π kn/N is unique only for n mod N )

X e = j 2π (k+m N )n/N NT

X e j = 2π kn/N (3.5.2) N

We can match this equation with the IDFS/IDFT formula (3.4.8)/(3.4.3) for ˜x N [n] = ˜x P (nT )

to write the relation between the CTFS coefficients X k of a periodic signal ˜x P (t ) (with period P = N T ) and the N -point DFT/DFS coefficients ˜X N (k) of ˜x N [n] =

˜x P (nT ) as

(3.5.3) This implies that the DFT/DFS of ˜x N [n] = ˜x P (nT ) is qualitatively the periodic

extension of the CTFS of ˜x P (t ) (with period N in frequency index k), i.e., the sum of infinitely many shifted version of CTFS. This explains how the DFT/DFS

3.5 Relationship Among CTFS, CTFT, DTFT, and DFT 161 strays from the CTFS because of frequency-aliasing unless the CTFS spectrum

X k is strictly limited within the low-frequency band of (−(N/2)ω 0 , (N /2)ω 0 )= (−π/T, π/T ) where the DFT size N equals the number of samples per period P

and the fundamental frequency is ω 0 = 2π/P = 2π/N T .

3.5.2 Relationship Between CTFT and DTFT

To investigate the relationship between CTFT and DTFT, suppose we have a continuous-time signal x(t ) and its discrete-time version x[n] = x(nT ). As a bridge between x(t ) and x[n], let us consider the sampled version of x(t ) with sampling interval or period T as

(3.5.4) Noting that x ∗ (t ) is still a continuous-time signal, we can use Eq. (2.2.1a) to write

x ∞ ∗ (t ) = x(t)δ T (t ) (δ T (t ) =

δ n=−∞ (t − nT ) : the impulse train)

its CTFT as

x (nT )e − jωnT

n=−∞

−∞ n=−∞ ∞

= (3.1.1) x [n] e − jΩn

=X d (Ω) | Ω =ωT (3.5.5) This implies that X d (Ω) = DTFT{x[n]} and X ∗ (ω) = CTFT{x ∗ (t )} are essentially

n=−∞

Ω =ωT

the same and that X d (Ω) can be obtained from X ∗ (ω) via a variable substitution ω = Ω / T . On the other hand, we recall from Eq. (E2.13.3) that X ∗ (ω) = CTFT{x ∗ (t )} is expressed in terms of X (ω) = CTFT{x(t)} as

(ω + mω s ) with ω s = T

X (3.5.6)

m=−∞

Combining these two equations (3.5.5) and (3.5.6), we can write the relation between the CTFT and the DTFT as

X d (Ω) =X ∗ (ω) | ω =Ω/T =

where ω and Ω are called the analog and digital frequency, respectively. This implies that the DTFT of x[n] = x(nT ) is qualitatively the periodic

extension of the CTFT of x(t ) (with period 2π/ T in analog frequency ω or 2π in digital frequency Ω), i.e., the sum of infinitely many shifted version of CTFT. This explains how the DTFT strays from the CTFT because of frequency-aliasing

162 3 Discrete-Time Fourier Analysis unless the CTFT spectrum X (ω) is strictly limited within the low-frequency band

of (−π/T, π/T ) where T is the sampling interval of x[n] = x(nT ). Fig. 3.7(b1) illustrates the deviation of the DTFT spectrum from the CTFT spectrum caused by

frequency-aliasing.

3.5.3 Relationship Among CTFS, CTFT, DTFT, and DFT/DFS

As stated in Remark 2.7(2) and illustrated in Fig. 2.8, the CTFS X k ’s of a periodic function ˜x P (t ) are the samples of the CTFT X (ω) of the one-period function x P (t )

at kω 0 = 2πk/P:

(3.5.8) Likewise, as discussed in Sect. 3.4 and illustrated in Fig. 3.11, the DFT/DFS X (k)’s

X (2.2.4)

= X(ω)| ω =kω 0 =2πk/P

of a periodic sequence ˜x P [n] are the samples of the DTFT X d (Ω) of the one-period sequence x P [n] at kΩ 0 = 2πk/N :

d (Ω) | Ω =kΩ 0 =2πk/N (3.5.9) Figure 3.13 shows the overall relationship among the CTFS, CTFT, DTFT, and

X (3.4.1)&(3.4.2) (k) = X

DFT/DFS based on Eqs. (3.5.3), (3.5.7), (3.5.8), and (3.5.9). Figure 3.14 shows the CTFT, DTFT, DFT/DFS, and CTFS spectra for a continuous-time/discrete- time rectangular pulse or wave, presenting us with a more specific view of their relationship. Some observations are summarized in the following remark:

Remark 3.7 Relationship among the CTFS, CTFT, DTFT, and DTFS (DFT/DFS) Figures 3.13 and 3.14 shows the overall relationship among the CTFS, CTFT, DTFT, and DTFS (DFT/DFS) from a bird’s-eye point of view. The following observations and comparisons are made.

(1) Among the four Fourier spectra, the CTFS and CTFT are more desired than the DTFS and DTFT since all physical signals are continuous-time signals.

Time-domain periodic extension with period P (2.2.1a) ∞

Frequency-domain

X(ω) =

– ∞ x(t) e ω t dt

∫ p 2π ∼ = x(t )e – j 2πkt / P dt

at ω = k ω 0 =k P CTFS (3.5.6) 1 ∞

CTFT

∞ Time-domain X d (Ω) =

∞X sampling Frequency-domain

=– ∞X(ω + m )⎮ T ω = Ω/T

Frequency-domain at t = nT periodic extension

at t = nT

DTFT Frequency-domain sampling

periodic extension

DTFS (DFT/DFS)

X (Ω) = (3.1.1) Σ ∞

–j Ωn 2π

(3.4.2) X (K) =

Σ N –1 x d ∼ n =– ∞ x [n]e –j 2πkn / N at Ω = k Ω 0 =k n = 0 [n]e

Time-domain periodic extension with period N

Fig. 3.13 Overall relationship among the CTFT, CTFS, DTFT, and DTFS (DFT/DFS)

3.5 Relationship Among CTFS, CTFT, DTFT, and DFT 163

(a0) A pulse signal x(t ) –4 π

–2 π – π 0 π 2 π 4 π (b0) The CTFT spectrum X( ω ) of x(t)

(a1) x [n] obtained by sampling x (t ) with T = 1 (b1) The DTFT spectrum X d (Ω) of x[n]

Period N N = 8 X d (Ω) X(k)

0 8 16 (a2) x 8 [n] – a periodic extension of x [n] with period N = 8 (b2) N = 8-point DFS/DFT spectrum X(k) of x ∼ 8 (n)

Period 0 p

0 8 16 (a3) x 8 (t) – a periodic extension of x (t ) with period p = 8 (b3) The CTFS spectrum X k of x ∼ 8 (t )

(a4) x (n) obtained by sampling x(t ) withT = 0.5 (b4) The DTFT spectrum X d (Ω) of x [n]

N Period = 16 X

d (Ω) X(k)

0 8 16 k (a5) x ∼ 16 (n) – a periodic extension of x [n] with period N = 16 (b5) N = 16- point DFS/DFT spectrum X(k) of x ∼ 16 [n]

Fig. 3.14 Examples of CTFT, DTFT, DFS/DFT, and CTFS spectra

Between the CTFS and CTFT, we prefer to have the CTFT because it has all the information contained in the CTFS on the assumption that the CTFS consists of the samples of CTFT (Eq. (3.5.8) and Fig. 3.14(b3)). Besides, the CTFS is not so practical because it is hard to find the period or even periodicity of periodic signals due to a noise. Therefore, we think of the CTFT as a standard when we need to compare the spectra in terms of how faithfully they describe the frequency contents of a given signal.

(2) The problem with the CTFS and CTFT is that they are difficult to compute due to the integration. Compared with them, the DTFT X d (Ω) is easier to deal

with since it has only multiplications and additions. However, the sampling of x(t ) (with sampling interval T ) to make x[n] = x(nT ) produces the peri-

odic extension of the CTFT spectrum X (ω) with period 2π/ T in ω, causing frequency -aliasing in the case of non-zero frequency components outside the principal analog frequency band [−π/T, π/T ]. This is the cost we pay in return for the computational convenience of the DTFT. This frequency-aliasing

164 3 Discrete-Time Fourier Analysis can be reduced by decreasing the sampling interval T so that more frequency

components can be contained in [−π/T, π/T ]. (Compare the DTFT spectra in Fig. 3.14(b1) (for T = 1) and (b4) (for T = 0.5) with the CTFT plotted in dotted lines.) Refer to the sampling theorem to be discussed in Sect. 5.3, which presents a criterion for selecting the sampling interval.

(cf.) To compare the DTFT X d (Ω) with the CTFT X (ω), we should divide X (ω) by the sampling interval T (refer to Eq. (3.5.7)).

(3) The DTFT X d (Ω) of x[n] is computationally advantageous over the CTFS or CTFT, but is still not so handy since it is continuous in the frequency Ω and thus requires an integration for IDTFT (inverse DTFT). That is why

we sample the DTFT in the frequency domain at kΩ 0 = 2πk/N for k =

0 : N − 1 to make an N -point DFT X(k) for more computational efficiency. However, it also costs us the (illusory) periodic extension of x[n] with period

N (the DFT size) irrespective of whether x[n] is originally periodic or not and no matter what the real period is even if x[n] is really periodic. This causes time-aliasing if the original signal is not sufficiently covered within the whole time interval [0, N − 1] (Example 3.15) and spectral leakage prob- lem when the DFT size does not conform to the real period of the signal (Example 3.16).

(4) The analog resolution frequency ω 0 =Ω 0 / T = 2π/N T = 2π/P can be improved by increasing the whole time interval P = N T . Increasing the DFT

size N (, say, by zero-padding) helps the DFT spectrum to become close to the DTFT spectrum. Decreasing the sampling interval T increases the period 2π/ T of periodic extension of the CTFT spectrum (Eq. (3.5.7)) or equivalently, expands the principal analog frequency band [−π/T, π/T ] so that the chance and degree of frequency aliasing can be reduced.

(5) Generally, we can choose the DFT size N and sampling interval T and thus, eventually P = N T (the product of N and T ) during which a signal is to be

measured, sampled, and collected as a set of N data points. Therefore, it is hard to imagine that N T happens to be identical with the real period of the signal. For this reason, it will be reasonable to call P = N T the whole time interval rather than the period that was originated from the definition of the CTFS.