Discrete-Time Fourier Transform (DTFT)

3.1 Discrete-Time Fourier Transform (DTFT)

3.1.1 Definition and Convergence Conditions of DTFT Representation

As the discrete-time counterpart of the CTFT (continuous-time Fourier transform)

X ( j ω) = F{x(t)} = x (t ) e − jωt dt

(2.2.1a)

we define the Fourier transform of a discrete-time sequence x[n] as

( j Ω) = X[e ∞ ] = F{x [n]} = x [n] e − j Ωn (3.1.1)

X jΩ

n=−∞

which is called the DTFT (discrete-time Fourier transform). From now on, we will more often use X (Ω) than X ( j Ω) or X [e jΩ ] for simplicity. Let us see the physical meaning of the DTFT.

Remark 3.1 Physical Meaning of the DTFT – Signal Spectrum and Frequency Response

(1) If a sequence x[n] represents a physical signal, its DTFT X (Ω) = F{x[n]} means the signal spectrum, which describes the frequency contents of the signal.

(2) In particular, if a time function g[n] represents the impulse response of a discrete-time LTI (linear time-invariant) system, its DTFT G(Ω) = F{g[n]}

means the frequency response, which describes how the system responds to a

3.1 Discrete-Time Fourier Transform (DTFT) 131 sinusoidal input sequence of digital (angular) frequency Ω (refer to Sect. 1.2.6

for the definition of frequency response). One striking difference of the DTFT from the CTFT is its periodicity (with period

2π ) in the (digital) frequency variable Ω, which results from the fact that it is a function of e jΩ periodic with period 2π in Ω, i.e., e j (Ω+2πn)

=e jΩ . Based on the periodicity of the DTFT, we are going to use the CTFS (for periodic functions) to

derive the IDTFT (inverse discrete-time Fourier transform) formula. To this end, we can use Eq. (2.1.5a) with P = 2π, ω 0 = 2π/P = 1, t = Ω, and k = −n to write

the (continuous-frequency) Fourier series representation of X (Ω) as

where the Fourier coefficients are x (2.1.5b)

X (Ω) e j Ωn dΩ (the integral over one period of length 2π )

P=2π, ω 0 =1,t=Ω,k=−n 2π

(3.1.2b) Noting that Eq. (3.1.2a) is the same as Eq. (3.1.1) multiplied by a scaling factor

1/2π , we can multiply Eq. (3.1.2b) by the same scaling factor 1/2π to write the IDTFT formula as

x [n] = F −1

X (Ω) e {X(Ω)} = j Ωn dΩ (3.1.3)

We call Eqs. (3.1.1) and (3.1.3) the DTFT pair where Eq. (3.1.1) is the analysis equation and Eq. (3.1.3) is the synthesis equation.

Like the convergence conditions (2.2.2a) and (2.2.2b) for the CTFT, it can be stated that the DTFT will exist if the sequence x[n] has finite energy, i.e.,

|x[n] | 2 < ∞

(3.1.4a)

n=−∞

or if it is absolutely summable, i.e.,

(3.1.4b) Remark 3.2 Frequency Response Existence Condition and Stability Condition of a

n=−∞ |x[n] | < ∞

System

Note that, for the impulse response g[n] of a discrete-time LTI system, the absolute-summability condition (3.1.4b) is identical with the stability condition (1.2.27b). This implies that a stable LTI system has a well-defined frequency response G(Ω) = F{g[n]}.

132 3 Discrete-Time Fourier Analysis

3.1.2 Examples of DTFT Analysis

Example 3.1 DTFT of a Rectangular Pulse Sequence For the rectangular pulse of duration 2M+1 from −M to M shown in Fig. 3.1(a1)

and described by

(E3.1.1) we can apply Eq. (3.1.1) to get its DTFT as

r ′ 2M+1 [n] = u s [n + M] − u s [n − M − 1]

R 2M+1 (Ω) =

r ′ n=−∞ 2M+1 [n] e − j Ωn

(E3.1.1)

e − j Ωn (D.23) j ΩM 1−e − j Ω(2M+1) =e

(Dirichlet kernel)

n=−M

1−e −jΩ

j ΩM e − j Ω(2M+1)/2 (e (2M+1)/2 −e − j Ω(2M+1)/2 ) =e

jΩ

e − j Ω/2 (e j Ω/ 2 −e − j Ω/2 )

(D.22) sin(Ω(2M + 1)/2) =

(E3.1.2) sin(Ω/2)

whose magnitude and phase are depicted in Fig. 3.1 (b1) and (c1), respectively. Likewise, for the rectangular pulse of duration 2M + 1 from 0 to 2M shown in

Fig. 3.1(a2) and described by r 2M+1 [n] = u s [n] − u s [n − 2M − 1] = r ′ 2M+1 [n − M]

(E3.1.3)

1 r 2M + 1 ' [n] r 2M + 1 [n ] 1 with M = 1 =r 2M + 1 ' [n–M ]

with M = 1

0 12 n (a1) r 2M + 1 ' [n]

–1 0 1 n

(a2) r 2M + 1 [n ]=r 2M + 1 ' [n – M ]

0 π 2 π (b1) R 2M + 1 ' (Ω)

(b2) R 2M + 1 (Ω) π

– π – π (c1) ∠R 2M + 1 ' (Ω)

(c2) ∠R 2M + 1 (Ω) Fig. 3.1 Two rectangular pulse sequences and their magnitude & phase spectra

3.1 Discrete-Time Fourier Transform (DTFT) 133 we can apply Eq. (3.1.1) to get its DTFT as

R 2M+1 (Ω) =

[n] e − j Ωn

n=−∞ 2M+1 (E3.1.3)

(D.23) 1−e − j Ω(2M+1)

=R ′ 2M+1 (Ω)e − j ΩM (E3.1.2) sin(Ω(2M + 1)/2)

sin(Ω(2M + 1)/2)

sin(Ω/2) ∠ − MΩ

sin(Ω/2)

(E3.1.4) whose magnitude and phase are depicted in Figs. 3.1 (b2) and (c2), respectively.

Note the following: – In contrast with the continuous-time case (Fig. 2.8(a)), the DTFT of a rectan-

gular pulse sequence is periodic, being no longer a pure sinc function, but an aliased sinc function. Still the magnitude spectra show that the rectangular pulses have low frequency components (around Ω = 0) more than high frequency ones around Ω = ±π.

– The spectrum (E3.1.2) of an even-symmetric pulse sequence is real-valued and

its phase is 0 or ±π depending on whether it is positive or negative. Especially for the frequency range such that R ′ 2M+1 (Ω) < 0, we set its phase to +π or −π

so that the overall phase curve is odd-symmetric to comply with the conjugate symmetry (3.2.4) of DTFT (see Fig. 3.1(c1)).

– Due to the fact that the phase of R 2M+1 (Ω) (Eq. (E3.1.4)) is proportional to the

(digital) frequency Ω as −MΩ, the (shifted) rectangular pulse sequence is said to have a linear phase despite the phase jumps (discontinuities) caused by the

change of sign or the wrapping (modulo 2π ) operation of angle (see the piecewise linear phase curve in Fig. 3.1(c2)).

Example 3.2 DTFT of an Exponential Sequence For an exponential sequence e 1 [n] = a n u s [n] with |a| < 1, we can apply Eq. (3.1.1) to get its DTFT as

E 1 (Ω) =

a u s [n]e − j Ωn =

(ae −jΩ )

n (D.23)

= 1−ae −jΩ

n=−∞

n=0

1 1−ae jΩ (D.20) 1 − a cos Ω − j a sin Ω =

2 (E3.2.1) 1−ae

1 − 2a cos Ω + a whose magnitude and phase are

1 −a sin Ω |E 1 (Ω)| = +

2 2 and ∠E 1 (Ω) = tan −1 (1 − a cos Ω) + (a sin Ω)

1 − a cos Ω (E3.2.2)

134 3 Discrete-Time Fourier Analysis : a = 0.5 : a = –0.5

: a = 0.8 : a = –0.8

(a1) (a2)

E 1 (Ω) a = –0.8 a = 0.8 E 4 1 (Ω) 2 2

a = 0.5 a = –0.5

(b2) a = 0.8 1 ∠E 1 (Ω)

a = –0.8 1 ∠E (Ω)

(c1) (c2) Fig. 3.2 Four exponential sequences and their magnitude & phase spectra

Fig. 3.2(a1–a2), (b1–b2), and (c2–c2) show the exponential sequences e 1 [n] =

a n u s [n] with a = ±0.5 and ±0.8 and their magnitude and phase spectra. Note the following:

- From Fig. 3.2(a1), we see that e 1 [n] with a = 0.8 is smoother than that with

a = 0.5. This is reflected in Fig. 3.2(b1) showing that the magnitude spectrum |E 1 [Ω]| with a = 0.8 is larger/smaller than that with a = 0.5 around Ω = 0 (low frequency)/ Ω = ±π (high frequency). Also from Fig. 3.2(a2), we see that e 1 [n]

with a = −0.8 changes more rapidly than that with a = −0.5. This is reflected

in Fig. 3.2(b2) showing that the magnitude spectrum |E 1 [Ω]| with a = −0.8 is

larger/smaller than that with a = −0.5 around Ω = ±π (high frequency)/Ω = 0 (low frequency).

- Comparing Fig. 3.2(a1) and (a2), we see that e 1 [n] with a < 0 changes much

more rapidly than that with a > 0. This is reflected in that the magnitude

spectrum |E 1 [Ω]| with a < 0 (Fig. 3.2(b2)) is large around Ω = ±π (high

frequency) and that with a > 0 (Fig. 3.2(b1)) is large around Ω = 0 (low frequency).

Example 3.3 DTFT of a Symmetric Exponential Sequence For the exponential sequence e 2 [n] = a |n| with |a| < 1, we can apply Eq. (3.1.1) to get its DTFT as

3.1 Discrete-Time Fourier Transform (DTFT) 135 :a = 0.5 :a = –0.5

:a = 0.8 :a = –0.8

n –1 0 1 0

(a1) (a2)

| E 2 (Ω) | | E 2 (Ω) |

a = 0.8 a = –0.8

a = 0.5 a = –0.5

(b2) Fig. 3.3 Four exponential sequences and their magnitude spectra

a −n e −jΩn

n=−∞

n=− ∞

+ n a e −jΩn = (ae ) + (ae −jΩ )

jΩ n

n=0 (D.23)

2 : real-valued 1−ae

jΩ +

1−ae −jΩ

1 − 2 a cos Ω + a (E3.3.1)

whose magnitudes with a = ±0.5 and ±0.8 are depicted in Fig. 3.3. Example 3.4 DTFT of the Unit Sample (Impulse) Sequence

For the impulse (unit sample) sequence δ[n], we can apply Eq. (3.1.1) to get its DTFT as

n=0 =1 As with the continuous-time case (Eq. (E2.6.1) in Example 2.6), this implies that a

D ∞ (Ω) =

δ [n] e − j Ωn

= δ[n] e

discrete-time impulse signal has a flat or white spectrum, which is evenly distributed over all digital frequencies.

(cf.) Very interestingly, applying the IDTFT formula (3.1.3) to (E3.4.1) yields a useful expression of the unit sample (impulse) sequence.

(D.33) e jπn −e − jπn (D.22) sin nπ =

(E3.4.1) 1 π

= (E3.4.2) 2π −π

j Ωn

1e dΩ =

2π j n

nπ

136 3 Discrete-Time Fourier Analysis

g [n]

G(Ω) B= 2

2 π – 2 π –B – 2 π + B – B 0 B – 2 π –B – 2 π + B Ω

B B B B (a) The impulse response of an ideal LPF

(b) The frequency response of an ideal LPF Fig. 3.4 The impulse response and frequency response of an ideal LPF

Example 3.5 IDTFT of an Ideal Lowpass Filter Frequency Response Let us consider the frequency response of an ideal lowpass filter (LPF) depicted in Fig. 3.4(b):

G (Ω) = for |Ω − 2mπ| ≤ B ≤ π (m : an integer) (E3.5.1)

0 elsewhere

We can take the IDTFT (inverse discrete-time Fourier transform) (3.1.3) to get the impulse response of the LPF, which is shown in Fig. 3.4(a), as follows:

(e [n] = F jBn { G(Ω) } = = −jBn )

(D.33) 2π j n sin(Bn)

−B

(E3.5.2) π n = sinc π π

3.1.3 DTFT of Periodic Sequences

In analogy with the arguments in Sect. 2.3 for deriving Eq. (2.3.2) as the generalized CTFT of continuous-time periodic functions, we begin with applying the IDTFT formula (3.1.3) to an impulse train type spectrum

i =−∞ 2π δ(Ω − Ω 0 − 2π i) to get a complex sinusoidal sequence as x (3.1.3) [n] = F −1 1 j Ωn

δ (1.1.25) )e j Ωn dΩ jΩ 0 = n (Ω − Ω

0 =e

This implies the following DTFT relation:

e jΩ 0 n F ↔ X(Ω) = ∞

i =−∞ 2π δ(Ω − Ω 0 − 2π i)

3.1 Discrete-Time Fourier Transform (DTFT) 137 Example 3.6 DTFT of a Constant Sequence

Let us consider a constant sequence c[n] = 1. Noting that this is a kind of periodic function obtained by substituting Ω 0 = 0 into the LHS of DTFT relation

(3.1.5), we can substitute Ω 0 = 0 into the RHS of that DTFT relation to obtain the DTFT of c[n] = 1 as

F c ∞ [n] = 1 ↔ C (Ω) =

2π δ(Ω − 2π i) (E3.6.1)

(3.1.5) with Ω 0 =0

i =−∞

Example 3.7 DTFT of a Sine/Cosine Sequence (a) For sin(Ω (D.22)

0 n ) jΩ 0 = (e n − jΩ 0 −e n )/j 2, we use Eq. (3.1.5) to get its DTFT as

sin(Ω 0 n ) ↔ jπ (δ(Ω + Ω i =−∞ 0 − 2π i) − δ(Ω − Ω 0 − 2π i)) (E3.7.1) (b) For cos(Ω (D.21)

0 n ) = (e jΩ 0 n

− jΩ +e n 0 )/2, we use Eq. (3.1.5) to get its DTFT as

i =−∞ (δ(Ω + Ω − 2πi) + δ(Ω − Ω − 2πi)) Example 3.8 DTFT of the Unit Step Sequence

F cos(Ω ∞

0 n ) ↔π

0 0 (E3.7.2)

Similarly to Example 2.8 for deriving the CTFT of the unit step function, we first decompose the unit step sequence u s [n] into the sum of an even sequence and an odd sequence as

(E3.8.1) where

u s [n] = u e [n] + u o [n]

u e [n] = (u s

2 [n] + u [−n]) = 2 (1 + δ [n] )

(E3.8.2)

u o [n] = (u s [n] − u s sign(n) (E3.8.3)

2 [−n]) = 2

Then we can take the DTFT of the even and odd parts as

U e (Ω) = F{ u e [n] } =

2 F{ δ [n] } + F 2

(E3.4.1),(E3.6.1) 1 ∞

2 F{u s [n]} − F{u s [−n]})

j Ωn n=1

= 2 n=1 −e (D.23) 1 e −jΩ

(E3.8.5) 1−e

2 −jΩ −

(D.20)

1−e

jΩ

2(1 − cos Ω)

138 3 Discrete-Time Fourier Analysis Now we add these two results to obtain the Fourier transform of the unit step

sequence as

(E3.8.6) 1−e

δ (Ω − 2π i)

i =−∞