EEL3135: Discrete-Time Signals and Systems The Discrete ... · EEL3135: Discrete-Time Signals and Systems The Discrete-Time Fourier Transform (DTFT) - 3 - Hence, the DTFT of any discrete-time

EEL3135: Discrete-Time Signals and Systems The Discrete-Time Fourier Transform (DTFT)

The Discrete-Time Fourier Transform (DTFT)

1. IntroductionIn these notes, we introduce the discrete-time Fourier transform (DTFT) and explore some of its properties.

2. Discrete-Time Fourier Transform (DTFT)

A. Introduction

Previously, we have defined the continuous-time Fourier transform (CTFT) as,

(1)

where is a continuous-time signal, is the CTFT of the continuous-time signal, denotes the timevariable (in seconds), and denotes the frequency variable (in Hertz). The discrete-time Fourier transform(DTFT) is defined by,

(2)

where is a discrete-time signal, is the DTFT of the discrete-time signal, denotes the timeindex, and denotes the frequency variable.

Comparing the definitions for the CTFT and the DTFT, we observe the following differences. For the DTFT,the integration over time has been replaced by a summation over the time index ; also, the frequency vari-able is unitless. In order to relate the frequency variable to a real frequency, we need to know the sam-pling frequency . That is, if the discrete-time sequence represents a sampled continuous-time signal,sampled at frequency , then and the corresponding real frequency are related as follows:

, . (3)

B. Analytic example

We begin our study of the DTFT by looking at the DTFT of a simple discrete-time signal, namely a pulse ofwidth centered at :

(4)

While in general it is much more difficult to derive an analytic expression for the DTFT than for the CTFT,for the example signal in (4) it is possible to derive a closed-form expression. Applying definition (2),

(5)

(6)

(7)

X f( ) x t( )e j2πft– td∞–

∞∫=

x t( ) X f( ) tf

X ejθ( ) x n[ ]e jnθ–

n ∞–=

∞

∑=

x n[ ] X ejθ( ) nθ

t nθ θ

fs x n[ ]fs θ f

fθfs2π-------= θ 2πf

fs--------=

2M 1+ n 0=

x n[ ]10

=n M≤n M>

X ejθ( ) e jnθ–

n M–=

M

∑=

X ejθ( ) e j n M–( )θ–

n 0=

2M

∑=

X ejθ( ) ejMθ e jnθ–

n 0=

2M

∑=

- 1 -


(8)

From (7) to (8) above, we used the following finite geometric sum identity:

, .1 (9)

where, for our case,

and . (10)

Equation (8) can be further simplified by symmetrizing the powers of the exponentials in the numerator anddenominator:

(11)

(12)

Note that,

(13)

so that equation (12) reduces to:

(14)

(15)

In Figure 1 below, we plot for a pulse with . Note that this DTFT appears to be periodicin the frequency variable with period . This is not only the case for this example, but for the DTFTof any discrete-time signal . The periodic property of the DTFT is easily shown by going back to thedefinition in equation (2):

(16)

Note that for integer ,

(17)

so that:

. (18)

1. Strictly speaking, equation (9) is not valid for , , since for those values of . For these values of , . Using L’Hopital’s Rule, this is, however, precisely the limiting value of the final expression for in equation (15). Therefore, we do not treat this case separately.

X ejθ( ) ejMθ 1 e j 2M 1+( )θ––1 e jθ––

------------------------------------- =

αn

n 0=

N

∑ 1 α N 1+( )–1 α–

---------------------------= α 1≠

θ 2nπ±= n 0 1 2 …, , ,{ }∈ α 1=θ θ X ejθ( ) 2M 1+=

X ejθ( )

α e jθ–= N 2M=

X ejθ( ) ejMθ e j 2M 1+( )θ 2⁄– ej 2M 1+( )θ 2⁄ e j 2M 1+( )θ 2⁄––( )e jθ 2⁄– ejθ 2⁄ e jθ 2⁄––( )

---------------------------------------------------------------------------------------------------------------=

X ejθ( ) ejMθ e j 2M 1+( )θ 2⁄–

e jθ 2⁄–----------------------------------

ej 2M 1+( )θ 2⁄ e j 2M 1+( )θ 2⁄––ejθ 2⁄ e jθ 2⁄––

----------------------------------------------------------------------- =

ejMθ e j 2M 1+( )θ 2⁄–

e jθ 2⁄–----------------------------------

1=

X ejθ( ) ej 2M 1+( )θ 2⁄ e j 2M 1+( )θ 2⁄––ejθ 2⁄ e jθ 2⁄––

----------------------------------------------------------------------- ej 2M 1+( )θ 2⁄ e j 2M 1+( )θ 2⁄––2j

----------------------------------------------------------------------- 2j

ejθ 2⁄ e jθ 2⁄––----------------------------------

= =

X ejθ( )

θ2--- 2M 1+( )

sin

θ2---

sin---------------------------------------=

X ejθ( ) x n[ ] M 4=θ T 2π=

x n[ ]

X ej θ 2π+( )( ) x n[ ]e jn θ 2π+( )–

n ∞–=

∞

∑ e j2πn– x n[ ]e jnθ–

n ∞–=

∞

∑ e j2πn– X ejθ( )= = =

n

e j2πn– 1=

X ej θ 2π+( )( ) X ejθ( )=

- 2 -


Hence, the DTFT of any discrete-time signal is periodic in wit period . This result is closelyrelated to our previous discussion on sampling and aliasing as we shall see shortly.

Now, let us see how the DTFT in equation (15) changes as a function of the pulse width parameter . In Fig-ure 2, we plot and for . Note the strong similarity of Figure 2 below, and Fig-ure 5 on page 7 of the Fourier Series to Fourier Transform notes. For the narrowest pulse ( ,

), the frequency content is uniformly distributed for , while for the wider pulses, thefrequency content becomes more and more concentrated around . This is the same phenomenon as forthe CTFT (Figure 5, Fourier Series to Fourier Transform notes).

3. Some additional DTFT illustrations

Below, we explore additional examples of the DTFT.

A. Finite-length cosine wave

Here we consider the DTFT of a discrete-time signal , sampled from the continuous-time signal ,

(19)

with sampling frequency for a total of 50 samples, such that,

. (20)

Applying definition (2),

(21)

In Figure 3, we plot and for the discrete-time sequence in (20) as a function of the dimen-sionless frequency variable , and as a function of the real frequency variable (in Hertz), where we make

-6 -4 -2 0 2 4 6-2

0

2

4

6

8

-10 -5 0 5 10-2

0

2

4

6

8

-20 -10 0 10 200

0.5

1

1.5

2

-3 -2 -1 0 1 2 3-2

0

2

4

6

8M 4=

X ejθ( ) X ejθ( )

X ejθ( )

θ

n θ

θ

Figure 1

x n[ ]

θ T 2π=

Mx n[ ] X ejθ( ) M 0 1 2 4 8, , , ,=

M 0=x n[ ] δ n[ ]= π– θ π≤ ≤

θ 0=

x n[ ] xc t( )

xc t( ) 2πt( )cos=

fs 10Hz=

x n[ ]xc n fs⁄( )

0

=n 0 1 … 48 49, , , ,{ }∈elsewhere

X ejθ( ) xc n fs⁄( )e jnθ–

n 0=

49

∑=

X ejθ( ) X ejθ( )∠θ f

- 3 -


-20 -10 0 10 200

0.5

1

1.5

2

-3 -2 -1 0 1 2 3

0

5

10

15

-20 -10 0 10 200

0.5

1

1.5

2

-3 -2 -1 0 1 2 3-2

0

2

4

6

8

-20 -10 0 10 200

0.5

1

1.5

2

-3 -2 -1 0 1 2 3

-1

0

1

2

3

4

5

-20 -10 0 10 200

0.5

1

1.5

2

-3 -2 -1 0 1 2 3-1

0

1

2

3

-20 -10 0 10 200

0.5

1

1.5

2

-3 -2 -1 0 1 2 30

0.5

1

1.5

2M 0=

n

x n[ ]

M 1=

n

x n[ ]

M 2=

n

x n[ ]

M 8=

n

x n[ ]

X ejθ( )

θ

X ejθ( )

θ

X ejθ( )

θ

X ejθ( )

θ

Figure 2

X ejθ( )

θ

x n[ ]

n

M 4=

- 4 -


the substitution in equation (3). That is, the bottom two plots in Figure 3 correspond to the two expressionsbelow:

and (22)

These plots are more meaningful to us, since they clearly shows two spikes in the DTFT centered at .Now, let us make a few observations. First, note that when plotted as a function of , one period of the DTFTcovers the frequency range ( ). Since the DTFT is periodic, thissame frequency spectrum will be repeated outside this frequency range. Recall from our discussion on sam-pling, this is exactly what we said would occur —namely, that the frequency spectrum of the original contin-uous-time waveform will be repeated at offsets of , . In Figure 4, for example, we plot

for , corresponding to .

At this point, the reader might object to the above discussion on the grounds that we have previously said thatthe continuous-time spectrum of the cosine wave in equation (19) is given by,

(23)

-4 -2 0 2 40

5

10

15

20

25

-4 -2 0 2 4-3

-2

-1

0

1

2

3

-3 -2 -1 0 1 2 30

5

10

15

20

25

-3 -2 -1 0 1 2 3-3

-2

-1

0

1

2

3

X ejθ( )∠ θ 2πf fs⁄=

X ejθ( )∠

f

θ

Figure 3

0 10 20 30 40 50-1

-0.5

0

0.5

1x n[ ]

X ejθ( )

n

X ejθ( ) θ 2πf fs⁄=

θ

f

X ejθ( ) θ 2πf fs⁄= X ejθ( )∠ θ 2πf fs⁄=

1Hz±f

fs– 2⁄ fs 2⁄,[ ] 5Hz– 5Hz,[ ]= fs 10Hz=

kfs± k 1 2 3 …, , ,{ }∈X ejθ( ) θ 2πf fs⁄= 25Hz– f 25Hz< < 5π– θ 5π< <

xc t( )

Xc f( ) 12---δ f 1+( ) 1

2---δ f 1–( )+=

- 5 -


That is, the CTFT has two distinct spikes at , and is zero everywhere else. The CTFT in (23) is, how-ever, correct only for a cosine wave that is not time-limited. Note that by restricting our sampling of thecosine wave to 5 periods of the 1Hz cosine wave (50 samples), we implicitly assumed that the discrete-timesignal is zero everywhere else. That is, the DTFTs in Figures 3 and 4 represent the discrete-time fre-quency spectrum of a time-limited cosine wave. For comparison, we can compute the CTFT for the followingtime-limited continuous-time function,

(24)

by applying definition (1):

(25)

Using the inverse Euler relations for the cosine function:

(26)

(27)

(28)

(29)

(30)

Figure 5 below plots the time-limited, continuous-time signal in (24) and its CTFT magnitude spectrum. Note that except for a scaling difference, the CTFT and the DTFT appear very similar over the fre-

quency range . Thus, the fact that the DTFT spectral representation of the sampled, time-lim-ited cosine wave is not entirely localized at (as might have been expected) is not a consequence of theDTFT itself, but rather a consequence of the finite-length sampling process. We refer to this phenomenon,namely, the spreading of frequency content from the idealized peaks to the entire frequency range due tofinite-length sampling, as spectral leakage.

The longer the sequence length, the more concentrated the frequency content of that discrete-time signal willbe in the neighborhood of its dominant frequencies. Below, we consider the DTFT for the discrete-time signal

,

-20 -10 0 10 200

5

10

15

20

25

Figure 4

n


f

1Hz±

x n[ ]

x t( ) 2πt( )cos u t( ) u t 5–( )–[ ]=

X f( ) x t( )e j2πft– td∞–

∞∫ 2πt( )e j2πft–cos td

0

5∫= =

X f( ) 12---ej2πt 1

2---e j2πt–+

e j2πft– td0

5∫=

X f( ) 12--- e j2πt f 1–( )– e j2πt f 1+( )–+[ ] td

0

5∫=

X f( ) 12--- e j2πt f 1–( )–

j2π f 1–( )–---------------------------- e j2πt f 1+( )–

j2π f 1+( )–----------------------------+

t 0=

t 5==

X f( ) 12--- e j10π f 1–( )–

j2π f 1–( )–---------------------------- e j10π f 1+( )–

j2π f 1–( )–----------------------------+

1j2π f 1–( )–

---------------------------- 1j2π f 1–( )–

----------------------------+ –=

X f( ) j2--- e j10π f 1–( )– 1–

2π f 1–( )------------------------------------- e j10π f 1+( )– 1–

2π f 1–( )-------------------------------------+

=

X f( )fs– 2⁄ fs 2⁄,[ ]

1Hz±

x n[ ]

- 6 -


(31)

where is again a 1Hz cosine wave, and the sampling frequency is again given by . Note thatthe difference between the discrete-time signals in equations (31) and (20) is that now consists of 200samples (20 cycles of the cosine wave), instead of 50 samples (5 cycles of the cosine wave). In Figure 6, weplot the DTFT as a function of frequency for in equation (31). Comparing the plots in Figure 6 tothose in Figure 3 (50 samples), note how much more tightly focused the frequency content is about the fre-quencies .

B. Finite-length sum of cosines

Here we consider the DTFT of a discrete-time signal , sampled from the continuous-time signal ,

(32)

with sampling frequency for a total of 50 samples, such that,

. (33)

Applying definition (2),

(34)

In Figure 7, we plot and for the discrete-time sequence in (33) as a function of the dimen-sionless frequency variable , and as a function of the real frequency variable (in Hertz), where we again

-20 -10 0 10 200

0.5

1

1.5

2

2.5

-1 0 1 2 3 4 5 6-1

-0.5

0

0.5

1

Figure 5

x t( )

X f( )

t

f

x n[ ]xc n fs⁄( )

0

=n 0 1 … 198 199, , , ,{ }∈elsewhere

xc t( ) fs 10Hz=x n[ ]

f x n[ ]

1Hz±

x n[ ] xc t( )

xc t( ) 1 2 2πt( )cos 4 4πt( )cos+ +=

fs 10Hz=

x n[ ]xc n fs⁄( )

0

=n 0 1 … 48 49, , , ,{ }∈elsewhere

X ejθ( ) xc n fs⁄( )e jnθ–

n 0=

49

∑=

X ejθ( ) X ejθ( )∠θ f

- 7 -


make the substitution in equation (3). That is, the bottom two plots in Figure 7 correspond to the two expres-sions below:

and (35)

Note the peaks in the frequency spectrum at , and , corresponding to thethree terms in equation (32). In Figure 8, we plot for , corresponding to

. Note again that the frequency spectrum of the time-limited continuous-time waveform is rep-licated at offsets of , for the time-limited sampled waveform. Note the similarity of thespectrum in Figure 8 to that of the idealized, infinite-time sampled waveform, plotted in Figure 9 below.

In the examples above, the sampling frequency was larger than the Nyquist frequency of . In thenext section, we will look at the DTFT when the sampling frequency is less than the Nyquist frequency.

4. Aliasing and the Discrete-Time Fourier Transform (DTFT)

A. Introduction

In the previous section, we saw an example of the discrete-time Fourier Transform (DTFT) for finite-lengthsequences ,

(36)

where,

(37)

and the sampling frequency is given as , which is greater than the Nyquist frequency,

. (38)

-4 -2 0 2 40

20

40

60

80

100

-4 -2 0 2 4-3

-2

-1

0

1

2

3

0 50 100 150 200-1

-0.5

0

0.5

1

Figure 6

x n[ ]

X f( )

n


f


f

X ejθ( ) θ 2πf fs⁄= X ejθ( )∠ θ 2πf fs⁄=

f 0Hz= f 1Hz±= f 2Hz±=X ejθ( ) θ 2πf fs⁄= 25Hz– f 25Hz< <

5π– θ 5π< <kfs± k 1 2 3 …, , ,{ }∈

fs 2fmax

x n[ ]

x n[ ]xc n fs⁄( )

0

=n 0 1 … 48 49, , , ,{ }∈elsewhere

xc t( ) 1 2 2πt( )cos 4 4πt( )cos+ +=

fs 10Hz=

2fmax 2 2Hz× 4Hz= =

- 8 -


Below, we explore the DTFT for the same continuous-time signal, but for two sampling frequencies belowthe Nyquist frequency.

B. Undersampled examples

Here we look at the DTFT of the following finite-length sequence:

(39)

where and is given by equation (37) above. That is, we sample the continuous-time signal for the same length of time, only at a lower sampling frequency (below the Nyquist frequency of 4Hz)

than before. In order to relate our present discussion to our previous discussion on sampling and aliasing, weplot the following functions in Figure 10 below: (1) the original signal ; (2) the sampled signal ;(3) the infinite-length, continuous-time frequency spectrum (CTFT) of the continuous-time signal

; (4) the sampled freq. spectrum (CTFT) for the infinite-length, continuous-time signal ,

(40)

-4 -2 0 2 40

20

40

60

80

100

-4 -2 0 2 4-3

-2

-1

0

1

2

3

-3 -2 -1 0 1 2 30

20

40

60

80

100

-3 -2 -1 0 1 2 3-3

-2

-1

0

1

2

3

0 10 20 30 40 50

-2

0

2

4

6


X ejθ( )∠

f

θ

Figure 7

x n[ ]

X ejθ( )

n


θ

f

x n[ ]xc n fs⁄( )

0

=n 0 1 … 14 15, , , ,{ }∈elsewhere

fs 3Hz= xc t( )xc t( )

xc t( ) x n[ ]Xc f( )

xc t( ) Xs f( ) xs t( )

xs t( ) x' n[ ]δ t nfs---–

n ∞–=

∞

∑=

- 9 -


where , ; and (5) the magnitude DTFT of the finite-length, dis

Figure 9: Sampled spectrum of infinite-length sampled sum of cosines.

x' n[ ] xc n fs⁄( )= n∀ X ejθ( ) θ 2πf fs⁄=
crete-time signal as a function of frequency . Note that except for the spectral leakage (defined previously)caused by the finite-length of , the DTFT has peaks of the same relative magnitude and at the same fre-quencies as .
In Figure 11, we plot the same functions as in Figure 10, except that now the finite-length sequence is givenby:

(41)

where and is again given by equation (37) above. Again, note the similarity between theDTFT and .

C. Oversampled example

Finally for comparison, we generate the same five plots as above for the oversampled case from the previoussection (Figure 12). The finite-length sequence is now the same as last time [equation (36)] with .In all three of these examples, the DTFT for the finite-length sampled sequences generates a similar fre-quency distribution as the CTFT for the infinite-length sampled sequences (represented in the continuous-time domain as ), with two main differences: (1) scaling and (2) spectral leakage caused by the finite-length sampling processes in equations (36), (39) and (41).

5. Conclusion

The Mathematica notebook “dtft.nb” was used to generate the examples in this set of notes. Next time, we willintroduce the discrete Fourier transform (DFT) and show how it is related to the DTFT.

-20 -10 0 10 200

0.5

1

1.5

2

n

f

-20 -10 0 10 200

20

40

60

80

100

Figure 8

n


f

x n[ ] fx n[ ]

Xs f( )

x n[ ]xc n fs⁄( )

0

=n 0 1 … 16 17, , , ,{ }∈elsewhere

fs 3.5Hz= xc t( )Xs f( )

fs 10Hz=

xs t( )

- 10 -


-6 -4 -2 0 2 4 60

0.5

1

1.5

2

2.5

3

-6 -4 -2 0 2 4 60

0.5

1

1.5

2

2.5

3

0 1 2 3 4 5

-2

0

2

4

6

0 2 4 6 8 10 12 14

-2

0

2

4

6

-6 -4 -2 0 2 4 60

10

20

30

40

50

Xc f( )

DTFT

f

t n

Figure 10

xc t( )

Xs f( )

f

x n[ ]


f

fs 3Hz=

- 11 -


-7.5 -5 -2.5 0 2.5 5 7.50

5

10

15

20

25

30

35

-7.5 -5 -2.5 0 2.5 5 7.50

0.5

1

1.5

2

2.5

3

-7.5 -5 -2.5 0 2.5 5 7.50

0.5

1

1.5

2

2.5

3

0 1 2 3 4 5

-2

0

2

4

6

0 2.5 5 7.5 10 12.5 15

-2

0

2

4

6

Xc f( )

DTFT

f

t n

Figure 11

xc t( )

Xs f( )

f

x n[ ]


f

fs 3.5Hz=

- 12 -


-20 -10 0 10 200

20

40

60

80

100

-20 -10 0 10 200

0.5

1

1.5

2

2.5

3

-20 -10 0 10 200

0.5

1

1.5

2

2.5

3

0 1 2 3 4 5

-2

0

2

4

6

0 10 20 30 40 50

-2

0

2

4

6

Xc f( )

DTFT

f

t n

Figure 12

xc t( )

Xs f( )

f

x n[ ]


f

fs 10Hz=

- 13 -

EEL3135: Discrete-Time Signals and Systems The Discrete ... · EEL3135: Discrete-Time Signals and Systems The Discrete-Time Fourier Transform (DTFT) - 3 - Hence, the DTFT of any discrete-time

Documents