10. The Discrete-Time Fourier Transform (DTFT)

10.1. Definition of the discrete-time Fourier transform

The Fourier representation of signals plays an important role in both continuous and discrete signal processing. In this section we consider discrete signals and develop a Fourier transform for these signals called the discrete-time Fourier transform, abbreviated DTFT. The discrete-time Fourier transform of a discrete sequence (mx ) is defined as follows:

∞ ~ ~ ( jω )= ∑ ()mxeX e jmω− (10.1) m −∞=

where ω~ is called the normalized frequency. ~ The notation X (e jω ) is justified by the observation that the frequency dependency is in exponential form. In order for the DTFT of a sequence to exist, the summation in (10.1) must converge. It will hold if ()mx is absolutely summable, that is

∞ ∑ ()mx ∞< . (10.2) m −∞=

Note that the DTFT of a discrete-time sequence is a function of a continuous variable ω~ .

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Since

∞ ∞ ~ ~ ~ X (e ()2j π+ω )= ∑ ()mx e m()2j π+ω− = ∑ ()mx m ee m2jj π−ω− = m −∞= m −∞= ∞ ~ ~ = ∑ () jmω− =Xmx (ee jω ) m −∞= then the DTFT is periodic in ω~ with a period of 2π . ~ Since X(e jω ) is periodic in ω~ with the period equal to 2π , we can express it by an exponential Fourier series in variable ω~ . Therefore

∞ ∞ jω~ ~ jmω~ ~ jmω− ~ X (e )= cm e = ∑∑ c−m e (10.3) m −∞= m −∞= where

π 1 ~ ~ c~ = X ( ω ) jj mωdee ω~ . (10.4) −m 2π ∫ π−

Comparison of equations (10.1) and (10.3) shows that the discrete signal (mx ) jω~ ~ corresponding to the spectrum X (e) is (mx ) = c−m . Therefore, from equation (10.4) the inverse discrete-time Fourier transform is

π 1 ~ ~ ()= Xmx ( ) jj mωω dee ω~ . (10.5) 2π ∫ π−

To shed more light on the subject we will derive the DTFT using an alternative approach. Let us consider the Fourier transform of a continuous-time signal ()tx

∞ ()j =ω ∫ ()jω− t de ttxX . ∞−

We approximate the integral as follows:

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∞ jω− mTs X ()j ≅ω ∑ (mTx s )e Ts (10.6) m −∞=

and replace the product ωTs by the normalized frequency

~ ω Ts =ω=ω . f s

If we ignore the scale factor Ts and replace (mTx s ) by (mx ), the resulting ~ summation, denoted by X (e jω ), is the DTFT

∞ ~ ~ X (e jω )= ∑ ()mx e jω− m . (10.7) m −∞=

Example 10.1

Let us consider the signal

( ) = m (muamx ) a <1 .

The DTFT of this signal is

∞ ∞ ~ ~ ~ m X (e jω )= ame jmω− = ∑∑ (ae jω− ) . m=0 m=0

Since the expression on the right hand side is the geometric series, we obtain

jω~ 1 X (e )= ~ − ae1 jω−

provided a <1.

Example 10.2

Let (mx ) be the unit sample δ(m) . The DTFT of this signal is

∞ ~ ~ X (e jω ) ∑ ()m jω− m =δ= 1e . m −∞= 224

10.2. Some properties of the DTFT

In this section we formulate some properties of the discrete time Fourier transform.

Periodicity

This property has already been considered and it can be written as follows

~ ~ X (e ( 2j π+ω ) )= X (e jω ). (10.8)

Linearity

The DTFT is a linear operator, i.e. the discrete-time Fourier transform of a signal

( ) = 11 ( )+ 22 (mxamxamx ) is

jω~ jω~ jω~ (e)= 11 (e )+ XaXaX 22 (e )

jω~ where X k (e ) is the DTFT of k ( ) ( = ,kmx 21 ).

Shifting

Let us consider a shifted signal

ˆ ( ) = ( − mmxmx 0 ).

The DTFT of this signal is (see (10.7))

∞ ∞ ~ ~ ~ ~ ˆ jω jmω− j()mm 0 jm0ω−ω−− X (e ) ∑ ()−= mmx 0 e ∑ ()−= mmx 0 ee . m −∞= m −∞=

Let −= mmk 0 , then

∞ ~ ~ ~ ~ ~ Xˆ ( jω )= ee jm0ω− ∑ ()kx jkω− = jm0ω− X (eee jω ). k −∞=

Thus, we conclude that shifting in time results in the multiplication of the DTFT ~ by a complex exponential e jm0ω− . 225

Example 10.3

Let us consider the shifted unit sample ( ) = δ( − mmmx 0 ). Using the shifting property and knowing that the DTFT of δ(m) is 1, we obtain

~ ~ X (e.jω )= e jm0ω−

Frequency shifting

Let us consider a signal (mx ) multiplied by e jmω0

ˆ ( ) = (mxmx )e jmω0 .

The DTFT of this signal is

∞ ∞ ~ ~ ~ ~ Xˆ (e jω )= ∑ ()mx jmω0 ee jmω− = ∑ ()mx e jm()ω−ω− 0 = X ()e j ()ω−ω 0 . m −∞= m −∞=

Thus, multiplying a sequence by a complex exponential e jmω0 results in shifting in frequency of the DTFT.

Convolution theorem

The convolution of signals (mx ) and (my ) is given by

∞ () ()=∗ ∑ ()(− kmykxmymx ). k −∞=

The DTFT of the convolution is

∞ ∞ ∞ ∞ ⎛ ⎞ ~ ⎛ ~ ⎞ ⎜ ⎟ jmω− ⎜ jmω− ⎟ ∑∑⎜ () (− kmykx )⎟ e = ∑∑⎜ () (− kmykx )e ⎟ = mk−∞= ⎝ −∞= ⎠ k −∞= ⎝ m −∞= ⎠ ∞ ∞ ∞ ∞ ⎛ ~ ~ ⎞ ⎛ ~ ~ ⎞ = ⎜ () (− kmykx ) j( km ) ee jkω−ω−− ⎟ = ⎜ ()kx e jkω− ()py e jpω− ⎟ = ∑∑⎜ ⎟ ∑∑⎜ ⎟ k −∞= ⎝ m −∞= ⎠ k −∞= ⎝ p −∞= ⎠ ~~ = ( )YX (ee jj ωω )

226 where = − kmp . Thus, the DTFT of a convolution of signals (mx ) and (my ) is the product of the DTFTs of (mx ) and (my ) .

Parseval’s theorem

Let us consider a discrete signal (mx ). Parseval’s theorem states that

∞ π 2 1 ~ 2 ()mx = X ( jω ) de ω~ . (10.9) ∑ 2π ∫ m −∞= π−

In Section 12 it will be shown that this equation gives signal energy in the time domain and in the frequency domain.

10.3. Comparing of the DTFT to the DFT

Recall that the DFT of a sequence { m = (mxx )} where = 10 " N,,,m −1 is

2π N −1 N −1 − j mn −mn N n = m = ∑∑ xwxX me = 10 "" N,,,n −1. (10.10) m=0 m=0

On the other hand, the DTFT of the same sequence is

N −1 jω~ jmω− ~ X (e )= ∑ xme . (10.11) m=0

Comparing (10.10) to (10.11) we find

jω~ n = XX (e ) 2πn = 10 "" N,,,n −1. (10.12) ~ =ω N

Equation (10.12) states that the coefficients of the DFT are samples of the 2π continuous spectrum given by the DTFT at ~ =ω n . N Note that the DFT coefficients correspond to N samples of the (zX ) taken at N equally spaced points around the unit circle

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2π j n z = e N = " N,,,n −110 .

10.4. Generalized DTFT

Some discrete-time signals do not have a DTFT but they have a generalized DTFT as explained below. ~ Let the DTFT of a signal (mx ) be X (e jω ) (ωδ= ~). To find this signal, we use the inverse DTFT:

π 1 ~ 1 ()mx = ()jmωde ~~ =ωωδ . 2π ∫ 2π π−

1 This result states that the constant signal ()mx = has the DTFT equal to ()ωδ ~ . 2π Hence, the constant signal ()mx =1 has the DTFT equal to 2πδ(ω~), or

~ ()mx 1 ↔= X ( jω ) 2e (ωπδ= ~) . (10.13)

Note that the signal ()mx =1 does not have the DTFT in the ordinary sense because the series

∞ ~ ∑e jmω− m −∞= is not convergent. Therefore we say that 2πδ(ω~) is a generalized DTFT of the signal ()mx =1. Now we consider a discrete signal (mx ) having the DTFT jω~ ~ X (e ) (ω+ωδ= 0 ). Then

π 1 ~ 1 ()mx = ()~ jmωde ~ =ωω+ωδ e jmω− 0 2π ∫ 0 2π π− or

~ jmω− 0 jω ~ ()mx = e ↔ X ( ) 2e ( ω+ωπδ= 0 ) (10.14)

holds. 228

Likewise, we find

~ jmω0 jω ~ ()mx = e ↔ X ( ) 2e ( ω−ωπδ= 0 ). (10.15)

Since

1 j( 0m α+ω ) 1 j( 0m α+ω− ) ()0mcos =α+ω e + e = 2 2 1 1 = jαee jω0m + jα− ee jω− 0m 2 2 then using (10.14) and (10.15) we determine the DTFT of the signal

()mx cos (0m α+ω= ) as follows:

~ 1 1 X (e jω ) jα 2e ()~ jα− 2e ()~ =ω+ωπδ+ω−ωπδ= 2 0 2 0 jα ~ jα− ~ ( 0 ω+ωδ+ω−ωδπ= 0 )(e)(e ) or

jω~ jα ~ jα− ~ ()mx cos (0m ) X ( ) ( 0 ω+ωδ+ω−ωδπ=↔α+ω= 0 )(e)(ee ). (10.16)

A similar approach leads to the DTFT of the signal (mx ) = sin(ω0m + α)

1 1 ()mx sin (m )=α+ω= e j( 0m α+ω ) − e j( 0m α+ω− ) = 0 j2 j2

1 1 = jαee jω0m − jα− ee jω− 0m . j2 j2

To determine the DTFT of (mx ), we apply (10.14) and (10.15)

jω~ 1 jα ~ 1 jα− ~ X (e ) 2e ()0 −ω−ωπδ= 2e ()0 =ω+ωπδ j2 j2 jα− ~ jα ~ ( 0 ω−ωδ−ω+ωδπ= 0 )(e)(ej ) or

jω~ jα− ~ jα ~ ()mx sin( 0m ) X ( ) ( 0 ω−ωδ−ω+ωδπ=↔α+ω= 0 )(e)(eje ). (10.17) 229

In the special case when =α 0 we have:

jω~ ~ ~ ()mx cos 0m ↔ω= X ( ) ( 0 ω+ωδ+ω−ωδπ= 0 )()(e ) (10.18)

jω~ ~ ~ () sin 0mmx ↔ω= X ( ) ( 0 ω−ωδ−ω+ωδπ= 0 )()(je ). (10.19)

10.5. Frequency response of LTI discrete systems

Let an LTI discrete system be represented by its unit sample response (mh ) (see Fig.10.1). The response of the system due to the input (mx ) is given by convolution

∞ () () ()=∗= ∑ ()(− kmxkhmxmhmy ). k −∞=

x(m) y(m) h(m)

Fig.10.1. An LTI discrete system

Convolution theorem states that

~ ~ ~ (e jω )= ( )XHY (ee jj ωω ) (10.20)

~ ~ where Y (e jω ) is the DTFT of the output (my ) , X (e jω ) is the DTFT of the input ~ (mx ) and H (e jω ) is called the frequency response function of the discrete system. ~ Note that H (e jω ) is, in general, a complex-valued function of the frequency ω~ and can be written in the polar representation

~ ~ jω~ ( jω )= HH ( jω ) eee φH (ej ) . (10.21)

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Thus, the frequency response function of the discrete system is the DTFT of the unit sample response (mh ) and is a continuous function of ω~ . Having the jω~ jω~ magnitude H (e ) and the phase φH (e ) we find:

~ ~ ~ ( jω ) = ( jω ) XHY (eee jω ) (10.22)

jω~ jω~ jω~ Y (e ) H ( ) φ+φ=φ X (ee ). (10.23)

Example 10.4

Let us consider the input signal

( ) = cos (ω0mAmx + α)

To find the output response of a system specified by a frequency response ~ function H (e jω ), we apply (10.20). The DTFT of the signal (mx ) is given by (10.16) repeated below

jω~ jα ~ jα− ~ ( ) AX ( 0 ω+ωδ+ω−ωδπ= 0 )(e)(ee ).

Hence, we obtain

jω~ j~ jαω ~ ~ jj α−ω ~ Y (e ) π= AH ( ) 0 π+ω−ωδ AH ( ) 0 )(ee)(ee =ω+ωδ

jα jω0 ~ jα− jω− 0 ~ π= ()HeA 0 +ω−ωδ H 0 )()(ee)()(e =ω+ωδ

jω0 ~ j( H α+φ ) jω0 ~ j( H α+φ− ) π= ()HA ()e ()ω−ωδ 0 e H ()e ()ω+ωδ+ 0 e .

To find (my ) , we use the inverse DTFT

π ~ Aπ jω0 ~ j( H α+φ ) ~ j( H α+φ− ) jmω ~ ()my = ()H ()()ω−ωδ 0 e(e ()ω+ωδ+ 0 de)e =ω 2π ∫ π− A = H ()ω0 ()ee jj ( 0m H ) + e j( 0m H α+φ+ω−α+φ+ω ) = 2

jω0 = HA ()cose ()0m H α+φ+ω .