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 x( m) is defined as follows: ∞ ~ ~ X( ejω )= ∑ x() m e −jm ω (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 x() m is absolutely summable, that is ∞ ∑ x() m < ∞ . (10.2) m=−∞ Note that the DTFT of a discrete-time sequence is a function of a continuous variable ω~ . 222 Since ∞ ∞ ~ ~ ~ X (e j()ω+ 2 π )= ∑ x() m e −jm() ω+ 2 π = ∑ x() m e−jm ω e − jm 2 π = m=−∞ m=−∞ ∞ ~ ~ = ∑ x() me−jm ω = X ( e 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~ ~ −jm ω~ X (e )= ∑cm e = ∑ c−m e (10.3) m=−∞ m=−∞ where π 1 ~ ~ c~ = X (ejω ) e jmω dω~ . (10.4) −m 2π ∫ −π Comparison of equations (10.1) and (10.3) shows that the discrete signal x( m) jω~ ~ corresponding to the spectrum X (e) ixs( m) = c−m . Therefore, from equation (10.4) the inverse discrete-time Fourier transform is π 1 ~ ~ x() m= X (ejω) e jm ω dω~ . (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 x() t ∞ X()jω = ∫ )x te (−j ωt d t . −∞ We approximate the integral as follows: 223 ∞ −j ωmTs X ()jω ≅ ∑ x ( mTs )e Ts (10.6) m=−∞ and replace the product ωTs by the normalized frequency ~ ω ω = ωTs = . f s If we ignore the scale factor Ts and replace x( mTs ) by x( m), the resulting ~ summation, denoted by X (e jω ), is the DTFT ∞ ~ ~ X (e jω )= ∑ x() m e−j ωm . (10.7) m=−∞ Example 10.1 Let us consider the signal x( m) = am u( m) a <1 . The DTFT of this signal is ∞ ∞ ~ ~ ~ m X (e jω )= ∑ame−jm ω = ∑(ae−j ω ) . m=0 m=0 Since the expression on the right hand side is the geometric series, we obtain jω~ 1 X (e )= ~ 1− a e−j ω provided a <1. Example 10.2 Let x( m) be the unit sample δ(m) . The DTFT of this signal is ∞ ~ ~ X (e jω )=∑ δ()m e−j ωm = 1. 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 j(ω+ 2 π) )= X (e jω ). (10.8) Linearity The DTFT is a linear operator, i.e. the discrete-time Fourier transform of a signal x( m) = a1 x 1( m)+ a2 x 2 ( m) is jω~ jω~ jω~ X(e)= a1 X 1(e )+ a2 X 2 (e ) jω~ where X k (e ) is the DTFT of xk ( m) ( k=1 , 2). Shifting Let us consider a shifted signal ˆx( m) = x( m− m0 ). The DTFT of this signal is (see (10.7)) ∞ ∞ ~ ~ ~ ~ ˆ jω −jm ω −j()m − m0 ω −jm0 ω X (e )=∑ x() m − m0 e =∑ x () m − m0 e e . m=−∞ m=−∞ Let k= m − m0 , then ∞ ~ ~ ~ ~ ~ Xˆ (ejω )= e−jm0 ω ∑ x() k e−jk ω = e−jm0 ω X ( e jω ). k =−∞ Thus, we conclude that shifting in time results in the multiplication of the DTFT ~ by a complex exponential e−jm0 ω . 225 Example 10.3 Let us consider the shifted unit sample x( m) = δ( m− m0 ). Using the shifting property and knowing that the DTFT of δ(m) is 1, we obtain ~ ~ X (e.jω )= e−jm0 ω Frequency shifting Let us consider a signal x( m) multiplied by e jmω0 ˆx( m) = x( m)e jmω0 . The DTFT of this signal is ∞ ∞ ~ ~ ~ ~ Xˆ (e jω )= ∑ x() m ejmω0 e−jm ω = ∑ x() m e −jm() ω−ω0 = X ()e j ()ω−ω0 . 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 x( m) and y( m) is given by ∞ x() m∗ y () m = ∑ x ()( k y m− k ). k =−∞ The DTFT of the convolution is ∞ ∞ ∞ ∞ ⎛ ⎞ ~ ⎛ ~ ⎞ ⎜ ⎟ −jm ω ⎜ −jm ω ⎟ ∑∑⎜ x() k y ( m− k )⎟ e = ∑∑⎜ x() k y ( m− k )e ⎟ = mk=−∞ ⎝ =−∞ ⎠ k =−∞ ⎝ m=−∞ ⎠ ∞ ∞ ∞ ∞ ⎛ ~ ~ ⎞ ⎛ ~ ~ ⎞ = ⎜ x() k y ( m− k )e−j(m − k ) ω e −jk ω ⎟ = ⎜ x() k e−jk ω y () p e−jp ω ⎟ = ∑∑⎜ ⎟ ∑∑⎜ ⎟ k =−∞ ⎝ m=−∞ ⎠ k =−∞ ⎝ p=−∞ ⎠ ~ ~ = XY(ejω) ( e j ω ) 226 where p= m− k . Thus, the DTFT of a convolution of signals x( m) and y( m) is the product of the DTFTs of x( m) and y( m) . Parseval’s theorem Let us consider a discrete signal x( m). Parseval’s theorem states that ∞ π 2 1 ~ 2 x() m = X (ejω ) dω~ . (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 {xm = x( m)} where m= 0 , 1 ," , N −1 is 2π N −1 N −1 − j mn −mn N Xn = ∑ xm w= ∑ xme n= 0 , 1 ,"" , N −1. (10.10) m=0 m=0 On the other hand, the DTFT of the same sequence is N −1 jω~ −jm ω~ X (e )= ∑ xme . (10.11) m=0 Comparing (10.10) to (10.11) we find jω~ XXn = (e ) 2πn n= 0 , 1 ,"" , 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 X( z) taken at N equally spaced points around the unit circle 227 2π j n z = e N n= 0 , 1 ," , N − 1. 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 x( m) be X (e jω )= δ( ω~). To find this signal, we use the inverse DTFT: π 1 ~ 1 x() m = δ() ω~ejmω d ω~ = . 2π ∫ 2π −π 1 This result states that the constant signal x() m = has the DTFT equal to δ() ω~ . 2π Hence, the constant signal x() m =1 has the DTFT equal to 2πδ(ω~), or ~ x() m =1 ↔ X (ejω )= 2 πδ( ω~) . (10.13) Note that the signal x() m =1 does not have the DTFT in the ordinary sense because the series ∞ ~ ∑e−jm ω m=−∞ is not convergent. Therefore we say that 2πδ(ω~) is a generalized DTFT of the signal x() m =1. Now we consider a discrete signal x( m) having the DTFT jω~ ~ X (e )= δ( ω + ω0 ). Then π 1 ~ 1 x() m = δ() ω~ + ωejmω d ω~ = e−jm ω0 2π ∫ 0 2π −π or ~ −jm ω0 jω ~ x() m = e ↔ X (e)= 2 πδ( ω + ω0 ) (10.14) holds. 228 Likewise, we find ~ jmω0 jω ~ x() m = e ↔ X (e)= 2 πδ( ω − ω0 ). (10.15) Since 1 j(ω0m +α) 1 −j( ω0m +α) cos()ω0 m + α = e + e = 2 2 1 1 = ejα e jω0m + e−j α e−j ω0m 2 2 then using (10.14) and (10.15) we determine the DTFT of the signal x() m =cos ( ω0m + α ) as follows: jω~ 1 jα ~ 1 −j α ~ X (e )=e 2 πδ() ω− ω0 +e 2 πδ() ω + ω0 = 2 2 jα ~ −j α ~ =π(e δω−ω (0 ) + e δω+ω (0 )) or jω~ jα ~ −j α ~ x() m =cos ( ω0m +α↔ ) X (e) =π( e δω−ω+ (0 ) e δω+ω (0 )). (10.16) A similar approach leads to the DTFT of the signal x( m) = sin(ω0m + α) 1 1 x() m =sin ( ωm + α ) = e j(ω0m +α) − e−j( ω0m +α) = 0 2j 2j 1 1 = ejα e jω0m − e−j α e−j ω0m . 2j 2j To determine the DTFT of x( m), we apply (10.14) and (10.15) jω~ 1 jα ~ 1 −j α ~ X (e )=e 2 πδ() ω− ω0 − e 2πδ() ω+ ω0 = 2j 2j −j α ~ jα ~ =πj( e δω+ω (0 ) − e δω−ω (0 )) or jω~ −j α ~ jα ~ x() m =sin( ω0m +α↔) X (e) =π j( e δω+ω− (0 ) e δω−ω (0 )). (10.17) 229 In the special case when α = 0 we have: jω~ ~ ~ x() m =cos ω0m ↔ X (e)=πδω−ω( (0 ) +δω+ω (0 )) (10.18) jω~ ~ ~ x() m=sin ω0 m ↔ X (e)= j πδω+ω( (0 ) −δω−ω (0 )). (10.19) 10.5. Frequency response of LTI discrete systems Let an LTI discrete system be represented by its unit sample response h( m) (see Fig.10.1).