Duality in Time and Frequency

10.2.2 Duality in Time and Frequency

In practice, there are many signals of interest that do not satisfy the absolute summability condition, and so we cannot find their DTFTs with the definition given in the previous section. Duality in the time and frequency representation of signals permits us to obtain the DTFT of those signals.

Consider the DTFT of the signal δ[n − k] for some integer k. Since Z[δ[n − k]] = z −k with ROC the whole z-plane except for the origin, the DTFT of δ[n − k] is e −jωk . By duality, as in the continuous-

time domain, we would expect that the signal e −jω 0 n , −π ≤ ω 0 <π , would have 2π δ(ω +ω 0 ) (where δ(ω) is the analog delta function) as its DTFT. Indeed, the inverse DTFT of 2π δ(ω +ω 0 ) gives

Using these results, we have the following dual pairs:

x[k]δ[n − k] ⇔

x[k]e −jωk

k =−∞

k =−∞

X[k]e −jω k n

⇔ X 2π X[k]δ(ω +ω

The top left equation is the generic representation of a discrete-time signal x[n] and the corresponding term on the right is its DTFT X(e jω ) , one more verification of Equation (10.1). The bottom pair is a

dual of the above. 1 Using Equation (10.9), we then have the following dual pairs as special cases:

x[n]

= Aδ[n] ⇔ X(e jω ) =A

y[n] = A, −∞ < n < ∞ ⇔ Y(e jω ) = 2πAδ(ω)

The signal y[n] is not absolutely summable, and as a constant it does not change from −∞ to ∞, so that its frequency is ω = 0, thus its DTFT Y(e jω ) is concentrated in that frequency. Consider then

1 Calling this a “dual” is not completely correct given that the ω k are discrete values of frequency instead of continuous as expressed by ω , and that the delta functions are not the same in the continuous and the discrete domains, but a duality of some sort exists in these

two pairs, which we would like to take advantage of.

C H A P T E R 10: Fourier Analysis of Discrete-Time Signals and Systems

a sinusoid x[n] = cos(ω 0 n + θ), which is not absolutely summable. According to the above pairs, we get

h i x[n] =

1 h j(ω n

e 0 +θ) +e −j(ω 0 +θ)

⇔ X(e jω ) =π e jθ δ(ω −ω 0 ) +e −jθ δ(ω +ω 0 )

The DTFT of the cosine indicates that its power is concentrated at the frequency ω 0 . The “dual” pairs

δ [n − k], integer k ⇔ e −jωk

(10.11) allow us to obtain the DTFT of signals that do not satisfy the absolutely summable condition. Thus, in general,

e −jω 0 n , −π≤ω 0 <π ⇔ 2πδ(ω + ω 0 )

we have

X X[k]e −jω k n

⇔ X 2π X[k]δ(ω +ω

The linearity of the DTFT and the above result give that for a signal that is not absolutely summable

x[n] X = A

ℓ cos(ω ℓ n +θ ℓ )

its DTFT is

If x[n] is periodic, the discrete frequencies are harmonically related, i.e., ω ℓ = ℓω 0 where ω 0 is the fundamental frequency of x[n].

■ Example 10.2

The DTFT of a signal x[n] is X(e jω ) = 1 + δ(ω − 4) + δ(ω + 4) + 0.5δ(ω − 2) + 0.5δ(ω + 2)

The signal x[n] = A + B cos(ω 0 n) cos(ω 1 n) is given as a possible signal that gives X(e jω ) . Determine whether you can find A, B, and ω 0 and ω 1 to obtain the desired DTFT. If not, provide a better x[n].

Solution

Using trigonometric identities or Euler’s identity, we have that cos(ω 0 n) cos(ω 1 n) = 0.5 cos((ω 0 +ω 1 ) n) + 0.5 cos((ω 1 −ω 0 ) n)

10.2 Discrete-Time Fourier Transform 577

so that x[n] = A + 0.5B cos((ω 0 +ω 1 ) n) + 0.5B cos((ω 1 −ω 0 ) n). Letting ω 2 =ω 0 +ω 1 and ω 3 = ω 1 −ω 0 , the DTFT of x[n] is

X(e jω ) = 2πA + πB(δ(ω − ω 2 ) + δ(ω + ω 2 )) + πB(δ(ω − ω 3 ) + δ(ω + ω 3 )) Comparing this DTFT with the given one, we find that

2π A = 1 ⇒ A = 1/(2π) ω 2 =ω 0 +ω 1 =4 ω 3 =ω 1 −ω 0 =2

π B = 1, 0.5 ⇒ no unique value for B

Although we find that A = 1/(2π) and ω 0 = 1 and ω 1 = 3, there is no unique value for B, so the given x[n] is not the correct answer. The correct signal should be x[n] = (1/π)(0.5 + cos(4n)) + ( 1/2π ) cos(2n), which has the desired DTFT (verify it!).