Fourier transform representation of CT aperiodic signals Section 4.1

Transcription

1 Fourier transform representation of CT aperiodic signals Section 4. A large class of aperiodic CT signals can be represented by the CT Fourier transform (CTFT). The (CT) Fourier transform (or spectrum) of x(t) is X(jω) = x(t)e jωt dt. x(t) can be reconstructed from its spectrum using the inverse Fourier transform x(t) = 2π X(jω)e jωt dω. The above two equations are referred to as the Fourier transform pair with the first one SM 07

2 being the analysis equation and the second being the synthesis equation. Notation: X(jω) = F{x(t)} x(t) = F {X(jω)} x(t) and X(jω) form a Fourier transform pair, denoted by x(t) F X(jω) Often, we will use the simpler notation x(t) X(jω) Example: rect(t) or Π(t) =, t < /2 0, t > /2 /2, t = /2 SM 08

3 Π( t) rect(t) t SM 09

4 F{ Π( t) } 6π 4π 2π 0 2π 4π 6π ω Fourier transform of rect(t) Example: x(t) = e at u(t),a > 0. We want to show that e at u(t) a+jω,a > 0. SM 0

5 Since the above FT is complex-valued, it is customary to plot its magnitude and phase, i.e., a+jω = a2 +ω 2 a+jω = tan ( ω a ) a + jω a a 0 a Plot of spectrum magnitude ω a jω π 2 a 0 a ω π 2 Plot of spectrum phase SM

6 Example: triangle function (t) = { t, t 0, elsewhere Λ( t) 0 t Exercise: Show that triangle function (t) F sinc 2 ( ω 2π ) SM 2

7 Properties of CT Fourier transform Section 4.3 Table 4. on p. 328 summarizes many CTFT properties.. Linearity Section 4.3. Let x(t) X(jω), y(t) Y(jω). Then, z(t) = ax(t)+by(t) Proof: Z(jω) = ax(jω)+by(jω). SM 3

8 Example: What is the CTFT of the signal, z(t), shown in the figure below? z( t) t Signal z(t) SM 4

9 2. Time shift Section If x(t) X(jω), then Proof: x(t t 0 ) e jωt 0 X(jω). Notethatwhenasignalisdelayedbyt 0, the spectrum amplitude is unchanged whereas the spectrum phase is changed by ωt 0. SM 5

10 Example: What are the CTFT s of the signals shown below? Π( t + 0.5) 0 t Π( t 0.5) 0 Shifted rect signals t SM 6

11 3. Time scaling Section If x(t) X(jω), then x(at) a X ( ) jω where a is a non-zero real constant. The time scaling property states that if a signal is compressed in time by a factor a, then its spectrum is expanded in frequency by the same factor a and vice-versa. This property is an example of the inverse relationship between the time and frequency domains. Example: Since ( ω Π(t) sinc, 2π) a then Π(2t) 2 sinc ( ω 4π ). SM 7

12 Note also that if a = (corresponding to a time reversal) in the time scaling property, we have x( t) X( jω) i.e. the spectrum is also reversed. Combining the time shift and scaling properties What is the Fourier transform of x(at b)? SM 8

13 4. Conjugation Section If x(t) X(jω), then x (t) X ( jω). As a result, if x(t) is real, we have X( jω) = X (jω) i.e. the spectrum magnitude is an even function of ω and the spectrum phase is an odd function of ω. More generally, we can summarize the relationship between a signal and its spectrum as follows: SM 9

14 5. Convolution Section 4.4 y(t) = x(t) h(t) Y (jω) = X(jω)H(jω). Proof: Application: Since (t) = Π(t) Π(t), F{ (t)} = F{Π(t)}F{Π(t)} = sinc 2( ω ) 2π SM 20

15 6. Differentiation& Integration Section If x(t) X(jω), then Proof: d x(t) jωx(jω). dt This result indicates that differentiation accentuates the high frequency components in the signal. Application: Consider y(t) as shown below y( t) 0 t SM 2

16 Since y(t) = d dt (t), we have F{y(t)} = jω F{ (t)} = jω sinc 2( ω ) 2π If x(t) X(jω), then t x(τ)dτ jω X(jω)+πX(0)δ(ω). We see that in contrast to differentiation, integration attenuates the high frequency components in the signal. 7. Area property If x(t) X(jω), then x(t)dt = X(0). SM 22

17 This result follows directly from the definition of the Fourier transform, i.e. Also, since X(jω) = x(t) = 2π x(t)e jωt dt. X(jω)dω = 2π x(0) X(jω)e jωt dω. Example: Evaluate sinc(x)dx. This can be obtained as follows: SM 23

18 8. Duality Due to the similarity between the FT analysis and synthesis equations, we have a duality relationship. If x(t) X(jω), then Proof: X(jt) 2πx( ω). SM 24

19 Example: Determine the inverse FT of Π(ω). Since Π(t) sinc ( ω, 2π) then sinc ( ) t 2π 2πΠ( ω) = 2πΠ(ω) or equivalently ( ) t 2π sinc 2π Π(ω). Duality can also be useful in suggesting new properties of the FT. SM 25

20 Example: 9. Parseval s relation If x(t) X(jω), then x(t) 2 dt = 2π X(jω) 2 dω. Remarks: (a) The LHS is the total energy in the signal x(t). (b) X(jω) 2 describes how the energy in x(t) is distributed as a function of frequency. It is commonly called the energy density spectrum of x(t). SM 26

21 Proof: Example: Evaluate sinc2 x dx. SM 27

22 CT unit impulse function pp , pp The CT unit impulse function is also known as the Dirac delta function, δ(t). Contrast with the Kronecker delta function, δ ij. The introduction of δ(t) allows us to look at the FT of periodic signals. δ(t) is defined by the sifting property, namely δ(t)f(t) dt = f(0) if f(t) is continuous at t = 0. Properties of the Dirac delta function. δ(t) dt = SM 28

23 δ( t) represents the strength of the impulse function 0 t 2. δ(t t 0 )f(t) dt = = f(t 0 ) δ(τ)f(τ +t 0 ) dτ The first line is obtained using τ = t t f(t) δ(t) = = f(t) δ(τ)f(t τ) dτ The above is referred to as the replication property of δ(t). SM 29

24 4. Proof: δ(at) = a δ(t) 5. What is the Fourier transform of δ(t)? F{δ(t)} = SM 30

25 6. What is the inverse FT of δ(ω)? F {δ(ω)} = 2π = 2π δ(ω)e jωt dω Thus, 2πδ(ω). 7. It follows from the last result that e jωt dt = 2πδ(ω) Therefore, 2πδ(ω ω c ) = = = F { e jω ct } e j(ω ωc)t dt e jωct e jωt dt SM 3

26 We thus have e jω ct 2πδ(ω ω c ). This result is useful in examining the FT of a periodic signal. 8. Using the last result, we can write cosω c t = 2 [ e jω c t +e jω ct ] 2 [2πδ(ω ω c)+2πδ(ω +ω c )] = πδ(ω ω c )+πδ(ω +ω c ) { cosω c t} π ω c 0 ω c ω SM 32

27 Similarly, we have sinω c t = 2j [ e jω c t e jω ct ] π j [δ(ω ω c) δ(ω +ω c )] { sin t} ω c π -- j ω c 0 ω c ω π -- j SM 33

28 Fourier transform of periodic signals Section 4.2 Recall from our discussion of FS representation: A periodic signal, x(t), with fundamental period T can be represented by x(t) = k= a k e jk ( 2π T )t where the (possibly complex) Fourier coefficients {a k } are given by a k = T T 2 T 2 x(t) e jk ( 2π T )t dt, k = 0,±,±2,... Let x(t) = { x(t), T 2 t T 2 0, otherwise. SM 34

29 Thus, x(t) is simply one basic period of x(t). Then, with ω 0 = 2π T a k = T, we can write = T X(jkω 0). The last line follows since X(jω) = x(t) e jkω 0t dt x(t)e jωt dt. In other words, a k is equal to T multiplied by the FT of x(t) evaluated at ω = kω 0. Therefore, x(t) = T k= k= T X(jkω 0) e jkω 0t X(jkω 0 ) 2πδ(ω kω 0 ) SM 35

30 In summary, the FT of a periodic signal consists of a series of impulses located at frequencies which are multiples of the fundamental frequency ω 0. The strength of the impulse at the kth harmonic frequency kω 0 is 2πa k. Example: What is the FT of the impulse train or comb function? Recall that the comb function is given by δ T (t) = k= δ(t kt) and is shown in the figure below. δ T ( t) 3T 2T T 0 T 2T 3T t SM 36

31 In this example, x(t) = δ(t) so that X(jω) =. Therefore, δ T (t) 2π T k= δ ( ω k 2π ). T { δ T ( t) } 2π T 6π T 4π T 2π T 0 2π T 4π T 6π T ω We will see that the comb function is very useful in discussing sampling. SM 37