Signal and systems. Linear Systems. Luigi Palopoli. Signal and systems p. 1/5

Transcription

1 Signal and systems p. 1/5 Signal and systems Linear Systems Luigi Palopoli

2 Wrap-Up Signal and systems p. 2/5

3 Signal and systems p. 3/5 Fourier Series We have see that is a signal is periodic, it can be conveniently expressed as a Fourier Series: X s(t) = s n e jnωt n= or, if s(t) is real, s(t) = X n=1 2 s n cos(nωt + Φ(s n )) This is essentially a representation of the signal in an appropriate base made of exponential orthonormal functions The convenience of this representation can easily be seen when we process a signal through a linear systems

4 Signal and systems p. 4/5 Fourier Series Because exponential functions are eigenfunctions, the response to a sinusoidal signal is: S e st = H(s)e st, H(s) = R h(t)e st dt S {cos(ωt + φ 0 )} = H(jω) cos(ωt + φ 0 + Φ(H(jω)) S {1} = H(0) Where the last expression is true iff H(s) converges for s = jω. Using the superimposition principle, for a generic periodic signal s(t), we get: S {s(t)} = H(0)s 0 + P n=1 2 s n H(jnω) cos(ωnt + Φ(s n ) + Φ(H(jnω))) The expression above reveals the power of our formalism. By simply knowing H(jω), we are able to compute the response to any periodic signal without solving differential equations. We would like to apply this to non-periodic signals as well.

5 Linear systems: Frequency Domain Signal and systems p. 5/5

6 Nonperiodic signals Signal and systems p. 6/5

7 Signal and systems p. 7/5 Fourier series of a non periodic signal A non-periodic signal can be obtained as the limit of a periodic signal: s(t) = lim T0 rep T0 s(t) where rep T0 s(t) = P h= s(t + ht 0) The coefficients of the periodic signal rept0 s(t) are given by: s n = 1 T 0 Z T0 /2 T 0 /2 = 1 Z T0 /2 X T 0 [rep T0 s(t)]e jnω 0t dt T 0 /2 h= s(t + ht 0 )e jnω 0t dt

8 Signal and systems p. 8/5 Fourier series of a non periodic signal Observing that e j2πnh = 1, we can write s n = 1 Z T0 /2 X T 0 T 0 /2 h= = 1 Z T0 /2 X T 0 = 1 T 0 X = 1 T 0 X T 0 /2 h= Z T0 /2 h= T 0 /2 Z T0 (h+1/2) h= T 0 (h 1/2) = 1 Z s(t )e jnω 0t dt T 0 s(t + ht 0 )e jnω 0t e jω 0nhT 0 dt s(t + ht 0 )e jnω 0(t+hT 0 ) dt s(t + ht 0 )e jnω 0(t+hT 0 ) dt s(t )e jnω 0t dt

9 Signal and systems p. 9/5 Fourier series of a nonperiodic signal Define S(ω) = R s(t)e jωt dt We get: sn = 1 S(nω T 0 ) 0 Notice that two coefficients are spaced out by ω = ω0. Drawing the spectrum in the (ω s n ) plan, the rows s n become closer and closer as T 0 : the spectrum tends to be continuous. Therefore, s n = ω 2π S(n ω) We can write: rept0 s(t) = P n= ω 2π S(n ω)ejnω 0t As T ( ω 0), we get the definition of the integral: s(t) = lim T 0 rep T 0 s(t) = 1 2π Z S(ω)e jω dω

10 Signal and systems p. 10/5 Fourier Transform Starting from the definition of Fourier series (that applies to periodic signal) we have [informally] derived the fourier transform, S(ω) = Z s(t)e jωt dt s(t) = 1 2π Z S(ω)e jωt dω It can be seen that if the signal is absolutely integrable and bounded, (plus other technical mathematical conditions) i.e., R s(t) dt = L <, then the fourier transform exists for almost all ω R also the Fourier is squared -integrable: S(ω) 2 dω < its inverse transform converges to s(t) (in squared error sense) Notice If a signal has finite is absolutely integrable, then it has finite energy: R s(t) 2 dt is finite. The above conditions are only sufficient (we will find the Fourier transform of several signals that do not meet these conditions).

11 Signal and systems p. 11/5 Example Compute the FT of a rectangular impulse: s(t) = AG τ (t) = 8 < A if τ/2 t τ/2 : 0 Otherwise By applying the definition: S(ω) = = A Z Z τ/2 s(t)e jωt dt τ/2 e jωt dt = A 1 jω e jωt τ/2 τ/2 = A 1 jω ( 2j sin(ω τ 2 ) = Aτ 1 ω τ 2 sin(ω τ 2 ) = AτSinc(ω τ 2 )

12 Signal and systems p. 12/5 Example Compute the FT of a dirac impulse: s(t) = δ(t) By applying the definition: S(ω) = = 1 Z δ(t)e jωt dt

13 Signal and systems p. 13/5 Properties of the Fourier transform Linearity: F(u 1 (t)) = U 1 (ω), F(u 2 (t)) = U 2 (ω) F(a 1 u 1 (t) + a 2 u 2 (t)) = a 1 U 1 (ω) + a 2 U 2 (ω) This is a fairly obvious consequence of the linearity of the integral operator. Time shifting: F(u 1 (t)) = U 1 (ω) F(u 1 (t t 0 )) = = Z Z = e jωt 0 U 1 (ω) u 1 (t t 0 )e jωt dt = u 1 (t t 0 )e jω(t t 0) e jωt 0 dt =

14 Signal and systems p. 14/5 Properties of the Fourier transform Shifiting in the frequency domain F(u 1 (t)) = U 1 (ω), F(e jω0t u 1 (t)) = = Z Z = U 1 (ω ω 0 ) e jω 0t u 1 (t)e jωt dt = u 1 (t)e j(ω ω 0)t dt =

15 Signal and systems p. 15/5 Properties of the Fourier transform Time scaling. Assume a positive F(u 1 (t)) = U 1 (ω) F(u 1 (at)) = = Z Z u 1 (at)e jωt dt = 1 a u 1(t )e j ω a t dt = = 1 a U 1( ω a ) For general a we can easily see F(u 1 (t)) = U 1 (ω) F(u 1 (at)) = 1 a U 1( ω a )

16 Properties of the Fourier transform Differentiation in the time domain: F(u 1 (t)) = U 1 (ω), u 1 (t) = 1 2π F( d dt u 1(t)) = F( d dt Z = F( 1 2π Z Z = jωu 1 (ω) U 1 (ω)e jωt dt = U 1 (ω)e jωt dt) = jωu 1 (ω)e jωt dt) = More generally F(u 1 (t)) = U 1 (ω), F( dn u 1 dt n ) = (jω)n U 1 (ω) Signal and systems p. 16/5

17 Signal and systems p. 17/5 Properties of the Fourier transform Duality F(f(t)) = F(ω) F(F(t)) = 2πf( ω) Proof: F(f(t)) = F(ω) f(t) = 1 2π Z F(ω)e jωt dt If we simply swap the two variables t and ω, we find: f(ω) = 1 2π Z F(t)e jωt dt = 1 2π F(F(t)) ω

18 Signal and systems p. 18/5 An example Consider the signal s(t) = 1. This is not absolutely integrable (it does not converge to 0). However, if we apply duality we get: F(δ(t)) = 1 F(1) = 2πδ(ω) This is extremely important because it shows that if we consider generalised functions (δ(.)) we can find the Fourier Transform of function that are not absolutely integrable

19 Signal and systems p. 19/5 Another (important) example Let us consider the signal (non absolutely integrable) 1(t) = 8 < 1 t 0 : 0 Otherwise We can see that: 1(t) = 1 sgn(t) = (1 + sgn(t)) 2 8 < 1 t 0 : 1 Otherwise Function sgn(t) is not an absolutely integrable function, but we can manage it with some trick...

20 Signal and systems p. 20/5 Another (important) example We can write: sgn(t) = lim S α (t) α 0 8 < e αt t 0 S α (t) = : e αt t < 0 Sα (t) is absolutely integrable, hence we can deal with it: F(S α (t)) = = Z Z 0 S α (t)e jωt dt e αt e jωt dt + Z 0 e αt e jωt dt = 1 α jω e(α jω)t 0 1 α + jω e (α+jω)t 0 = 1 α jω + 1 α + jω

21 Signal and systems p. 21/5 Another (important) example Hence: We can conclude: F(sgn(t)) = lim α 0 1 α jω + 1 α + jω = 2 jω F(1(t)) = 1 «2 2 jω + 2πδ(ω) = 1 jω + πδ(ω)

22 Signal and systems p. 22/5 Properties of the Fourier transform Convolution Proof F(u1(t) u2(t)) = F(u 1 (t)) = U 1 (ω), F(u 2 (t)) = U 2 (ω), F(u 1 (t) u 2 (t)) = U 1 (ω)u 2 (ω) = = = Z Z Z Z Z Z u 1 (τ)u 2 (t τ)dτe jωt dt u 1 (τ)u 2 (t τ)dτe jωt dt»z u 1 (τ) = U 1 (ω)u 2 (ω) u 1 (τ)u 1 (ω)e jωτ dτ u 2 (t τ)e jωt dt dτ

23 Signal and systems p. 23/5 Properties of the Fourier transform Product F(u 1 (t)) = U 1 (ω), F(u 2 (t)) = U 2 (ω), F(u 1 (t)u 2 (t)) = 1 2π U 1(ω) U 2 (ω) Proof: It descends from duality + convolution

24 Signal and systems p. 24/5 Properties of the Fourier transform Integration F(u 1 (t)) = U 1 (ω), Z t F( u 1 (τ)dτ) = 1 τ= jω U 1(ω) + πu 1 (0)δ(ω) Proof Z t τ= u 1 (τ)dτ = u 1 (t) 1(t) Z t F( u 1 (τ)dτ) = F (u 1 (t) 1(t)) τ= = ( 1 jω + πδ(ω))u 1(ω) = U 1(ω) jω + πδ(ω))u 1(0)

25 Signal and systems p. 25/5 Example Consider the f(t) = B cos ω0 t F(f(t)) = B 2 F ``e jω0t jω + e 0t We have seen that F(δ(t)) = 1; applying the duality property: F (1) = 2πδ ( ω) Now, we apply frequency shifting property: F `e jω 0t = 2πδ (ω ω 0 ) Therefore, we get: F (B cos ω 0 t) = Bπ (δ(ω ω 0 ) + δ(ω + ω 0 ))

26 Signal and systems p. 26/5 Example We have seen that Let us find: F ( AG τ/2 ) F (AG τ (t)) = AτSinc ( ω τ ) 2 We can apply time scaling rule: F ( AG τ/2 ) = Aτ 2 Sinc ( ω 2 τ ) 2

27 Signal and systems p. 27/5 Spectrum Also for non periodic signals g(t) we can associate a frequency domain spectrum G(ω) It is typically depicted by giving its norm G(ω) and its phase (G(ω)) For real signals the following hold: G(ω) = G( ω), (G(ω)) = (G( ω)) If the signal is even, then G(ω) is real If the signal is odd, then G(ω) is imaginary An interesting example is the following ideal filter. Its spectrum is given by: H(ω) = 1GateB (ω) and (H(ω) = ωt 0

28 Signal and systems p. 28/5 Why is an ideal filter ideal? Consider, for simplicity, t0 = 0 Applying duality, we get: h(t) = 2πBSinc t B 2 As we can see this filter is not a causal system... Therefore the system is not pysically implementable

29 Signal and systems p. 29/5 Fourier Transform of periodic signals We have seen that the fourier transform of a cosine is the sum of two δ The same applies also to other periodic signals For a periodic signal, we have seen that it is possible to write them in terms of Fouries series: X s(t) = s n e jnω 0t n= we have seen that F `ejnω 0 t = 2πδ(ω nω 0 ) Therefore we get: F (s(t)) = X n= 2πs n δ(ω nω 0 )

30 Signal and systems p. 30/5 Fourier Transform of periodic signals What if we construct a periodic repeating a non periodic signal? s(t) = X i= s c (t + it 0 ) The signal can be expressed as s(t) = s c (t) X i= δ(t + it 0 ) We can compute the Fourier series of signal sr (t) = P i= δ(t + it 0): s n = 1 T 0 Z T0 /2 s r (t) = T 0 /2 X n= s n e jnω 0t δ(t)e jnω 0t dt = 1 T 0

31 Signal and systems p. 31/5 Fourier Transform of periodic signals Therefore: s(t) = s c (t) 1 T 0 X n= e jnω 0t Which, corresponds, in the frequency domain, to: S(ω) = S c (ω) 2π T 0 X n= δ(ω nω 0 )

32 Signal and systems p. 32/5 Fourier Transform of periodic signals Example:

33 Mathematical Complements Signal and systems p. 33/5

34 Signal and systems p. 34/5 Discussion We have seen that for signals compying with the following conditions: s(t) limited Finite number of minima and maxima and of singularities Absolutely integrable The Fourier transform exists and the inverxe transform converges to s(t). For these signals we have the Parseval equality: Z s(t) 2 dt = 1 2π Z S(ω) 2 dω We can compute the integral in the easier domain (for instance for a low pass filter it is much easier in the frequency domain)

35 Signal and systems p. 35/5 Discussion If we consider signals with finite power (e.g. periodic signals) we can still compute the fourier transform using generalized functions (δ) We have derived the Fourier transform from the Fourier series, but we also have seen that the Fourier series is a special case of the Fourier transform.

36 Two interesting applications Signal and systems p. 36/5

37 Signal and systems p. 37/5 Amplitude Modulation We want to use the same medium (e.g., the air), to transmit multiple signals (e.g., different channels) Assume that each transmission can have a limited bandwidth One of the oldest ways for doing this was to modulate the amplitude of the signal by multiplying it by a sinusoidal oscillation: x AM (t) = x(t)cos(5t)

38 Signal and systems p. 38/5 Amplitude Modulation 1.0 x(t) K10 K t * cos(5t) K10 K K0.5 K1.0 = x(t)cos(5t) K10 K5 K t 10 K0.4 K0.6

39 Signal and systems p. 39/5 Amplitude Modulation - Frequency domain It is important to see what happens in the frequency domain. Remember cos ω 0 t = ejω 0 t +e jω 0 t 2 Therefore x AM (t) = x(t)cos(ω 0 t) = x(t) ejω0t + e jω 0t 2 X AM (ω) = X(ω ω 0) + X(ω + ω 0 ) 2

40 Signal and systems p. 40/5 Fourier Transform of periodic signals Example:

41 Signal and systems p. 41/5 FDM The idea outlined above can be used to do a Frequency Division Demultiplexing In practice, the spectrum of each singnal is translated to a different frequency range: X i (ω) X i (ω ω i ) In order for the idea to work, the frequency used to translate the signal must be sufficiently spaced out so as to avoid interference: ω i+1 ω > B/2, where B is the bandwidth To demodulate the signal, we first isoltate the part of the spectrum we are interested in, translate the spectrum by ω i and then elimnate spurious component by a low pass filter.

42 FDM Signal and systems p. 42/5

43 FDM - demodulation Signal and systems p. 43/5

44 Sampling Signal and systems p. 44/5

45 Signal and systems p. 45/5 Ideal sampler Ideal sampling can intuitively be seen as generated by multiplying a signal by a sequence of dirac s δ

46 Signal and systems p. 46/5 Properties of δ + f(t)δ(t a)dt = f(a) t δ(τ)dτ = 1 F (δ(t)) = 1 f(t) δ(t t 0 ) = f(t t 0 )

47 Signal and systems p. 47/5 F-trasform of r Using the above properties it is possible to write: F (r (t)) = F r(t) P n= δ(t nt) We can express the sampling signal using the Fourier series: X n= δ(t nt) = 1 T X h= jh 2π e T t Hence F (r (t)) = F r(t) 1 T P t h= ejh2π T Applying the frequency shifting property we get: F (r (t)) = 1 P h= R(ω h 2π T T )

48 Signal and systems p. 48/5 Example R(j ω) 0.5 R * (jω) π /T ω ω

49 Signal and systems p. 49/5 Aliasing The spectrum might be altered (i.e., signal not attainable from samples!) sin(2 π t/3) samples collected with T = 3/ sin(2 π t)

50 Signal and systems p. 50/5 Shannon theorem If the signal has a finite badwidth then the signal can be reconstructed from samples collected with a period such that 1 T 2B Band-limited signals have infinite duration; many signals of interest have infinite bandwidth Typically a low-pass filter is used to de-emphasize higher frequencies

51 Signal and systems p. 51/5 Data Extrapolation If the following hypotheses hold the signal has limited bandwidth B (i.e., the spectrum is not null in the range [ B, B]. the signal is sampled at frequency fs = 1 T 2B then the signal can be reconstructed using an ideal lowpass filter L(s) with L(ω) = 8 < T if ω [ π T, π T ] : 0 elsewhere. Signal l(t) is given by: l(t) = 1 2π Z pi/t pi/t Te jωt dω = sin(πt/t) πt/t = sinc(πt/t)

52 Signal and systems p. 52/5 Data Extrapolation I The reconstructed signal is computed as follows: r(t) = r (t) l(t) = R + r(τ) P δ(τ kt)sinc π(t τ = P + r(kt)sinc π(t kt) T The function sinc is not causal and has infinite duration T dτ = In communication applications The duration problem can be solved truncating the signal The causality problem can be solved introducing a delay and collecting some sample before the reconstruction Not viable in control applications since large delays jeopardise stability