4 The Continuous Time Fourier Transform

Transcription

1 96 4 The Continuous Time ourier Transform ourier (or frequency domain) analysis turns out to be a tool of even greater usefulness Extension of ourier series representation to aperiodic signals oundation of frequency-domain methods for continuous-time signals and systems Outline 4.1 The ourier Transform 4.2 The ourier Transform for Periodic Signals 4.3 Properties of the Continuous Time ourier Transform 4.4 The Convolution Property 4.5 The Multiplication Property 4.6 requency Response and Linear Constant Coefficient Differential Equations

2 The ourier Transform Motivation/Background Goal: ourier (frequency-domain) representation of aperiodic signals Idea: view aperiodic signals as periodic signals with infinite period. Example: ourier representation of the aperiodic square wave x(t) with x(t) = 0 for t > T 1 1 x(t) T T 1 T 1 T t Periodic square wave with period T ( 0 = /T) 1 x(t) 2T T T 1 T 1 T 2T t Observe: x(t) x(t) for T

3 98 requency domain: or with envelope x(t) S a k = 2sin(k 0T 1 ) k 0 T Ta k = 2sin(T 1) = X(j) =k0 X(j) = 2sin(T 1) =k0 T = 4T 1 Ta k T = 8T 1 Ta k T = 16T 1 Ta k Observe: {Ta k } X(j) for T (complex exponentials occur over a continuum of frequencies)

4 99 More generally, when x(t) = 0 for t > T/2 Ta k = and and T/2 T/2 x(t) = ourier transform x(t)e jk 0t dt = k= a k e jk 0t = 1 X(j) = and inverse ourier transform x(t) = 1 x(t)e jt dt = X(j) =k0 k= x(t) T x(t) x(t)e jt dt X(j)e jt d X(jk 0 )e jk 0t 0 =k0 X(j): referred to as the ourier transform, the ourier integral or the spectrum of signal x(t) Short-hand notation x(t) X(j) X(j) = {x(t)} x(t) = 1 {X(j)}

5 100 Convergence of the ourier Transform Given X(j) = x(t)e jt dt when is ˆx(t) = 1 a valid representation of x(t)? X(j)e jt d Similar to the ourier series representation (Section 3.2.1): two sets of conditions and forms of convergence of the ourier integral inite Energy Condition or signals x(t) with finite energy x(t) 2 dt < the ourier transform X(j) is finite and the energy of the error ˆx(t) x(t) is zero ˆx(t) x(t) 2 dt = 0

6 101 Dirichlet Conditions If the Dirichlet Conditions 1. x(t) is absolutely integrable x(t) dt <. 2. x(t) has a finite number of maxima and minima within any finite interval. 3. x(t) has a finite number of discontinuities within any finite interval. urthermore, these discontinuities must be finite. are fulfilled, ˆx(t) is equal to x(t) except at discontinuities, where ˆx(t d ) = x(t d+) + x(t d ) 2 Remark: The above conditions guarantee that the ourier transform exists. However, there are signals that do not meet those conditions but still have a ourier transform. One important example are periodic signals..

7 102 Example: 1. Let then x(t) = e at u(t), Re{a} > 0 X(j) = x(t)e jt dt = e at e jt dt = 1 a + j e (a+j)t 0 0 = 1 a + j 2. Unit impulse ourier transform X(j) = x(t) = δ(t) δ(t) δ(t)e jt dt = 1 1 δ(t) {δ(t)} 1 t

8 Shifted unit impulse ourier transform X(j) = x(t) = δ(t t 0 ) δ(t t 0 )e jt dt = e jt 0 δ(t t 0 ) e jt 0 X(j) δ(t t 0 ) 1 X(j) t 0 t 4. Rectangular pulse signal ourier transform X(j) = x(t) = rect(t/t 1 ) = T 1 /2 T 1 /2 e jt dt = 2 sin(t 1/2) { 1, t < T1 /2 0, t > T 1 /2 = T 1 sinc ( ) T1

9 104 rect(t/t 1 ) T 1 sinc T 1 ( ) T1 {rect(t/t 1 )} 1 rect(t/t 1 ) T 1 /2 T 1 /2 t 4π T 1 T 1 0 T 1 4π T 1 6π T 1 Remark: Two frequently used functions Rectangular pulse of length T r and unit amplitude { 1, t < Tr /2 rect(t/t r ) = 0, t > T r /2 Sinc function Note: sinc(0) = 1 sinc(t) dt = 1 sinc(t) dt = sinc(t) = sin(πt) πt

10 The ourier Transform for Periodic Signals Periodic signals Example: do not meet the Convergence Conditions of Section 4.1, but also have ourier transforms, which can directly be constructed from their ourier series. Consider x(t) with ourier transform Inverse ourier transform Accordingly x(t) = 1 X(j) = δ( 0 ) δ( 0 )e jt d = e j 0t e j 0t δ( 0 ) x(t) = k= a k e jk 0t S a k X(j) = k= a k δ( k 0 ) ourier transform of a periodic signal train of impulses occurring at harmonically related frequencies area of impulse at kth harmonic frequency k 0 is times the kth ourier series coefficient a k

11 106 Example: 1. Sine function ourier series coefficients a k = ourier transform x(t) = sin( 0 t) 1 2j, k = 1 1 2j, k = 1 0 k 1 X(j) = 1 2j δ( 0) 1 2j δ(+ 0) = jπδ( 0 )+jπδ(+ 0 ) jπ X(j) sin( 0 t) jπ 2. Cosine function x(t) = cos( 0 t) ourier transform X(j) = 1 2 δ( 0)+ 1 2 δ(+ 0) = πδ( 0 )+πδ(+ 0 ) π X(j) π cos( 0 t) 0 0 0

12 Impulse train x(t) = ourier series coefficients T/2 k= δ(t kt) a k = 1 T δ(t)e jk 0t dt = 1 T T/2 ourier transform X(j) = T k= δ ( k ) T x(t) 1 2T T 0 T 2T t X(j) T 4π T T 0 T 4π T

13 Properties of the Continuous Time ourier Transform Linearity ourier transform {ax(t) + by(t)} = a{x(t)} + b{y(t)} inverse ourier transform 1 {cx(j) + dy (j)} = c 1 {X(j)} + d 1 {Y (j)} Time Shifting If then Proof: x(t) x(t t 0 ) {x(t t 0 )} = X(j) e jt 0 X(j) x(t t 0 )e jt dt = x(τ)e j(τ+t 0) dτ = e jt 0 x(τ)e jτ dτ = e jt 0 X(j)

14 109 Time and requency Scaling If then (a 0) x(at) x(t) X(j) 1 a X ( ) j a Observe: Scaling in time by a factor a corresponds to scaling in frequency by a factor 1/a a > 1: Time-domain signal x(at) compressed, spectrum X(j) expanded a = 1: Time inversion x( t) Conjugation and Conjugate Symmetry If then x(t) x (t) Real x(t): conjugate symmetry X( j) X(j) X ( j) X( j) = X (j) Imaginary x(t): X(j) = X ( j)

15 110 Conjugation + time reversal properties x(t) = Re{Ev{x(t)}} + Re{Od{x(t)}} + jim{ev{x(t)}} + jim{od{x(t)}} X(j) = Re{Ev{X(j)}} + Re{Od{X(j)}}+ jim{ev{x(j)}} + jim{od{x(j)}} Example: Desired: ourier transform of the signal Already known Express x(t) as x(t) = e a t x(t) = e a t, a > 0 e at u(t) 1 a + j = e at u(t) + e at u( t) ( e at u(t) + e at ) u( t) = 2 2 = 2Ev{e at u(t)} e at u(t) real valued and symmetry properties of ourier transform { } Ev{e at u(t)} 1 Re a + j Therefore { } 1 X(j) = 2Re = 2a a + j a 2 + 2

16 111 Differentiation If then Proof: dx(t) dt x(t) dx(t) dt = d 1 dt = 1 X(j) jx(j) = {jx(j)} X(j)e jt d (jx(j))e jt d Integration Property If then t x(τ) dτ x(t) X(0)δ(): dc component of x(t) X(j) 1 X(j) + πx(0)δ() j

17 112 Example: Unit step function u(t) Known and δ(t) 1 = G(j) u(t) = t δ(τ) dτ Integration property X(j) = {u(t)} = G(j) j + πg(0)δ() = 1 j + πδ() u(t) 1 j + πδ() Duality ormal similarity between ourier and inverse ourier transform Let x(t) X(j) and changing roles of and t in inverse ourier transform then x( ) = 1 X(jt) X(jt)e jt( ) dt x( )

18 113 Duality of properties requency Shift e j 0t x(t) Differentiation of Spectrum X(j( 0 )) jtx(t) Integration of Spectrum 1 x(t) + πx(0)δ(t) jt dx(j) d x(η) dη Example: 1. ourier transform of ( ) t/t1 x(t) = sinc Rectangular pulse z(t) = rect(tt 1 ) Exploiting duality we find and x(t) = T 1 Z(jt) Z(j) = 1 T 1 sinc ( ) t/t1 T 1 Z(jt) = sinc ( ) t/t1 sinc ( ) /T1 T 1 z( ) = X(j) T 1 rect(t 1 )

19 ourier transform of Already known x(t) = te at u(t) Re{a} > 0 z(t) = e at u(t) 1 a + j = Z(j) Exploiting dual differentiation property we find jtz(t) = jx(t) te at u(t) dz(j) d = d d 1 (a + j) 2 ( 1 ) a + j Parseval s Relation If then x(t) x(t) 2 dt = 1 X(j) X(j) 2 d

20 115 Proof: x(t) 2 dt = x(t)x (t) dt = = 1 = 1 x(t) 1 X (j) X (j)e jt d dt X(j) 2 d x(t)e jt dt d Calculate energy of signal x(t) either in time or in frequency domain X(j) 2 : Energy density spectrum of x(t). Example: ( ) Energy E of x(t) = sinc t/t1 Use ( ) t/t1 sinc T 1 rect(t 1 ) and calculate E from spectrum E = 1 X(j) 2 d = 1 1/(2T 1 ) 1/(2T 1 ) (T 1 ) 2 d = T 1

21 The Convolution Property Consider LTI system Recall: x(t) y(t) e jt is eigenfunction with eigenvalue H(j) (frequency response) Periodic signals: ourier series coefficients of output signal are product of ourier series coefficients of input signal and system s frequency response evaluated at corresponding harmonic frequencies. Guess for aperiodic signals and LTI system: ourier transform of output signal is product of ourier transform of input signal and system s frequency response. Derivation from ourier series Interpretation of aperiodic signal as periodic signal with infinite period x(t) = 1 1 = lim 0 0 X(j)e jt d k= X(jk 0 )e jk 0t 0 LTI system with impulse response h(t) and frequency response H(j) = {h(t)} 1 X(jk 0 )e jk0t 0 1 X(jk 0 )H(jk 0 )e jk0t 0 k= k=

22 117 Limit 0 0 Direct derivation y(t) = lim 0 0 = 1 1 k= X(jk 0 )H(jk 0 )e jk 0t 0 X(j)H(j)e jt d = 1 {X(j)H(j)} Y (j) = X(j)H(j) Convolution integral x(t) y(t) = x(τ)h(t τ) dτ ourier transform of system output Y (j) = x(τ)h(t τ) dτ e jt dt = x(τ) h(t τ)e jt dt dτ = H(j) x(τ)e jτ dτ = H(j)X(j)

23 118 Convolutional property y(t) = h(t) x(t) Important to memorize Convolution in time domain domain Y (j) = H(j)X(j) multiplication in frequency requency response H(j) of LTI system is ourier transform of impulse response h(t). Impulse response completely characterizes LTI system. requency response completely characterizes LTI system. Properties of LTI systems in terms of their frequency response E.g., cascading of systems x(t) H 1 (j) H 2 (j) y(t) x(t) H 1 (j)h 2 (j) y(t) x(t) H 2 (j) H 1 (j) y(t)

24 119 Existence of frequency response convergence of ourier transform Example: LTI system is BIBO stable h(t) dt <, first Dirichlet condition fulfilled Practically important systems: second and third Dirichlet conditions also fulfilled Stable LTI systems have frequency response H(j) = {h(t)}. 1. LTI system with impulse response requency response h(t) = δ(t t 0 ) H(j) = e jt 0 Input-output relation in frequency domain Y (j) = H(j)X(j) = e jt 0 X(j) Time shifting property of ourier transform y(t) = x(t t 0 ) Note: time shift t 0 in time domain unit magnitude and linear phase characteristic t 0 in frequency domain

25 Differentiator y(t) = dx(t) dt Differentiation property of ourier transform Y (j) = jx(j) requency response of differentiator H(j) = j 3. requency response of an ideal lowpass filter H(j) = rect(/(2 c )) = { 1, < c 0, > c 1 H(j) c c Stopband Passband Stopband Impulse response h(t) = c π sinc ( ) c t Observe: Impulse response is not causal and of infinite duration ideal lowpass not realizable π

26 121 Often convenient: Computing response of LTI system via frequency domain x(t) h(t) y(t) 1 X(j) H(j) Y (j) Example: Computational bottleneck : inverse ourier transform Solutions: Look up table of (basic) ourier transform pairs (e.g., Table 4.3 in text book) Partial fraction expansion for ratio of polynomials 1. LTI system with impulse response and input signal Output y(t)? h(t) = e at u(t), a > 0 x(t) = e bt u(t), b > 0 ourier transforms of x(t) and h(t) and X(j) = 1 b + j H(j) = 1 a + j

27 122 ourier transform of y(t) Y (j) = X(j)H(j) = Inverse ourier transform (a) a b 1 (a + j)(b + j) Write with (b) a = b Y (j) = A a + j + B b + j A = (a + j)y (j) = 1 b a j= a B = (b + j)y (j) = 1 b a j= b Y (j) = y(t) = ( 1 1 b a a + j 1 ) b + j 1 ( e at e bt) u(t) b a We have Y (j) = 1 (a + j) 2 y(t) = te at u(t)

28 Ideal lowpass filter impulse response h(t) = ( ) c π sinc c t π and input signal Output y(t)? x(t) = i π sinc ( ) i t Time-domain approach: Convolution of two sinc functions very involved requency-domain approach π where and Y (j) = H(j)X(j) H(j) = rect(/(2 c )) X(j) = rect(/(2 i )) Y (j) = rect(/(2 0 )), 0 = min{ c, i } y(t) = ( ) 0 π sinc 0 t, 0 = min{ c, i } π Note: Convolution of two sinc functions is again a sinc function.

29 The Multiplication Property Dual to convolution property multiplication property r(t) = s(t)p(t) R(j) = 1 S(j) P(j) Multiplication in time domain convolution in frequency domain Multiplication of p(t) with s(t) = scaling or modulating amplitude of p(t) with amplitude of s(t) Multiplication often referred to as amplitude modulation Multiplication property also referred to as modulation property Example: 1. Multiplication with Cosine (Modulation) Signal s(t) with spectrum S(j) S(j) A 1 1

30 125 Carrier p(t) = 2cos 0 t, i.e., P(j) = δ( 0 ) + δ( + 0 ) P(j) 0 0 Modulated signal r(t) = s(t)p(t) with spectrum R(j) Graphically R(j) A/ Mathematically R(j) = 1 1 (S(j) P(j)) = S(j( 0 ))+ 1 S(j(+ 0 )) 2 2 Assumption: 0 > 1 no overlap r(t) uniquely represents s(t) sinusoidal amplitude modulation

31 Multiplication with Cosine (Demodulation) Modulated signal r(t) from previous example and p(t) = 2cos0 t Let Spectrum G(j) of g(t) g(t) = r(t)p(t) A G(j) A/ Note: g(t) = r(t)p(t) = s(t)p 2 (t) = (1 + cos(2 0 t))s(t) Reconstruction of s(t) from g(t): 1, < 1 Lowpass filter H(j) 0, > (2 0 1 ) don t care, else

32 Spectrum of signal x(t) = sin(t)sin(t/2) πt 2 Write x(t) as product of two sinc functions x(t) = 1 ( ) ( ) t t sinc sinc π Use multiplication property of ourier transform X(j) = 1 { ( )} { ( )} t t 4π 2 sinc sinc π Use correspondence sinc() rect() X(j) = 1 rect(/2) rect() 2 Convolution of rectangular pulses X(j) 1/

33 128 Multiplication property (dual frequency-shift property) Realization of bandpass filter with variable center frequency x(t) e j ct y(t) 0 H(j) 1 0 w(t) e j ct f(t) Multiplication with complex exponential e j ct requency shift Y (j) = X(j( c )) requencies near c shifted into passband of lowpass filter Multiplication with e j ct requency shift (j) = W(j( + c )) Example: X(j) H(j) Y (j) 0 0 c W(j) 0 0 (j) c

34 requency Response and Linear Constant Coefficient Differential Equations Recall: System description by linear constant coefficient differential equation N d k y(t) M d k x(t) a k = b dt k k dt k k=0 requency response H(j)? k=0 Apply x(t) = e jt as input output y(t) = H(j)e jt Alternatively Convolution property H(j) = Y (j)/x(j) Apply ourier transform to differential equation + differentiation property { N } { d k M } y(t) d k x(t) a k = b dt k k dt k k=0 k=0 N { d k } y(t) M { d k } x(t) a k = b dt k k dt k k=0 N a k (j) k Y (j) = k=0 Consequently H(j) = Y (j) X(j) = k=0 M b k (j) k X(j) k=0 M k=0 N k=0 b k (j) k a k (j) k

35 130 Observe Example: H(j) obtained from differential equation by inspection H(j) is a rational function in j inverse ourier transform by partial-fraction expansion 1. Stable LTI system characterized by dy(t) + ay(t) = x(t), a > 0 dt requency response H(j) = 1 j + a Impulse response h(t) = e at u(t) 2. Stable LTI system characterized by d 2 y(t) + 4 dy(t) + 3y(t) = dx(t) + 2x(t) dt 2 dt dt requency response by inspection j + 2 H(j) = (j) 2 + 4j + 3 Partial fraction expansion H(j) = Impulse response j + 2 (j + 1)(j + 3) = 1/2 j /2 j + 3 h(t) = 1 2 e t u(t) e 3t u(t)

36 System of the last example and input Spectrum of output Y (j) = H(j)X(j) = = x(t) = e t u(t) ( j + 2 (j + 1) 2 (j + 3) Partial fraction expansion j + 2 (j + 1)(j + 3) Y (j) = A 11 j A 12 (j + 1) 2 + A 21 j + 3, with A 11 = 1/4, A 12 = 1/2, and A 21 = 1/4. Inverse ourier transform ( 1 y(t) = 4 e t te t 1 ) 4 e 3t u(t) )( 1 ) j + 1