Chapter 4 Fourier transforms

Chapter 4 Fourier transforms - ECSE-303A Signals and...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ECSE-303A Signals and Systems I Wednesday, March 03, 2010 LECTURE 17 Section 4.1-Continuous-time Fourier transform Assignment #2 due Wednesday Chapter 4: The continuous-Time Fourier transform 4.0 Introduction 4.1 The continuous-time Fourier transform 4.2 The Fourier transform for periodic signals 4.3 Properties of the continuous-time Fourier transform 4.4 The convolution property 4.5 The multiplication property 4.6 Tables of Fourier properties and of basic Fourier transform pairs 4.7 Systems characterized by linear constantcoefficient differential equations 4.1 The Continuous-Time Fourier Transform At the limit of a Fourier Series where the periodicity is infinitely large ⇒ Fourier Transform To illustrate this, consider the Fourier series representation of the periodic rectangular wave The spectral coefficients are ak = η sinc ( kη ) ECSE303 Chap. 4 MR - 3/3/2010 2t0 ⎫ ⎧ ⎨η = ⎬ T⎭ ⎩ 3 2t0 ⎛ 2t0 ⎞ sinc ⎜ k ak = ⎟ T T⎠ ⎝ T=2*2t0 η=0.5 Let’s increase the period T by a factor of 4… T=8*2t0 η=0.125 ECSE303 Chap. 4 MR - 3/3/2010 4 The spectrum of the Fourier coefficients ak could be plotted as a function of ω=ω0k instead than k alone 2t0 ⎛ 2t0 ⎞ ak = sinc ⎜ k ⎟ T ⎝ T⎠ 2t0 ⎛ k ω0 t 0 ⎞ sinc ⎜ = ⎟ T ⎝π⎠ 2t0 ⎛ ω t0 ⎞ sinc ⎜ = T π⎟ ⎝ ⎠ This eliminates the dependence of the sinc on the variable k ECSE303 Chap. 4 MR - 3/3/2010 5 Now ak as a function of ω 2t0 ⎛ ω t0 ⎞ sinc ⎜ ak = ⎟ T ⎝π ⎠ T=2*2t0 η=0.5 Let’s increase the period T by a factor of 4… T=8*2t0 η=0.125 ECSE303 Chap. 4 MR - 3/3/2010 The sinc envelope remain unchanged along the horizontal 6 axis And another way to see this solution is to plot Tak as a function of ω 2t0 ⎛ ωt0 ⎞ sinc ⎜ ak = ⎟ T ⎝π ⎠ ⎛ ω t0 ⎞ Tak = 2t0sinc ⎜ ⎟ ⎝π ⎠ Then the sinc envelope becomes independent of the signal period along both the amplitude and the frequency axes ECSE303 Chap. 4 MR - 3/3/2010 7 Now Tak as a function of ω ⎛ ω t0 ⎞ Tak = 2t0sinc ⎜ ⎟ ⎝π ⎠ T=2*2t0 η=0.5 Let’s increase the period T by a factor of 4… T=8*2t0 η=0.125 ECSE303 Chap. 4 MR - 3/3/2010 8 By expressing the spectral coefficients of the square wave function with ωt ⎛ 0⎞ Tak = 2t0sinc ⎜ ⎟ ⎝π ⎠ Then the spectrum becomes a sinc envelope that has an amplitude and width that remain unchanged by T. This envelope is sampled at a frequency spacing given by the periodicity of the square function T=8*2t0 η=0.125 ECSE303 Chap. 4 MR - 3/3/2010 9 Let’s define a signal x(t) equal to the periodic square function over only one period and zero elsewhere. x (t ) x (t ) This signal with an infinite fundamental period is in fact aperiodic ECSE303 Chap. 4 MR - 3/3/2010 10 Let's define X(jkω0) to represent the envelope of the spectral coefficients of x(t) +∞ x (t ) = X ( jkω0 ) ≡ Tak +∞ = ∫ k =−∞ x(t )e − jkω0t dt −∞ = t0 ∫ ∑ ak = ak e jkω0t 1 x ( t ) e − jkω0t dt ∫ TT e − jkω0t dt − t0 = 2t0sinc( t0ω π ) For a periodic square function, this spectral envelope Tak is independent of T. Are we going to get this for any periodic function? Let’s develop the expression of the spectral coefficient envelope X(jkω0) in the general case. ECSE303 Chap. 4 MR - 3/3/2010 11 ~ A periodic signal x ( t ) has the Fourier series representation +∞ +∞ 1 jkω0t x ( t ) = ∑ ak e = ∑ X ( jkω0 ) e jkω0t k =−∞ k =−∞ T Or, equivalently with T=2π/ω0 +∞ ω0 x(t ) = ∑ X ( jkω0 )e jkω0t k =−∞ 2π In the limit of T⇒∞, we get +∞ 1 k ω0 = ω x(t ) = X ( jω )e jω t d ω 2π −∞ ω0 → d ω It is a T-independent →∫ ∑ spectrum x (t ) → x (t ) representation of x(t) ECSE303 Chap. 4 ∫ MR - 3/3/2010 12 We thus have formed a set of two equations which will be called the Fourier transform pair. 1 x(t ) = 2π X ( jω ) = +∞ ∫ +∞ ∫ X ( jω )e jωt d ω Inverse FT −∞ x(t )e − jωt dt Fourier transform −∞ ECSE303 Chap. 4 MR - 3/3/2010 13 Example Spectrum of the impulse x (t ) = δ (t ) X ( jω ) = 1 x (t ) = 2π X ( jω ) = +∞ +∞ ∫ X ( jω )e jωt d ω −∞ +∞ ∫ x (t )e − jωt dt −∞ δ ( t ) e − jωt dt ∫ −∞ +∞ = ∫ δ ( t ) dt −∞ =1 An impulse is made of the contribution of all frequencies with an amplitude of 1 ECSE303 Chap. 4 MR - 3/3/2010 14 Example Spectrum of the delayed impulse X ( jω ) = x ( t ) = δ ( t − t0 ) X ( jω ) = 1 x (t ) = 2π +∞ ∫ X ( jω )e jωt d ω −∞ +∞ ∫ x (t )e − jωt dt −∞ +∞ δ ( t − t0 ) e − jωt dt ∫ −∞ +∞ = δ ( t − t0 ) e − jωt dt ∫ 0 −∞ = e − jωt0 A delayed impulse is made of the contribution of all frequencies with an amplitude of 1 and varying phase-shift ECSE303 Chap. 4 MR - 3/3/2010 15 Example The square function T1 X ( jω ) = ∫e − jω t 1 x (t ) = 2π X ( jω ) = dt = x1 ( t ) FT t ECSE303 Chap. 4 MR - 3/3/2010 +∞ ∫ x (t )e − jωt dt ω 2 ⎛ ωT ⎞ sin c ⎜ 1 ⎟ T1 ⎝π ⎠ T1 −∞ sin ωT1 1 −T1 ∫ X ( jω )e jωt d ω −∞ −T1 =2 +∞ ⇔ X1 ( jω ) 2 T1 Surf. area ω −π T1 π T1 π/half-width 16 1 x (t ) = 2π Example The Fourier transform of x ( t ) = e − at u ( t ) X ( jω ) = +∞ ∫ X ( jω )e jωt d ω −∞ +∞ ∫ x (t )e − jωt dt −∞ {a ≥ 0, real} is given by X ( jω ) = +∞ e − at u ( t ) e − jωt dt ∫ −∞ +∞ = e −( a + jω )t dt = ∫ 0 = ECSE303 Chap. 4 MR - 3/3/2010 1 jω + a 17 1 X ( jω ) = jω + a The magnitude of X(jω) is X ( jω ) 1/a 1 1 X ( jω ) = ω 2 + a2 a2 -a a ω The 3dB cutoff frequency of this filter is ωc=a ECSE303 Chap. 4 MR - 3/3/2010 18 Phase of X(jω) 1 X ( jω ) = jω + a ∠X ( jω ) ⎛ ω⎞ ∠X ( jω ) = a tan ⎜ − ⎟ ⎝ a⎠ π/2 π/4 -a −π/4 −π/2 a ω The maximum phase added to the input signal is ±π/2 for |ω | >> ωc ECSE303 Chap. 4 MR - 3/3/2010 19 Remarks For the case where e-atu(t) is the impulse response h(t) of a first-order differential LTI system: • The system is a lowpass filter with DC gain of 1/a • High frequencies in the input signal x(t) are attenuated • The 3 dB (or 50% power, or 70.7% amplitude) attenuation cutoff is ωc=a • The maximum phase added to the input signal is ±π/2 for ω →±∞ ECSE303 Chap. 4 MR - 3/3/2010 20 1 x (t ) = 2π +∞ ∫ X ( jω )e jωt d ω −∞ Example +∞ X ( jω ) = ∫ x (t )e − jωt dt The complex exponential −∞ − at x2 ( t ) = e u (t ) {a = α + j β , α , β ≥ 0, real} X 2 ( jω ) = +∞ e − at u ( t ) e− jωt dt ∫ −∞ +∞ = ∫e 0 −(α + j β + jω )t 1 dt = j (ω + β ) + α Magnitude of X2(jω) X 2 ( jω ) = 1 α 1 2 (ω + β ) + α 2 1 α2 −α−β −β α−β ECSE303 Chap. 4 MR - 3/3/2010 ω 21 Phase of X2(jω) 1 X 2 ( jω ) = j (ω + β ) + α ⎛ (ω + β ) ⎞ ∠X 2 ( jω ) = a tan ⎜ − ⎟ α⎠ ⎝ π/2 π/4 -α−β α −β −β −π/4 ω −π/2 ECSE303 Chap. 4 MR - 3/3/2010 22 Example Find the Fourier transform of x3 (t ) = e −α t sin(ω0t )u (t ) −1 − (α + jω0 )t = (e u (t ) − e − (α − jω0 )t u (t )) 2j We have just seen that the Fourier transform of a complex exponential is x2 ( t ) = e − at u (t ) X 2 ( jω ) = ECSE303 Chap. 4 MR - 3/3/2010 {a = α + j β , α , β ≥ 0, real} 1 j (ω + β ) + α 23 x2 ( t ) = e − at u (t ) X 2 ( jω ) = by linearity {a = α + j β , α , β ≥ 0, real} 1 j (ω + β ) + α ⎞ 1 1 −1 ⎛ X 3 ( jω ) = − ⎜ ⎟ 2 j ⎝ j (ω + ω0 ) + α − j (−ω + ω0 ) + α ⎠ ⎞ 1 1 −1 ⎛ = − ⎜ ⎟ 2 j ⎝ j (ω + ω0 ) + α j (ω − ω0 ) + α ⎠ 1 ⎛ j (ω + ω0 ) + α − [ j (ω − ω0 ) + α ] ⎞ = ⎜ ⎟ 2 j ⎝ [ j (ω − ω0 ) + α ][ j (ω + ω0 ) + α ] ⎠ ω0 = 2 ( jω + α ) 2 + ω0 ECSE303 Chap. 4 MR - 3/3/2010 24 x3 ( t ) = e −α t ω0 sin (ω0t ) u ( t ) ⇔ X 3 ( jω ) = 2 2 jω + α ) + ω0 ( FT | X 3 ( jω )| 1 2α ω0 ω2 +α2 0 2 − ω0 − 2 ECSE303 Chap. 4 MR - 3/3/2010 2 ω0 − α 2 ω 25 X 3 ( jω ) = ω0 2 ( jω + α ) 2 + ω0 ⎡ ⎤ 2αω ∠X 3 ( jω ) = − atan ⎢ 2 ⎥ α + ω02 − ω 2 ⎦ ⎣ π π/2 2 − ω0 + α 2 2 ω0 + α 2 −π/2 ω −π ECSE303 Chap. 4 MR - 3/3/2010 26 1 x (t ) = 2π Example Find the Fourier Transform of the following signal X ( jω ) = X ( jω ) = ∫ x(t )e − jω t dt 0 = ∫ −T1 T1 −∞ +∞ ∫ x (t )e − jωt dt T1 T1 -T1 −∞ ∫ X ( jω )e jωt d ω −∞ x (t ) +∞ +∞ (−T1 − t )e − jω t dt + ∫ (T1 − t )e − jω t dt t -T1 0 ⎛ 2sin(ω T1 ) 2T1 ⎞ = j⎜ − 2 ω⎟ ω ⎝ ⎠ ECSE303 Chap. 4 MR - 3/3/2010 27 Spectrum of the function ⎛ 2sin(ωT1 ) 2T1 ⎞ X ( jω ) = j ⎜ − 2 ω ω⎟ ⎝ ⎠ T1=1 sec X ( jω ) = ∠X ( jω ) = 2sin(ωT1 ) − ω π 2 π 2 ECSE303 Chap. 4 MR - 3/3/2010 2 − ω>0 ω<0 28 2T1 ω Chapter 4: The continuous-Time Fourier transform 4.0 Introduction 4.1 The continuous-time Fourier transform 4.2 The Fourier transform for periodic signals 4.3 Properties of the continuous-time Fourier transform 4.4 The convolution property 4.5 The multiplication property 4.6 Tables of Fourier properties and of basic Fourier transform pairs 4.7 Systems characterized by linear constantcoefficient differential equations x (t ) = +∞ ∑ k =−∞ ak = ak e jkω0t 1 x ( t ) e − jkω0t dt ∫ TT Question A periodic signal x(t) is expressed in Volts. What are the units of ak? ECSE303 Chap. 3 MR - 3/3/2010 30 1 x (t ) = 2π X ( jω ) = +∞ ∫ X ( jω )e jωt d ω −∞ +∞ ∫ x (t )e − jωt dt −∞ Question A non periodic signal x(t) is expressed in Volts. What are the units of X(jω)? ECSE303 Chap. 4 MR - 3/3/2010 31 ak = 1 v(t )e − jkω0t dt ∫ TT v (t ) = = 1 δ (t )e − jkω0t dt T −T∫/ 2 -2T -T ∑ δ ( t − nT ) n =−∞ 1 T /2 +∞ T 2T 3T 1 = The spectrum of an impulse train is a real, constant T t sequence. ak 1/T -16 -8 8 16 k ECSE303 Chap. 4 MR - 3/3/2010 32 4.2 Fourier Transforms for Periodic Signals What happens if we use the Fourier transform to represent periodic signals? Consider a signal x(t) with Fourier transform that is a single impulse of area 2π at frequency kω0. This can be represented as X ( jω ) = 2πδ (ω − kω 0 ) Taking the inverse Fourier transform yields 1 x (t ) = 2π =e ECSE303 Chap. 4 MR - 3/3/2010 z +∞ 2πδ (ω − kω 0 )e jωt dω −∞ jkω 0t 1 x (t ) = 2π X ( jω ) = +∞ ∫ X ( jω )e jωt d ω −∞ +∞ ∫ x (t )e − jωt dt −∞ 33 Therefore, a periodic signal for which the Fourier transform can +∞ expressed as a sum of impulses be X ( jω ) = ∑ 2πa δ (ω − kω k 0 ) k =−∞ has the time-domain form x (t ) = +∞ a k e jkω 0t ∑ k =−∞ which in fact is the Fourier series representation of x(t) Therefore the Fourier transform of a periodic signal with Fourier series coefficients ak, is a train of impulses of surface area 2πak occurring at frequencies ω=kω0. ECSE303 Chap. 4 MR - 3/3/2010 34 Example The two impulses in the continuous frequency domain X ( jω ) = 2π [δ (ω − ω0 ) + δ (ω + ω0 ) ] In fact represent a periodic cosine in the timedomain +∞ 1 x(t ) = 2π [δ (ω − ω0 ) + δ (ω + ω0 ) ] e jωt d ω ∫ 2π −∞ =e jω0t +e − jω0t = 2 cos (ω0t ) Similarly, 1 x (t ) = 2π X ( jω ) = +∞ ∫ X ( jω )e jωt d ω −∞ +∞ ∫ x (t )e − jωt dt −∞ FT 2sin (ω0t ) ⇔ ( 2π j ) [δ (ω − ω0 ) − δ (ω + ω0 )] ECSE303 Chap. 4 MR - 3/3/2010 35 Fourier Transform of a Periodic Impulse Train Consider the impulse train signal +∞ x (t ) = ∑ δ (t − kT ) −∞ x (t ) We know that the spectral coefficients of the impulse train are ak =1/T. 11 -2T -T T 2T 3T t The Fourier spectrum is then 2π X ( jω ) = T X ( jω ) +∞ 2 1π ∑ δ (ω − kω ) k =−∞ 0 −2ω0 −ω0 -2T ECSE303 Chap. 4 MR - 3/3/2010 -T ω0 T 2T 2ω0 3T 3ω0 tω 36 Reading/exercises Read sections 4.1-4.2 Problems 4.1-4.3, 4.21 ECSE303 Chap.4 MR - 3/3/2010 42 ECSE-303A Signals and Systems I Wednesday, March 03, 2010 LECTURE 18 Section 4.3-Properties of the continuous-time Fourier transform x(α t ) ↔ Remark The Fourier-transform pair 1 x (t ) = 2π X ( jω ) = +∞ ∫ 1 α X ( jω / α ) X ( jω )e jωt d ω −∞ +∞ ∫ x (t )e − jωt dt −∞ Can also be rewritten as +∞ x (t ) = ∫ X ( jf )e j 2π ft df −∞ +∞ X ( jf ) = ∫ x (t )e − j 2π ft dt −∞ ECSE303 Chap. 4 MR - 3/3/2010 44 Remark The transform X ( jω ) e jω0t FT ↔ 2πδ (ω − ω0 ) 2π ω ω0 Can in fact be rewritten as X ( jω ) e jω0t FT ↔ δ ( f − f0 ) 1 f f0 ECSE303 Chap. 4 MR - 3/3/2010 45 Chapter 4: The continuous-Time Fourier transform 4.0 Introduction 4.1 The continuous-time Fourier transform 4.2 The Fourier transform for periodic signals 4.3 Properties of the continuous-time Fourier transform 4.4 The convolution property 4.5 The multiplication property 4.6 Tables of Fourier properties and of basic Fourier transform pairs 4.7 Systems characterized by linear constantcoefficient differential equations 4.3 Properties of the Fourier Transform We denote the relationship between a signal and its Fourier transform as FT x(t ) ↔ X ( jω ) The Fourier transform has the following properties Linearity The Fourier transform is a linear operation FT ax(t ) + by (t ) ↔ aX ( jω ) + bY ( jω ) ECSE303 Chap. 4 MR - 3/3/2010 47 Time Shifting A time shift in the time domain turns into a phase shift in the frequency domain FT x (t ) ↔ X ( jω ) FT x (t − t0 ) ↔ e − jωt0 X ( jω ) ECSE303 Chap. 4 MR - 3/3/2010 48 1 x (t ) = 2π Example Spectrum of the delayed impulse X ( jω ) = x (t ) x ( t ) = δ ( t − t0 ) +∞ ∫ X ( jω )e jω t dω −∞ +∞ ∫ x (t )e − jωt dt −∞ 1 t t0 X ( jω ) = +∞ δ ( t − t0 ) e − jωt dt ∫ −∞ 1 +∞ = δ ( t − t0 ) e − jωt dt ∫ 0 X ( jω ) ∠X ( jω ) ω ω −∞ = e − jωt0 A delayed impulse is made of the contribution of all frequencies with an amplitude of 1 and varying phase-shift ECSE303 Chap. 4 MR - 3/3/2010 49 Time/Frequency Scaling Scaling the time variable either expands or compresses the spectrum FT x(t ) ↔ X ( jω ) FT x(α t ) ↔ 1 α X ( jω / α ) For α>1, the signal x(αt) is sped up (or compressed in time), so its spectrum expands in the frequency domain. 1 e u (t ) ↔ jω + a FT 1 at e u (−t ) ↔ jω − a − at Example e-atu(t) and eatu(-t), a>0 ECSE303 Chap. 4 MR - 3/3/2010 FT 50 Example The square function T1 X ( jω ) = ∫ e − jωt dt =2 1 x (t ) = 2π = x1 ( t ) −∞ +∞ ∫ ω 2 ⎛ ωT ⎞ sin c ⎜ 1 ⎟ T1 ⎝π ⎠ FT t T1 x (t )e − jωt dt −∞ 1 ECSE303 Chap. 4 MR - 3/3/2010 ∫ X ( jω )e jωt d ω sin ωT1 X ( jω ) = −T1 −T1 +∞ ↔ X1 ( jω ) 2T1 Surf. area ω −π T1 π T1 π/half-width 51 Conjugation and Conjugate Symmetry Taking the general case of a complex signal, the conjugate of the signal has a Fourier transform FT x(t ) ↔ X ( jω ) FT x (t ) ↔ X ∗ (− jω ) ∗ Proof ⎡∞ ⎤ ∗ − jωt [ X ( jω )] = ⎢ ∫ x ( t ) e dt ⎥ ⎣ −∞ ⎦ X ∗ ( jω ) = ∞ ∫ ∗ x∗ ( t ) e jωt dt −∞ X ∗ (− jω ) = ECSE303 Chap. 4 MR - 3/3/2010 ∞ ∫ −∞ x∗ ( t ) e − jωt dt 52 FT x(t ) ↔ X ( jω ) Then for a real signal with x(t) = x*(t), Considering a cartesian notation, FT x (t ) ↔ X ∗ (− jω ) ∗ X ( jω ) = Re { X ( jω )} + j Im { X ( jω )} the spectrum takes the form of: an even function of ω (represented by cosines), Re{ X ( jω )} = Re{ X ( − jω )} or an odd function of ω (represented by sinuses), Im{ X ( jω )} = − Im{ X ( − jω )} Considering a phasor notation X ( jω ) = X ( jω ) e j∠X ( jω ) The magnitude is an even function of ω | X ( jω )| =| X ( − jω )| The phase, an odd even function of ω ∠X ( jω ) = −∠X ( − jω ) ECSE303 Chap. 4 MR - 3/3/2010 53 FT x(t ) ↔ X ( jω ) FT x (t ) ↔ X ∗ (− jω ) ∗ If x(t) is real and even (combination of cosines), then the spectrum is real and even X ( jω ) = X ( − jω ) If x(t) is real and odd (sum of sinus), then the spectrum is purely imaginary and odd X ( jω ) = − X ( − jω ) = − X ∗ ( jω ) ECSE303 Chap. 4 MR - 3/3/2010 54 Differentiation Differentiating a signal in the time domain is equivalent to a multiplication of the Fourier transform by jω FT x(t ) ↔ X ( jω ) FT d x (t ) ↔ jω X ( jω ) dt 1 x (t ) = 2π X ( jω ) = +∞ ∫ X ( jω )e jωt d ω −∞ +∞ ∫ x (t )e − jωt dt −∞ ECSE303 Chap. 4 MR - 3/3/2010 55 Integration Integrating a signal in the time domain is equivalent to a division of the Fourier transform by jω. Also, an infinite DC term πX(0)δ(ω) must be added to the Fourier transform. FT x(t ) ↔ X ( jω ) t 1 ∫ x(τ )dτ ↔ jω X ( jω ) + π X (0)δ (ω ) −∞ FT 1 x (t ) = 2π X ( jω ) = +∞ ∫ X ( jω )e jωt d ω −∞ +∞ ∫ x (t )e − jωt dt −∞ ECSE303 Chap. 4 MR - 3/3/2010 56 t 1 ∫ x(τ )dτ ↔ jω X ( jω ) + π X (0)δ (ω ) −∞ FT Example Find the Fourier transform of the running integral of x(t)=δ(t) FT δ (t ) ↔ 1 t 1 u (t ) ↔ + πδ (ω ) jω FT 1 ∫ δ (τ )dτ ↔ jω + πδ (ω ) −∞ FT Note that FT{u(t)} cannot be found from the Fourier transform integral alone because X ( jω ) = +∞ ∫ e − jωt dt = undefined 0 u(t) is not integrable. Even then, the spectrum found for u(t) is often used. ECSE303 Chap. 4 MR - 3/3/2010 57 FT u (t ) ↔ t 1 + πδ (ω ) jω 1 X ( jω ) + π X (0)δ (ω ) Example ∫ jω Find the Fourier transform of the running integral of x(t)=u(t)-u(t-t0) FT x(τ )dτ ↔ −∞ FT u (t ) − u (t − t0 ) ↔ ⎡1 ⎤ 1 + πδ (ω ) − e − jωt0 ⎢ + πδ (ω ) ⎥ jω ⎣ jω ⎦ (1 − e u (t ) − u (t − t ) ↔ FT − jωt0 ) jω 0 FT u (t ) − u (t − t0 ) ↔ e − jωt0 2 sin (ωt0 2 ) ω2 FT u (t ) − u (t − t0 ) ↔ t0 e − jωt0 2 sinc (ωt0 2π ) t FT ∫ u (τ ) − u (τ − t0 )dτ ↔ −∞ ECSE303 Chap. 4 MR - 3/3/2010 1 ⎡t0 e − jωt0 2 sinc (ωt0 2π ) ⎤ + π t0δ (ω ) ⎦ jω ⎣ 59 Parseval's Relation Just like the total average power of a periodic signal is equal to the sum of the powers of all its harmonics, ∞ 2 1 P = ∫ x ( t ) dt = ∑ ak TT k =−∞ 2 the total energy in an aperiodic signal is equal to the total energy in its spectrum. +∞ ∫ x (t ) −∞ ECSE303 Chap. 4 MR - 3/3/2010 2 1 dt = 2π +∞ ∫ X ( jω ) 2 dω −∞ 60 +∞ ∫ x (t ) Example Find the energy contained in 2 −∞ 1 dt = 2π +∞ ∫ X ( jω ) 2 −∞ 1 e u (t ) ↔ jω + a FT − at In time +∞ ∫ x ( t ) dt = 2 −∞ +∞ ∫ e − at u ( t ) dt −∞ +∞ = e −2 at dt ∫ 0 1 = 2a ECSE303 Chap. 4 MR - 3/3/2010 In frequency 2 1 2π +∞ ∫ X ( jω ) −∞ 2 1 dω = 2π 1 = 2π 1 = 2π = 1 2a +∞ ∫ −∞ 2 1 dω jω + a +∞ 1 ∫ ω 2 + a 2 dω −∞ +∞ ⎡1 ⎛ ω ⎞⎤ ⎢ a atan ⎜ a ⎟ ⎥ ⎝ ⎠ ⎦ −∞ ⎣ 61 dω Energy-density spectrum The energy density spectrum of an aperiodic signal is defined as the magnitude squared of its spectrum, i.e., 2 X ( jω ) We can find the energy of a signal in a given frequency band by integrating E[ωa ,ωb ] ECSE303 Chap. 4 MR - 3/3/2010 1 = 2π ωb ∫ ω X ( jω ) dω 2 a 62 Note that it is customary to include the negative frequency band as well. For example, to estimate the energy contained in a signal between, say, f=5 kHz and f=10 kHz E[10000π ,20000π ] −10000π ⎡ 20000π ⎤ 1 2 2 = ⎢ ∫ X ( jω ) d ω + ∫ X ( jω ) dω ⎥ 2π ⎣10000π −20000π ⎦ And if the signal is real | X ( jω )| =| X ( − jω )| E[10000π ,20000π ] = ECSE303 Chap. 4 MR - 3/3/2010 1 π 20000π ∫π X ( jω ) d ω 2 10000 63 Duality The Fourier transform pair is symmetric. This results in a duality between the time domain and the frequency domain. For example, x1 ( t ) X1 ( jω ) FT x1 ( t ) ↔ X 1 ( jω ) 1 2T1 Surf. area t −T1 ω −π T1 T1 ωc π x2 ( t ) π/half-width T1 X 2 ( jω ) FT x2 ( t ) ↔ X 2 ( jω ) π 1 ω t −π ECSE303 Chap. 4 MR - 3/3/2010 ωc π ωc −ωc ωc 64 Example Find the frequency derivative of X(jω)=1/(jω+a). What does it represents in the time domain? We know that e − at 1 u (t ) ↔ jω + a FT From the definition of the Fourier transform X ( jω ) = +∞ ∫ x (t )e − jωt dt ⇒ −∞ +∞ dX ( jω ) = ∫ − jtx (t )e − jωt dt dω −∞ Taking the derivative of X(jω) thus leads to ⎛1⎞ − jte u ( t ) ↔− j ⎜ ⎟ jω + a ⎠ ⎝ − at ECSE303 Chap. 4 MR - 3/3/2010 FT ⎛1⎞ te u ( t ) ↔ ⎜ ⎟ jω + a ⎠ ⎝ − at FT 2 2 First derivative in frequency is similar to a multiplication by –jt in the time domain 65 Example The impulse response of a system for which H(jω)=e-a|ω| {Re(a)>0} is 1 h (t ) = 1 2π +∞ ∫ e − aω e jωt d ω + 0 1 2π 0 ∫ x (t ) = e aω e jωt d ω −∞ +∞ 0 2π X ( jω ) = ⎡ eω ( jt − a ) ⎤ 1 ⎡ eω ( jt + a ) ⎤ ⎢ ⎥+ ⎢ ⎥ 2π ⎣ jt + a ⎦ −∞ jt − a ⎦ 0 ⎣ FT 1 2a 1 2a −a ω ↔e = 2π ( t 2 + a 2 ) 2π t 2 + a 2 +∞ ∫ X ( jω )e jωt d ω −∞ +∞ ∫ x (t )e − jωt dt −∞ 1 = 2π ( Compare with ) e −a t FT ↔ 2a ω 2 + a2 {Re ( a ) > 0} {Re ( a ) > 0} Duality leads to equivalent profiles when converting from one domain (t, ω) to the opposite one (ω, t) with ECSE303 Chap. 4 a factor of 2π MR - 3/3/2010 66 Suggested exercises 4.1, 4.2, 4.4, 4.6, 4.8, 4.12, 4.17 Note: for those attending the Thursday tutorial session, please contact Vahid at vahid.raissidehkordi@mcgill.ca for an appointment ECSE-303A Signals and Systems I Wednesday, March 03, 2010 LECTURE 19 Sections 4.4-4.7 Convolution, multiplication, tables and systems characterized by linear differential equations Mid term #2 Friday Feb 5th Chapter 4: The continuous-Time Fourier transform 4.0 Introduction 4.1 The continuous-time Fourier transform 4.2 The Fourier transform for periodic signals 4.3 Properties of the continuous-time Fourier transform 4.4 The convolution property 4.5 The multiplication property 4.6 Tables of Fourier properties and of basic Fourier transform pairs 4.7 Systems characterized by linear constantcoefficient differential equations 4.4 Convolution The convolution of two signals in the time domain is equivalent to a multiplication of their Fourier transforms in the frequency domain. FT a (t ) ∗ b(t ) ↔ A( jω ) B( jω ) A direct application is the calculation of the response of an LTI system to an arbitrary input signal: Y ( jω ) = H ( jω ) X ( jω ) The output signal in the time-domain is obtained by taking the inverse Fourier transform of its spectrum. ECSE303 Chap. 4 MR - 3/3/2010 70 Proof ∞ y (t ) = ∫ x(τ )h(t − τ )dτ −∞ ⎡∞ ⎤ − jωt Y ( jω ) = ∫ ⎢ ∫ x (τ ) h(t − τ ) dτ ⎥ e dt −∞ ⎣ −∞ ⎦ ∞ ⎡∞ ⎤ − jω t = ∫ x (τ ) ⎢ ∫ h(t − τ )e dt ⎥ dτ −∞ ⎣ −∞ ⎦ ∞ ∞ = ∫ −∞ x (τ ) ⎡ e − jωτ H ( jω ) ⎤ dτ ⎣ ⎦ = H ( jω ) X ( jω ) ECSE303 Chap. 4 MR - 3/3/2010 71 Example Find the spectrum of a system with impulse response h(t)=e-atsin(ω0t)u(t) to an input x(t)=rectangle (width=2t0,center=t0) H ( jω ) = 2 ( jω + a ) 2 + ω0 X ( jω ) = e ECSE303 Chap. 4 MR - 3/3/2010 ω0 − jωt0 ⎛ t0ω ⎞ 2t0 sinc ⎜ π⎟ ⎝ ⎠ Y ( jω ) = X ( jω ) H ( jω ) ⎛t ω ⎞ 2ω0t0 e − jωt0 sinc ⎜ 0 ⎟ ⎝π ⎠ = 2 ( jω + a ) 2 + ω0 72 Chapter 4: The continuous-Time Fourier transform 4.0 Introduction 4.1 The continuous-time Fourier transform 4.2 The Fourier transform for periodic signals 4.3 Properties of the continuous-time Fourier transform 4.4 The convolution property 4.5 The multiplication property 4.6 Tables of Fourier properties and of basic Fourier transform pairs 4.7 Systems characterized by linear constantcoefficient differential equations 4.5 Multiplication This property is the dual of the convolution property. The multiplication of two signals in the time domain is equivalent to a convolution of their spectra. 1 a (t )b(t ) ↔ A( jω ) ∗ B ( jω ) 2π FT Modulation is based on this property. ECSE303 Chap. 4 MR - 3/3/2010 74 Example Consider the amplitude-modulated signal x(t) described by y (t ) = c(t ) x(t ) The Fourier transform of the carrier c(t)=cos(ω0t) is C ( jω ) = πδ (ω − ω0 ) + πδ (ω + ω0 ) Spectrally, it is represented by two impulses of surface area π, one at -ω0 and the other at +ω0. ECSE303 Chap. 4 MR - 3/3/2010 75 Suppose that the spectra of x(t) and c(t) look like this Then the Fourier transform of the output signal FT 1 x(t)c(t) looks like this c(t ) x(t ) ↔ C ( jω ) ∗ X ( jω ) 2π Hence multiplication of a signal by a sinusoid duplicates the spectrum at the sinusoid’s frequency band. ECSE303 Chap. 4 MR - 3/3/2010 76 Application Voice has a bandwidth of less than 5kHz . It is transmitted over typical AM radio at carrier modulation frequencies in the range 500kHz to 1.5 MHz. ECSE303 Chap. 4 MR - 3/3/2010 77 Example Knowing that a triangular function (=convolution between two identical square functions) in the frequency domain is represented as a sinc2 in the time domain, and using the properties of Fourier transforms, evaluate the following integral ∞ ∫ ⎡ sinc ( t )⎤ ⎡3 sinc ( 3t )⎤ dt ⎣ ⎦⎣ ⎦ −∞ We first modify the argument of the integral ∞ ∞ −∞ −∞ ⎣ ⎦⎣ ⎦ ∫ ⎡ sin c ( t )⎤ ⎡3 sin c ( 3t )⎤ dt = ∫ x ( t ) dt x ( t ) = ⎡ sin c ( t ) ⎤ ⎡3 sin c ( 3t ) ⎤ ⎣ ⎦⎣ ⎦ ECSE303 Chap. 4 MR - 3/3/2010 78 ωc π x2 ( t ) X 2 ( jω ) FT x2 ( t ) ↔ X 2 ( jω ) 1 ω t −π ωc π ωc −ωc ωc ⎡1, ω < π ⎤ ⎡1, ω < 3π ⎤ ⎢ ⎥* ⎢ ⎥ 0 , ω > π ⎦ ⎣0 , ω > 3π ⎦ ⎣ ⎡ω + 4π −4π < ω < −2π ⎤ 1⎢ 2π = −2π < ω < 2π ⎥ ⎥ 2π ⎢ ⎢ 4π − ω 2π < ω < 4π ⎥ ⎣ ⎦ 1 X ( jω ) = 2π 1 ⎡ − ω + 4π ω < 4π ⎤ 1 ⎡ − ω + 2π ω < 2π ⎤ = ⎢ ⎥− ⎢ ⎥ 0 0 ω > 4π ⎦ 2π ⎣ ω > 2π ⎦ 2π ⎣ 1 ⎡1 ω < 2π ⎤ ⎡1 ω < 2π ⎤ 1 ⎡1 ω < π ⎤ ⎡1 = ⎢ ⎥* ⎢ ⎥− ⎢ ⎥* ⎢ 2π ⎣0 ω > 2π ⎦ ⎣0 ω > 2π ⎦ 2π ⎣0 ω > π ⎦ ⎣0 ECSE303 Chap. 4 MR - 3/3/2010 ω <π⎤ ⎥ ω >π⎦ 79 x ( t ) = ⎡ 2 sin c ( 2t ) ⎤ − ⎡ sin c ( t ) ⎤ ⎣ ⎦⎣ ⎦ 2 ∞ ∞ −∞ −∞ ∫ x ( t ) dt = ∫ ⎡2 sin c ( 2t )⎤ ⎣ ⎦ 2 2 − ⎡ sin c ( t ) ⎤ dt ⎣ ⎦ 2 This integral can easily be performed using Parseval’s relation 1 ⎡1 ⎣ ⎦⎣ ⎦ ∫ ⎡2 sin c ( 2t )⎤ − ⎡ sin c ( t )⎤ dt = 2π −∞ ⎢0 ∫⎣ −∞ ∞ 2 ∞ 2 1 = 2π 2 2 ω < 2π ⎤ ⎡1 ω < π ⎤ ⎥ −⎢ ⎥ dω 2π ⎦ ⎣0 ω > π ⎦ ω> 2π 1 12 d ω − ∫ 2π −2π π 12 d ω ∫ −π 1 1 4π − 2π 2π 2π =1 = ECSE303 Chap. 4 MR - 3/3/2010 80 Example Problem 4.21f: Calculate the Fourier transform of the signal ⎡ sin (π t ) ⎤ ⎡ sin ( 2π ( t − 1) ) ⎤ ⎥ ⎢ ⎥⎢ ⎣ π t ⎦ ⎢ π ( t − 1) ⎥ ⎣ ⎦ ECSE303 Chap. 4 MR - 3/3/2010 81 ωc π x2 ( t ) X 2 ( jω ) FT x2 ( t ) ↔ X 2 ( jω ) 1 ω t −π ωc π ωc −ωc ωc ⎡ sin (π t ) ⎤ ⎡ sin ( 2π ( t − 1) ) ⎤ ⎥ = ⎡ sinc ( t ) ⎤ ⎡ 2 sin c ( 2t ) ∗ δ ( t − 1) ⎤ ⎢ ⎥⎢ ⎦⎣ ⎦ π t ⎦ ⎢ π ( t − 1) ⎥ ⎣ ⎣ ⎣ ⎦ FT ⎧ 1, ω < π δ ( t − 1) ↔ e − jω sinc ( t ) ↔ ⎨ ⎩0 , otherwise FT ⎧ 1, ω < 2π 2 sinc ( 2t ) ↔ ⎨ ⎩0 , otherwise FT ECSE303 Chap. 4 MR - 3/3/2010 82 1 ⎡ ⎤⎡ ⎤ ⎣ sinc ( t ) ⎦ ⎣ 2 sin c ( 2t ) ∗ δ ( t − 1) ⎦ ↔ 2π FT Sq4π ( Ω ) ⎡ e − jΩ , Ω < 2π ⎤ Sq4π ( Ω ) = ⎢ ⎥ 0 , otherwise ⎦ ⎣ 1 -2π -π π 2π Sq2π (ω − Ω ) 1 ω-π ECSE303 Chap. 4 MR - 3/3/2010 ω+π ⎡ 1, ω < π ⎤ ⎡ e − jω , ω < 2π ⎤ ⎥ ⎢ ⎥* ⎢ 0 , otherwise ⎦ ⎣ 0 , otherwise ⎦ ⎣ Ω ⎡1, ω − Ω < π ⎤ Sq2π (ω − Ω ) = ⎢ ⎥ 0 , otherwise ⎦ ⎣ Ω 83 Pattern 1: No overlap Pattern 2: Partial overlap ω-π ω+π -2π ω +π ∫π −2 -3π<ω<-π -π π Sq4π ( Ω ) Sq2π (ω − Ω )d Ω = Ω 2π ω +π ∫π e − jΩ d Ω −2 e − j ( ω +π ) − e j 2π = −j ( ) = j −e − jω − 1 ECSE303 Chap. 4 MR - 3/3/2010 84 Pattern 3: Full overlap ω-π -2π ω +π ∫π −2 -π -π<ω<π ω+π π 2π Sq4π ( Ω ) Sq2π (ω − Ω )d Ω = ω +π ∫π ω Ω e − jΩ d Ω − e − j ( ω + π ) − e − j ( ω −π ) = −j ( = j −e − jω + e − jω ) =0 ECSE303 Chap. 4 MR - 3/3/2010 85 Pattern 4: Partial overlap Pattern 5: No overlap π<ω<3π ω-π -2π ω +π ∫π −2 -π π Sq4π ( Ω ) Sq2π (ω − Ω )d Ω = ω+π Ω 2π 2π ∫π ω e − jΩ d Ω − e − j ( 2 π ) − e − j ( ω −π ) = −j ( = j 1 + e − jω ECSE303 Chap. 4 MR - 3/3/2010 ) 86 1 ⎡ sinc ( t ) ⎤ ⎡ 2 sin c ( 2t ) ∗ δ ( t − 1) ⎤ ↔ ⎣ ⎦⎣ ⎦ 2π FT ⎡ 1, ω < π ⎤ ⎡ e − jω , ω < 2π ⎤ ⎥ ⎢ ⎥* ⎢ 0 , otherwise ⎦ ⎣ 0 , otherwise ⎦ ⎣ ( ) ( ) ⎡ − j 1 + e − jω FT 1 ⎢ ⎡ sinc ( t ) ⎤ ⎡ 2 sin c ( 2t ) ∗ δ ( t − 1) ⎤ ↔ 0 ⎣ ⎦⎣ ⎦ 2π ⎢ ⎢ j 1 + e − jω ⎢ ⎣ ECSE303 Chap. 4 MR - 3/3/2010 −3π < ω < −π ⎤ ⎥ −π < ω < π ⎥ ⎥ π < ω < 3π ⎥ ⎦ 87 Chapter 4: The continuous-Time Fourier transform 4.0 Introduction 4.1 The continuous-time Fourier transform 4.2 The Fourier transform for periodic signals 4.3 Properties of the continuous-time Fourier transform 4.4 The convolution property 4.5 The multiplication property 4.6 Tables of Fourier properties and of basic Fourier transform pairs 4.7 Systems characterized by linear constantcoefficient differential equations ECSE303 Chap. 4 MR - 3/3/2010 89 +Duality: but be careful as ω and t can be interchanged by including a multiplication factor of 2π ECSE303 Chap. 4 MR - 3/3/2010 90 ECSE303 Chap. 4 MR - 3/3/2010 91 ECSE303 Chap. 4 MR - 3/3/2010 92 The Inverse Fourier Transform 1) By direct calculation of the integral Example Consider the ideal lowpass filter with cutoff frequency ωc and given the frequency spectrum R1, H ( jω ) = S T0, ω < ωc ω > ωc The corresponding impulse response is calculated from the frequency spectrum ECSE303 Chap. 4 MR - 3/3/2010 •93 R1, H ( jω ) = S T0, 1 x (t ) = 2π ω < ωc ω > ωc X ( jω ) = h(t ) = 1 2π 1 = 2π +∞ ∫ +∞ ∫ X ( jω )e jωt d ω −∞ +∞ ∫ x (t )e − jωt dt −∞ H ( jω )e jωt d ω −∞ + ωc ∫ ω − e jωt d ω c 1 jωt ωc ⎡e ⎤ = ⎦ −ωc 2π ( jt ) ⎣ ωc ωc sin( π tπ ) = ωcπ t π π ωc ⎛ ωc ⎞ = sinc ⎜ t ⎟ π ⎝π ⎠ ECSE303 Chap. 4 MR - 3/3/2010 94 Thus the impulse response of an ideal lowpass filter is a sinc function extending from t=-∞ to t=+∞ ωc π ωc ⎛ ωc ⎞ h(t ) = sinc ⎜ t ⎟ π ⎝π ⎠ t π − ωc π ωc Take note that this function is real but noncausal ECSE303 Chap. 4 MR - 3/3/2010 95 2) Using transforms pairs from tables The Fourier transform is often in the form of a rational function of jω (a ratio of polynomials). 2 − jω Y ( jω ) = 2 ω − 5 jω − 6 To determine the inverse Fourier transform: • Perform a partial fraction expansion to simplify the expression as a sum of first order polynomial terms. • Apply the Inverse Fourier transform to the simplified terms and use the linear property of FT. ECSE303 Chap. 4 MR - 3/3/2010 96 Example Consider the response of an LTI system with impulse response h(t)=e-2tu(t) to the input x(t)=e-3tu(t) using Fourier transforms. The solution is given from multiplying the Fourier transforms of the input and the impulse response. 1 H ( jω ) = jω + 2 1 X ( jω ) = jω + 3 1 Y ( jω ) = X ( jω ) H ( jω ) = ( jω + 3)( jω + 2) ECSE303 Chap. 4 MR - 3/3/2010 97 The partial fraction expansion consists in expressing the function as a sum of first-order terms. Y ( jω ) = A B 1 = + ( jω + 2)( jω + 3) ( jω + 2) ( jω + 3) To solve for A and B: (1) Equate the transform with its sum of partial fractions, and let s=jω 1 A B = + ( s + 2)( s + 3) ( s + 2) ( s + 3) (2) To obtain the coefficient A, multiply both sides of the equation by (s+2) and evaluate at s=-2 1 ( s + 2) B = A+ ( s + 3) s = −2 ( s + 3) ECSE303 Chap. 4 MR - 3/3/2010 ⇒ 1 A= =1 −2 + 3 s =−2 98 Applying again step (2) for the constant B, we obtain 1 A B = + ( s + 2)( s + 3) ( s + 2) ( s + 3) 1 ( s + 3) A = +B ( s + 2) s =−3 ( s + 2) s =−3 1 ⇒ B= = −1 −3 + 2 The coefficients A and B are called residues at poles ECSE303 Chap. 4 MR - 3/3/2010 99 FT x (t ) = e u (t ) ↔ − at X ( jω ) = 1 jω + a Finally, the partial fraction expansion of the Fourier transform of the output is given by 1 1 − Y ( jω ) = ( jω + 2) ( jω + 3) Which leads to y (t ) = e −2t u (t ) − e −3t u (t ) ECSE303 Chap. 4 MR - 3/3/2010 100 Chapter 4: The continuous-Time Fourier transform 4.0 Introduction 4.1 The continuous-time Fourier transform 4.2 The Fourier transform for periodic signals 4.3 Properties of the continuous-time Fourier transform 4.4 The convolution property 4.5 The multiplication property 4.6 Tables of Fourier properties and of basic Fourier transform pairs 4.7 Systems characterized by linear constantcoefficient differential equations 4.7 Systems characterized by linear differential equations The output y(t) of a stable LTI system with impulse response h(t) and subjected to an input x(t) is given by +∞ y ( t ) = x ( t ) * h ( t ) = ∫ x (τ ) h ( t − τ ) dτ −∞ We can also calculate the output by taking the inverse Fourier transform of Y ( jω ) = H ( jω ) X ( jω ) Represented in a block diagram as X ( jω ) ECSE303 Chap. 4 MR - 3/3/2010 H ( jω ) Y ( jω ) 102 For a cascade of two stable LTI systems with impulse responses h1(t), h2(t), we have X ( jω ) H1 ( jω ) H 2 ( jω ) Y ( jω ) y ( t ) = x ( t ) * h1 ( t ) * h2 ( t ) Y ( jω ) = H2 ( jω ) H1 ( jω ) X ( jω ) ECSE303 Chap. 4 MR - 3/3/2010 103 For a parallel connection of two stable LTI systems with impulse responses h1(t), h2(t), we have H1 ( jω ) X ( jω ) + Y ( jω ) + H 2 ( jω ) y ( t ) = ⎡ h1 ( t ) + h2 ( t ) ⎤ * x ( t ) ⎣ ⎦ Y ( jω ) = [ H1 ( jω ) + H2 ( jω )] X ( jω ) ECSE303 Chap. 4 MR - 3/3/2010 104 For a feedback interconnection of two stable LTI systems with impulse responses h1(t), h2(t), we have E ( jω ) = X ( jω ) − Y ( jω ) H 2 ( jω ) Y ( jω ) = E ( jω ) H1 ( jω ) H1 ( jω ) X ( jω ) Y ( jω ) = 1 + H1 ( jω ) H 2 ( jω ) ECSE303 Chap. 4 MR - 3/3/2010 105 LTI Differential Systems Consider the stable LTI system defined by an Nthorder linear constant-coefficient differential equation initially at rest: d k y (t ) M d k x (t ) ∑0 ak dt k = ∑0 bk dt k k= k= N Assume that FT FT x ( t ) ↔ X ( jω ) y ( t ) ↔ Y ( jω ) Recall that differentiation in the time domain is equivalent to a multiplication in the Fourier domain by jω. Thus, . N M k =0 k =0 a k ( jω ) k Y ( jω ) = ∑ bk ( jω ) k X ( jω ) ∑ ECSE303 Chap. 4 MR - 3/3/2010 106 N M k =0 k =0 a k ( jω ) k Y ( jω ) = ∑ bk ( jω ) k X ( jω ) ∑ Because Y ( jω ) H ( jω ) = X ( jω ) the frequency response of the system is given by M Y ( jω ) H ( jω ) = = X ( jω ) bk ( jω ) k ∑ k =0 N a k ( jω ) k ∑ k =0 ECSE303 Chap. 4 MR - 3/3/2010 107 Example Consider the LTI differential system dy ( t ) dt + ay ( t ) = Ax ( t ) Applying the Fourier transform operation leads to ( jω + a ) Y ( jω ) = AX ( jω ) The frequency response of the system is Y ( jω ) A H ( jω ) = = X ( jω ) jω + a And thus it has an impulse response h ( t ) = Ae − at u ( t ) ECSE303 Chap. 4 MR - 3/3/2010 108 Example The frequency response of the second-order LTI differential system d 2 y (t ) dy (t ) dx (t ) +3 + 2 y (t ) = − x (t ) 2 dt dt dt is calculated as follows: [( jω ) 2 + 3 jω + 2]Y ( jω ) = ( jω − 1) X ( jω ) Y ( jω ) jω − 1 H ( jω ) = = X ( jω ) ( jω ) 2 + 3 jω + 2 ECSE303 Chap. 4 MR - 3/3/2010 109 H ( jω ) = jω − 1 ( jω ) 2 + 3 jω + 2 Suppose we want to obtain the step response of this system. The Fourier transform of the step function is X ( jω ) = 1 + πδ (ω ) jω Then the output Y ( jω ) = H ( jω ) X ( jω ) ⎡ ⎤⎡ 1 ⎤ jω − 1 =⎢ + πδ (ω ) ⎥ ⎥⎢ 2 ⎢ ( jω ) + 3 jω + 2 ⎥ ⎣ jω ⎦ ⎣ ⎦ jω − 1 1 = − πδ (ω ) Partial fraction 2 ⎡( jω ) + 3 jω + 2 ⎤ jω 2 expansion ⎣ ⎦ ECSE303 Chap. 4 MR - 3/3/2010 110 Y ( jω ) = jω − 1 1 − πδ ( jω ) ⎡( jω )2 + 3 jω + 2 ⎤ jω 2 ⎣ ⎦ Expanding the rational function on the right-hand side into partial fractions, we get (s=jω) s −1 s −1 A B C = =+ + s 2 + 3s + 2 s ( s + 1)( s + 2) s s s + 1 s + 2 and the coefficients are calculated as follows A= s −1 1 =− ( s + 1)( s + 2 ) s =0 2 s −1 B= =2 s ( s + 2 ) s =−1 C= ECSE303 Chap. 4 MR - 3/3/2010 s −1 3 =− 2 s ( s + 1) s =−2 s −1 = 2 ⎡ s + 3s + 2 ⎤ s ⎣ ⎦ −1 2 3 + − 2s s + 1 2 ( s + 2 ) 111 Hence, jω − 1 1 Y ( jω ) = − πδ (ω ) 2 ⎡( jω ) + 3 jω + 2 ⎤ jω 2 ⎣ ⎦ ⎛ 1 ⎞ 3⎛ 1 ⎞ 1 11 2⎜ =− + ⎟− ⎜ ⎟ − πδ (ω ) 2 jω ⎝ jω + 1 ⎠ 2 ⎝ jω + 2 ⎠ 2 ⎤ ⎛ 1 ⎞ 3⎛ 1 ⎞ 1⎡ 1 =− ⎢ + πδ (ω ) ⎥ + 2 ⎜ ⎟− ⎜ ⎟ 2 ⎣ jω jω + 1 ⎠ 2 ⎝ jω + 2 ⎠ ⎦ ⎝ We find the output by inspection LM− 1 + 2e y (t ) = N2 ECSE303 Chap. 4 MR - 3/3/2010 −t OP Q 3 −2 t − e u( t ) 2 112 Example Find h(t) for the following stable second-order LTI differential system d 2 y (t ) dy (t ) + + y (t ) = x (t ) 2 dt dt The frequency response of this system is given by H ( jω ) = = ECSE303 Chap. 4 MR - 3/3/2010 Y ( jω ) X ( jω ) 1 ( jω ) 2 + jω + 1 113 Letting s=jω, and expanding the right-hand side into partial fractions, we get 1 1 = s 2 + 1s + 1 ⎡ ⎛ 1 3 ⎞⎤ ⎡ ⎛ 1 3 ⎞⎤ ⎢s + ⎜ + j ⎟⎥ ⎢ s + ⎜ − j ⎟⎥ 2 2 ⎠⎥ ⎢ ⎝ 2 2 ⎠⎥ ⎢⎝ ⎣ ⎦⎣ ⎦ A B = + ⎛1 ⎛1 3⎞ 3⎞ s+⎜ + j ⎟ s+⎜ − j ⎟ 2⎠ 2⎠ ⎝2 ⎝2 the residues at poles are then A= 1 ⎛1 3⎞ s+⎜ − j ⎟ 2 2 ⎠ ⎛1 ⎝ s =−⎜ + j ⎜ ⎝2 ECSE303 Chap. 4 MR - 3/3/2010 =j 3⎞ ⎟ 2⎟ ⎠ 1 3 B= 1 ⎛1 3⎞ s+⎜ + j ⎟ 2 2 ⎠ ⎛1 ⎝ s =−⎜ − j ⎜ ⎝2 =−j 1 3 3⎞ ⎟ 2⎟ ⎠ 114 Hence, 1 ⎡ ⎛1 ⎛1 3 ⎞⎤ ⎡ 3 ⎞⎤ ⎢ jω + ⎜ + j ⎟ ⎥ ⎢ jω + ⎜ − j ⎟⎥ 2 2 ⎠⎥ ⎢ 2 2 ⎠⎥ ⎢ ⎝ ⎝ ⎣ ⎦⎣ ⎦ j 1 j 1 = − ⎛1 ⎛1 3 3 3⎞ 3⎞ jω + ⎜ + j jω + ⎜ − j ⎟ ⎟ 2 2⎠ 2 2⎠ ⎝ ⎝ H ( jω ) = we find h(t) ⎡ j −1t− j 3t j − 1 t + j 23 t ⎤ h(t ) = ⎢ e2 2 − e2 ⎥ u (t ) 3 ⎢3 ⎥ ⎣ ⎦ 3 j t⎤ j − 1 t ⎡ − j 23 t = − e 2 ⎥ u (t ) e 2 ⎢e 3 ⎢ ⎥ ⎣ ⎦ ECSE303 Chap. 4 MR - 3/3/2010 2 −1t 3 = e 2 sin( t )u (t ) 2 3 115 Example Find the differential equation that leads to ( jω ) 2 + ωc 2 jω H hp ( jω ) = ( jω ) 2 + ωc 2 jω + ωc2 Y ( jω ) H hp ( jω ) = X ( jω ) ⎡( jω ) 2 + ωc 2 jω + ωc2 ⎤ Y ( jω ) = ⎡( jω ) 2 + ωc 2 jω ⎤ X ( jω ) ⎣ ⎦ ⎣ ⎦ d 2 y (t ) dy (t ) d 2 x(t ) dx(t ) 2 + ωc 2 + ω c y (t ) = + ωc 2 2 2 dt dt dt dt ECSE303 Chap. 4 MR - 3/3/2010 116 Suggested exercises 4.18-4.20, 4.25, 4.26, 4.33 ECSE303 Chap. 4 MR - 3/3/2010 121 ...
View Full Document

Ask a homework question - tutors are online