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6: Fourier Transform Periodic Signals Duality Time Shifting and Scaling Gaussian Pulse Summary

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6: Fourier Transform Periodic Signals Duality Time Shifting and Scaling Gaussian Pulse Summary 6: Fourier ⊲ Transform Fourier Series as T → ∞ Fourier Transform Fourier Transform Examples Dirac Delta Function Dirac Delta Function: Scaling and Translation Dirac Delta Function: Products and Integrals 6: Fourier Transform Periodic Signals Duality Time Shifting and Scaling Gaussian Pulse Summary E1.10 Fourier Series and Transforms (2014-5559) Fourier Transform: 6 – 1 / 12 Fourier Series as T → ∞ 6: Fourier Transform i2πnF t Fourier Series as Fourier Series: u(t) = n∞= Une ⊲ T → ∞ −∞ Fourier Transform 1 Fourier Transform The harmonic frequenciesP are nF ∀n and are spaced F = T apart. Examples Dirac Delta Function As T gets larger, the harmonic spacing becomes smaller. Dirac Delta Function: Scaling and e.g. T = 1 s ⇒ F = 1Hz Translation Dirac Delta Function: T = 1 day ⇒ F = 11.57 µHz Products and Integrals Periodic Signals If T → ∞ then the harmonic spacing becomes zero, the sum becomes an Duality Time Shifting and integral and we get the Fourier Transform: Scaling + i πft Gaussian Pulse ∞ 2 u(t) = f= U(f)e df Summary −∞ Here, U(f), is the spectralR density of u(t). • U(f) is a continuous function of f . • U(f) is complex-valued. • u(t) real ⇒ U(f) is conjugate symmetric ⇔ U(−f) = U(f)∗. • Units: if u(t) is in volts, then U(f)df must also be in volts ⇒ U(f) is in volts/Hz (hence “spectral density”). E1.10 Fourier Series and Transforms (2014-5559) Fourier Transform: 6 – 2 / 12 Fourier Transform 6: Fourier Transform i2πnF t Fourier Series as Fourier Series: u(t) = n∞= Une T → ∞ −∞ ⊲ Fourier Transform The summation is over a set of equally spaced frequencies Fourier Transform P 1 Examples fn = nF where the spacing between them is ∆f = F = T . Dirac Delta Function Dirac Delta Function: i2πnF t 0.5T i2πnF t Scaling and Un = u(t)e− = ∆f t= 0.5T u(t)e− dt Translation − Dirac Delta Function: Products and Spectral Density: If u(t) has finite energy,R Un → 0 as ∆f → 0. So we Un Integrals define a spectral density, U(fn) = , on the set of frequencies {fn}: Periodic Signals ∆f Duality . T Un 0 5 i2πfnt Time Shifting and U(fn) = ∆f = t= 0.5T u(t)e− dt Scaling − Gaussian Pulse we can write [Substitute Un = U(fn)∆f] Summary R i2πfnt u(t) = n∞= U(fn)e ∆f −∞ Fourier Transform:PNow if we take the limit as ∆f → 0, we get u(t) = ∞ U(f)ei2πftdf [Fourier Synthesis] −∞ ∞ i2πft U(f) =R t= u(t)e− dt [Fourier Analysis] −∞ R Un For non-periodic signals Un → 0 as ∆f → 0 and U(fn) = ∆f remains finite. However, if u(t) contains an exactly periodic component, then the corresponding U(fn) will become infinite as ∆f → 0. We will deal with it. E1.10 Fourier Series and Transforms (2014-5559) Fourier Transform: 6 – 3 / 12 Fourier Transform Examples 6: Fourier Transform 1 Example 1: a=2 Fourier Series as 0.5 T → ∞ u(t) at 0 Fourier Transform e− t ≥ 0 -5 0 5 Time (s) Fourier Transform u(t) = 0.5 ⊲ Examples 0 t < 0 0.4 ( 0.3 0.2 Dirac Delta Function |U(f)| 0.1 Dirac Delta Function: i2πft -5 0 5 Scaling and U(f) = ∞ u(t)e− dt Frequency (Hz) Translation −∞ 0.5 Dirac Delta Function: at i2πft 0 Products and = ∞ e− e− dt R 0 -0.5 Integrals (rad/pi) <U(f) -5 0 5 ( a i2πf)t Frequence (Hz) Periodic Signals = ∞ e − − dt Duality R0 1 ( a i2πf)t 1 Time Shifting and = − e − − ∞ = Scaling Ra+i2πf 0 a+i2πf 1 Gaussian Pulse a=2 0.5 Summary v(t) Example 2: 0 -5 0 5 v(t) = e a t Time (s) − | | 1 0.5 i2πft |V(f)| V (f) = ∞ v(t)e dt 0 − -5 0 5 −∞ Frequency (Hz) 0 R at i2πft ∞ at i2πft = e e− dt + 0 e− e− dt −∞ 0 1 e(a i2πf)t 1 e( a i2πf)t ∞ = Ra i2πf − R + a+−i2πf − − 0 −1 1 −∞ 2a = + = 2 2 2 [v(t) real+symmetric a i2πf a+i2πf a +4π f − ⇒ V (f) real+symmetric] E1.10 Fourier Series and Transforms (2014-5559) Fourier Transform: 6 – 4 / 12 Dirac Delta Function 6: Fourier Transform δ (x) Fourier Series as We define a unit area pulse of width w as 4 0.2 T → ∞ δ (x) 1 0.5 Fourier Transform 2 −0.5w ≤ x ≤ 0.5w δ (x) Fourier Transform w 3 dw(x) = 0 Examples 0 otherwise -3 -2 -1 0 1 2 3 Dirac Delta ( x Function ⊲ 1 Dirac Delta Function: This pulse has the property that its integral equals δ(x) Scaling and 0.5 Translation 1 for all values of w: Dirac Delta Function: 0 Products and ∞ -3 -2 -1 0 1 2 3 Integrals x= dw(x)dx = 1 x Periodic Signals −∞ Duality If we makeR w smaller, the pulse height increases to preserve unit area. Time Shifting and Scaling We define the Dirac delta function as δ(x) = limw 0 dw(x) Gaussian Pulse → Summary • δ(x) equals zero everywhere except at x = 0 where it is infinite. • However its area still equals 1 ⇒ ∞ δ(x)dx = 1 −∞ • We plot the height of δ(x) as its areaR rather than its true height of ∞. δ(x) is not quite a proper function: it is called a generalized function. E1.10 Fourier Series and Transforms (2014-5559) Fourier Transform: 6 – 5 / 12 Dirac Delta Function: Scaling and Translation 6: Fourier Transform Translation: δ(x − a) Fourier Series as 1 δ(x) δ(x-2) T → ∞ Fourier Transform δ(x) is a pulse at x = 0 0 Fourier Transform δ Examples δ(x − a) is a pulse at x = a -0.5 (x+2) -1 Dirac Delta Function -3 -2 -1 0 1 2 3 Dirac Delta bδ x x Function: Scaling Amplitude Scaling: ( ) ⊲ and Translation 1 Dirac Delta Function: δ(x) has an area of 1 ⇔ ∞ δ(x)dx = 1 Products and δ(4x) = 0.25δ(x) −∞ 0 Integrals Periodic Signals bδ(x) has an area of b sinceR -3δ(-4x-8) = -0.75δ(x+2) -1 Duality -3 -2 -1 0 1 2 3 ∞ (bδ(x)) dx = b ∞ δ(x)dx = b x Time Shifting and Scaling −∞ −∞ Gaussian Pulse b canR be a complex numberR (on a graph, we then plot only its magnitude) Summary Time Scaling: δ(cx) dy ∞ ∞ c > 0: x= δ(cx)dx = y= δ(y) c [sub y = cx] −∞ −∞ 1 ∞ 1 1 = c y= δ(y)dy = c = c R R −∞ | | dy ∞ −∞ c < 0: x= δ(cx)dx = y=+R δ(y) c [sub y = cx] −∞ ∞ + 1 ∞ 1 1 R R= −c y= δ(y)dy = −c = c −∞ | | 1 In general, δ(cx) = c δ(x) for cR6= 0 | | E1.10 Fourier Series and Transforms (2014-5559) Fourier Transform: 6 – 6 / 12 Dirac Delta Function: Products and Integrals 6: Fourier Transform 6 Fourier Series as If we multiply δ(x − a) by a function of x: y=x2 T → ∞ y = x2 × δ(x − 2) 4 Fourier Transform 2 Fourier Transform The product is 0 everywhere except at x = 2. Examples 0 Dirac Delta Function -1 0 1 2 Dirac Delta Function: So δ(x − 2) is multiplied by the value taken by x Scaling and 2 6 δ Translation x at x = 2: y= (x-2) Dirac Delta 2 2 4 Function: x × δ(x − 2) = x x × δ(x − 2) Products and =2 2 ⊲ Integrals = 4 × δ(x − 2) 0 Periodic Signals -1 0 1 2 Duality x Time Shifting and 6 2 In general for any function, f(x), that is y=2 ×δ(x-2) = 4δ(x-2) Scaling 4 Gaussian Pulse continuous at x = a, Summary 2 f(x)δ(x − a) = f(a)δ(x − a) 0 -1 0 1 2 x Integrals: ∞ f(x)δ(x − a)dx = ∞ f(a)δ(x − a)dx −∞ −∞ = f(a) ∞ δ(x − a)dx R R = f(a) −∞ [if f(x) continuous at a] R ∞ x2 x δ x dx x2 x Example: 3 − 2 ( − 2) = 3 − 2 x=2 = 8 −∞ R E1.10 Fourier Series and Transforms (2014-5559) Fourier Transform: 6 – 7 / 12 Periodic Signals 6: Fourier Transform i2πft Fourier Series as Fourier Transform: u(t) = ∞ U(f)e df [Fourier Synthesis] T → ∞ −∞ ∞ i2πft Fourier Transform U(f) = t= u(t)e− dt [Fourier Analysis] Fourier Transform R −∞ 1.5 Examples U f . δ f . δ f 1.5δ(f+2) 1.5δ(f-2) Dirac Delta Function Example: ( ) = 1 5 ( +R 2) + 1 5 ( − 2) 1 Dirac Delta Function: i2πft 0.5 Scaling and u(t) = ∞ U(f)e df Translation 0 −∞ i2πft -3 -2 -1 0 1 2 3 Dirac Delta Function: = ∞ 1.5δ(f + 2)e df Frequency (Hz) Products and R −∞ Integrals + ∞ 1.5δ(f − 2)ei2πftdf 4 3cos(4πt) ⊲ Periodic Signals R 2 i πft−∞ i πft Duality = 1.5 e 2 + 1.5 e 2 0 f= 2 f=+2 -2 Time Shifting and R − Scaling i4πt i4πt 0 1 2 3 4 5 = 1.5 e + e− = 3 cos 4πt Time (s) Gaussian Pulse Summary If u(t) is periodic then U(f) is a sum of Dirac delta functions: i2πnF t u(t) = n∞= Une ⇒ U(f) = n∞= Unδ (f − nF ) −∞ −∞ Proof:P u(t) = ∞ U(f)ei2πftdf P −∞ ∞ i2πft =R n∞= Unδ (f − nF ) e df −∞ −∞ ∞ i2πft = R n∞=P Un δ (f − nF ) e df −∞ −∞ i2πnF t = Pn∞= UneR −∞ E1.10 Fourier Series and Transforms (2014-5559) P Fourier Transform: 6 – 8 / 12 Duality 6: Fourier Transform i2πft Fourier Series as Fourier Transform: u(t) = ∞ U(f)e df [Fourier Synthesis] T → ∞ −∞ ∞ i2πft Fourier Transform U(f) = t= u(t)e− dt [Fourier Analysis] Fourier Transform R −∞ Examples Dirac Delta Function R Dirac Delta Function: Dual transform: Scaling and Translation Suppose v(t) = U(t), then Dirac Delta Function: i πft Products and ∞ 2 V (f) = t= v(t)e− dτ Integrals −∞ Periodic Signals ∞ i2πgt V (g) = t= U(t)e− dt [substitute f = g, v(t) = U(t)] ⊲ Duality R −∞ Time Shifting and ∞ i2πgf Scaling =R f= U(f)e− df [substitute t = f] Gaussian Pulse −∞ Summary = Ru(−g) So: v(t) = U(t) ⇒ V (f) = u(−f) Example: t 2 u(t) = e−| | ⇒ U(f) = 1+4π2f 2 [from earlier] 2 f f v(t) = 1+4π2t2 ⇒ V (f) = e−|− | = e−| | E1.10 Fourier Series and Transforms (2014-5559) Fourier Transform: 6 – 9 / 12 Time Shifting and Scaling 6: Fourier Transform i2πft Fourier Series as Fourier Transform: u(t) = ∞ U(f)e df [Fourier Synthesis] T → ∞ −∞ ∞ i2πft Fourier Transform U(f) = t= u(t)e− dt [Fourier Analysis] Fourier Transform R −∞ Examples Dirac Delta Function R Dirac Delta Function: Time Shifting and Scaling: Scaling and Translation Suppose v(t) = u(at + b), then Dirac Delta Function: Products and ∞ i2πft Integrals V (f) = t= u(at + b)e− dt [now sub τ = at + b] Periodic Signals −∞ − i2πf τ b 1 Duality =Rsgn(a) ∞ u(τ)e− ( a ) dτ Time Shifting and τ= a ⊲ Scaling −∞ Gaussian Pulse R 1 a > 0 Summary note that ±∞ limits swap if a < 0 hence sgn(a) = (−1 a < 0 2 1 i πfb i π f τ a ∞ 2 a = a e τ= u(τ)e− dτ | | −∞ 2 1 πfb f i a R = a e U a | | 2πfb f 1 i a So: v(t) = u(at + b) ⇒ V (f) = a e U a | | E1.10 Fourier Series and Transforms (2014-5559) Fourier Transform: 6 – 10 / 12 Gaussian Pulse 2 6: Fourier Transform t 1 2 Fourier Series as Gaussian Pulse: u(t) = e− 2σ T → ∞ √2πσ2 Fourier Transform ∞ Fourier Transform This is a Normal (or Gaussian) probability distribution, so u(t)dt = 1.
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