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Sinusoidal Signals Sinusoidal Signals

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Sinusoidal Signals Sinusoidal Signals SINUSOIDAL SIGNALS SINUSOIDAL SIGNALS x(t) = cos(2¼ft + ') = cos(!t + ') (continuous time) x[n] = cos(2¼fn + ') = cos(!n + ') (discrete time) f : frequency (s¡1 (Hz)) ! : angular frequency (radians/s) ' : phase (radians) 2 Sinusoidal signals A cos(ω t) A sin(ω t) A 0 −A 0 T/2 T 3T/2 2T xc(t) = A cos(2¼ft) xs(t) = A sin(2¼ft) = A cos(2¼ft ¡ ¼=2) - amplitude A -amplitude A - period T = 1=f -period T = 1=f - phase: 0 -phase: ¡¼=2 3 WHY SINUSOIDAL SIGNALS? ² Physical reasons: - harmonic oscillators generate sinusoids, e.g., vibrating structures - waves consist of sinusoidals, e.g., acoustic waves or electromagnetic waves used in wireless transmission ² Psychophysical reason: - speech consists of superposition of sinusoids - human ear detects frequencies - human eye senses light of various frequencies ² Mathematical (and physical) reason: - Linear systems, both physical systems and man-made ¯lters, a®ect a signal frequency by frequency (hence low- pass, high-pass etc ¯lters) 4 EXAMPLE: TRANSMISSION OF A LOW-FREQUENCY SIGNAL USING HIGH-FREQUENCY ELECTROMAGNETIC (RADIO) SIGNAL - A POSSIBLE (CONVENTIONAL) METHOD: AMPLITUDE MODULATION (AM) Example: low-frequency signal v(t) = 5 + 2 cos(2¼f¢t); f¢ = 20 Hz High-frequency carrier wave vc(t) = cos(2¼fct); fc = 200 Hz Amplitude modulation (AM) of carrier (electromagnetic) wave: x(t) = v(t) cos(2¼fct) 5 8 6 4 2 v 0 −2 −4 −6 −8 8 6 4 2 x 0 −2 −4 −6 −8 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 t Top: v(t) (dashed) and vc(t) = cos(2¼fct). Bottom: transmitted signal x(t) = v(t) cos(2¼fct). 6 Frequency contents of transmitted signal x(t) = v(t) cos(2¼fct) = (5 + 2 cos(2¼f¢t)) cos(2¼fct) = 5 cos(2¼fct) + 2 cos(2¼f¢t) cos(2¼fct) Trigonometric identity: 1 1 cos ® cos ¯ = cos(® ¡ ¯) + cos(® + ¯) 2 2 ) 2 cos(2¼f¢t) cos(2¼fct) = cos (2¼(fc ¡ f¢)t)+cos (2¼(fc + f¢)t) 7 ) x(t) = 5 cos(2¼fct) + cos (2¼(fc ¡ f¢)t) + cos (2¼(fc + f¢)t) 5 Spectrum Spectrum of v(t) of vc(t) 2 1 - 0 20 200 Frequency Spectrum of x(t) 5 1 1 - 0 180 200 220 Frequency 8 We see that a low-frequency signal in frequency range 0 · fs · fmax (baseband signal) can be transmitted as a signal in the frequency range fc ¡ fmax · f · fc ¡ fmax ("RF" (radio frequency) signal). Another analog modulation technique is frequency modulation (FM) 9 Representation of sinusoidal signal A. Amplitude and phase B. Sine and cosine components C. Complex exponentials 10 A. Amplitude and phase x(t) = A cos(2¼ft + ') or x(t) = A cos(!t + ') where ! = 2¼f is the angular frequency (radians/second) The phase ' describes translation in time compared to cos(!t): x(t) = A cos(!t + ') = A cos(!(t + '=!)) 11 B. Sine and cosine components x(t) = a cos(!t) + b sin(!t) Follows from trigonometric identity: cos(® + ¯) = cos(®) cos(¯) ¡ sin(®) sin(¯) ) x(t) = A cos(!t + ') = A cos(') cos(!t) ¡ A sin(') sin(!t) Hence: a = A cos('); b = ¡A sin(') and, conversely, p A = a2 + b2;' = ¡ arctan(b=a) 12 EXAMPLE x(t) 1 0 −1 −2 0 2 4 6 8 10 12 Signal x(t) = cos(!t + ') Period: T = 10 (s) Frequency: f = 1=T = 0:1 (s¡1 =Hz (periods/second)) Angular frequency: ! = 2¼f = 0:628 (radians/s) 13 A. Representation using amplitude and phase Amplitude: A = 1 From ¯gure we see that x(t) = cos(!(t ¡ 2)). But x(t) = cos(!t + ') = cos(!(t + '=!) ) Phase: ' = ¡2! = ¡0:4¼ radians or ¡0:4 ¢ 180=¼ = ¡72± 14 B. Sine and cosine components x(t) = a cos(!t) + b sin(!t) where a = cos(') = cos(¡0:4¼) = 0:3090 b = ¡ sin(') = ¡ sin(¡0:4¼) = 0:9511 1 x(t) 0 −1 −2 0 2 4 6 8 10 12 1 Cosine component Sine component 0 −1 −2 0 2 4 6 8 10 12 x(t) = 0:3090 cos(!t) + 0:9511 sin(!t) 15 C. Representation using complex exponentials Background. A common operation is to combine sinusoidal components with di®erent phases: x(t) = A1 cos(!t + '1) + A2 cos(!t + '2) One way: ¯rst write A1 cos(!t + '1) = a1 cos(!t) + b1 sin(!t) A2 cos(!t + '2) = a2 cos(!t) + b2 sin(!t) Then x(t) = (a1 + a2) cos(!t) + (b1 + b2) sin(!t) 16 Streamlining the above procedure We have: cos(!t + ') = cos(') cos(!t) ¡ sin(') sin(!t) and sin(!t + ') = sin(') cos(!t) + cos(') sin(!t) Instead of working with the two signals cos(!t) and sin(!t) it is convenient to working with the single complex-valued signal g(!t) = cos(!t) + j sin(!t) p where j = ¡1 (imaginary unit number) 17 Im 6 g(!t) = cos(!t) + j sin(!t) j sin(!t) ©©* ©© ©© ©© 1 ©© ©© ©© ©© ©© ©© !t ©© - cos(!t) Re Then: g(!t + ') = cos(!t + ') + j sin(!t + ') = cos(') cos(!t) ¡ sin(') sin(!t) ¡ ¢ +j sin(') cos(!t) + cos(') sin(!t) ¡ ¢¡ ¢ = cos(') + j sin(') cos(!t) + j sin(!t) = g(')g(!t) 18 In terms of the complex-valued signal, phase shift with angle ' is equivalent to multiplication by the complex number g('): g(!t + ') = g(')g(!t)(A) ) cos(!t + ') = Re (g(')g(!t)) sin(!t + ') = Im (g(')g(!t)) We will see that using the multiplication (A) to describe phase shifts results in signi¯cant simpli¯cation of formulae and calculations 19 Properties of the complex function g(θ) = cos(θ) + j sin(θ): - g(0) = 1 - g(θ1 + θ2) = g(θ1)g(θ2) dg(θ) - dθ = ¡ sin(θ) + j cos(θ) = jg(θ) The only function which satis¯es these relations is g(θ) = ejθ e = 2:71828182845904523536 ::: (Euler's number) 20 Hence we have: Euler's relation ejθ = cos(θ) + j sin(θ) e = 2:71828182845904523536 ::: (Euler's number) As e¡jθ = cos(θ) ¡ j sin(θ) we have 1 ¡ ¢ cos(θ) = ejθ + e¡jθ 2 1 ¡ ¢ sin(θ) = ejθ ¡ e¡jθ 2j 21 It follows that x(t) = a cos(!t) + b sin(!t) µ ¶ µ ¶ a b a b = + ej!t + ¡ e¡j!t 2 2j 2 2j 1 1 = (a ¡ jb) ej!t + (a + jb) e¡j!t 2 2 or x(t) = c ej!t + c¤e¡j!t 1 ¤ 1 where c = 2 (a ¡ jb) and c = 2 (a + jb) (complex conjugate) Thus the cosine and sine components are parameterized as the real and imaginary components of the complex number c 22 PREVIOUS EXAMPLE: The signal x(t) = cos(!t ¡ 0:4¼) = a cos(!t) + b sin(!t) = 0:3090 cos(!t) + 0:9511 sin(!t) can be represented using complex exponentials as x(t) = c ej!t + c¤ e¡j!t where 1 c = (a ¡ jb) = 0:1545 ¡ j0:4755 2 1 c¤ = (a + jb) = 0:1545 + j0:4755 2 Hence: x(t) = (0:1545 ¡ j0:4755) ej!t + (0:1545 + j0:4755) e¡j!t 23 SUMMARY Let us summarize the results so far. We have the following alternative representations of a sinusoidal signal: ² Amplitude and phase: x(t) = A cos(!t + ') ² Sine and cosine components: x(t) = a cos(!t) + b sin(!t) where a = A cos('); b = ¡A sin(') or p A = a2 + b2;' = ¡ arctan(b=a) 24 ² Complex exponentials: x(t) = c ej!t + c¤ e¡j!t where 1 1 c = (a ¡ jb) ; c¤ = (a + jb) 2 2 or a = 2Re(c); b = ¡2Im(c) 25 EXAMPLE - Signal transmission with reflected component Transmitted signal: x(t) = cos(!t) Received signal: y(t) = x(t) + rx(t ¡ ¿) PROBLEM: Determine amplitude and phase of y(t) We have y(t) = cos(!t) + r cos(!(t ¡ ¿)) The problem can be solved in a straightforward way by introducing complex exponential: 1 ¡ ¢ x(t) = ej!t + e¡j!t 2 26 Then: y(t) = cos(!t) + r cos(!(t ¡ ¿)) 1 ¡ ¢ 1 ³ ´ = ej!t + e¡j!t + r ej!(t¡¿) + e¡j!(t¡¿) 2 2 1 ¡ ¢ 1 ¡ ¢ = 1 + re¡j!¿ ej!t + 1 + rej!¿ e¡j!t 2 2 Here, 1 ¡ ¢ 1³ ´ 1 + re¡j!¿ = 1 + r cos(!¿) ¡ jr sin(!¿) 2 2 and 1 ¡ ¢ 1³ ´ 1 + rej!¿ = 1 + r cos(!¿) + jr sin(!¿) 2 2 27 Then, de¯ning a = 1 + r cos(!¿); b = r sin(!¿) we have (cf. above) 1 1 y(t) = (a ¡ jb) ej!t + (a + jb) e¡j!t 2 2 = a cos(!t) + b sin(!t) = Ay cos(!t + 'y) where p 2 2 Ay = a + b ;'y = ¡ arctan(b=a) 28 SOME SIMPLIFICATIONS As 1 1 a cos(!t) + b sin(!t) = (a ¡ jb) ej!t + (a + jb) e¡j!t 2 2 it follows that it is su±cient to consider the complex-valued signal 1 (a ¡ jb) ej!t 2 Here, the complex number 1 c = (a ¡ jb) 2 determines the signal's amplitude and phase. 29 To see how this works, consider a linear dynamical system with input x(t) and output y(t) x(t) - G - y(t) For the sinusoidal input x(t) = cos(!t) the output takes the form y(t) = A cos(!t + ') = a cos(!t) + b sin(!t) and we want to determine the amplitude A and phase '. 30 As the phase shift introduces a combination of cosines and sines, the problem can be simpli¯ed by embedding the input signal into a larger class of signals involving both a cosine and a sine component. It turns out that a convenient way to do this is to consider the complex input signal xc(t) = cos(!t) + j sin(!t) = ej!t The output then also has real and imaginary components: yc(t) = A cos(!t + ') + jA sin(!t + ') = Aej(!t+') = Aej'ej!t 31 or j!t yc(t) = c e = c xc(t) where c = Aej'.
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  • Multiple-Wavelength Phase-Shifting Interferometry
    Multiple-wavelength phase-shifting interferometry Yeou-Yen Cheng and James C. Wyant This paper describes a method to enhance the capability of two-wavelength phase-shifting interferometry. By introducing the phase data of a third wavelength, one can measure the phase of a very steep wave front. Experiments have been performed using a linear detector array to measure surface height of an off-axis pa- rabola. For the wave front being measured the optical path difference between adjacent detector pixels was as large as 3.3 waves. After temporal averaging of five sets of data, the repeatability of the measurement is better than 25-Å rms (l = 6328 Å). I. Introduction alent wavelength l eq according to Eq, (1) with the as- Conventional single-wavelength phase-shifting in- sumption that the difference of OPD between any ad- 1-3 terferometry (PSI) is a technique which can perform jacent pixels is <l eq/2: direct phase measurement by first obtaining three or more intensity readings of fringe patterns with some known amounts of phase shifts and then using these intensity readings to calculate the phase value at each data point. Since the phase distribution across the interferogram is a measured modulo 2p one needs an Once these D OPDs between pixels were obtained, the assumption to remove the 2p discontinuities in the relative wave-front (OPD) plot or relative surface height measured phase data. The fundamental assumption plot could be obtained by integrating all these D OPD for single-wavelength PSI is that the difference of op- values. The problem for the first method of the two- tical path difference (OPD) between any adjacent pixels wavelength phase-shifting interferometer (TWLPSI) is £l/2.
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