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Contents1. Finite Waves 12. Finite Waves to Infinite Waves 23. Negative Waves and Eternal Sinusoidals 44. Negative Frequencies Analogy 65. Phase of Eternal Sinusoidals 86. Phase Sign and Value 97. Sinusoidals in Frequency Domain 118. Polar form 149. Periodic Signals 1510. Harmonics 1611. The Almost Square Wave (ASqW) 1712. Fourier Analysis 1813. The Orthogonal Property 21 2
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1. Finite WavesAny finite wave propagating in the positive direction in the time domain can be completelyidentified by three pieces of information (parameters): amplitude, frequency and phase. Weshall denote these by the triple (A, +ω, ∅). Time domain one cycle phase ∅ peak +A 0 To +t T0 = fundamental period ƒ = frequency = 1/T0 Hz ω0 = angular frequency = 2π/T0 [rads/sec] ∅ = phase [rads] = fraction of a cycleWith the triple (A, +ω, ∅) we can describe the wave. More precisely, to completely describe thisfinite wave we will need to know the number of full cycles as well. The first cycle is generallyassumed to start at time t = 0 as shown in the diagram. We may embed n, the number of cycles,in the phase: ∅ = {n(2π) + fraction of a cycle}. This is known as continuous phase. Frequency domaina mm a pl +A g n ∅ = 3(2π) + 2i i •t tu ut de e 0 +ɷo +ɷ 0 +ɷo +ɷ Amplitude spectrum Continuous phase spectrumWith the triple (A, +ω, ∅) we can describe the wave and depict the information using thefrequency axis in the frequency domain as shown above. 3
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2. Finite Waves to Infinite WavesNow we extend this finite wave to +∞. The wave now propagates in the positive direction from0 to +∞. So the number of cycles becomes irrelevant. But what happens to the phase? Whenthe wave extends to infinity there is no fraction of a cycle. Also the number of cycles is infinite.Do we need the phase information? We can have two waves from 0 to +∞ with the same peakamplitude +A and the same frequency +ɷo as shown below. period To period To +A +A ••• ••• 0 t+ +t 0 t+ +t 2 2 phase ∅ = = t+ [rads] phase ∅ = = t+ [rads] 2 4 How can we differentiate between these two infinite waves?Both the waves have the same peak amplitude A and frequency +ωo. To distinguish between thetwo waves we need an additional piece of information. This information is t+ , the first positive 2peak in [0, To]. Hence we define phase as phase ∅ = t+ [rads]. This is known as modulo 2π phase. 4
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So the two infinite waves may now be described as (A, +ωo, π/2) and (A, +ωo, π/4). In thefrequency domain we have separate spectrums for the phase information. Frequency domain Frequency domain +π +π +π/2 +π/4 0 ɷo +ɷ 0 ɷo +ɷ − π − π Phase spectrum Phase spectrum 5
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3. Negative Waves and Eternal SinusoidalsLet us now look at infinite waves propagating in the negative direction: from 0 to −∞.Angular displacement: If an object is moving in a circular path (unit radius) with constantangular speed ωs and initial angular displacement ∅, then in terms of variable t: y angular displacement ϴ = ωs t + ∅ horizontal displacement = cos(ϴ) ωs t vertical displacement = sin(ϴ) ∅ -1 0 +1 xWhen a body is in uniform circular motion we know that its horizontal co-ordinate or x co-ordinate describes a cosine wave and its vertical co-ordinate or y co-ordinate describes a sinewave.Counter-clockwise direction of rotation is defined as positive. In uniform circular motion thiscorresponds to angular speed +ωs. In wave motion this corresponds to angular frequency +ω orto cos(t) and sin(t) with t ≥ 0 progressing to the right. +ωs +ω +ω • • • • • • 0 0 +t +t 0 ≤t → + ∞ 0 ≤t → + ∞ counter-clockwise cosine wave sine wave 6
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Clockwise direction of rotation is defined as negative. In uniform circular motion this corresponds to angular speed −ωs. In wave motion this corresponds to angular frequency –ω or to cos(t) and sin(t) with t ≤ 0 progressing to the left. −ω −ω • • • • • • −t 0 −t 0 −∞ ← t ≤0 −∞ ← t ≤0 −ωs clockwise cosine wave sine wave Two frequencies: If we combine the intervals t ≥ 0 and t ≤ 0 we get the eternal (san-antana) cosine wave and eternal (san-antana) sine wave from −∞ t +∞. −ω +ω −ω +ω • • • • • • • • •• • • 0 0 −∞ ← t ≤0 0 ≤t → + ∞ −∞ ← t ≤0 0 ≤t → + ∞ eternal cosine wave eternal sine wave Infinite waves from 0 to +∞ are called sinusoidals. Infinite waves from −∞ to +∞ are also called sinusoidals, more precisely eternal sinusoidals. 7
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4. Negative Frequencies AnalogyBefore we proceed further let us see an analogy. We are so used to working with numbers in thedecimal digital form. We can express numbers in analog form or digital form. The numbertwenty nine in analog form may be expressed using twenty nine strokes. In digital form theexpression depends on the digital system we use. The number twenty nine in the . . . 102 101 100DECIMAL DIGITAL SYSTEM is 2 9 . . . 24 23 22 21 20BINARY DIGITAL SYSTEM is 1 1 1 0 1 3 •••8 82 81 80OCTAL DIGITAL SYSTEM is 3 5We associate analog with the time domain and digital with the frequency domain.To transform numbers from time domain to frequency domain we need a base. The exponentsof this base act as the digital frequencies. This set of frequencies has to be ℤ, the set of integers.All the frequencies should be present in order to express or represent any number (integer,rational or irrational) in digital form. This is the familiar place and place-value system welearned in class four. Here, we can think of the places: 100, 101, 102, • • • as the decimal digitalfrequencies and the values in each place as the amplitudes. 3 • • •10 102 101 100632= 6 3 26 3 2 is decomposed into the sum of its frequency components. 6 3 2 = 6 x 102 + 3 x 101 + 2 x 100The values 6, 3 and 2 are the magnitudes or amplitude coefficients an at these frequencies. 8
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We have no difficulty in writing down the rational 129/16 in decimal form +2 • • •10 10+1 100 10-1 10-2 10-3 10-4 • • • 129/16 = 8 • 0 6 2 5 +∞ ••• a+2 a+1 a0 a-1 a-2 a-3 a-4 • • • = ∙ 10 =−∞ Note the presence of negative decimal frequencies: 10-1, 10-2, 10-3, 10-4, • • •. Again, 8.0625 is decomposed into the sum of its frequency components. So for the sinusoidals in the time domain with t ≥ 0 the corresponding frequencies in the frequency domain are denoted by +ω. And analogous to positive frequencies for discrete numbers: 100, 101, 102, 103, • • • we have positive frequencies: +ω0, +2ω0, +3ω0, • • • for sinusoidals. Likewise, for the sinusoidals in the time domain with t ≤ 0 the corresponding frequencies in the frequency domain are denoted by –ω. So analogous to negative frequencies for discrete numbers: 10-1, 10-2, 10-3, • • • we have negative frequencies: –ω0, –2ω0, –3ω0, • • • for sinusoidals. ••• +A ••• +A−t t+ 0 −t t+ 0 3 2 7 2 phase ∅ = 2 = t+ [rads] phase ∅ = 4 = t+ [rads] Because the waves propagate in the negative directions we say they have negative frequencies denoted by –ω. Again we can completely describe these waves by the triples (A, −ω0, 3π/2) and 2 (A, −ω0, 7π/4) where t+ is the first positive peak in [0, −To] and the phase is ∅ = t+ [rads]. Later we shall see that, analogous to representing analog numbers in the decimal digital domain, we will need a discrete set, dense set or complete (continuous) set of frequencies to represent a signal in the frequency domain. The type of set depends on the type the signal. 9
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5. Phase of Eternal SinusoidalsLet us now combine the corresponding positive and negative waves as depicted below: ••• ••• −t t+ 0 t+ +t 3 2 2 phase ∅ = 2 = t+ [rads] phase ∅ = 2 = t+ [rads] ••• ••• −t t+ 0 t+ +t 7 2 2 phase ∅ = = t+ [rads] phase ∅ = 4 = t+ [rads] 4 Now we have what are called eternal sinusoidals (or sometimes just referred to as sinusoidals),that is to say: waves extending from –∞ to +∞. We can see in the diagrams that each sinusoidalhas two phases. Which phase do we use? 2 − +In this situation we define the phase ∅ = t+ [rads] where t+ is the first positive peak in , 2 2and so the phase lies between –π and +π. See diagrams on page 3.Adding an integer multiple of 2π to this form of the phase does not change the sinusoidal.However, it can lead to discontinuities when depicting phase spectrum. For more information onthis please refer to DSP by Steven Smith pages 164 to 168 and page 188. 10
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6. Phase Sign and Value We still have more work to do regarding the phase. Phase has both sign and value. Let us see some examples.ADVANCED ADVANCED ADVANCED • • • t+ t+ t+ 2 2 2 ∅ = +|t+| [rads] = +π ∅ = +|t+| [rads] = + ∅ = +|t+| [rads] =+ 2 4 • t+ = 0 2 ∅ = |t+| [rads] = 0 DELAYED DELAYED • • • t+ t+ t+ DELAYED 2 2 2∅ = −|t+| [rads] =− ∅ = −|t+| [rads] =− ∅ = −|t+| [rads] = −π 4 2 11
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Given a sinusoidal cos(ω0t + ∅) the phase value ∅ = |t+| 2π/T0 [rads] where t+ = first positivepeak closest to t = 0.The phase sign in the single-sided phase spectrum (rectangular form) is defined by convention: DELAY is negative when t+ 0 ADVANCE is positive when t+ 0The phase sign in the double-sided phase spectrum (rectangular form) and in the phasespectrum (polar form) is defined by convention: (sign of frequency ωi) ∙ {phase sign in the single-sided phase spectrum(rectangular form)} + 2 phase value = |t | [rads] − +where t+ is the first positive peak in , and so the phase lies between –π and +π. 2 2For the time being this is sufficient information on phase. We shall go into more detail on phasein another module. 12
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7. Sinusoidals in Frequency DomainWe have two forms of representing sinusoidals and signals in the frequency domain: rectangularform and polar form. In the rectangular form there are two ways of representation: single-sidedamplitude spectrum which has only the positive frequency axis (+ɷ axis) and double-sidedamplitude spectrum which has both the negative and positive frequency axii (−ɷ axis and +ɷaxis).In the single-sided amplitude spectrum (as shown below) the symbol a is usually used torepresent the peak amplitude of the cosine sinusoidal as in ai.cos(ɷit). The amplitude co-efficients ai may be positive or negative. Likewise the symbol b is usually used to represent thepeak amplitude of the sine sinusoidal as in bi.sin(ɷit). The amplitude co-efficients bi may bepositive or negative. Time domain Frequency domain +a.cos(ɷot) +a +a 0 +t 0 +ɷo +ɷ single-sided amplitude spectrum (rectangular form) −a.cos(ɷot) +ɷo 0 +ɷ 0 +t −a −a single-sided amplitude spectrum (rectangular form) 13
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Time domain Frequency domain +b.sin(ɷot) +b +b 0 +t 0 +ɷo +ɷ single-sided amplitude spectrum (rectangular form) +ɷ o 0 +ɷ −b −b −b.sin(ɷot) single-sided amplitude spectrum (rectangular form)In the double-sided amplitude spectrum the amplitude is equally divided over both the positiveand negative frequencies as shown below. +a.cos(ɷot) +a + + 2 2 0 +t −ɷo 0 +ɷo double-sided amplitude spectrum (rectangular form) 14
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Time domain Frequency domain −ɷo +ɷo 0 +t 0 −a.cos(ɷot) −a − − 2 2 .cos(ɷot) double-sided amplitude spectrum (rectangular form) + 2 +.sin(ɷot) .cos(ɷot) −ɷo 0 +t 0 +ɷo − 2 double-sided amplitude spectrum (rectangular form)−b.sin(ɷot) + 2.cos(ɷot) +ɷo 0 +t −ɷo 0 − 2 double-sided amplitude spectrum (rectangular form)In the double-sided phase spectrum (rectangular form) the phase value is not divided because itrepresents the position of the first positive peak. Whereas the amplitude co-efficients in thedouble-sided amplitude spectrum (rectangular form) ai and bi represent energy which can beequally divided over both +ɷi and −ɷi. 15
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8. Polar formIn the polar form we have a magnitude spectrum and phase spectrum. The magnitude spectrumhas both the negative and positive frequency axii (−ɷ axis and +ɷ axis). The magnitude m of afrequency component ɷ is m = sqrt(a2 + b2). Hence the magnitude m is always positive.However, the magnitude mi is the same for both the +ɷi and −ɷi. In other words it is not equallydivided like in the double-sided amplitude spectrum (rectangular form).We have two types of eternal sinusoidals sine sinusoidal from -∞ to +∞ cosine sinusoidals from -∞ to +∞In rectangular form we need both types: sine as well as cosine. And the phase information ishidden in the amplitudes of the sinusoidals ai.cos(ɷit) and bi.sin(ɷit). This can be extracted: by tan ∅ = arctan(tan∅) = ∅ = arctan( )In polar form we need only cosine type sinusoidals because when we add a cosine sinusoidal anda sine sinusoidal of the same frequency we get a cosine sinusoidal of the same frequency butwith the different phase. The phase information is explicit. ai.cos(ɷit) + bi.sin(ɷit) = mi.cos(ɷit+∅) where mi = sqrt(ai2 + bi2)Here the addition is of the sine and cosine parts of a frequency component present in a signal.We may extract each frequency component as described in Convolution LTI Systems page82 http://www.nitte.ac.in/nmamit/articles.php?linkId=131parentId=20mainId=20facId=131#For more information on polar form please read DSP by Steven W Smith pages 161 to 164. 16
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9. Periodic SignalsLet us see how we can form a periodic signal using sinusoidals. Are they any restrictions orconditions?Let ƒ1(t) = sin(ɷ0t) where ɷ0 = 2π/1, here the fundamental period T0 = 1. Observe that ƒ1(t) is periodic. 2 (t) = sin(ɷ0t) where ɷ0 = 2π , here T0 = 2. Observe that 2 (t) is also periodic. 2 ƒ1(t) (t) 2To 0 To To= 2 2To 3To t - axisIs ƒ(t) = ƒ1(t) + (t) periodic?If ƒ(t) is periodic, then at some point in time, say Tf secs, the intercepts of ƒ1(t) and 2 (t) on the t-axiswill coincide for the first time. This will happen again at 2Tf secs, 3Tf secs and so on. So Tf will be thefundamental period of ƒ(t). But this cannot happen. Let us see why.Proof by Way of Contradiction.Suppose ƒ1(t) and 2 (t) do coincide at some instant Tf on the time axis. This means that: m cycles of ƒ1(t) = n cycles of 2 (t) 2 2 = 1 2 Therefore 2 = implying 2 is rational. But 2 is an irrational number. So this cannot happen.Let us look at the ratio of the periods of ƒ1(t) and 2 (t). period of ƒ1 (t):T 0 =1 1 period of ƒ 2 (t):T 0 = 2 = 2 is irrational 17
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10. HarmonicsIf the ratio of the periods of the periodic functions is irrational, then their sum is NOT a periodicfunction.Hence we can say: Given ƒ(t) = g(t) + h(t) where both g(t) and h(t) are continuous and periodic ƒ(t) is periodic The ratio of the periods of g(t) and h(t) are rationalIn other words: the fundamental frequencies of g(t) and h(t) must be rational multiples ofeach other.In general, we can form a periodic signal ƒ(t) by summing two or more periodic functions whosefundamental frequencies are rational multiples of each other.Composing a periodic signal from sinusoidals whose fundamental frequencies are rationalmultiples of each other is known as synthesis. These sinosoidals are called harmonics. Periodic ƒ(t) HARMONICSHence we may write continuous periodic ƒ(t) as: ∞ ƒ(t) = =0 cos(0 ) + sin(0 )This is known as the Fourier Series or more precisely the CT-FS.Let us now look at a periodic signal (the simplest type of signal) and see how we can extractthe information of the frequency components present in it, that is to say the amplitude co-efficients ai and bi of the sinusoidals of the frequencies i present in the signal. 18
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11. The Almost Square Wave (ASqW) Let us now compose the periodic signal ASqW from its three components. ASqW = b1sin(+1sinɷ0t ) + b3sin(+3ɷ0t) + b5sin(+5sinɷ0t) A1 almost square wave a m p l T0 i t 0 - axis u d e s 10 a m 1 p 1 b1 sin +ωo t l T0 i t u 0 - axis 10 d e T 1 s 1 a 3 b3 sin +3ωo t m p T0 l i 0 - axis t 10 u 3 d e s b5 sin +5ωo t 1 a 5 3 5 5 5 7 5 9 5 m p T0 l i t 0 - axis u d 2 4 6 8 10 e 5 5 5 5 5 s••• t0 t1 t2 t3 t4 t5 t6 t7 t8 t9 t10 • • • [t1] • • • 19
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12. Fourier AnalysisGiven the periodic signal/function ƒ(t) = ASqW (in the TIME DOMAIN) how can wedetermine the components (sinusoidals) present and the (A, , ∅) of each component?The amplitude coefficients ai (of cosine sinusoidals) and bi (of sine sinusoidals) are called theFourier Coefficients.The operation of converting an operand from analog form in the TIME DOMAIN to digital form inthe FREQUENCY DOMAIN is known as a transform. Finding the amplitude coefficients in theFREQUENCY DOMAIN is knows as Fourier analysis. The conversion of an operand from digitalform in the FREQUENCY DOMAIN to analog form in the TIME DOMAIN by the inverse transformoperation is known as synthesis. The transform and its inverse transform are called a transformpair. ∞FOURIER SERIES: HARMONICS = =0 cos(0 ) + sin(0 ) − + 0 ≤ ≤ ≡ 2 ≤ ≤ 2 sin(0) , sin 30 , sin(50) {ƒ(t) ⊛ sin(0 ),Periodic ƒ(t) CARRIERS ƒ(t) ⊛ sin(30 ), ASqW ⊛ Linear Time Invariant System ƒ(t) ⊛ sin(50 )} LINEAR: HOMOGENEITY AND EXPANSION AS SUM OF SINUSOIDALS TIME/SHIFT INVARIANT: INPUT PERIOD = OUTPUT PERIODWe shall find the Fourier coefficients in five steps. The first two steps are not usually shown intext books. It is important to see the central role of the circular convolution operation and theorthogonal property of sinusoidals in determining the coefficients. 20
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1. CIRCULAR CONVOLUTION ƒ(t) ⊛ sin(0 ) = ƒ(t) . sin(−0 ) ƒ(t) ⊛ sin(30 ) = ƒ(t) . sin(−30 ) ƒ(t) ⊛ sin(50 ) = ƒ(t) . sin(−50 ) 2. EXPANSION ƒ(t) . sin(−0 ) = b1 sin(0 ) + b3 sin 30 + b5 sin(50 ) . sin(−0 ) The minus sign in sin(−0 ) on the L.H.S. will cancel out with the minus sign in sin(−0 ) onthe R.H.S. The amplitude coefficients b1, b3, b5 are unknown. 3. ORTHOGONAL Due to the ORTHOGONAL PROPERTY, on R.H.S. we have: L.H.S. R.H.S. ƒ(t) . sin(1. 0 ) = 1 . sin2 (10 ) 4. TRIGOMETRIC MANIPULATION 1 sin2 (0 ) = 2 1 − cos(20 ) 5. COMPLETE the ∫ on R.H.S. The ∫ came from the CIRCULAR CONVOLUTION. over one period ∫ ∫ ∫ cos(2 ) b1 b1 b1 b1 sin2 (0 ) = 1 − cos(20 ) = − 0 2 2 2 over To over To over To b 1 T0 b 1 T0 = − 0 = 2 2 2 2 So: b1 = 0 b1 sin2 (0 ) = 0 ∫ ƒ(t). sin(1. ) over To 0 2 CT-FS coefficient bk = 0 ∫ƒ(t). sin(. ) over To 0 21
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Likewise we get b3 b3 sin2 (+30 ) = cos(20 ) − cos(40 ) 2 ∫ ƒ(t). sin(+3 )dt 2 So: b3 = 0 0 over Toand b5 b5 sin2 (+50 ) = cos(40 ) − cos(60 ) 2 ∫ ƒ(t). sin(+5 )dt 2 So: b5 = 0 0 over ToThis is what the picture looks like in the FREQUENCY DOMAIN. b1 Note: periodic signals have discrete amplitude spectrums a m p l i t u d b3 e s b5 0 +10 + 30 +50 - axis single-sided amplitude spectrum (rectangular form)Usually for frequencies 1 2 3 , we have amplitude coefficients 1 2 3 . Thatis to say: the lower the frequency, the larger the amplitude. This is exactly what happens innature. Because high frequency with large amplitude requires more power.In the equation for period of the pendulum: = 2 , we see that 2 1 implies2 1 . Now, 2 1 implies frequency 1 2 . But 1 2 means amplitude 1 2 .Hence frequency 1 2 implies amplitude 1 2 . The higher the frequency the smaller the amplitude. 22
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13. The Orthogonal Property cos(A+B) + cos(A−B) = 2cosA. cosBLet A = m0 and B = n0 . Then 1 cos(m0 ).cos(n0 ) = 2 cos (m + n)0 + cos (m − n)0 1 ∫ cos(m0 ).cos(n0 ) dt = 2 ∫ cos (m + n) 0 + cos (m − n)0 dt ∫cos (m + n) dt = 0 when m ≠ n 0 ∫cos (m − n) dt = 0 when m ≠ n 0 ∫sinusoidal over one period = 0 because of equal area above and below the time axis. 1When m = n : ∫ cos (m + n) + cos (m − n) dt 2 0 0 1 1 = 2 ∫ ∫ cos 2m. 0 dt + 2 cos 0. 0 dt ∫1.dt 1 = 0 + 2 = 2In short, the Orthogonal Property states that: the circular convolution of two sinusoidals ofdifferent frequencies is zero and of the same frequency is non-zero. Recall that convolution bydefinition is the combined effect of each on each summed up over an interval. So from thePhysics point of view we may say that when the frequencies of the sinusoidals are different theyhave no effect on each other. And when they are same there is some effect or response likeresonance. This is what allows us to extract the individual frequency components present in acomposite continuous periodic signal like the ASqW. 23
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