Chapter 2 signals and spectra,
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The document discusses Fourier analysis techniques. It covers topics like line spectra and Fourier series, including periodic signals and average power. Key aspects covered include phasor representation of sinusoids, convergence conditions of Fourier series, and Parseval's power theorem relating signal power to Fourier coefficients.
![Phasors and Line Spectra
11/30/2012 8:18 AM
v(t) = A cos(ω0t + φ)
The phasor representation of a sinusoidal signal comes from Euler’s theorem
𝒆±𝒋𝜽 = 𝐜𝐨𝐬 𝜽 ± 𝒋 𝐬𝐢𝐧 𝜽
Any sinusoid as the real part of a complex exponential
4
𝒋(𝝎 𝟎 𝒕+𝝋) 𝒋𝝋 𝒋𝝎 𝟎 𝒕
𝑨 𝐜𝐨𝐬 𝝎 𝟎 𝒕 + 𝝋 = 𝑨 𝑹𝒆 𝒆 = 𝑹𝒆 [𝑨𝒆 𝒆 ]](https://image.slidesharecdn.com/chapter2-signalsandspectra-121130101800-phpapp02/85/Chapter-2-signals-and-spectra-4-320.jpg)
![Any sinusoid as the real part of a complex exponential
𝑨 𝐜𝐨𝐬 𝝎 𝟎 𝒕 + 𝝋 = 𝑨 𝑹𝒆 𝒆 𝒋(𝝎 𝟎 𝒕+𝝋) = 𝑹𝒆 [𝑨𝒆 𝒋𝝋 𝒆 𝒋𝝎 𝟎 𝒕 ]
11/30/2012 8:18 AM
This is called a phasor representation
Only three parameters completely specify a phasor: amplitude, phase angle, and
rotational frequency 5](https://image.slidesharecdn.com/chapter2-signalsandspectra-121130101800-phpapp02/85/Chapter-2-signals-and-spectra-5-320.jpg)
![Recalling that Re[z] = ½ (z + z*)
If 𝒛 = 𝑨𝒆 𝒋𝝋 𝒆 𝝎 𝟎 𝒕
11/30/2012 8:18 AM
then 𝐴 𝑗𝜑 𝑗 𝜔0 𝑡
𝐴 −𝑗𝜑 −𝑗 𝜔0 𝑡
𝐴 cos 𝜔0 𝑡 + 𝜑 = 𝑒 𝑒 + 𝑒 𝑒
2 2
Two sided spectra
7](https://image.slidesharecdn.com/chapter2-signalsandspectra-121130101800-phpapp02/85/Chapter-2-signals-and-spectra-7-320.jpg)
![trigonometric Fourier Series a one-sided spectrum
∞
𝑣 𝑡 = 𝑐0 + |2𝑐 𝑛 |cos
(2𝜋𝑛𝑓0 𝑡 + arg 𝑐 𝑛 )
11/30/2012 8:18 AM
𝑛=1
or ∞
𝑣 𝑡 = 𝑐0 + 𝑎 𝑛 cos 2𝜋𝑛𝑓0 𝑡 + 𝑏 𝑛 sin 2𝜋𝑛𝑓0 𝑡
𝑛=1
an = Re[cn] and bn = Im[cn]
These sinusoidal terms represent a set of orthogonal basis functions,
Functions vn(t) and vm(t) are orthogonal over an interval from t1 to t2 if
𝑡2
0 𝑛≠ 𝑚
𝑣 𝑛 𝑡 𝑣 𝑚 𝑡 𝑑𝑡 = 𝑤𝑖𝑡ℎ 𝐾 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑡1 𝐾 𝑛= 𝑚
13](https://image.slidesharecdn.com/chapter2-signalsandspectra-121130101800-phpapp02/85/Chapter-2-signals-and-spectra-13-320.jpg)
![v(λ) has the same shape as v(t)
11/30/2012 8:18 AM
But obtaining the picture of w(t - λ) as a function of requires two steps:
First, we reverse w(t) in time and replace t with λ to get w(λ);
Second, we shift w(λ) to the right by t units to get w[-(λ-t)] = w(t-λ) for a given
value of t.
82](https://image.slidesharecdn.com/chapter2-signalsandspectra-121130101800-phpapp02/85/Chapter-2-signals-and-spectra-82-320.jpg)
Chapter 2 signals and spectra,
- 2. Roadmap 11/30/2012 8:18 AM 1. Line Spectra and Fourier Series 2. Fourier Transforms and Continuous Spectra 3. Time and Frequency Relations 4. Convolution 5. Impulses and Transforms in the Limit 6. Discrete Time Signals and the Discrete Fourier Transform 2
- 3. 11/30/2012 8:18 AM LINE SPECTRA AND FOURIER SERIES • Phasors and Line Spectra • Periodic Signals and Average Power • Fourier Series • Convergence Conditions and Gibbs Phenomenon • Parseval’s Power Theorem 3
- 4. Phasors and Line Spectra 11/30/2012 8:18 AM v(t) = A cos(ω0t + φ) The phasor representation of a sinusoidal signal comes from Euler’s theorem 𝒆±𝒋𝜽 = 𝐜𝐨𝐬 𝜽 ± 𝒋 𝐬𝐢𝐧 𝜽 Any sinusoid as the real part of a complex exponential 4 𝒋(𝝎 𝟎 𝒕+𝝋) 𝒋𝝋 𝒋𝝎 𝟎 𝒕 𝑨 𝐜𝐨𝐬 𝝎 𝟎 𝒕 + 𝝋 = 𝑨 𝑹𝒆 𝒆 = 𝑹𝒆 [𝑨𝒆 𝒆 ]
- 5. Any sinusoid as the real part of a complex exponential 𝑨 𝐜𝐨𝐬 𝝎 𝟎 𝒕 + 𝝋 = 𝑨 𝑹𝒆 𝒆 𝒋(𝝎 𝟎 𝒕+𝝋) = 𝑹𝒆 [𝑨𝒆 𝒋𝝋 𝒆 𝒋𝝎 𝟎 𝒕 ] 11/30/2012 8:18 AM This is called a phasor representation Only three parameters completely specify a phasor: amplitude, phase angle, and rotational frequency 5
- 6. A suitable frequency-domain description would be the line spectrum 11/30/2012 8:18 AM One sided spectra Phase angles will be measured with respect to cosine waves. Hence, sine waves need to be converted to cosines via the identity sin ωt = cos (ωt – 90o) We regard amplitude as always being a positive quantity. When negative signs appear, they must be absorbed in the phase using 6 - A cos ωt = A cos (ωt ± 180o)
- 7. Recalling that Re[z] = ½ (z + z*) If 𝒛 = 𝑨𝒆 𝒋𝝋 𝒆 𝝎 𝟎 𝒕 11/30/2012 8:18 AM then 𝐴 𝑗𝜑 𝑗 𝜔0 𝑡 𝐴 −𝑗𝜑 −𝑗 𝜔0 𝑡 𝐴 cos 𝜔0 𝑡 + 𝜑 = 𝑒 𝑒 + 𝑒 𝑒 2 2 Two sided spectra 7
- 8. consider the signal w(t) = 7 – 10 cos(40πt – 60o) + 4 sin 120πt 11/30/2012 8:18 AM w(t) = 7 cos 2π0t + 10 cos(2π20t + 120o) + 4 cos (2π60t – 90o) 8
- 10. Periodic Signals and Average Power 11/30/2012 8:18 AM The average value of any function v(t) is defined as 𝑇/2 1 𝑣(𝑡) = lim 𝑣 𝑡 𝑑𝑡 𝑇→∞ 𝑇 −𝑇/2 In case of periodic signal 𝑡 1 +𝑇 𝑜 1 1 𝑣(𝑡) = 𝑣 𝑡 𝑑𝑡 = 𝑣 𝑡 𝑑𝑡 𝑇𝑜 𝑡1 𝑇𝑜 𝑇𝑜 The average power (normalized) 1 𝑃 = |𝑣 𝑡 |2 = |𝑣 𝑡 |2 𝑑𝑡 𝑇𝑜 𝑇𝑜 10 The average value of a power signal may be positive, negative, or zero.
- 11. Fourier Series Let v(t) be a power signal with period T0=1/f0. Its exponential Fourier series expansion is 11/30/2012 8:18 AM ∞ 𝑣 𝑡 = 𝑐 𝑛 𝑒 𝑗 2𝜋𝑛𝑓0 𝑡 𝑛 = 0, 1, 2, … 𝑛=−∞ The series coefficients are related to v(t) by 1 𝑐𝑛 = 𝑣 𝑡 𝑒 −𝑗 2𝜋𝑛𝑓0 𝑡 𝑑𝑡 𝑇0 𝑇0 The coefficients are complex quantities in general, they can be expressed in the polar form 𝑐 𝑛 = 𝑐 𝑛 𝑒 𝑗 arg 𝑐𝑛 the nth term of the Fourier series equation being 11 𝑗 2𝜋𝑛𝑓0 𝑡 𝑐𝑛 𝑒 = 𝑐𝑛 𝑒 𝑗 arg 𝑐 𝑛 𝑒 𝑗 2𝜋𝑛𝑓0 𝑡
- 12. |c(nf0)| represents the amplitude spectrum as a function of f, and arg c(nf0) represents the phase spectrum. Three important spectral properties of periodic power signals are listed below. 11/30/2012 8:18 AM 1. All frequencies are integer multiples or harmonics of the fundamental frequency f0=1/T0. Thus the spectral lines have uniform spacing f0. 2. The DC component equals the average value of the signal, since setting n = 0 1 𝑐 0 = 𝑣 𝑡 𝑑𝑡 = 𝑣(𝑡) 𝑇𝑜 𝑇𝑜 3. If v(t) is a real (noncomplex) function of time, then 𝑐−𝑛 = 𝑐 ∗ = 𝑐 𝑛 𝑒 𝑗 arg 𝑛 𝑐𝑛 With replacing n by - n 𝑐 −𝑛𝑓0 = 𝑐 −𝑛𝑓0 arg 𝑐 −𝑛𝑓0 = − arg 𝑐 −𝑛𝑓0 which means that the amplitude spectrum has even symmetry and the phase spectrum has odd 12 symmetry.
- 13. trigonometric Fourier Series a one-sided spectrum ∞ 𝑣 𝑡 = 𝑐0 + |2𝑐 𝑛 |cos (2𝜋𝑛𝑓0 𝑡 + arg 𝑐 𝑛 ) 11/30/2012 8:18 AM 𝑛=1 or ∞ 𝑣 𝑡 = 𝑐0 + 𝑎 𝑛 cos 2𝜋𝑛𝑓0 𝑡 + 𝑏 𝑛 sin 2𝜋𝑛𝑓0 𝑡 𝑛=1 an = Re[cn] and bn = Im[cn] These sinusoidal terms represent a set of orthogonal basis functions, Functions vn(t) and vm(t) are orthogonal over an interval from t1 to t2 if 𝑡2 0 𝑛≠ 𝑚 𝑣 𝑛 𝑡 𝑣 𝑚 𝑡 𝑑𝑡 = 𝑤𝑖𝑡ℎ 𝐾 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑡1 𝐾 𝑛= 𝑚 13
- 14. The integration for cn often involves a phasor average in the form 𝑇 2 1 1 1 𝑒 𝑗 2𝜋𝑓 𝑡 𝑑𝑡 = 𝑒𝑗 𝜋𝑓𝑇 − 𝑒 −𝑗 𝜋𝑓𝑇 = sin 𝜋𝑓𝑇 𝑇0 −𝑇 2 𝑗2𝜋𝑓𝑇 𝜋𝑓𝑇 11/30/2012 8:18 AM we’ll now introduce the sinc function defined by sin 𝜋𝜆 𝑠𝑖𝑛𝑐 𝜆 ≜ 𝜋𝜆 sinc λ is an even function of λ having its peak at λ = 0 and zero crossings at all other integer values of λ, so 0 𝜆=0 𝑠𝑖𝑛𝑐 𝜆 = 1 𝜆 = ±1, ±2, … 14
- 15. EXAMPLE: Rectangular Pulse Train 𝐴 𝑡 < 𝜏/2 11/30/2012 8:18 AM 𝑣 𝑡 = 0 𝑡 > 𝜏/2 To calculate the Fourier coefficients 𝑇0 /2 𝜏/2 1 −𝑗 2𝜋𝑛 𝑓0 𝑡 1 𝑐𝑛 = 𝑣 𝑡 𝑒 𝑑𝑡 = 𝐴𝑒 −𝑗 2𝜋𝑛 𝑓0 𝑡 𝑑𝑡 𝑇0 𝑇0 /2 𝑇0 𝜏/2 𝐴 𝐴 sin 𝜋𝑛𝑓0 𝜏 = (𝑒 −𝑗 𝜋𝑛𝑓0 𝜏 − 𝑒 +𝑗 𝜋𝑛𝑓0 𝜏 ) = −𝑗𝜋𝑛𝑓0 𝑇0 𝑇0 𝜋𝑛𝑓0 Multiplying and dividing by t finally gives 𝐴𝜏 15 𝑐𝑛 = 𝑠𝑖𝑛𝑐 𝑛𝑓0 𝜏 𝑇0
- 16. The amplitude spectrum obtained from |c(nf0)| = |cn| = Af0 τ|sinc nf0 τ| 11/30/2012 8:18 AM for the case of τ/T0 = τf0 =1/4 We construct this plot by drawing the continuous function Af0 τ|sinc nfτ| as a dashed curve, which becomes the envelope of the lines. The spectral lines at 4f0, 8f0, and so on, are ―missing‖ since they fall precisely at multiples of 1/τ where the envelope equals zero. The dc component has amplitude c(0) = Aτ/T0 which should be recognized as the average value of v(t). 16 Incidentally, τ/T0 equals the ratio of ―on‖ time to period, frequently designated as the duty cycle in pulse electronics work
- 17. The phase spectrum is obtained by observing that cn is always real but sometimes negative. Hence, arg c(nf0) takes on the values 0o and 180o, depending on the sign of sinc nf0 τ. 11/30/2012 8:18 AM Both +180o and -180o were used here to bring out the odd symmetry of the phase. Having decomposed the pulse train into its frequency components, let’s build it back up again. For that purpose, we’ll write out the trigonometric series, still taking τ/T0 = f0 τ = 1/4, so c0=A/4 and |2cn|=(2A/4) |sinc n/4| = (2A/πn) |sin πn/4|. Thus 𝐴 2𝐴 𝐴 2𝐴 𝑣 𝑡 = + cos 𝜔0 𝑡 + cos 2𝜔0 𝑡 + cos 3𝜔0 𝑡 + … 4 𝜋 𝜋 3𝜋 17
- 20. Convergence Conditions and Gibbs Phenomenon 11/30/2012 8:18 AM The Dirichlet conditions for Fourier series expansion are as follows: If a periodic function v(t) has •a finite number of maxima, minima, and discontinuities per period, •and if v(t) is absolutely integrable, so that v(t) has a finite area per period, then the Fourier series exists and converges uniformly wherever v(t) is continuous. These conditions are sufficient but not strictly necessary. 20
- 21. An alternative condition is that v(t) be square integrable, so that |v(t)|2 has finite area per period—equivalent to a power signal. 11/30/2012 8:18 AM Under this condition, the series converges in the mean such that if 𝑁 𝑣𝑁 𝑡 = 𝑐 𝑒 𝑗 2𝜋𝑛 𝑓0 𝑡 𝑛=−𝑁 then 2 lim 𝑣 𝑡 − 𝑣𝑁 𝑡 𝑑𝑡 = 0 𝑁→∞ 𝑇0 In other words, the mean square difference between v(t) and the partial sum vN(t) vanishes as more terms are included. 21
- 22. Regardless of whether v(t) is absolutely integrable or square integrable, the series exhibits a behavior known as Gibbs phenomenon at points of discontinuity. 11/30/2012 8:18 AM Gibbs ears, Height is independent of N 22
- 23. Parseval’s Power Theorem 11/30/2012 8:18 AM Parseval’s theorem relates the average power P of a periodic signal to its Fourier coefficients. 1 1 𝑃= |𝑣 𝑡 |2 𝑑𝑡 = 𝑣 𝑡 𝑣 ∗ (𝑡)𝑑𝑡 𝑇𝑜 𝑇𝑜 𝑇𝑜 𝑇𝑜 ∞ ∗ ∞ 𝑣∗ 𝑡 = 𝑐 𝑛 𝑒 𝑗 2𝜋𝑛 𝑓0 𝑡 = 𝑐 ∗ 𝑒 −𝑗 2𝜋𝑛 𝑛 𝑓0 𝑡 𝑛=−∞ 𝑛=−∞ ∞ 1 𝑃= 𝑣 𝑡 𝑐 ∗ 𝑒 −𝑗 2𝜋𝑛 𝑛 𝑓0 𝑡 𝑑𝑡 𝑇𝑜 𝑇𝑜 𝑛 =−∞ ∞ 1 = 𝑣 𝑡 𝑒 −𝑗 2𝜋𝑛 𝑓0 𝑡 𝑑𝑡 𝑐∗ 𝑛 𝑇0 𝑇0 𝑛 =−∞ ∞ ∞ 23 𝑃= 𝑐 𝑛 𝑐∗ = 𝑛 |𝑐 𝑛 |2 𝑛=−∞ 𝑛=−∞
- 24. DRILL PROBLEMS 11/30/2012 8:18 AM 24
- 26. 11/30/2012 8:18 AM But and its fundamental frequency is just 26
- 27. 11/30/2012 8:18 AM Solution Amplitude 27
- 28. 11/30/2012 8:18 AM Why? with Fourier coefficients Applying Parseval’s power theorem then yields 28
- 29. Problem 1.3 Sketch the two-sided amplitude spectrum of the even symmetric triangular wave listed in Table T.2 in the textbook. 11/30/2012 8:18 AM Solution The periodic signal waveform is given by 29
- 30. with complex Fourier series coefficients 11/30/2012 8:18 AM by even symmetry How? Amplitude 30
- 31. Problem 1.4 Calculate the average power of the periodic signal of Problem 1.3. 11/30/2012 8:18 AM 31
- 32. 11/30/2012 8:18 AM FOURIER TRANSFORMS AND CONTINUOUS SPECTRA • Fourier Transforms • Symmetric and Causal Signals • Rayleigh’s Energy Theorem • Duality Theorem • Transform Calculations 32
- 33. Fourier Transforms 11/30/2012 8:18 AM Non-periodic signals … average v(t) or |v(t)|2 over all time you’ll find that these averages equal zero. normalized signal energy ∞ 2 𝐸 ≜ 𝑣 𝑡 𝑑𝑡 −∞ 33 When the integral exists and yields 0 < E < ∞, the signal v(t) is said to have well- defined energy and is called a nonperiodic energy signal.
- 34. To introduce the Fourier transform, we’ll start with the Fourier series representation of a periodic power signal 11/30/2012 8:18 AM Let the frequency spacing approach zero and the index n approach infinity such that the product nf0 approaches a continuous frequency variable f Fourier transform 34
- 35. The Fourier transform of v(t) 11/30/2012 8:18 AM The time function v(t) is recovered from V( f ) by the inverse Fourier transform a nonperiodic signal will have a continuous spectrum rather than a line spectrum. In the periodic case we return to the time domain by summing discrete-frequency phasors, while in the nonperiodic case we integrate a continuous frequency function 35
- 36. Three major properties of V( f ) are listed below 1. The Fourier transform is a complex function, so |V(f)| is the amplitude spectrum 11/30/2012 8:18 AM of v(t) and arg V(f) is the phase spectrum. 2. The value of V(f) at f = 0 equals the net area of v(t), since which compares with the periodic case where c(0) equals the average value of v(t) 3. If v(t) is real, then 36
- 37. EXAMPLE: Rectangular Pulse 11/30/2012 8:18 AM and so V(0) = Aτ, which clearly equals the pulse’s area. 37
- 38. 11/30/2012 8:18 AM We may take 1/τ as a measure of the spectral ―width.‖ If the pulse duration is reduced (small τ), the frequency width is increased, whereas increasing the duration reduces the spectral width Thus, short pulses have broad spectra, and long pulses have narrow spectra. This 38 phenomenon, called reciprocal spreading,
- 39. Symmetric and Causal Signals 11/30/2012 8:18 AM Using 𝑒 −𝑗 2𝜋𝑓𝑡 = cos 𝜔𝑡 − 𝑗 sin 𝜔𝑡 even part of V(f) odd part of V(f) 39
- 40. When v(t) has time symmetry, 11/30/2012 8:18 AM where w(t) stands for either v(t) cos ωt or v(t) sin ωt If v(t) has even symmetry then v(t) cos ωt is even whereas v(t) sin ωt is odd, then Vo(f) = 0 and if v(t) has odd symmetry, then Ve(f) = 0 and 40
- 41. Consider the case of a causal signal, defined by the property that 11/30/2012 8:18 AM One-sided Laplace transform is a function of the complex variable s = σ + jω defined by you can get V( f ) from the Laplace transform by letting s = jω = j2πf. 41
- 42. EXAMPLE: Causal Exponential Pulse 11/30/2012 8:18 AM which is a complex function in unrationalized form. Multiplying numerator and denominator by b - j2πf yields the rationalized expression 42
- 43. 11/30/2012 8:18 AM Conversion to polar form then gives the amplitude and phase spectrum 43
- 45. Rayleigh’s Energy Theorem 11/30/2012 8:18 AM Rayleigh’s energy theorem states that the energy E of a signal v(t) is related to the spectrum V(f) by Integrating the square of the amplitude spectrum over all frequency yields the total energy. 45
- 46. The energy spectral density of a rectangular pulse, whose spectral width was claimed to be |f|=1/τ. The energy in that band is the shaded area in the figure, namely 11/30/2012 8:18 AM the total pulse energy is E ≈ A2τ 46
- 47. Rayleigh’s theorem is actually a special case of the more general integral relationship 11/30/2012 8:18 AM 47
- 48. Duality Theorem 11/30/2012 8:18 AM The theorem states that if v(t) and V( f ) constitute a known transform pair, and if there exists a time function z(t) related to the function V( f ) by then 48
- 49. EXAMPLE: Sinc Pulse 11/30/2012 8:18 AM z(t) = A sinc 2Wt We’ll obtain Z(f) by applying duality to the transform pair Re-writing 49
- 50. Transform Calculations 11/30/2012 8:18 AM • 50
- 51. Fourier transform table. 11/30/2012 8:18 AM 51
- 52. Fourier transform table. (continued) 11/30/2012 8:18 AM 52
- 53. DRILL PROBLEMS 11/30/2012 8:18 AM 53
- 54. 11/30/2012 8:18 AM Solution The signal waveform may be sketched according to the following table after noting its even symmetry, as shown below. 54
- 55. 11/30/2012 8:18 AM Ponder! How would you elaborate on reciprocal spreading for this signal? and 55
- 57. 11/30/2012 8:18 AM TIME AND FREQUENCY RELATIONS • Superposition • Time Delay and Scale Change • Frequency Translation and Modulation • Differentiation and Integration 57
- 58. Superposition 11/30/2012 8:18 AM If a1 and a2 are constants and then generally 58
- 59. Time Delay and Scale Change Replacing t by t - td in v(t) produces the time-delayed signal v(t - td). 11/30/2012 8:18 AM The amplitude spectrum remains unchanged in either case, since proof λ = t - td 59
- 60. Scale change, produces a horizontally scaled image of v(t) by replacing t with αt. The scale signal v(αt) will be expanded if |α| < 1 or compressed if |α| > 1; a 11/30/2012 8:18 AM negative value of α yields time reversal as well as expansion or compression. 60
- 61. proof, for case α < 0 α = - |α | λ = - |α|t 11/30/2012 8:18 AM t = λ/α dt = -dλ/|α| 61
- 62. EXAMPLE: Superposition and Time Delay 11/30/2012 8:18 AM 62
- 63. 11/30/2012 8:18 AM we have : θ1 = - πftd θ2 = - πf (td + T) θ1 – θ2 = πfT θ1 – θ2 = πft0 where t0 = td + T/2 63
- 64. with T=τ t0 = 0 11/30/2012 8:18 AM 64
- 65. Frequency Translation and Modulation 11/30/2012 8:18 AM Frequency translation or complex modulation 65
- 66. 1. The significant components are concentrated around the frequency fc. 11/30/2012 8:18 AM 2. Though V(f) was bandlimited in W, V (f - fc) has a spectral width of 2W. Translation has therefore doubled spectral width. Stated another way, the negative-frequency portion of V(f) now appears at positive frequencies. 3. V(f - fc) is not hermitian but does have symmetry with respect to translated origin at f = fc. These considerations are the basis of carrier modulation, we have the following modulation theorem: 66
- 67. EXAMPLE: RF Pulse 11/30/2012 8:18 AM 67
- 68. Differentiation and Integration 11/30/2012 8:18 AM Differentiation theorem 68
- 69. Integration theorem 11/30/2012 8:18 AM Differentiation enhances the high-frequency components of a signal, since |j 2πfV(f )| > |V(f)| for |f | > 1/2π. Conversely, integration suppresses the high-frequency components. 69
- 70. EXAMPLE: Triangular Pulse 11/30/2012 8:18 AM Has zero net area, and integration produces a triangular pulse shape 70
- 71. Applying the integration theorem to Zb (f) 11/30/2012 8:18 AM 71
- 72. The triangular function 11/30/2012 8:18 AM 72
- 73. DRILL PROBLEMS 11/30/2012 8:18 AM 73
- 74. 11/30/2012 8:18 AM (a) (b) Solution 74 How?
- 76. 11/30/2012 8:18 AM Ponder! Is it always possible to interchange the order of the mathematical operations as with this problem? 76
- 77. The modulating signal 11/30/2012 8:18 AM Signal waveform Its spectrum 77
- 78. Modulated signal (a) 11/30/2012 8:18 AM Signal waveform Its spectrum 78
- 79. Modulated signal (b) 11/30/2012 8:18 AM Signal waveform Its spectrum 79 Note: The phase spectrum in this sketch is not to scale.
- 80. 11/30/2012 8:18 AM CONVOLUTION • Convolution Integral • Convolution Theorems 80
- 81. Convolution Integral 11/30/2012 8:18 AM The convolution of two functions of the same variable, say v(t) and w(t), is defined by take the functions 81
- 82. v(λ) has the same shape as v(t) 11/30/2012 8:18 AM But obtaining the picture of w(t - λ) as a function of requires two steps: First, we reverse w(t) in time and replace t with λ to get w(λ); Second, we shift w(λ) to the right by t units to get w[-(λ-t)] = w(t-λ) for a given value of t. 82
- 83. As v(t) * w(t) is evaluated for -∞ < t < + ∞, w(t - λ) slides from left to right with respect to y(λ) 11/30/2012 8:18 AM when t < 0 w(t - λ) t-T t functions don’t overlap 83
- 84. when 0 < t < T 11/30/2012 8:18 AM w(t - λ) t-T t functions overlap for 0 < λ < t 84
- 85. when t > T 11/30/2012 8:18 AM w(t - λ) t-T t functions overlap for t - T < λ < t 85
- 87. Convolution Theorems 11/30/2012 8:18 AM proof 87
- 88. EXAMPLE: Trapezoidal Pulse 11/30/2012 8:18 AM If τ1 > τ2, the problem breaks up into three cases: no overlap, partial overlap, and full overlap. 88
- 89. case 1: no overlap 11/30/2012 8:18 AM or |t| > (τ1 + τ2)/2 89
- 90. case 2: partial overlap The region where there is partial overlap 11/30/2012 8:18 AM and 90
- 91. case 2: total overlap 11/30/2012 8:18 AM 91
- 92. 11/30/2012 8:18 AM The spectrum becomes (A1τ sinc fτ) (A2τ sinc fτ) = Aτ sinc2fτ 92
- 93. EXAMPLE: Ideal Lowpass Filter Rectangular function v(t) = AΠ (t/τ) whose transform, V(f) = Aτ sinc fτ, 11/30/2012 8:18 AM We can lowpass filter this signal at f = 1/τ by multiplying V(f) by the rectangular function The output function is 93
- 94. Review Questions 11/30/2012 8:18 AM 94
- 95. Problems to Ponder 11/30/2012 8:18 AM 95
- 96. 11/30/2012 8:18 AM IMPULSES AND TRANSFORMS IN THE LIMIT • Properties of the Unit Impulse • Impulses in Frequency • Step and Signum Functions • Impulses in Time 96
- 97. Properties of the Unit Impulse 11/30/2012 8:18 AM The unit impulse or Dirac delta function δ(t) 97
- 100. Impulses in Frequency 11/30/2012 8:18 AM Knowing 100
- 101. Direct application of the frequency-translation and modulation theorems yields 11/30/2012 8:18 AM 101
- 102. 11/30/2012 8:18 AM then its Fourier transform is 102
- 103. EXAMPLE: Impulses and Continuous Spectra 11/30/2012 8:18 AM 103
- 105. Step and Signum Functions 11/30/2012 8:18 AM The unit step function the signum function (also called the sign function) 105
- 106. The signum function is a limited case of the energy signal z(t), where v(t) = e-btu(t) and 11/30/2012 8:18 AM so that z(t) → sgn t if b → 0 106
- 107. The step and signum functions are related by 11/30/2012 8:18 AM Hence, 107
- 108. An impulsive DC term appears in the integration theorem when the signal being integrated has nonzero net area. 11/30/2012 8:18 AM since u(t - λ) = 0 for λ > 0 108
- 109. Impulses in Time 11/30/2012 8:18 AM let τ → 0 An impulsive signal with ―zero‖ duration has infinite spectral width, whereas a constant signal with infinite duration has ―zero‖ spectral width. 109
- 110. 11/30/2012 8:18 AM show that the inverse transform does, indeed, equal v(t). But the bracketed integral equals δ(t - λ) 110
- 111. We relate the unit impulse to the unit step by means of the integral 11/30/2012 8:18 AM Repeatedly differentiate the signal in question until one or more stepwise discontinuities first appear. The next derivative, say the nth, then includes an impulse Ak δ(t - tk) for each discontinuity of height Ak at t = tk, so where w(t) is a nonimpulsive function 111
- 112. EXAMPLE: Raised Cosine Pulse 11/30/2012 8:18 AM We’ll use the differentiation method to find the spectrum V(f ) and the high- frequency rolloff 112 has no discontinuities
- 113. d2v(t)/dt2 is discontinuous at t = τ so 11/30/2012 8:18 AM the first term of d3v(t)/dt3 can be written as w(t) = -(π/τ)2dv(t)/dt 113
- 115. Review Questions 11/30/2012 8:18 AM 115
- 116. Problems to Ponder 11/30/2012 8:18 AM 116
- 117. 11/30/2012 8:18 AM DISCRETE TIME SIGNALS AND THE DISCRETE FOURIER TRANSFORM • Foundation • Convolution using the DFT 117
- 118. Foundation 11/30/2012 8:18 AM If we sample a signal at a rate at least twice its bandwidth, then it can be completely represented by its samples sampled at rate fs = 1/Ts 118 where x(k) is a discrete-time signal
- 119. 11/30/2012 8:18 AM Function X(n) is the Discrete Fourier transform (DFT) The DFT spectrum repeats itself every N samples or every fs Hz The DFT is computed only for positive N Note the interval from n→(n+1) represents fs /N Hz 119 the discrete frequency
- 120. Inverse discrete Fourier transform (IDFT) 11/30/2012 8:18 AM 120
- 121. Convolution using the DFT 11/30/2012 8:18 AM 121
- 122. Review Questions 11/30/2012 8:18 AM 122
- 123. Problems to Ponder 11/30/2012 8:18 AM 123