(updated 2016-Nov-4,7:40pm) MEDE2500 (2016-2017) Tutorial 3 MEDE2500 Tutorial 3 2016-Nov-7 Content 1. The Dirac Delta Function, singularity functions, even and odd functions 2. The sampling process and aliasing 3. A simple filtering system 1a. Dirac Delta Function The following comes from chapter 1 of my other document on Delta function. (Other parts of the pdf is related to integration on delta function, which is not the main concern here http://www.eee.hku.hk/~msang/DiracDeltaFunction_Ang.pdf) For simplicity, we can say that the Unit Delta Function has the form 1 푖푓 푡 = 0 훿(푡) = { 0 푒푙푠푒 The following shows some delta functions with different shift The other name of Delta function is impulse function, Dirac pulse function. 1 MEDE2500 (2016-2017) Tutorial 3 Impulse is very useful. It can be used to generate other function. The unit step function (Heaviside Unit Step function) 푡 푢(푡) = ∫ 훿(푠)푑푠 −∞ For simplicity, we can say that the Unit Step Function has the form 1 푖푓 푡 ≥ 0 푢(푡) = { 0 푒푙푠푒 The following shows some step functions The ramp function 푡 푟(푡) = ∫ 푢(푠)푑푠 −∞ For simplicity, we can say that the ramp function has the form 푡 푖푓 푡 ≥ 0 푟(푡) = { 0 푒푙푠푒 Relationship between impulse, step and ramp 푑 푑2 푑 훿 = 푢 = 푟 , 푢 = 푟 = ∫ 훿푑푠 , 푟 = ∫ 푢푑푠 = ∫ ∫ 훿푑푠2 푑푡 푑푡2 푑푡 (훿, 푢, 푟) together are called singular functions Singular function are useful for expressing other function. 2 MEDE2500 (2016-2017) Tutorial 3 EXAMPLE 1. Write down the expression of the following signal. Answer Note. 푥(푡) = 2(푢(푡) − 푢(푡 − 2)) + (푢(푡 − 2) − 푢(푡 − 4)) + 2(푢(푡 − 4) − 푢(푡 − 6)) + 2(푢(푡 − 6) − 푢(푡 − 8))(1 − 푟(푡 − 6)) After simplification 푥(푡) = 2푢(푡) − 푢(푡 − 2) + 푢(푡 − 4) − 2푢(푡 − 8) − 2(푢(푡 − 6) − 푢(푡 − 8))푟(푡 − 6) 3 MEDE2500 (2016-2017) Tutorial 3 EXAMPLE 2. Write down the expression of the following signal. Answer: 2 [푢(푡) − 푢(푡 − 1.5)] [1 − 푟(푡)] 1.5 2 +[푢(푡 − 1.5) − 푢(푡 − 3)] 푟(푡 − 1.5) 1.5 −2훿(푡 − 4) EXAMPLE 3. Write down the expression of the following signal. (Called impulse train or Daric comb). ∞ Answer: 푠(푡) = ∑푛=−∞ 훿(푡 − 푛푇) 1b. Even and Odd function A function is even if it is symmetric along the y-axis. A function is even if it is anti-symmetrical along the y-axis. 푓 is even ↔ 푓(푥) = 푓(−푥) 푓 is odd ↔ 푓(푥) = −푓(−푥) Example of even and odd function 4 MEDE2500 (2016-2017) Tutorial 3 Even and odd function decomposition theorem. Any function can be written as a sum of even and odd functions. 푓(푥) = 푓푒(푥) + 푓표(푥) where 1 푓푒(푥) = (푓(푥) + 푓(−푥)) 2 1 푓 (푥) = (푓(푥) − 푓(−푥)) 표 2 Decomposition of function as even and odd function. EXAMPLE 4. Decompose the function below into even and odd function. Answer: EXAMPLE 5. Decompose the function below into even and odd function. Answer: 5 MEDE2500 (2016-2017) Tutorial 3 Why care about even and odd function: speed up calculation. 1 EXAMPLE 6. Find the spectrum of 푥(푡) = [푢(푡 − 1) − 푢(푡 + 1)]. (The 푥 in example 4). 2 푒 The standard way: ∞ 푋(휔) = ∫ 푥(푡)푒−푗휔푡푑푡 −∞ 1 1 = ∫ 푒−푗휔푡푑푡 2 −1 −1 = 푒−푗휔푡|1 2푗휔 −1 −1 = (푒−푗휔 − 푒푗휔) 2푗휔 1 = sin 휔 휔 = sinc 휔 The “fast” way: ∞ 푋(휔) = ∫ 푥(푡)푒−푗휔푡푑푡 −∞ Expand the 푒−푗휔푡 using Euler formula ∞ 푋(휔) = ∫ 푥(푡)[cos 휔푡 − 푗 sin 휔푡]푑푡 −∞ ∞ ∞ 푋(휔) = ∫ 푥(푡) cos 휔푡 푑푡 − 푗 ∫ 푥(푡) sin 휔푡 푑푡 −∞ −∞ Note that (i) 푥(푡) is even and sin 휔푡 is odd, (ii) 푒푣푒푛 × 표푑푑 = 표푑푑. (iii) the integration of odd function is zero. Thus we can ignore the second term. ∞ 푋(휔) = ∫ 푥(푡) cos 휔푡 푑푡 −∞ 1 1 = ∫ cos 휔푡 푑푡 2 −1 1 = sin 휔푡 |1 2휔 −1 1 = sin 휔 휔 = sinc 휔 6 MEDE2500 (2016-2017) Tutorial 3 EXAMPLE 7. Find the spectrum of 푥0(푡) in example 4. 푡 푥 (푡) = [푢(푡 − 1) − 푢(푡 + 1)] 표 2 “Normal way” “Fast way” ∞ ∞ 푋(휔) = ∫ 푥(푡)푒−푗휔푡푑푡 푋(휔) = ∫ 푥(푡)푒−푗휔푡푑푡 −∞ −∞ 1 Expand the exp term 1 = ∫ 푡푒−푗휔푡푑푡 ∞ 2 −1 = ∫ 푥(푡)[cos 휔푡 − 푗 sin 휔푡]푑푡 −∞ Can ignore the first term since it is zero. 1 −푗 푋(휔) = ∫ 푡 sin 휔푡 푑푡 2 −1 Perform integration by parts 1 푗 = ∫ 푡푑푐표푠휔푡 2휔 −1 푗 2 = [2 cos 휔 − sin 휔] 2휔 휔 sin 휔 − 휔 cos 휔 = −푗 휔2 Message from this example: both are clumsy. But making use of even-odd function may help reduce some of the step!!! 7 MEDE2500 (2016-2017) Tutorial 3 2a. Sampling Sampling turns a continuous time signal x(t) into discrete time signal x[n]. There are many sampling schemes but usually uniform sampling will be used: the sampling interval is fix (constant T). Discrete signal is obtained by multiplying a sampling signal s(t) with the original function. s(t) is an impulse train ∞ 푠(푡) = ∑ 훿(푡 − 푛푇) 푛=−∞ Therefore, the multiplication between 푥(푡) and 푠(푡) creates a signal 푥(푛푇) 푥푐(푡) = 푥(푛푇) = 푥(푡)푠(푡) ∞ = 푥(푡) ∑ 훿(푡 − 푛푇) 푛=−∞ ∞ = ∑ 푥(푛푇)훿(푡 − 푛푇) 푛=−∞ Now, consider the spectrum. Recall that Fourier Series of 푣(푡) if 푣(푡)is periodic Spectrum of a signal 푣(푡) = { Fourier Transform of 푣(푡) if 푣(푡)is aperiodic Let’s assume the spectrum of x(t) is 푋(휔). (or 푋(푗휔) as shown below.) The spectrum is band-limited. That means the frequency content (complex spectrum) of the signal x(t) is within a frequency range [−휔푀 + 휔푀] (subscript M means max). In “drawing” a spectrum, actually any kind of shape is possible. But usually triangle will be used. A triangle spectrum means “strong low frequency, weak high frequency”. 8 MEDE2500 (2016-2017) Tutorial 3 Now assumes we have the spectrum 푆(휔) of s(t). To find the spectrum 푋푐(휔) of 푥푐(푡), there is a theorem: Convolution theorem on spectrum. The relationship between 푋(휔), 푆(휔) and 푋푐(휔) is 푋푐(휔) = 푋(휔) ∗ 푆(휔) where * is convolution. Some technical details 1. The complete theorem (i). Two signal multiplied together in the time domain is equivalent to the two signal convoluted together in the frequency domain. (ii). Two signal convoluted in the time domain is equivalent to the two signal multiplied together in the frequency domain. i.e. 푥(푡) ∗ 푠(푡) ↔ 푋(휔)푆(휔) 푥(푡)푠(푡) ↔ 푋(휔) ∗ 푆(휔) 2. Normalization constant Sometime there will be a normalization constant before the expression. For example 1 1 푥(푡)푠(푡) ↔ 푋(휔) ∗ 푆(휔) 표푟 푥(푡) ∗ 푠(푡) ↔ 푋(휔)푆(휔) 2휋 √2휋 These are just different notations used in different books. The spectrum - the Fourier Series of 풔(풕). Since 푠(푡) is periodic. So we compute the Fourier series. Fourier series of 푠(푡) is ∞ 푗푚휔푡 푠(푡) = ∑ 푐푚푒 푚=−∞ where 푇/2 1 푐 = ∫ 푠(푡)푒푗푚휔푡푑푡 푚 푇 −푇/2 Plug in the equation of s(t) into the integration 푇/2 ∞ 1 푐 = ∫ ∑ 훿(푡 − 푛푇) 푒푗푚휔푡푑푡 푚 푇 −푇/2 푛=−∞ 푇 푇 Since integral range is [− + ] and 훿(푡 − 푛푇) = 0 for 푡 ≠ 푛푇, so all the terms except 훿(푡) 2 2 in the summation vanish. 푇/2 1 푐 = ∫ 훿(푡)푒푗푚휔푡푑푡 푚 푇 −푇/2 9 MEDE2500 (2016-2017) Tutorial 3 Mathematics trick: since 훿(푡) = 0 for 푡 ≠ 0, we can extend the integration range ∞ 1 푐 = ∫ 훿(푡)푒푗푚휔푡푑푡 푚 푇 −∞ Now we can use the “screening property” of Delta function (page 1 property 2) 1 푐 = 푒푗푚휔0 푛 푇 1 = 푇 Thus the Fourier series of 푠(푡) is ∞ 1 푠(푡) = ∑ 푒푗푚휔푡 푇 푚=−∞ Now, recall that spectrum of a periodic signal are the coefficient of the Fourier series (tutorial 2). So the spectrum is As the spectrum is also an impulse train, so it can be expressed mathematically as ∞ ∞ 1 푛 2휋 푆(푓) = ∑ 훿 (푓 − ) 표푟 푆(휔) = ∑ 훿(휔 − 푛휔) 푇 푇 푇 푛=−∞ 푛=−∞ Note. If s(t) is not infinite pulse train but finite pulse train. Then it is NOT periodic function and the process above changed from Fourier series to Fourier Transform. What we have now: (i) All the expressions of 푥(푡), 푠(푡) , 푥푐(푡), 푋(휔) and 푆(휔) (ii) The relationships between them. (The convolution theorem on spectrum). From (ii), what we need to do is to convolute 푋(휔) and 푆(휔) to get the spectrum of the sampled signal. 10 MEDE2500 (2016-2017) Tutorial 3 Convolution with impulse train illustration How to “think” about convolution “When f is only one impulse. Then convolution is just translation” “Amplitude is just magnification” 11 MEDE2500 (2016-2017) Tutorial 3 “Multiple impulses? Just add them together!” “Don’t forget the shape of g matters!” EXAMPLE 8. 푓 = [1 2 3 4], 푔 = [1 − 1 1 ]. Find 푓 ∗ 푔. Answer Consider 푓 ∗ 푔 Result: 푓 ∗ 푔 = 푓 ∗ 푔[1] + 푓 ∗ 푔[2] + 푓 ∗ 푔[3] = [1, 1,2,3, −1,4] Consider 푔 ∗ 푓 Also give you [1, 1,2,3, −1,4], since 푓 ∗ 푔 = 푔 ∗ 푓 12 MEDE2500 (2016-2017) Tutorial 3 Hence, the overall picture of the sampling process is shown below (I) and (II) produce (III) through multiplication.