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Chapter 7 Fourier analysis for Continuous-Time systems In preceding chapter, we have discussed the linear time-invariant Systems and have seen that linear time-invariant systems can be described by constant coefficient differential equation and may be characterized by its impulse response. So that, the linear time invariant system may be analysis in time domain with both of the convolution operation and classical Solving method. In this chapter，we will discuss the Analysis for continuous Time invariant system in frequency domain. 7.1 System Analysis by Fourier Transform 1. Principle of the Analysis in Frequency domain The impulse response of the system is the solution to the differential equation when forcing signals a Dirac delta function. Tat is, r t ht , when et t . This impulse response may be used to obtain the system output when input is not an impulse, In this case, the r t can obtained by convolution operation, r t et * ht 7.1-1 that is , the impulse response can be used to characterize the response of the system in time domain. The response of the system in frequency domain can also be obtained by taking Fourier Transform of both sides of eq7.1-1. Using convolution theorem, we get R E H 7.1-2 Where R F r t E F et H F ht H is said to be the transfer function or frequency response of the system. Of course, the transfer function H is, in general, a complex quantity and can be written in polar form H H e jh Where H is known as amplitude-frequently response and h is known as 203 phase-frequently response. Since ht is a real function of time, so that H f is an even function of frequency and h is an odd function of frequency. We can, also writing 7.1-2 as follow R e jr E e je H e jh or R E H 7.1-3 r e h 7.1-4 eq7.1-3, 7.1-4 implied that output spectrum, both Amplitude-frequently and phase-frequency, has been modified by transfer function of the system. The input to system spectrum has changed into output spectrum. And the out Amplitude spectrum is weighted by H . The out phase spectrum is weighted by n . 2. Zero-state Response As we discussed in chapter 6, a particular important and useful class of continuous-time LTI (Linear-Time Invariant) systems are those for which the input and output satisfy a linear constant-coefficient differential equation of the form n d k r t m d k et k 0 ak dt k bk k 0 dt k 7.1-5 From R E H 7.1-6 or, equivalently R H 7.1-7 E Here, of course, we are assuming that these three Fourier transforms all exist. Next, consider applying the Fourier transforms to both sides of eq7.1-5 to obtain n d k r t m d k et F ak F bk k 0 dt k k 0 dt k From the linearly properly this becomes n d k r t m d k et ak F dt k bk F dt k k 0 k 0 and from the differentiation property 204 n m k 0 ak j R k b j E k 0 k k or, equivalently, n k m k R ak j E bk j k 0 k 0 Thus, from eq7.1-7 get m R b j k k H k 0 7.1-8 E n a j k 0 k k From eq7.1-8 we observe that H is a rational function, that is, it is a ratio of polynomials in j . The coefficients of the numerator polynomial are the same coefficients as those that appear on the right hand side of differential equation, and the coefficients of denominator polynomial are the same coefficients as appear on the left side of differential equation. Thus, we see that the frequency response given in eq7.1-8 for the LTI system characterized by eq7.1-5 can be written down directly by inspection. Example 7.1-1 Consider the LTI system that is initially at rest and that is characterized by drt a r t et dt with a 0 , from eq7.1-8, the frequency response is H 1 j a inspection of H ，we see that is the Fourier transform of e -atu t . Thus, the impulse response of the system is recognized as ht e -atu t Example 7.1-2 Consider an LTI system initially at rest that is characterized by the differential equation d 2 r t dr t det 2 4 3r t 2et dt dt dt From eq7.1-8 the frequency response is 205 H j 2 j 2 4 j 3 To determine the corresponding impulse response, we use the method of partial fraction to obtain the following expressions for H j 2 H A1 A2 j 1 j 3 j 1 j 3 j 2 1 and A1 j 3 j 1 2 j 2 1 A2 j 1 j 3 2 The inverse transform of each term can be recognized by inspection with the result that 1 ht e t e 3t u t 1 2 2 In addition, if the Fourier transform E of input to such a system is also of polynomials in j , then so is R E H . In this case we can use the same technique to solve the differential equation, in other word, we can use the same technique to find the r t the system response for arbitrary input et . That is, to find the response r t to the input et , we shall illustrate it in the next examples. Example 7.1-3 Consider the system that as same as Example r 4r 3r e 2e and suppose that the input is et e t u t Then, we have that j 2 R H E 1 j 1 j 3 j 1 In this case we seek a partial fraction expansion of form R A11 A12 A2 j 1 j 12 j 3 206 A11 d j 1 R 2 j 3 j 2 1 dj j 1 j 32 j 1 4 j 2 A12 j 1 R 1 2 j 1 j 3 j 1 2 j 2 1 A2 j 3R j 3 j 12 j 3 4 R 1 1 1 So that 4 2 4 j 1 j 1 2 j 3 The inverse transform is then found to be t 1 4 e t 1 te t 1 e 3t u t 2 4 Example 7.1-4: The RC lower pass filter is illustrated shown in figure 7-1 and input is et Eut -ut- . Find the vc t on the capacitor C, where the circuit is at rest. Figure 7-1 Solution: (1) find the E E F E u t u t 1 1 j E e j j E j 1 e j (2) find the H The H can derived directly from the frequently model as illustrated as follow Figure 7-2 207 1 we see: E I R jc R I 1 jc 1 1 R jc H 1 RC by using E R 1 jRc 1 j 1 j jc RC 1 where RC (3) find the Vc Vc E H E 1 j j 1 e j A A2 here 1 j j j j A1 1 j j 0 A2 1 j j a so that 1 1 Vc E j j 1 e j 1 E 1 e j 1 1 e j j j (4) find the uc t from above we see E u t u t — E F 1 1 e j j so that E 1 j E u t e t u t e t therefore uc t E u t u t E u t e at u t e t 208 E 1 e at u t 1 e t u t 3. Response of System encouraged by Periodic Signals Assuming et is a periodic signal with T1 period. We represent (one of) first period of et as et T1 t e1 t 2 0 otherwise and E1 F e1 t then, et expressed as exponential Fourier series et c e n n jn 1t so that E F cn e jn1t n 2 c n n n 1 E1 n1 thus 1 because cn T1 2 E T1 E n n n 1 1 1 1 E n n n 1 1 1 by using of R H E get R H E 7.1-9 1 H n E n n n 1 1 1 1 The response of the system, then, let we take the inverse of Fourier Transform to both sides of eq7.1-9 209 t F 1R 1 H n1 E1 n1 F 1 n1 n 1 H n1 E1 n1 e jn1t T1 n Example 7.1-5 Find the response of the lower pass filter that the input Signal is periodic triangular pulse with T1 as the period. The input et and low pass filter is shown in figure 7-3 (a) and (b) respectivilly. Figure 7-3 Solution: (1) from former example the frequency response of the low pass filter is H j 1 where RC ET1 2 T1 (2) E1 F e1 t Sa 2 4 1 (3) t T1 n H n1 E1 n1 e jn1t 1 ET 2 n T 1 S a 1 1 e jn1t T1 n jn 1 2 4 n 1 E n j n1t artg 2 1 Sa e 2 n n1 2 2 2 7.2 Distortion-less Transmission In communication systems a distortion-less channel is often desired. This implies that the channel output is just proportional to a delayed version of the input r t Ket td 7.2-1 210 where K is the gain (which may be less than one) and t is the delay. The corresponding requirement in the frequency domain specification is obtained by taking the Fourier transform of both sides of (7.2-1) R KE e j t d 7.2-2 Thus, for distortion less transmission, we require that the transfer function of the channel be given by R H Ke j t d E This implies that for no distortion at the output of a linear time-invariant system, two requirements must be satisfied: (1) The amplitude response is flat. That is H constamt K 7.2-3(a) (2) The phase response is a linear function of frequency. That is, td 7.2-3(b) The two requirements can figured as bellow Figure 7-4 When the first condition is satisfied, it is said that there is no amplitude distortion. When the second condition is satisfied, there is no phase distortion. For distortion-less transmission both conditions must be satisfied. The second requirement is often specified in an equivalent way using the time delay. Define the time delay of the system by d Td 7.2-4 d for distortion-less transmission, the time delay is d Td t constant d 211 If Td is not a constants as a function of frequency, there is phase distortion because the phase response is not a linear function of frequency. 7.3 Ideal Low pass Filters An ideal filter is a system that transmits without distortion all the frequencies in a certain band. The amplitude spectrum over the band is a constant and the phase spectrum over the band is linear. Figure 7-5 depicts the filter characteristics for the ideal low pass and band pass filters Figure 7-5 1. Ideal low pass filter and impulse Response The ideal low-pass filter is define by the transfer function H given by e j t 0 ; m H 7.3-1 0; m The frequency m is often called the cutoff frequency of the fitter. We can find the impulse response of this system by taking the inverse Fourier transform of 7.3-1. Thus ht F 1 H H e j t d 1 2 e j t t 0 d 1 2 m cos t t0 d 1 0 sin t t0 0 m 1 1 t t0 212 sin m t t0 2 1 t t0 m s t t0 7.3-2 a m A sketch of the impulse response is shown in Figure 7-6. We conclude from the impulse-response function that the peak value of the response, m , is proportional to the cut-off frequency. Notice that Figure 7-6 as m the filter becomes an all-pass filter and the output response peak . In other words, the output response approaches the input, an impulse. That is m lim s t t0 t t0 7.3-3 m a m m so that s t t0 dt 1 a m or sa x dx 7.3-4 Again, the impulse response is not causal, because of ht 0 , t 0 . Why are the ideal low pass filter non causal？They have finite bandwidths and linear phase characteristics. 2. Constraint Condition for physically realizable filter. It can be shown that, for all H satisfying H d , a necessary and 2 sufficient condition on the magnitude spectrum for a filter to be physically realizable is that ln H 1 2 d This is called the Paley-Wiener criterion. From this result, we can conclude that for a realizable filter, the magnitude of H may not fall off toward zero faster than a function 213 k k 2 of simple exponential order (for example, e , but not as e , and the attenuation may not be infinite over any band of frequencies of nonzero width. For example, let H e , then 2 ln H 2 11 2 1 2 d 1 2 d 2 1 d 1 1 2 d 1 So that H e 2 cannot realize. Therefore we often attempt to built realizable filters to approximate the ideal filter characteristic as closely as possible. Example 7.3-1, A realizable Low pass filter can be built with RLC Network depict in Figure 7-7. Figure 7-7 Solution: Applied model of frequency in above circuit, we have 1 R j c 1 H R E jL 1 1 R j c 1 j 2 LC j L 1 R L L L Assume R then LC C R L C 1 Let c then LC H 1 1 7.3-5 j 2 2 1 j 1 j c c c c 214 H 1 7.3-6 2 2 2 1 2 c c c tg1 2 7.3-7 1 c The amplitude-frequency and phase-frequency response are depicted in Figure 7-8 Figure 7-8 The impulse response take can inverse of Fourier transform to Eq. 7.3-5. We first change H slightly. H 1 1 2 2 S S 1 j j 1 c c 2 c c j s c 2 c 2 S c c 2 2 2 2 2 S c c 2 4 3 3 2 c 2 2 c c 2 2 2 2 c 3 c 3 2 2 3 3 S c S c 2 2 2 2 then the 215 3 c 1 2c ht F 2 2 3 2 3 j c c 2 2 2c 2c t ct u t 3 e sin 3 2 The impulse response of the RLC network is depicted in Figure 7-9 Figure 7-9 3. Step Response of the Ideal low pass filter The concepts of impulse function and impulse response are valuable analytically, but more practical system test consists in applying a unit step to the input of a system and observing its response. Using the results obtained above, the step response follows using the convolution property: g t ht * ut h d t 7.3 - 8 Combining Eq 7.3-2, and 7.3-8, we have m g t Sa m t0 d t Changing variable, let x m t0 so that m t t 0 g t Sa x dx 1 Using the fact that Sa x dx 0 2 we have m t t 0 g t Sa x dx 1 1 2 0 7.3-9 216 S a x dx is a tabulated function known as the sine-integral function y The integral 0 and denoted by Si y S a x dx y 0 Form the properties of the Sa x function, we can deduce the fallowing: (1) Si y is an odd function: that is, Si y Si y ; (2) Si 0 0 ; (3) Si , Si 2 2 A sketch of Si y is shown in Figure 7-10 (a) Using the Si y the step response g t is shown in Figure 7-10 (b) Figure 7-10 Several important observations can be made from Figure 7-10 (b). Note that as the bandwidth W of LPF is decreased, the filter output g t rises to its peak value more slowly. A measure of the time it takes for this rise is called the “rise time” the filter. A convenient of choice for the LPF is the time required for the output ware form to rise from its minimum to 2 1 its maximum. From Fig 7-10 (b), tr m fm 217 Example 7.3-2 H1 and H 2 are distinct system they are all the LPF and expressed, individually, as H1 u 1 u 1 H 2 u 2 u 2 Where 2 1 find the impulse response of cascade system with H1 and H 2 as follow form as Figure 7-11 shown Figure 7-11 Solution: from Figure 7-11 we have E E H1 H 2 R or H H 2 H 2 H1 because 2 1 , so that H 2 H1 H1 thus H H 2 H1 therefore ht F 1H 2 H1 2 Sa 2t 1 Sa 1t Example 7.3-3 The H and input et given as shown as Figure 7-12 find the response of the system. 218 Figure 7-12 Solution: The et can be expanded to Fourier-Series form 4 et 1 1 sin 1t 3 sin 31t 5 sin 51t 2 8 2 Where 1 consider the bandwidth of the system m 4 41 . So that T T T only 1th and 3th harmonic can going through the LPF. Thus, the response of the LPF is 4 r t sin 1 t t0 3 sin 31 t t0 1 219 Problems 7-1 Find the zero state response by use of Fourier analysis method. The system function and encourage signal is given as following. H et e 3t u t 1 j 2 7-2 Find the frequency response and the impulse response of the systems having the output r t for the input et . (a) et e u t r t e t 2 t e 3t u t (b) et e u t r t e u t 2 3t 3t 2 7-3 Determine the frequency response and the impulse response for the systems described by the following differential equation. dr t (a) 3r t et dt d 2 r t dr t det (b) 2 5 6r t dt dt dt 7-4 In order to distortion-less transmission for given system (or circuit) determine the relation of R1 , R2 and C1 C2 . , 220 7-5 The system function of ideal low-pass filler is 2 1 H and the E Sa . Determine the response r t by 0 2 2 using the time convolution theorem. r t et ht . 7-6 The system is depicted in Fig 7-1. Fig 7-1 e Where H i is ideal Low-pass filter let t u t determine the r t . e jt 0 1 H i 0 1 7-7 Find the system function and the impulse response for given systems. 7-8 Find the response of the system that depicted in Fig 7-2. the input et S t c o 1 0 0 t sin 2t cos1000t t s 0 2t Fig 7-2 221