Tài liệu Signals And Systems P2 pdf

Signals and Systems with MATLAB Applications, Second Edition 1-19 Orchard Publications Summary int(d) % Integrate the delta function ans = Heaviside(t-a)*k 1.8 Summary • The unit step function that is defined as • The unit step function offers a convenient method of describing the sudden application of a volt- age or current source. • The unit ramp function, denoted as , is defined as • The unit impulse or delta function, denoted as , is the derivative of the unit step . It is also defined as and • The sampling property of the delta function states that or, when , • The sifting property of the delta function states that • The sampling property of the doublet function states that u 0 t() u 0 t() 0t0< 1t0> ⎩ ⎨ ⎧ = u 1 t() u 1 t() u 0 τ()τd ∞– t ∫ = δ t() u 0 t() δτ()τd ∞– t ∫ u 0 t()= δ t() 0 for all t0≠= f t()δta–()fa()δt()= a0= f t()δt() f0()δt()= ft()δt α–()td ∞– ∞ ∫ f α()= δ' t() f t()δ' ta–()fa()δ' ta–()f ' a()δta–()–= Chapter 1 Elementary Signals 1-20 Signals and Systems with MATLAB Applications, Second Edition Orchard Publications 1.9 Exercises 1. Evaluate the following functions: a. b. c. d. e. f. 2. a. Express the voltage waveform shown in Figure 1.24, as a sum of unit step functions for the time interval . b. Using the result of part (a), compute the derivative of , and sketch its waveform. Figure 1.24. Waveform for Exercise 2 tδsin t π 6 – ⎝⎠ ⎛⎞ 2tδcos t π 4 – ⎝⎠ ⎛⎞ t 2 δ t π 2 – ⎝⎠ ⎛⎞ cos 2tδtan t π 8 – ⎝⎠ ⎛⎞ t 2 e t– δ t2–()td ∞– ∞ ∫ t 2 δ 1 t π 2 – ⎝⎠ ⎛⎞ sin vt() 0t7 s<< vt() −10 −20 10 20 12345 6 7 0 vt() ts() e 2t– V() vt() Signals and Systems with MATLAB Applications, Second Edition 1-21 Orchard Publications Solutions to Exercises 1.10 Solutions to Exercises Dear Reader: The remaining pages on this chapter contain the solutions to the exercises. You must, for your benefit, make an honest effort to solve the problems without first looking at the solutions that follow. It is recommended that first you go through and solve those you feel that you know. For the exercises that you are uncertain, review this chapter and try again. If your results do not agree with those provided, look over your procedures for inconsistencies and computational errors. Refer to the solutions as a last resort and rework those problems at a later date. You should follow this practice with the exercises on all chapters of this book. Chapter 1 Elementary Signals 1-22 Signals and Systems with MATLAB Applications, Second Edition Orchard Publications 1. We apply the sampling property of the function for all expressions except (e) where we apply the sifting property. For part (f) we apply the sampling property of the doublet. We recall that the sampling property states that . Thus, a. b. c. d. We recall that the sampling property states that . Thus, e. We recall that the sampling property for the doublet states that Thus, f. 2. a. or δ t() f t()δta–()fa()δta–()= tδsin t π 6 – ⎝⎠ ⎛⎞ t t π 6⁄= δ t π 6 – ⎝⎠ ⎛⎞ sin π 6 δ t π 6 – ⎝⎠ ⎛⎞ sin 0.5δ t π 6 – ⎝⎠ ⎛⎞ === 2tδcos t π 4 – ⎝⎠ ⎛⎞ 2t t π 4⁄= δ t π 4 – ⎝⎠ ⎛⎞ cos π 2 δ t π 4 – ⎝⎠ ⎛⎞ cos 0=== t 2 δ t π 2 – ⎝⎠ ⎛⎞ cos 1 2 12tcos+() t π 2⁄= δ t π 2 – ⎝⎠ ⎛⎞ 1 2 1 πcos+()δt π 2 – ⎝⎠ ⎛⎞ 1 2 11–()δt π 2 – ⎝⎠ ⎛⎞ 0==== 2tδtan t π 8 – ⎝⎠ ⎛⎞ 2t t π 8⁄= δtan t π 8 – ⎝⎠ ⎛⎞ π 4 δ t π 8 – ⎝⎠ ⎛⎞ tan δ t π 8 – ⎝⎠ ⎛⎞ === ft()δt α–()td ∞– ∞ ∫ f α()= t 2 e t– δ t2–()td ∞– ∞ ∫ t 2 e t– t2= 4e 2– 0.54=== f t()δ' ta–()fa()δ' ta–()f ' a()δta–()–= t 2 δ 1 t π 2 – ⎝⎠ ⎛⎞ sin t t π 2⁄= 2 δ 1 t π 2 – ⎝⎠ ⎛⎞ sin d dt t t π 2⁄= 2 δ t π 2 – ⎝⎠ ⎛⎞ sin–= 1 2 12tcos–() t π 2⁄= δ 1 t π 2 – ⎝⎠ ⎛⎞ 2t t π 2⁄= δ t π 2 – ⎝⎠ ⎛⎞ sin–= 1 2 11+()δ 1 t π 2 – ⎝⎠ ⎛⎞ πδ t π 2 – ⎝⎠ ⎛⎞ sin– δ 1 t π 2 – ⎝⎠ ⎛⎞ == vt() e 2t– u 0 t() u 0 t2–()–[]10t 30–()u 0 t2–()u 0 t3–()–[]+= + 10– t50+()u 0 t3–()u 0 t5–()–[]10t 70–()u 0 t5–()u 0 t7–()–[]+ Signals and Systems with MATLAB Applications, Second Edition 1-23 Orchard Publications Solutions to Exercises b. (1) Referring to the given waveform we observe that discontinuities occur only at , , and . Therefore, and . Also, by the sampling property of the delta function and with these simplifications (1) above reduces to The waveform for is shown below. vt() e 2t– u 0 t() e 2t– u 0 t2–()10tu 0 t2–()30u 0 t2–()10tu 0 t3–()30u 0 t3–()+––+–= 10tu 0 t3–()– 50u 0 t3–()10tu 0 t5–()50u 0 t5–()10tu 0 t5–()+–++ 70u 0 t5–()10tu 0 t7–()70u 0 t7–()+–– e 2t– u 0 t() e 2t– 10t 30–+–()u 0 t2–()20t 80+–()u 0 t3–()20t 120–()u 0 t5–()+++= + 10t 70+–()u 0 t7–() dv dt 2e 2t– u 0 t() e 2t– δ t() 2e 2t– 10+()u 0 t2–()e 2t– 10t 30–+–()δt2–()++ +–= 20u 0 t3–() 20t– 80+()δt3–()20u 0 t5–()20t 120–()δt5–()+++– 10u 0 t7–() 10t– 70+()δt7–()+– t2= t3= t5= δ t() 0= δ t7–()0= e 2t– 10t 30–+–()δt2–() e 2t– 10t 30–+–() t2= δ t2–()= 10δ t2–()–≈ 20t– 80+()δt3–() 20t– 80+() t3= δ t3–()= 20δ t3–()= 20t 120–()δt5–()20t 120–() t5= δ t5–()= 20– δ t5–()= dv dt⁄ 2e 2t– u 0 t() 2e 2t– u 0 t2–()10u 0 t2–()10δ t2–()–++–= 20u 0 t3–()20δ t3–()20u 0 t5–()20δ t5–()10u 0 t7–()––++– 2e 2t– u 0 t() u 0 t2–()–[]10δ t2–()10 u 0 t2–()u 0 t3–()–[]20δ t3–()++––= 10 u 0 t3–()u 0 t5–()–[]– 20δ t5–()10 u 0 t5–()u 0 t7–()–[]+– dv dt⁄ dv dt⁄ 20 10 Vs⁄() t s() 20– 10– 1 2 3 4 5 6 7 10δ t2–()– 20δ t 3–() 20δ t 5–()– 2e 2t– – Chapter 1 Elementary Signals 1-24 Signals and Systems with MATLAB Applications, Second Edition Orchard Publications NOTES Signals and Systems with MATLAB Applications, Second Edition 2-1 Orchard Publications Chapter 2 The Laplace Transformation his chapter begins with an introduction to the Laplace transformation, definitions, and proper- ties of the Laplace transformation. The initial value and final value theorems are also discussed and proved. It concludes with the derivation of the Laplace transform of common functions of time, and the Laplace transforms of common waveforms. 2.1 Definition of the Laplace Transformation The two-sided or bilateral Laplace Transform pair is defined as (2.1) (2.2) where denotes the Laplace transform of the time function , denotes the Inverse Laplace transform, and is a complex variable whose real part is , and imaginary part , that is, . In most problems, we are concerned with values of time greater than some reference time, say , and since the initial conditions are generally known, the two-sided Laplace transform pair of (2.1) and (2.2) simplifies to the unilateral or one-sided Laplace transform defined as (2.3) (2.4) The Laplace Transform of (2.3) has meaning only if the integral converges (reaches a limit), that is, if (2.5) To determine the conditions that will ensure us that the integral of (2.3) converges, we rewrite (2.5) T L ft(){}Fs()= ft() ∞– ∞ ∫ e st– dt= L 1– Fs(){}ft()= 1 2πj Fs() σ jω– σ jω+ ∫ e st ds= L ft(){} f t() L 1– Fs(){} s σω s σ jω+= t tt 0 0== L ft(){}F= s() ft() t 0 ∞ ∫ e st– dt f t() 0 ∞ ∫ e st– dt== L 1– Fs(){}f= t() 1 2πj Fs() σ jω– σ jω+ ∫ e st ds= ft() 0 ∞ ∫ e st– dt ∞< Chapter 2 The Laplace Transformation 2-2 Signals and Systems with MATLAB Applications, Second Edition Orchard Publications as (2.6) The term in the integral of (2.6) has magnitude of unity, i.e., , and thus the condition for convergence becomes (2.7) Fortunately, in most engineering applications the functions are of exponential order * . Then, we can express (2.7) as, (2.8) and we see that the integral on the right side of the inequality sign in (2.8), converges if . Therefore, we conclude that if is of exponential order, exists if (2.9) where denotes the real part of the complex variable . Evaluation of the integral of (2.4) involves contour integration in the complex plane, and thus, it will not be attempted in this chapter. We will see, in the next chapter, that many Laplace transforms can be inverted with the use of a few standard pairs, and therefore, there is no need to use (2.4) to obtain the Inverse Laplace transform. In our subsequent discussion, we will denote transformation from the time domain to the complex frequency domain, and vice versa, as (2.10) 2.2 Properties of the Laplace Transform 1. Linearity Property The linearity property states that if have Laplace transforms * A function is said to be of exponential order if . ft()e σt– 0 ∞ ∫ e jωt– dt ∞< e jωt– e jωt– 1= ft()e σt– 0 ∞ ∫ dt ∞< f t() f t() ft() ke σ 0 t for all t0≥< ft()e σt– 0 ∞ ∫ dt ke σ 0 t e σt– 0 ∞ ∫ dt< σσ 0 > f t() L ft(){} Re s{} σ σ 0 >= Re s{} s f t() Fs()⇔ f 1 t()f 2 t()…f n t(),,, Signals and Systems with MATLAB Applications, Second Edition 2-3 Orchard Publications Properties of the Laplace Transform respectively, and are arbitrary constants, then, (2.11) Proof: Note 1: It is desirable to multiply by to eliminate any unwanted non-zero values of for . 2. Time Shifting Property The time shifting property states that a right shift in the time domain by units, corresponds to mul- tiplication by in the complex frequency domain. Thus, (2.12) Proof: (2.13) Now, we let ; then , and . With these substitutions, the second integral on the right side of (2.13) becomes 3. Frequency Shifting Property The frequency shifting property states that if we multiply some time domain function by an exponential function where a is an arbitrary positive constant, this multiplication will produce a shift of the s variable in the complex frequency domain by units. Thus, F 1 s()F 2 s()…F n s(),,, c 1 c 2 … c n ,, , c 1 f 1 t() c 2 f 2 t() … c n f n t()+++ c 1 F 1 s() c 2 F 2 s() … c n F n s()+++⇔ L c 1 f 1 t() c 2 f 2 t() … c n f n t()+++{}c 1 f 1 t() c 2 f 2 t() … c n f n t()+++[] t 0 ∞ ∫ dt= c 1 f 1 t() t 0 ∞ ∫ e st– dt c 2 f 2 t() t 0 ∞ ∫ e st– dt … + c n f n t() t 0 ∞ ∫ e st– dt++= c 1 F 1 s() c 2 F 2 s() … c n F n s()+++= f t() u 0 t() f t() t0< a e as– ft a–()u 0 ta–()e as– Fs()⇔ L ft a–()u 0 ta–(){}0 0 a ∫ e st– dt f t a–() a ∞ ∫ e st– dt+= ta– τ= t τ a+= dt dτ= f τ() 0 ∞ ∫ e s τ a+()– dτ e as– f τ() 0 ∞ ∫ e sτ– dτ e as– Fs()== f t() e at– a Chapter 2 The Laplace Transformation 2-4 Signals and Systems with MATLAB Applications, Second Edition Orchard Publications (2.14) Proof: Note 2: A change of scale is represented by multiplication of the time variable by a positive scaling factor . Thus, the function after scaling the time axis, becomes . 4. Scaling Property Let be an arbitrary positive constant; then, the scaling property states that (2.15) Proof: and letting , we get Note 3: Generally, the initial value of is taken at to include any discontinuity that may be present at . If it is known that no such discontinuity exists at , we simply interpret as . 5. Differentiation in Time Domain The differentiation in time domain property states that differentiation in the time domain corresponds to multiplication by in the complex frequency domain, minus the initial value of at . Thus, (2.16) Proof: e at– ft() Fs a+()⇔ L e at– ft(){}e at– ft() 0 ∞ ∫ e st– dt f t() 0 ∞ ∫ e sa+()t– dt F s a+()== = t a f t() f at() a fat() 1 a F s a ⎝⎠ ⎛⎞ ⇔ L fat(){} fat() 0 ∞ ∫ e st– dt= t τ a⁄= L fat(){} f τ() 0 ∞ ∫ e s τ a⁄()– d τ a ⎝⎠ ⎛⎞ 1 a f τ() 0 ∞ ∫ e sa⁄()τ– d τ() 1 a F s a ⎝⎠ ⎛⎞ == = f t() t0 − = t0= t0 − = f 0 − () f 0() s f t() t0 − = f ' t() d dt ft()= sF s() f0 − ()–⇔ L f ' t(){} f ' t() 0 ∞ ∫ e st– dt= [...]... by the equation F ( α ) = (2.23) b ∫a f ( x, α ) dx where f is some known function of integration x and the parameter α , a and b are constants independent of x and α , and the pardF tial derivative ∂f ⁄ ∂α exists and it is continuous, then = dα 2-6 b ∂( x, α ) ∫a - dx ∂( α ) Signals and Systems with MATLAB Applications, Second Edition Orchard Publications Properties of the Laplace Transform... τ = λ ; then, t = λ + τ , and dt = dλ By substitution into (2.35), * Convolution is the process of overlapping two signals The convolution of two time functions f 1 ( t ) and f 2 ( t ) is denoted as f 1 ( t )*f 2 ( t ) , and by definition, f 1 ( t )*f 2 ( t ) = ∞ ∫–∞ f1 ( τ )f2 ( t – τ ) dτ where τ is a dummy variable We will discuss it in detail in Chapter 6 Signals and Systems with MATLAB Applications,... s We let t ∫0 f ( τ ) dτ g(t) = then, g' ( t ) = f ( τ ) and g( 0) = 0 ∫0 f ( τ ) dτ = 0 Now, − L { g' ( t ) } = G ( s ) = sL { g ( t ) } – g ( 0 ) = G ( s ) – 0 sL { g ( t ) } = G ( s ) G(s) L { g ( t ) } = s ⎧ L ⎨ ⎩ t ⎫ ∫0 f ( τ ) dτ ⎬ = ⎭ F(s) -s (2.26) and the proof of (2.23) follows from (2.25) and (2.26) Signals and Systems with MATLAB Applications, Second Edition Orchard Publications... ) – f ' ( 0 ) Relations (2.19) and (2.20) can be proved by similar procedures We must remember that the terms f ( 0 − ), f ' ( 0 − ), f '' ( 0 − ) , and so on, represent the initial conditions Therefore, when all initial conditions are zero, and we differentiate a time function f ( t ) n times, this corresponds to F ( s ) multiplied by s to the nth power Signals and Systems with MATLAB Applications,... We apply the definition L { f( t)} = F(s) = ∞ ∫0 f ( t ) e – st dt and for this example, L { u1 ( t ) } = L { t } = ∞ ∫0 t e – st dt We will perform integration by parts recalling that ∫ u dv ∫ = uv – v du (2.39) We let u = t and dv = e – st then, – st -du = 1 and v = – e s * This condition was established in (2.9) 2-14 Signals and Systems with MATLAB Applications, Second Edition Orchard Publications... initial value f ( 0 − ) of the time function f ( t ) can be found from its Laplace transform multiplied by s and letting s → ∞ That is, − lim f ( t ) = lim sF ( s ) = f ( 0 ) t→0 s→∞ (2.32) Proof: From the time domain differentiation property, d f ( t ) ⇔ sF ( s ) – f ( 0 − ) dt or Signals and Systems with MATLAB Applications, Second Edition Orchard Publications 2-9 Chapter 2 The Laplace Transformation... 2 ( s ) 2πj For easy reference, we have summarized the Laplace transform pairs and theorems in Table 2.1 2.3 The Laplace Transform of Common Functions of Time In this section, we will present several examples for finding the Laplace transform of common functions of time Example 2.1 Find L { u 0 ( t ) } 2-12 Signals and Systems with MATLAB Applications, Second Edition Orchard Publications The Laplace... dt ⎭ ∞ ∫0 d f ( t ) e – st dt dt Taking the limit of both sides by letting s → 0 , we get 2-10 Signals and Systems with MATLAB Applications, Second Edition Orchard Publications Properties of the Laplace Transform T s→0 d ∫ - f ( t ) e T → ∞ ε dt − lim [ sF ( s ) – f ( 0 ) ] = lim lim s→0 – st dt ε→0 and by interchanging the limiting process, we get T d ∫ - f ( t ) dt T→∞ ε − lim [ sF ( s ) – f... 2T ∫T f( t)e – st dt + 3T ∫ 2T f ( t ) e – st dt + … In the first integral of the right side, we let t = τ , in the second t = τ + T , in the third t = τ + 2T , and so on The areas under each period of f ( t ) are equal, and thus the upper and lower limits of integration are the same for each integral Then, L {f(t)} = T ∫0 f(τ)e – sτ dτ + T ∫0 f(τ + T )e –s ( τ + T ) dτ + T ∫0 f ( τ + 2T ) e – s (... Final Value Theorem lim f ( t ) t→∞ lim sF ( s ) = f ( ∞ ) s→0 12 Time Convolution f 1 ( t )*f 2 ( t ) F 1 ( s )F 2 ( s ) 13 Frequency Convolution f 1 ( t )f 2 ( t ) 1 - F 1 ( s )*F 2 ( s ) 2πj Signals and Systems with MATLAB Applications, Second Edition Orchard Publications − lim sF ( s ) = f ( 0 ) s→∞ 2-13 Chapter 2 The Laplace Transformation Solution: We start with the definition of the Laplace . 5–()– 2e 2t– – Chapter 1 Elementary Signals 1-24 Signals and Systems with MATLAB Applications, Second Edition Orchard Publications NOTES Signals and Systems with MATLAB. function of integration x and the parameter , a and b are constants independent of x and , and the par- tial derivative exists and it is continuous, then