Ebook Engineering circuit analysis (8th edition): Part 2

10 Sinusoidal Steady-State Analysis KEY CONCEPTS INTRODUCTION The complete response of a linear electric circuit is composed of two parts, the natural response and the forced response. The natural response is the short-lived transient response of a circuit to a sudden change in its condition. The forced response is the long-term steady-state response of a circuit to any independent sources present. Up to this point, the only forced response we have consid-ered is that due to dc sources. Another very common forcing function is the sinusoidal waveform. This function describes the voltage available at household electrical sockets as well as the volt-age of power lines connected to residential and industrial areas. In this chapter, we assume that the transient response is of little interest, and the steady-state response of a circuit (a television set, a toaster, or a power distribution network) to a sinusoidal voltage or current is needed. We will analyze such circuits using a powerful technique that transforms integrodifferential equations into algebraic equations. Before we see how that works, it’s useful to quickly review a few important attributes of general sinusoids, which will describe pretty much all currents and voltages through-out the chapter. 10.1 CHARACTERISTICS OF SINUSOIDS Consider a sinusoidally varying voltage v(t) = Vm sinωt shown graphically in Figs. 10.1a and b. The amplitude of the sine waveisVm,andtheargumentisωt.Theradian frequency,orangular frequency, is ω. In Fig. 10.1a, Vm sinωt is plotted as a function of the argument ωt, and the periodic nature of the sine wave is evident. Characteristics of Sinusoidal Functions Phasor Representation of Sinusoids Converting Between the Time and Frequency Domains Impedance and Admittance Reactance and Susceptance Parallel and Series Combinations in the Frequency Domain Determination of Forced Response Using Phasors Application of Circuit Analysis Techniques in the Frequency Domain 371 372 CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS v(t) v(t) Vm Vm – p 0 p p 3p 2p t (rad) 2 2 – T 0 T T 3T T t (s) 4 2 4 –Vm –Vm (a) (b) FIGURE 10.1 The sinusoidal function v(t) = Vm sin ωt is plotted (a) versus ωt and (b) versus t. The function repeats itself every 2π radians, and its period is therefore 2π radians. In Fig. 10.1b, Vm sinωt is plotted as a function of t and the period is nowT.AsinewavehavingaperiodTmustexecute1/T periodseachsecond; itsfrequencyf is1/T hertz,abbreviatedHz.Thus, f = 1 and since ωT = 2π weobtainthecommonrelationshipbetweenfrequencyandradianfrequency, ω = 2πf Lagging and Leading Amore general form of the sinusoid, v(t) = Vm sin(ωt +θ) [1] includes a phase angle θ in its argument. Equation [1] is plotted in Fig. 10.2 as a function of ωt, and the phase angle appears as the number of radians by which the original sine wave (shown in green color in the sketch) is shifted to the left, or earlier in time. Since corresponding points on the sinusoid Vm sin(ωt +θ) occur θ rad, or θ/ω seconds, earlier, we say that Vm sin(ωt +θ) leads Vm sinωt by θ rad. Therefore, it is correct to describe v Vm Vm sin t p 2p t –Vm Vm sin (t + ) FIGURE 10.2 The sine wave Vm sin(ωt + θ) leads Vm sin ωt by θ rad. SECTION 10.1 CHARACTERISTICS OF SINUSOIDS sin ωt as lagging sin(ωt +θ) by θ rad, as leading sin(ωt +θ) by −θ rad, or as leading sin(ωt −θ) by θ rad. In either case, leading or lagging, we say that the sinusoids are out of phase. If the phase angles are equal, the sinusoids are said to be in phase. In electrical engineering, the phase angle is commonly given in degrees, rather than radians; to avoid confusion we should be sure to always use the degree symbol. Thus, instead of writing µ ¶ v = 100sin 2π1000t − 6 we customarily use v = 100sin(2π1000t −30◦) In evaluating this expression at a specific instant of time, e.g., t = 10−4 s, 2π1000t becomes 0.2π radian, and this should be expressed as 36° before 30° is subtracted from it. Don’t confuse your apples with your oranges. 373 Recall that to convert radians to degrees, we simply multiply the angle by 180/π. Two sinusoidal waves whose phases are to be compared must: 1. Both be written as sine waves, or both as cosine waves. 2. Both be written with positive amplitudes. 3. Each have the same frequency. Converting Sines to Cosines The sine and cosine are essentially the same function, but with a 90° phase difference. Thus, sin ωt = cos(ωt −90◦). Multiples of 360° may be added to or subtracted from the argument of any sinusoidal function without changing the value of the function. Hence, we may say that v1 = Vm1 cos(5t +10◦) = Vm1 sin(5t +90◦ +10◦) = Vm1 sin(5t +100◦) leads v2 = Vm2 sin(5t −30◦) by 130°. It is also correct to say that v1 lags v2 by 230°, since v1 may be written as v1 = Vm1 sin(5t −260◦) Note that: −sinωt = sin(ωt ±180◦) −cosωt = cos(ωt ±180◦) ∓sinωt = cos(ωt ±90◦) ±cosωt = sin(ωt ±90◦) v1 1008 08 –308 –2608 v2 We assume that Vm1 and Vm2 are both positive quantities. A graphical representation is provided in Fig. 10.3; note that the frequency of both sinu-soids (5 rad/s in this case) must be the same, or the comparison is meaning-less. Normally, the difference in phase between two sinusoids is expressed by that angle which is less than or equal to 180° in magnitude. The concept of a leading or lagging relationship between two sinusoids will be used extensively, and the relationship is recognizable both mathe-matically and graphically. FIGURE 10.3 A graphical representation of the two sinusoids v1 and v2. The magnitude of each sine function is represented by the length of the corresponding arrow, and the phase angle by the orientation with respect to the positive xaxis. In this diagram, v1 leads v2 by 100° + 30° = 130°, although it could also be argued that v2 leads v1 by 230°. It is customary, however, to express the phase difference by an angle less than or equal to 180° in magnitude. 374 CHAPTER 10 SINUSOIDAL STEADY-STATE ANALYSIS PRACTICE 10.1 Findtheanglebywhichi1 lagsv1 ifv1 = 120cos(120πt −40◦)V and i1 equals (a) 2.5cos(120πt +20◦) A; (b) 1.4sin(120πt −70◦) A; (c) −0.8cos(120πt −110◦) A. 10.2 FindA,B,C,andφif40cos(100t −40◦)−20sin(100t +170◦) = Acos100t + B sin100t = C cos(100t +φ). Ans: 10.1: −60◦; 120°; −110◦. 10.2: 27.2; 45.4; 52.9; −59.1◦. 10.2 FORCED RESPONSE TO SINUSOIDAL FUNCTIONS Now that we are familiar with the mathematical characteristics of sinusoids, we are ready to apply a sinusoidal forcing function to a simple circuit and obtain the forced response. We will first write the differential equation that applies to the given circuit. The complete solution of this equation is com-posed of two parts, the complementary solution (which we call the natural response) and the particular integral (or forced response). The methods we plan to develop in this chapter assume that we are not interested in the short-lived transient or natural response of our circuit, but only in the long-term or “steady-state” response. i R vs (t) = Vm cos t + L FIGURE 10.4 A series RL circuit for which the forced response is desired. The Steady-State Response The term steady-state response is used synonymously with forced response, and the circuits we are about to analyze are commonly said to be in the “sinusoidal steady state.” Unfortunately, steady state carries the connota-tion of “not changing with time” in the minds of many students. This is true for dc forcing functions, but the sinusoidal steady-state response is defi-nitely changing with time. The steady state simply refers to the condition that is reached after the transient or natural response has died out. The forced response has the mathematical form of the forcing function, plus all its derivatives and its first integral. With this knowledge, one of the methods by which the forced response may be found is to assume a solution composed of a sum of such functions, where each function has an unknown amplitude to be determined by direct substitution in the differential equa-tion. As we are about to see, this can be a lengthy process, so we will be sufficiently motivated to seek out a simpler alternative. Consider the series RL circuit shown in Fig. 10.4. The sinusoidal source voltage vs = Vm cosωt has been switched into the circuit at some remote time in the past, and the natural response has died out completely. We seek the forced (or “steady-state”) response, which must satisfy the differential equation L di + Ri = Vm cosωt obtained by applying KVLaround the simple loop. At any instant where the derivative is equal to zero, we see that the current must have the form i ∝ cosωt. Similarly, at an instant where the current is equal to zero, the SECTION 10.2 FORCED RESPONSE TO SINUSOIDAL FUNCTIONS 375 derivative must be proportional to cos ωt, implying a current of the form sinωt. We might expect, therefore, that the forced response will have the general form i(t) = I1 cosωt + I2 sinωt where I1 and I2 are real constants whose values depend upon Vm, R, L, and ω. No constant or exponential function can be present. Substituting the assumed form for the solution in the differential equation yields L(−I1ωsinωt + I2ωcosωt)+ R(I1 cosωt + I2 sinωt) = Vm cosωt If we collect the cosine and sine terms, we obtain (−LI1ω + RI2)sinωt +(LI2ω + RI1 − Vm)cosωt = 0 This equation must be true for all values of t, which can be achieved only if the factors multiplying cos ωt and sin ωt are each zero. Thus, −ωLI1 + RI2 = 0 and ωLI2 + RI1 − Vm = 0 and simultaneous solution for I1 and I2 leads to RVm 1 R2 +ω2L2 Thus, the forced response is obtained: ωLVm 2 R2 +ω2L2 i(t) = R2 +ω2L2 cosωt + R2 +ω2L2 sinωt [2] A More Compact and User-Friendly Form Although accurate, this expression is slightly cumbersome; a clearer picture of the response can be obtained by expressing it as a single sinusoid or cosinusoid with a phase angle. We choose to express the response as a cosine function, i(t) = Acos(ωt −θ) [3] At least two methods of obtaining the values of A and θ suggest them-selves. We might substitute Eq. [3] directly in the original differential equation, or we could simply equate the two solutions, Eqs. [2] and [3]. Selecting the latter method, and expanding the function cos(ωt −θ): Acosθ cosωt + Asinθ sinωt = R2 +ω2L2 cosωt + R2 +ω2L2 sinωt All that remains is to collect terms and perform a bit of algebra, an exercise left to the reader. The result is Several useful trigonometric identities are provided on the inside cover of the book. θ = tan−1 ωL and A = √R2 +ω2L2 and so the alternative form of the forced response therefore becomes i(t) = √R2 +ω2L2 cosµωt −tan−1 ωL¶ [4] ... - tailieumienphi.vn