Lecture Notes on Sinusoidal Steady, State Analysis | ECE 2300, Study notes of Electrical Circuit Analysis

Download Lecture Notes on Sinusoidal Steady, State Analysis | ECE 2300 and more Study notes Electrical Circuit Analysis in PDF only on Docsity! Circuit Analysis - ECE 2300 LECTURE NOTES DAVE SHATTUCK SET #5 Chapter 9 - Sinusoidal Steady-State Analysis Now we take on two major paradigm shifts, which are closely related. What we do in this chapter is: 1. Define what we mean by “steady state” when we have sinusoidal sources. 2. Define “phasors.” 3. Introduce a new solution technique, called phasor analysis. 4. Introduce and define “RMS.” First, I wish to introduce the first engineering paradigm, Fourier’s Theorem. Before we go on, the pronunciation. Everyone repeat, after me, 4-E-A. Furriers are people who make coats. Fourier’s Theorem - Any physically realizable signal can be represented by, and is equivalent to, a summation of sinusoids of different frequency, phase and amplitude. This theorem has profound implications, and has effectively shaped the way that we look at almost all of electrical engineering. It is remarkably powerful. It is profound. I am still not sure that I believe it. But, apparently, it is true. Now, through special mathematical techniques, called phasor analysis, the solution of complex differential equations, that describe circuits and their responses, can be found easily (relatively speaking), as long as the sources are sinusoidal. By Fourier’s Theorem, ultimately, all sources are sinusoidal. Thus, we have a simple technique with universal application. This is almost too good to be true. There are two hitches. It may require an infinite number of sinusoidal components to sinusoids. Fortunately, we do not do this very often in this course. We now define RMS. This means Root- Mean-Squared. It is a value that is associated with periodic functions. It has some particularly valuable information with regard to average power calculations using periodic functions. However, for the time being, we will simply say that we can calculate the RMS value of any periodic function by taking the square root of the mean value of the squared function. Some important notes: 1. In order to get this value, we take the inverse order. We square the function, then take the mean value of that squared function, and then take the square root of that mean value. So, we do SMR. 2. We take the RMS value of periodic functions, that is, functions that vary with time. However, since the RMS function includes a mean value, the RMS value is not a function of time. 3. The RMS value of a sinusoid, if we do the math, turns out to be Vrms = Vm / 2 4. Note that this formula does not apply to other waveforms. Read Section 9.1. Make sure that you understand it. Section 9.2 - The Sinusoidal Response Let us consider the total response of a simple circuit to a sinusoidal source. Consider a series RL circuit connected to a sinusoidal source by a switch, which closes at t=0. KVL gives us L (di/dt) + iR = Vm cos (t + ); for t>0. I am going to pull the solution out of a hat. i(t) = {-Vm / (R2 + 2L2)1/2 } cos () e - (R/L) t + {Vm / (R2 + 2L2)1/2 } cos (t + ) where  = arctan (L / R) Let us examine this solution. 1. The first term is not sinusoidal (with time). Rather, it is a decaying exponential. It dies out with time. After several ’s, it is gone. This is called the transient response. 2. The second term is sinusoidal (with time). It does not die out. This called the steady- state response. We will only be able to use phasor analysis to get the steady-state solution, but this may be all we want. Some fun facts to know and tell about the steady-state solution. 1. The steady-state solution is a sinusoidal function. We will always use an m subscript when we use magnitude based phasors. We can also define the Inverse Phasor Transform. Here, if we put in the phasor, we get the sinusoid out. Remember that we knew the frequency from the beginning. When we are in the phasor domain, it is understood that the frequency is known. Also, we assume that when we take the inverse transform, that we know whether the magnitude of the phasor is the zero to peak, or rms value, and convert accordingly. We will take phasors with magnitude values sometimes, and with rms values sometimes. We will use notation to make it clear which one we are using. Let us solve a problem the hard way, to make a point. I am going to work through all the steps. Often, in this class, we skip this process. However, here we will try to meet two goals: 1. Several steps in the solution will make clear what is going on in the simpler approach. 2. We will see why we never want to do this again. Take the RL circuit from before. (Draw it again, showing the series voltage source, resistor and inductor.) We want the steady-state solution for i(t), which we will call iss(t). From our previous analysis, we know that iss(t) = Im cos (t + ) = Im Re {e j(tej( We want Im and . Now, vS = Ri + L di/dt, and since iss(t) has got to work in this equation, we can plug it in to get: Vm Re {e j(tej( = R Im Re {e j(tej( + L d/dt [Im Re {ej(tej( We state, without proof, that we can put the derivative inside the Re statement. The magnitudes are real constants, and they can be moved inside as well. Then we get Re {Vm e j(tej( = Re {R Im e j(tej( + Re {L Im j e j(tej( Next, we claim that if the real parts of a general expression are equal, the whole quantities must be equal. This means that Vm e j(tej( = R Im e j(tej( + L Im j ej(tej( Next, we note that ej(t ≠ 0. Therefore we can divide through by ej(t and get Vm e j( = R Im e j( + L Im j e j( Vm / (R + jL) = Im Equation #1, phasor form Draw the solution by transform scheme on the board. Section 9.4 - The Passive Circuit Elements in the Phasor Domain Next, we want to find the relationship between phasor voltage and phasor current for passive elements. Resistor: Suppose that iR(t) = Imax cos (t + ) then, by Ohms Law, we have vR(t) = R Imax cos (t + ) Now, _ IR,m = Imax  and _ VR,m = R Imax  . If we take the ratio of these two, we get _ _ VR,m / IR,m = R which is called Complex Ohm’s Law. Inductor: Suppose that iL(t) = Imax cos (t + ), then, by the defining equation of an inductor, we have vL(t) = -L  Imax sin (t + ). By basic trig relationships, we can write vL(t) = -L  Imax cos (t +  - 90°). Now, _ IL,m = Imax  and _ VL,m = - L  Imax  - 90°. Putting this back in its exponential form, we see that VL,m = - L  Imax e jej(-90°) which is equal to VL,m = - L  Imax e j(-j) = jL Imax e j If we take the ratio of these two, we get _ _ VL,m / IL,m = jL. If this has a name, I don’t know what it is. Please be careful. It is not useful, nor typical for us to take the ratio for an inductor in the time domain. If we take vL(t) / iL(t) = -L  Imax sin(t + ) / Imax cos(t + ) = -L  tan(t + ). This function is sometimes zero, sometimes , and sometimes -. It is not equal to jL. By a similar approach, we can find that _ _