Circuit Analysis ECE 2300
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This 21 page Class Notes was uploaded by Karolann Wiegand on Saturday September 19, 2015. The Class Notes belongs to ECE 2300 at University of Houston taught by David Shattuck in Fall. Since its upload, it has received 16 views. For similar materials see /class/208286/ece-2300-university-of-houston in Electrical Engineering at University of Houston.
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Date Created: 09/19/15
Circuit Analysis ECE 2300 LECTURE NOTES DAVE SHATTUCK SET 5 Chapter 9 Sinusoidal SteadyState 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 wish to introduce the first engineering paradigm Fourier s Theorem Before we go on the pronunciation Everyone repeat after me 4EA 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 represent a real voltage This can make the solutions lengthy Second the technique only gives us the steadystate part of the response the response after a long time The transient response is not available with this technique It turns out in practice that the concept that this sinusoidal analysis could be done that allows us to approach problems in a different way Here is our approach to Chapter 9 Sinusoid Review RMS Definition Steady State Definition Phasor Definition Example Problem done the hard way Transform Concepts Impedance Definition AC Circuit Analysis Example Problem done the easy way Section 91 The Sinusoidal Source Sinusoids have the following form vt Vm cos rot 4 where Vm is the amplitude 03 is the frequency and I is the phase of the sinusoid Note that if T is the period of the sinusoid or any periodic source then f1T and 0321Tf The frequency ftypically has units of cycless or Hertz or Hz 31 The variable 0 is called the angular frequency and typically has units of radianss or rads It is very important to keep these two quantities straight Thus pt has typical units of radians The phase shift I has typical units of degrees or deg This can be awkward We need to be very careful especially when we evaluate sinusoids Fortunately we do not do this very often in this course We now define RMS This means Root MeanSquared 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 noz a function of time 3 The RMS value of a sinusoid if we do the math turns out to be vrmSVm l 2 4 Note that this formula does not apply to other waveforms Read Section 91 Make sure that you understand it Section 92 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 t0 KVL gives us L didt iR Vm cos rot 4 for tgt0 I am going to pull the solution out of a hat it vm R2 2L212 cos e e 39 RL t vm R2 2L212 cos oat e where 9 arctan 03L R Let us examine this solution 1 The first term is not sinusoidal with time Rather it is a decaying exponential lt dies out with time After several Is 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 steadystate response We will only be able to use phasor analysis to get the steadystate solution but this may be all we want Some fun facts to know and tell about the steadystate solution 1 The steadystate solution is a sinusoidal function 2 The frequency of the steadystate solution is the same as that of the source 3 The amplitude of the steadystate solution can be different from that of the source 4 The phase of the steadystate solution can be different from that of the source Thus when we want to solve for the steady state solution with sinusoidal sources we want to get two things the amplitude and the phase Everything else is already known Section 93 The Phasor Remember Euler s relation ej9cos0jsin9 So we can say that vt vm cos oat 4 vm Re elm ltlgt where Re means the real part of the expression that follows The Phoenician says The Phasor Transform of vt Vm Vmejd The phasor is a complex number whose magnitude is the magnitude of the sinusoid and whose phase is the phase of the sinusoid We will indicate phasors by using a bar over the variable which must absolutely must be uppercase The book uses boldface to indicate phasors but it is prohibitively difficult to draw boldface distinctively by hand Thus we will use the bar VmVmej Vmcos ijsin Vm 4 We also define the phasors in terms of the rms value Thus we will also use v vIrms vm I eJ39ltlgt vm 1 24 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 steadystate solution for it which we will call iSSt From our previous analysis we know that iSSt Im cos oat B Im Re eJ teJB We want lm and 3 Now vs Ri L didt and since iSSt has got to work in this equation we can plug it in to get vm Re ej03tej R Im Re eJ teJB L ddt Im Re ej 3tejB 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 ej tejltgt Re R m ej tejl3 Re L lm jo elwtejB Next we claim that if the real parts of a general expression are equal the whole quantities must be equal This means that vm ej tejltlgt R Im ej teJB Im jg ei teJB Next we note that ej 3t 0 Therefore we can divide through by ej 3t and get vm 914 R m 918 m jg 918 Note that this equation is not a function oft Next we factor out lm and eKB out of the terms and write vm 914 Im 913 R M We solve for lm eKB to get Vm R ij Im Equation 1 Now how many equations do we have here Ans There only appears to be one equation but it is a complex equation Thus the real parts must be equal 1 equation and the imaginary parts must be equal 1 equation This is 2 equations It is also true that the magnitudes must be equal and the phases must be equal which give us two more equations but they are not independent We can pick either pair and solve for our two unknowns So we can solve Equation 1 for lm and B and we get the solution for iSSt Im cos rot B Our goal now is to find a quicker way to get to Equation 1 That s what phasor analysis will give us Notice that we can always divide through by ej Dt This will always give us an equation which is independent of time We can rewrite Equation 1 in terms of phasors as Vm R ij Im Equation 1 phasor form Draw the solution by transform scheme on the board Section 94 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 iRt lmax cos rot 4 then by Ohms Law we have vRt R lmax cos rot 4 Now IRm Imax 44 and VR m R max 41 If we take the ratio of these two we get 7R m Rm R which is called Complex Ohm s Law Inductor Suppose that iLt max cos rot lt then by the defining equation of an inductor we have vLt L 0 max sin rot 4 By basic trig relationships we can write vLt L 0 max cos rot I 90 Now ILm Imax 44 and v1m L Imam 90 Puttin this back in its exponential form we see that vLm L 0 Imax eJltlgteJ90 which is equal to VL m L 0 Imax eJltlgt j ij Imax ej l If we take the ratio of these two we get 7Lm IEm ij lfthis 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 vLt iLt L 0 lmax Sin03t 4 lmax COS0t ltlgt L 0 tan03t This function is sometimes zero sometimes 00 and sometimes oo It is not equal to joaL By a similar approach we can find that vaI5m1jmc In general when we take the ratio of a phasor voltage to a phasor current we get what we call an impedance C m6 mZRjX The impedance Z is a complex number but is not a phasor The real part of the impedance R is called the resistance The imaginary part of the impedance X is called the reactance Now we can solve problems by doing the following 1 Convert voltages and currents to their phasors 2 Convert resistors inductors and capacitors to their impedances 3 Solve the resulting circuit using the impedances in the same way that you would use resistances 4 Convert the phasor voltagecurrent that you obtain back to a time domain voltagecurrent Do not skip step 4 Take special care not to mix domains at any time There should never be a t for time in the phasor domain There should never be aj square root of 1 in the time domain A single diagram should never have components or voltages or currents labelled from both domains Let s do some examples Questions Comments and Answers from the 3x5 cards September 2 2003 Dave Shattuck Question Was your circuit class as difficult as this one not counting the slide ruleabacus factor Answer It is hard to say The textbook was much worse and very hard to read There were no notes on the web or anyplace else There were no old exams available anywhere except at fraternity houses and at my college there were no fraternities I don t have the exams that I took at hand so I am not sure how difficult they were I do remember that lots of people failed to earn a passing grade in the course I took Comment When people have complimented you by saying the problems were not the same as the homework I think they mean to say the level of difficulty was not the same Answer I assume that you alleging that the exam and quiz problems are more difficult than the homework It is regularly alleged that the homework is easier than the exams However we have included several exam questions in the homework In addition there are several problems in the textbook that we assign which are old exam problems from ECE 2300 at the University of Houston Really I was a consultant on the 4th Edition of the textbook and I submitted several problems and they are still being used Unless you are contending that the exam problems have continuously increased in difficulty over the past 54 semesters each semester harder than the one before I don t think this comment can be valid I think there is an issue of perception here Comment You are losing too much weight What is going on Answer Ithink this is another perception thing I just weighed myself last night and I am not losing weight Unless my scale has been continuously under weighing me more and more over the past 54 semesters Question Is there any current in the wire Answer Sometimes Question When will Siemens be put to use in this course How can they be used Answer The term Siemens is the unit for conductance which is the inverse of resistance If one chose to one could label all resistors in terms of their conductance instead of their resistance There are some fields where this is done We will not do this in this course We will sometimes talk about conductance in this course and we also use the units when we have voltagedependent current sources Question Where can we see the schematic for negative resistance Answer It is available on the web in the Questions and Answers directory in a file called QAil4mar02doc Comment Nice haircut Answer Thanks Comment The book sucks Answer Maybe But it is the best book on the market in my opinion Just imagine how bad all the other ones are This is why I am trying to get a contract to write one myself It might turn out to be bad as well but it shouldn t be hard to beat all these Question What s your favorite cigar Answer It was the Ranger by White Owl However I gave up smoking when my daughter was born 17 years ago Question What are we doing for the next 4 weeks besides KCL and KVL Answer We will be learning how to use KCL KVL and Ohm s Law in new and more powerful more efficient ways Question How old are you Answer I am 49 years old Question If you hold a wire in the air does it still have current through it Answer I assume you mean if you hold one end of a wire in the air and that there is no thunderstorm in the area In this case there is no current through the wire Question Can we go over homework problem 120 after the homework has been turned in please Answer Yes we can However I need to have you ask for this in class You all may have figured it out in the two days between asking for this on the 3x5 card and the next class so it would be a waste of time Ask me to solve it at the beginning of class and I will Comment You did not define nodes clearly Answer Hmmmm It seemed clear to me Why didn t you challenge me to say it again in a different way I need you to do this in class Question Why are nodes not defined as locations with the same potential Answer Because while all nodes have the same potentials there can be locations with the same potential that are not the same node Think of having two identical batteries connected at one end but not at the other The two notconnected ends will have the same potential with respect to any other point but since they are not connected they are not the same node Comment Another example on nodes please Answer Of course There will literally be hundreds of them
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