ELECTRONIC INSTRUMENTATION ENGR 4300
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Electronic Instrumentation ENGR 4300 AC Steady State Analysis What if we are given a circuit and we wish to find the output of that circuit for a range of inpum We could wire the circuit attach the function generator to the input and hook up the scope to the output Then we could alter the input systematically and observe how the output responds to these changes If we could use this information to find a mathematical function that relates the input to the output then we could use the function to predict how the circuit will behave for any given input as shown in figure 1 below What if we could use the components in the circuit to derive this function Then we could predict the output of any circuit for any given input ml Cl Vout fVz39rz A What is a Transfer Function A transfer function relates the output and the input of a circuit In order for a transfer function to be useful it must be simple to use and easy to find using the circuit diagram Therefore let us define a transfer function H as the ratio of the output to the input of a circuit H Vout V equation 1 m In a circuit containing only resistors transfer functions can be easily found using voltage dividers In figure 2 a voltage divider is used to find the transfer function at Vout 39 anl 39 Vm Vm amp an39Rz RlR2R3 HEW w 2K 3K V 6 V V Vm I R3 1K 2K 3K 0 Figure 2 Susan Bonner l Revised 1162006 Rensselaer Polytechnic Institute Troy New York USA Electronic Instrumentation ENGR 4300 Since H is 56 for this circuit We can nd the output at Vout by multiplying any input Vin by 56 What if the circuit contains components other than resistors I V ldt C vnm dl VL LE V R 0 Figure 3 The equations that govern the behavior of capacitors and inductors are not linear like those for resistors Voltage dividers are based on these simple linear relationships How can We nd a simple transfer inction for a circuitWith voltage relationships that have derivatives and integrals We could use differential equations but that Would not be the simple solution We are looking for We need to find a Way to get rid of these integrals and derivatives so thatWe can get back to a simple linear relationship This is the indamental motivation for steady state analysis B A Madelfar Steady State Analysis Steady state analysis is a Way to analyze AC circuits mathematically We can do this using linear transfer inctions and Without the use of differential equations ifWe do two important things The first is to take advantage of the fact that AC signals are sinusoids that do not change in frequency The second is to map the input from the time domain into a domain based on the amplitude and frequency of the signals involved B I Steady State Sinuxoids The term rteady rtate refers to the concept that a circuit When attached to an AC input signal Will a er a nite period of time reach a state in Which the signal across each device in the circuit behaves like a sinusoid ofthe same frequency as the input The signal across each device Will have a constant amplitude and phase relative to the input that does not change once steady state has been reached Once a circuit has reached steady state the output at any point in the circuit Will look like a sinusoidWith the same frequency as e input a constant amplitude and a phase that does not change relative to the input phase It is this feature that We Will exploit to find transfer inctions Sinusoids have a special mathematical property in that the derivative of a sinusoid is just another sinusoid With the same frequency but shifted in phase The same is true for the S14 anBormer 2 Revised 1162006 Remselaer Polytechnic Innitute Troy New York USA Electronic Instrumentation ENGR43 00 integral os a sinusoid So when we need to take the derivative or integral of a sinusoidal signal we can think of it simply as the application of some change in amplitude and some shift in phase Mathematically the derivative of a sine wave of frequency 03 and amplitude A Vin A sinut is given by a cosine wave of frequency 03 and amplitude A03 alVin Au cosut The output of the derivative operation is a new sine wave with a alt different amplitude and a phase shifted by and additional 90 degrees Au sinut 7139 2 Mathematically this means that alVin dt A s1nut where A Au anal 2 The derivative ofa cos1ne or any other sinusoid results in the same changes For example if Vin A cosut then Ausinutg Ausinut 7239Aucosut 7239 7Aucosut 2 alt alVin and therefore d A cosut whereA Au anal 7139 2 Similarly the t integral of a sinusoid has a new amplitude of N03 and an additional phase shift of 90 degrees So if Vin Asinut then Vin alt A sinut 39 whereA39 anal u If we can somehow model our input and output waves so that we are concerning ourselves just with the changes in amplitude and phase brought about by the integrals and derivatives involved then we can eliminate the need to use differential equations Since the phase is an angle it seems logical to begin with polar coordinates B 2 Polar Coordinates Polar coordinates are a simple remapping of the traditional rectangular Xy coordinates on a plane into another set of variables r and 9 where r is the distance from the origin and 9 is the angle measured from the positive X aXis in counterclockwise direction This is shown in figure 4 To convert a point between the polar and Xy coordinate systems trigonometric identities must be used The distance to the origin can be found using the Pythagorean theorem r lxz y2 and the angle from the X aXis is simply found by tan 9 X or x 9 tan3911 To go the other way we can use x r cos 9 andy r sin 9 x Susan Bonner 3 Revised 1162006 Rensselaer Polytechnic Institute Troy New York USA Electronic Instrumentation ENGR4300 l lv rlsln l X1 mouse Figure 4 Ifwe think ofour signal as having aphase offset oftb and an amplitude ofA we can model 39 quot i 5 t t nequency begins at o and cycles around the origin one time for each cycle wt The ray has a lenhofAyt rlct um we have only look at eitherthe x coordinate A mstzz orthe y coordinate y Ammw st t l t c radians per second This corresponds to the constant cycling ofthe sinusoid in time We could just as easily drop the wt since it is asimple multiple of27r and we would have an x coordinate x A 0054 and ay coordinate y A 51714 Y Figure 5 Susan 8mm 4 27 Revzsed 1152005 Rznsselaer Patyterhmr Inmmte T my New Ym U Electronic Instrumentation ENGR4300 ma 39 39 39 sinu oiu For this we will need to rethink our model in terms of complex numbers 33 Camplzx Numbers 39 39 39 real part and an imaginary part In other words a complex number is a point on a plane called the complex plane A complex numberz can be written as z a y In a complex number x is called the real part y is called the imaginary part and J71 Note that in common mathematical notation l is used for J7 1 however in electrical engineering sincel is commonly used for current we denoteirl with j The term imaginary comes from the fact that J7 1 is not a real number Complex Thex A They 71 L L a part and the y coordinate is the imaginary part iv i Z W M yl Figure 5 dUUuL 345ml Bonnzr 5 mm 1152005 Remszlazr Palytzchrllc Institute Tray New York USA Electronic Instrumentation ENGR 43 00 Why do we need complex numbers What good could having a signal defined in terms of something that doesn t exist possibly do What we are working towards is de ning our sinusoid as a combination of a real part x Acoswt W and an imaginary part y jAsinwt This will enable us to work easily with sinusoids that have different phases relative to one another To do this we need to define complex polar coordinates B 4 Complex Numbers in Polar Format As we mentioned above poinm on a plane can be represented in polar coordinates Similarly any complex number can be written in the polar format shown in figure 7 imaginary m zrcosejrsine imaginary Figure 7 Now we have a complex number 2 defined in terms of a real part x rcosS and an imaginary part y j rsinS It is a simple matter to translate this so that it represenm the phase and amplitude of our sinusoid shown in figure 8 Note that the constant cycling of wt 21 for each cycle does not really affect the value of x and y in the model iv 2 Alos j Asm l A l v Asm l l Susan Bonner 6 Revised 1162006 Rensselaer Polytechnic Institute Troy New York USA Electronic Instrumentation ENGR4300 In this representation we can translate from the time domain to the frequency domain by usingx Acorg andy Arimi and back to the time domain using A x2 y2 and 57 tan391 1 This is the nal representation we will use We will call this model a X phasor and base steady state analysis upon this representation Before we go into details however let s look at a brief example Hope Jlly this will help clarify how the representation works 85 An Example iAdding Two Cosine Waves know that a complex polar number is given by z Acos j Asind The whole signal assuming it is a cosine can be represented only by the x part of this complex ar number What will we use y for Let s look at the example in figure 9 V2 v0 7 Frequency Domain Figure 9 Figure 9 shows two sinusoids in the time domain that we want to add together The sum with the largest amplitude is also shown The rst input sinusoid is Vl 65 cosoat 027 and the second sinusoid is V2 6 cosoat 077 If we use complex polar notation adding the imaginary part these two signals become V1 65 cosoat 027 j 65 sinoat 027 and V2 6 cosoat 077 j 6 sinoat 077 To translate into the frequency domain we use x Acorg and y Arimi This gives us the components ofthe two input vectors zl 49 j43 and 22 35 j47 We can add the x and y coordinates ofthese vectors to get the vector sum zom 14 j 90 This represents the sum of the sinusoids The real part can be found by determining the Swan Bonner 7 Revised 1162006 Renrrelaer Polytechnic Inr mte T ray New York USA Electronic Instrumentation ENGR43 00 amplitude using A 1le y2 wl142 902 9 l and the phase angle by using 90 tan391 tan391 0457239 rad This means that the sum is x Vout 91 cos oat 04511 Since the amplitude and phase along with the original frequency de ne the new signal the vectors have given us the sum Although the imaginary part is inserted at the beginning and discarded at the end it is instrumental in determining how much of the two signals overlap in time This representation is even more useful when dealing with multiplication and transfer functions C Phasors A phasor is a representation of a sinusoidal wave in complex polar form The variables are expressed in terms of the phase offset and amplitude Therefore if we have a sinusoidal signal vt A cosat the phasor of this signal would be expressed as l7 A cosat A sinat Since we know that the cat will simply return a multiple of 211 this can be further simplified to V A cos A sin equation 2 Phasors have special terminology The amplitude of the sine wave is called the magnitude of the phasor and is denoted I l7 l The phase offset of the sine wave is called the phase of the phasor and is denoted Al The magnitude and phase of a phasor given in terms of Cartesian coordinates z x j y can be determined using the equations for polar coordinates 2 2 l VI l x y equation 3 and z I tanil eQuation 4 C 1 Euler 3 Formula In order to make it easier to manipulate our phasors we must understand Euler s formula Euler s formula relates the complex polar definition we have defined to the natural logarithm e Euler s formula can be stated as follows Susan Bonner 8 Revised 1162006 Rensselaer Polytechnic Institute Troy New York USA Electronic Instrumentation ENGR4300 e cost9jsint9 This means that our de nition of phasors can be extended as follows 17 Aej Acos jAsin Along with Euler s formula come several shortcuts that we can use to manipulate phasors By a natural extension of Euler s formula it can be shown that if we have two jg complex polar numbers 21 1 16 and 22 7 26 then the product ofthese two numbers is given by 1399 f39913992 Z3 2122 ie 6 1399 1rze Z rer and the quotient of the two numbers is given by r ej39g1 r Z 1 161091792 Z2 rze r2 We can use these rules to multiply phasors If I71 Alejljl and V2 Azej z then I73 2171172 AIA2ejmlm We find that the magnitude of the product is the product of the magnitudes l73 A1142 and the phase of the product is the sum of the phases 4173 LIZ 4V2 1 2 We can also use these rules to divide phasors fl1 Alej l and V2 Azej Z then V 1 2eM 7l z 2 hFill Vii We find that the magnitude of the quotient is the quotient of the magnitudes A1 W A and the phase of the quotient is the difference of the phases 2 2 zV3 4V1 4V2 1 2 If the two phasors are expressed as a ratio of two Cartesian complex numbers l7 x 39 V1 x1jy1 and V2 x2jy2 and V3 27121 19 V2 x2 Jyz Susan Bonner 9 Revised 1162006 Rensselaer Polytechnic Institute Troy New York USA Electronic Instrumentation ENGR43 00 then the magnitude and phase of the quotient can be found as follows 2 2 e l 1 x1y1 in i V 2 2 equation5 2 x2 y2 and 11 11 g 1 1 y2 A 71 4V3 4V1 4V2 tan tan equation 6 C2 Phasors and Transfer Functions Recall that we are looking for a transfer function that relates the output to the input signal of a circuit using the relationship in equation 1 We have an input signal of arbitrary amplitude and phase and a corresponding output signal Vin t 2 Am 0050 m and Vow t Aouf 0050 ouf By defining the input and output in terms of phasors iwt m iwt m Iin Aine and Vout Acute we can express the function H as follows i 601 0 jwt j aux 1 H Vout Acute Amie e 2 Acute Aout ejwam m d i 601 1quot jwt Mm j m Iin Ame Aine e Aine A in If we have the phasor for the input and a compleX polar expression for H we can calculate the phasor for the output as follows V0 2 H 39 Vin equation 7 By defining H we have also defined a new type of phasor The magnitude of H is not the amplitude of any sinusoidal signal it is a factor that we multiply by the input amplitude to determine the output amplitude Likewise the phase of H is not a phase angle but the change in phase between the input and the output wave Thus given the transfer function H the amplitude and phase of the output phasor can be determined by Susan Bonner 10 Revised 1162006 Rensselaer Polytechnic Institute Troy New York USA Electronic Instrumentation ENGR43 00 Aout 39 Ai n equation 8 and out 411 in equation 9 Now we need to nd a way to use the components in the circuit itself to derive some phasor H that allows us to map the input phasor of the circuit to an appropriate output phasor D Finding Transfer Functions We know that by using phasors and complex polar coordinates we can greatly simplify the problem of determining the output of an AC circuit for any input It still remains however to develop a strategy for nding the transfer function phasor D The In uence of Resistors Capacitors and Inductors Conceptually H defines how the circuit alters the phase and amplitude of the input wave to produce the output wave Each component in the circuit is in uencing the amplitude and the phase in some way We can use the equations that govern the behavior of each of the three basic passive components resistors capacitors and inductors to determine the nature of the in uence of each In order to do this we will consider a sinusoidal input Vin I Am sinat m and examine the way in which each device alters its phase and amplitude Then we will examine what this in uence means in the complex polar coordinate system We will define the in uence of each component in terms of a complex quantity called impedance and denote it by the letter Z The fundamental equation that governs the behavior of a resistor is VR 1 iR t R If we let V R I Am sinat m then the output signal will be V0 I R 39 Am Sil l0t Qquot Note that the resistor changes the amplitude of the input by a factor of R but it does not affect the phase at all Since AW I Alquot and 0w LII in the magnitude of a phasor showing the in uence ofa resistor ZR must be R and the phase must be zero In complex polar form this would mean ZR Re Rcos0jR sin0 R Therefore the impedance of a resistor is given simply by ZR 2 R equation 10 Susan Bonner ll Revised 1162006 Rensselaer Polytechnic Institute Troy New York USA Electronic Instrumentation ENGR4300 61110 The fundamental equation that governs the behav1or of an 1nductor is V L l L7 Ifwe let VL t Am sinat m then the output signal will be vow t L Ain cocosat in coLAin sinat in 7 The inductor changes the amplitude of the input by a factor of 03L and shifts the phase by 1t2 Recall that the return from sine to cosine shifts the phase by 1t 2 Since AW Alquot and 0w A in the magnitude of a phasor showing the in uence of a inductor ZL must be 03L and the phase must be 1t 2 In complex polar form this would mean 21 60L 61 UL 39 COS jcoL sin ij Therefore the impedance of a capacitor is given by ZL equation 11 dv t The fundamental equation that governs the behavior of a capacitor is iC t C j 1 Ifwe solve this for the voltage we get that VC 0 EIZC 1 dl If we let vC t 2 Am sinat m then the output signal will be Va 0 39 Air 39 005a t in Am ocjsin ot in 7 The capacitor changes the amplitude of the input by a factor of lnC and shifts the phase by 112 The return from cosine to sine shifts the phase by 1t 2 and the negative sign shifts it by 11 The net phase shift caused by the integration is 11 2 Since AM Am and UM A m the magnitude of a phasor showing the in uence ofa capacitor Zc must be lnC and the phase must be 112 In complex polar form this would mean A 1 NJV 1 l l 1 Z e 2 cos s1n C WC WC ij ij j39wC Therefore the impedance of a capacitor is given by Z l C QC equation 12 We can now use these complex expressions to define transfer functions of RLC circuits Susan Bonner 12 Revised 1162006 Rensselaer Polytechnic Institute Troy New York USA Electronic Instrumentation ENGR4300 D2 Transfer Functions and Complex Impedance Recall that in gure 2 we used a voltage divider to nd the transfer function of a resistive circuit We can eXtend the use of voltage dividers to complex phasors in order to nd transfer functions for passive circuits containing capacitors inductors and resistors In gure 10 we have taken the circuit in gure 3 and represented the in uence of each component by its compleX impedance To this we can apply the voltage divider rule in the same way as we did in gure 2 to nd H zC JmC 1y V 2V 95 zLij ZCZLZR J 3 HI W Vquot ZCZLZR g ZRR 0 Figure 10 Let s look at some simpler examples D3 An RC Circuit Example Let s see how transfer functions work by considering the following circuit Vout R Fu V C C n 1 First note the impedances of the two devices Z R R and Z c Using the J a ZC voltage divider rule we can write the output as 17 ZRZC Vin or equivalently Susan Bonner l3 Revised 1162006 Rensselaer Polytechnic Institute Troy New York USA Electronic Instrumentation ENGR43 00 l Iyou H J m 2 I7 1 To make working with the transfer function easier it is in R JwC ij best to Simplify it by multiplying by j39wC l a j wC Vow ij 1 How ijC 1 NV Therefore the transfer function of the circuit is H 39a j 1jaJRC Now using equation 5 we can nd an expression for the magnitude Hjco1j0 my 1 1jcoRC J12mRC2 J1mRC2 and using equation 6 the phase RC AHja 41 j0 41 ijC tan 1 tan 1 tan 1aRC What is the output of this circuit if we let RlKQ and CluF and if the input is Vin Amcosat m 2Vcos27th 7V4 We know from the input signal that n 211K radsec We can substitute the angular frequency and the component values into the equations for the magnitude and phase of H003 to find 1 11272K1K1u2 We can now find the amplitude phase and function for the output of this circuit at 211K radsec using equations 8 and 9 Susan Bonner l4 Revised 1162006 Rensselaer Polytechnic Institute Troy New York USA Hj2K7r 0157 and mom tan39127zKlKlu 141 Electronic Instrumentation ENGR4300 Am11jcoA7w 01572V 0314v q o 4Hjw m 0785141 0625 rad Therefore the output of the circuit for this input is Vamp 0314Veos27K t7 0625 D4 An RLC Circuit Example Let s look at an example With all three types of componenw C wwmn kA V0ut First We Will nd the transfer inction of this circuit R j to RC j to RC 1 RijZLC1ijC 1701ijch ij Hja ij Since this is a more complex example let s substitute in the numerical Values before We nd expressions for the magnitude and the phase IfWe let LlmH Cl pF and R1KQ and We use the same input in the previous examp e Vmt Ameoswt 57 2Veos2IZK t 714 The transfer function becomes j27tK1Kly j63 j63 H0 7 14271021m1yj27r1lt11lt1 7 17395mj63 7 096j63 Now We can nd the magnitude and phase using equations 5 and 6 Susan Bonner 15 Revised 1162006 Rensselaer Polytechnic Institute T my New Y ork USA Electronlc Instrumentatlon ENGR743 00 Hj o and zHom W483 tanquot IrZ 7142 015 rad Thls rneans that the output othls crrcurt for the gryen rnput ls Vm 210Vca27r1ltt 785015ZVcax27th 935 For detarls on flndlng phases consult appendlx A E Filter Fllters are a fundamental concept ln electronlcs Many RL RC and RLc clrcults act as at nttelt p Flgure 11 sh nl Luau ll low pass lter hlgh pass lter band pass lter and band reject lter 1 11 1M m n frequency 00 n frequency 00 3 frequmcy 00 n equ ry Inwyau ku high pas rrltsr hurl pass lter hand rejact lter The names ofthe lters rndrcate exactly what they do A low pass lter for example passes low frequencles Thls rneans that lfthe rnputhas a low e uency the outputwrll beLhesm as n t No a e Lhel yalue thetra sferfunctron ofalowpass lter at lo equencres ls 1 Ifth puthasahlgh equ n y theo utwlllbemultlplle by the value ofthe trans ctron here uencleswhlc ro Hence output are erfun at lg q lsze e for a hlgh frequency rnput wlll be ltered out The lter haspassed low quencles and cles rejected hlgh frequen An tu t a lr ultl ltbehaves Hqunl lnuanll Susan Banner 7 16 e Revtsed 1162006 Renxxelaer Polytecth Institute Tray N w A Electronic Instrumentation ENGR43 00 E 1 Transfer Functions atLow Frequencies First let s consider how to nd the magnitude and phase of a transfer function at low frequencies If we have a transfer function with a general form all a 2 a1 a a0 b2 a 2 bla be frequencies To do this you must find the single term in the numerator and the denominator which dominates the behavior of the function as 03 gets very small This term will be the one with the smallest power of 03 Remembering of course that no H j39m then we first find how this function looks at very low To understand why this works consider 03103 Since 03210396 01103 and 001 azmza1ma0 azx10396 a1x10393 a0 which is approximately a0 If you assume all the ai are 1 then the term becomes 1001001 which rounds to 1 As 03 gets even smaller the number just gets closer to 1 The same holds true for the denominator For more on taking limits see Appendix B In our first example from the last section H j39m 4 the lowest power of n in 1 jch both the numerator and denominator is no or 1 and the function simplifies to HLOWUm 1 This is the complex number 1 j 0 We can use equations 5 and 6 to find the magnitude and phase 1 0 IHLOWaasa gt011 AHLOWatan T 0 rad Note that when we calculate the magnitude at low frequencies we want to take the magnitude of HLome and then take its limit as n approaches 0 The phase should be found by applying the phase equation to HLOWGm NOT to its magnitude Taking the phase of the magnitude is meaningless 39 RC Let s look at our other example Hja ij The term With the l a LCja RC lowest power of n in the numerator is ijC The term with the lowest power of n in the denominator is 1 Therefore the transfer function at low frequencies is 39wRC HLOWa jch This is complex number 0 joaRC To find the magnitude we take the limit of the magnitude of this function as n approaches zero Susan Bonner 17 Revised 1162006 Rensselaer Polytechnic Institute Troy New York USA Electronic Instrumentation ENGR43 00 IHLOWj39aasa gt0aRCasa gt00 To nd the phase go back to the general function for HLowGoa and nd its phase ZHLOW Jim tan71 7139 2 rod why In this particular case we have run into a problem with the phase equation We are dealing with a point where the tangent approaches infinity However since HLome ijC is a positive imaginary number we know its phase must be n2 The special cases when either the real or imaginary term is zero are fairly easy to deal with Decide whether the complex number you found for HLome is positive real negative real positive imaginary or negative imaginary Then use the following chart Although finding the phase for the limits of transfer functions is fairly straight forward because the limit always ends up one of the four cases in the chart finding phases in the general case can be quite complicated Please see appendix A for more information E 2 Transfer Functions at High Frequencies For high frequencies we must consider what happens when 03 gets very large Let us use the general form again and let to be 10 This means that m2106 05103 and 001 and that the numerator is azx106 a1x103 a0 If you assume all the ai are 1 than the term becomes 1001001 which rounds to 1 million As 03 gets even larger the number just gets closer to a2 Clearly the contribution of a2 is much more than the others Therefore at high frequencies the dominant term in both the numerator and denominator has the highest order of n For additional information on taking limits see appendix B 1 In our first example H j39m the numerator has only one term so the 1 jch highest power of n is 0 In the denominator we have two terms and we use the 31 term joaRC Therefore at high frequencies the function that governs the behavior is HHIGH j39w This is imaginary number 0 ij oaRC because 1j j jch Now we can find the magnitude and phase Susan Bonner 18 Revised 1162006 Rensselaer Polytechnic Institute Troy New York USA Electronic Instrumentation ENGR43 00 IHHIGHQ39w I as a gt oo as a gt oo i 0 AHHIGHUQ quot imagina1y rad OO mRC Since the transfer function is a negative imaginary number the phase from the chart must be 11 2 For more information about determining phase see appendix A jch In our other example Hja l szCjaRC At high frequencies this simplifies to 39 RC R Hja L L This is imaginary number 0 ij WmL a 2LC mL The magnitude and phase of this function at high frequencies is IHHIGH j39m as a gt oo L as a gt oo 0 and AHHIGH j39m quot imaginary rad a E3 Filters andLimits Once we have the limits of a transfer function it is simple to determine the type of filter quot J atlowf 39 isl it is In our first example Hja he 1 1 1 jch a t and the magnitude at high frequencies is 0 It must be a low pass filter At low frequencies this filter will not change the phase of the signal at all and at high frequencies there will be a 112 phase shift jch l mzLC jch frequencies and at high frequencies This does not mean that it is zero at all frequencies It means that it is a band pass filter It rejects both very low and very high frequencies but passes some band of frequencies in between The phase of this function shifts from 1t2 to 112 over the frequency range from zero to infinity Our second example Hja has a magnitude of0 both at low In order to understand how a filter functions we must know how it behaves at high and low frequencies but we must also know what else it does Where does it transition from high to lowfrom low to high Where is the pass or reject band on a band lter How wide is the band To answer these questions we must know how to find the comer frequency and the resonant frequency E 4 The Corner Frequency We have established that our first example Hja is a low pass filter We l jch still don t know however the frequency at which the filter switches from passing the Susan Bonner l9 Revised 1162006 Rensselaer Polytechnic Institute Troy New York USA Electronic Instrumentation ENGR43 00 input to rejecting it This is called the comer frequency In an ideal filter the transfer function would switch instantaneously from 1 to 0 In reality there is a range over which the value of the transfer function goes from 1 to 0 We assign a single frequency to define the approximate location of this area and call it the comer frequency By definition the comer frequency is the location at which the value of the transfer function 1s T or about 0707 This pomt is chosen Since the power that goes through the circuit 2 at this frequency is half of the input power For our first example we can use the definition of the comer frequency to derive an equation that we can use to find it for any simple RC circuit Hjw Hjw L l 1ijC 01302 45 1wRC2 2 21aRC2 1 2 1 g mo RC2 RC Dc is the frequency which defines about where our simple RC low pass filter switches from high to low As a matter of fact the comer frequency for any simple RC filter is given by a l e uation l3 5 R C l q You can go through a similar process to prove that the comer frequency for a simple RL circuit is R 0 I equation 14 E 5 The Resonant Frequency For the circuits that include capacitors and inductors such as our second example How 2ja2RC I l a LCa RC is defined We already know that this circuit is a band pass filter The resonant frequency of a simple RLC circuit like this one is the frequency at which resonance between the capacitor and the inductor occurs For the practical purposes of this course the resonant frequency occurs when the lnzLC term in the denominator goes to zero This occurs at another special frequency called the resonant frequency 1 0 0 equation 15 I LC Susan Bonner 20 Revised 1162006 Rensselaer Polytechnic Institute Troy New York USA Electrame Insmxnentahm ENGRemn The hweeeme amummeeam equemywe we mums cmxse us an appxaxxmauanta me acm xesammfnquency a my ample RLC mm mm meme ufmme camplzx mounts engneexs 1m fax pales acmms a am a a mgr man as nunexa39m am transfer nmmxs equalta zem We dan39thaveume m gm mm lms amwunt a sum saw me mermxy sauna and very ample sppmxxmmm mm use ufmx maul me xesmmtfxequmcy meme eemee uwee passbmd This mm alsahaswxa cmmxsmnl dame Lb wxd39h uwee band These canbe rm mg 1 me cmmxequ u RLC lms em be lww pass we pass RC and 1 L band pass a band 12 act In em band lms we xesamnt equancy gves wuthz lacahm uf me eenm arm band In law and we pass lms n gves ywu me appxaxxmate 1mm arm Cuzan EA Man Camplkahd Exampl InLhIshSL example we wxll exammz abandpass ex wah same campanzn39smpan zl Ths 5 we must cmnplex type uf mgr funcan memqu are hkelyta have m Amiga In was class R vm Recs um wxuge mam un y wk m senes cucuts H mce m azan m determine me meagre nchmfm mes urcuL we mm m cambxne thetwa parallel campamms Thsxs am umgme cambxnahmxulesfmpau el masters J Z zkzz Jam 1 1m 2M2 a 1 mum IrmeC J Jac Umgme valng dwde rule me kansfexfmmanxs gvmby Susan Eomv e21 e Revised 1152006 Rimselmv Fobmom 1mm My New York USA Electronic Instrumentation ENGR43 00 ij Z 2 2 39 L Hjw mumplybyw Humz RZCL RL l a LC Rl a LCaL l mzLC Now that we have H003 we can look at the behavior at low and high frequencies 39wL I n HLOWwjT IHLOWww gt00 ZHLOWJWErad ij j 1 7239 HHIGHUmzmz leHWWm 0 0 zHHEHUmFEmd This lter although it approaches 0 at both high and low frequencies does not block all frequencies It will pass a band around the resonant frequency which by definition is 1 ch 39 More useful information about a filter can be found by finding the transfer function magnitude and phase at the resonant frequency To do this you simply substitute the expression for the resonant frequency 030 into the equation for H003 LL HOW 2 J39woL ijLC 1 R1 mgLCjmoL R1 1 ZLC 1 1L ll JL C Hja0l 4Hja00rad You can see that this filter does indeed pass a band of frequencies around the resonant frequency because the magnitude of the transfer function at the resonant frequency is 1 Therefore for some band of frequencies centered around the resonant frequency the output will be equal to the input The width of the band can be found using the comer frequency equations Further information can be found in the Gingrich online notes at httpwww nhvs llnthI ta ca Ngingrichpth39J J 7 html G Conclusion Transfer functions relate the output to the input of a circuit For AC circuits we can use phasors to easily find and manipulate transfer functions Phasors are defined in terms of the compleX polar coordinate system because AC signals are sinusoids which are easily Susan Bonner 22 Revised 1162006 Rensselaer Polytechnic Institute Troy New York USA Electronic Instrumentation ENGR43 00 manipulated using this representation Transfer functions are useful for nding the output of a circuit for any given input They can also be used to determine if a circuit is behaving as a lter and enable us to find important features of that filter Other useful features of filters are the comer frequency and the resonant frequency Appendix A More about Phases Phases can be determined by looking at the real and imaginary parts of the H003 function The general equation for phase is given by 417 tan391Z when l7 x jy x Calculating phases using the inverse tangent function If the transfer function is given as a ratio of two complex numbers then the phase is given by the difference between the phases of the numerator and denominator if Hja then 4H tan391 tan391amp y2 x x xi 1 2 If X1 y1 X2 and y are all positive then the phase changes are all in the first quadrant and the equation can be applied directly with a calculator If one or more of them is negative then one must worry about which quadrant the phase angle is in The most reliable way to determine a phase change is to take the absolute value of the X and y coordinates of a complex number calculate tan391lyXl to find the reference angle use the signs of X and y to determine the quadrant and find the phase based on the reference angle and the quadrant I ll III IV Susan Bonner 23 Revised 1162006 Rensselaer Polytechnic Institute Troy New York USA Electronic Instrumentation ENGR43 00 In the gure above 5 is the reference angle for 0 We want to nd 0 the actual phase tan391lyxl will always give us the reference angle 5 We can nd 0 based on the sign and the quadrant X Note that all angles in the above chart represent a phase shift between 11 and 11 radians between 180 and 180 degrees 3j4 4 13 numerator l tan391l43l 093 xgt0 ygt0 Q1 Anum 093 denominator l tan391l34l 054 xgt0 ygt0 Q1 Aden 054 A H Anum Aden 093054 039 rad some examples Hja 3 j 4 4 j3 numerator l tan391l43l 093 xlt0 ygt0 Q2 Anum 314093 221 denominator l tan391l34l 054 xgt0 ylt0 Q3 Aden 054314 260 A H Anum Aden 221260 481 rad 147 rad HOW 3 j4 4 13 numerator l tan391l43l 093 xlt0 ylt0 Q4 Anum 093 denominator l tan391l34l 054 xgt0 ygt0Q1 Aden 054 A H Anum Aden 093054 147 rad HOW Special cases for nding phases Note that a calculator can be used to nd phases where the real and imaginary parts of a complex number have a nonzero value What does one do when the real or imaginary part of a complex number is zero These cases are best determined by examination using the complex plane Most functions we deal with in this class can be found using this simple method which avoids the use of the calculator entirely Susan Bonner 24 Revised 1162006 Rensselaer Polytechnic Institute Troy New York USA Electronlc Insh umentauon ENGR743 00 um l Very low frequencles there wlll be only one lemn m the numaator and one term m the n n l l a ltherthe real or lmaglnary part othe complex transt functlon wlll always be zero Thls makes four axes the fmctlon he on we can delemme Its phase by lnspecuon In the above gure any pomt on the axls llsted has the lndlcated phase r al ex R gt Oradlans x lans ereal ex mic gt n or 71 radlans 1 ex lJmC or 1mm gt rm mdlans Or one can remember the slmple relatlonshlps m the followlng table Anolhec common case 15 when the absolute value othereal and lmagmaryparts othe mpl nunl anal l l orby lnspecuon because the transt functlon lles along ether y x or ygtrx Susan Banner 7 25 e Remed 1162006 Renxxelaer Polytecth Inmate Tray N w A Electronic Instrumentation ENGR4300 1 yx yx 3 49539 495 real 39 real 6 In the above figure any point on yx or yx when yx has the indicated phase xgt0 and ygt0 ex 1j 9 n4 radians xlt0 and ygt0 ex lj 9 3754 radians xlt0 and ylt0 ex lj 9 374 radians xgt0 and ylt0 ex lj 9 7c4 radians The following chart contains some other useful values of tan39l39 Note that these values can be determined simply by nding the quadrant in the realimaginary plane Case real part x Imaginary tan391xjy in tan391xjy in part y degrees radians a 0 0 0 0 D A1j A A 45 754 g A1j A A 135 3754 h A1j A A 45 454 1 A1j A A 135 3754 Some examples Susan Bonner Rensselaer Polytechnic Institute H003 R j03L R j03L lj03C at high frequencies H003 003RC 032LC 003RC 1 032LC H003 gt 032LC032LC gt 1 at 03 gt oo 4 H003 gt tan3910l or zreal at 03 gt 00 this is caseb 4H030at03 gtoo H003 j03L R j03L lj03C at low frequencies H003 032LC 003RC 1 032LC H003 gt 032LC at 03 gt 0 4 H003 gt tan3910032LC or 4 real at 03 gt 0 this is cased 4H037at03 gt0 H0 03 j03RC 003RC l at the comer frequency 26 Revised 1162006 T my New York USA Electronic Instrumentation ENGR43 00 4 H003 4numerator 4denominator 4 H003 tan39IGmRCO tan39lg39oaRCl 4jszRc1 this is casec 7 casef iff Dc lRC 4 Hg39oac 112 tan391ll 112 n4 at m lRC A H mc n4 at Dc 1RC Appendix B More on Taking Limits You must be able to take limits in order to use transfer functions effectively Basically to take a limit as n gt 0 or n gt 00 you must determine the dominant term in both the numerator and denominator and then consider the value of the ratio as the function approaches the limit Note that when you first write out a transfer function by looking at the circuit it is often not in the best form for taking a limit Multiplying all terms by 003C usually puts it in a form where there are no fractions in the numerator and denominator More complex circuits may require more reduction When you take the limit try considering which of the two forms makes it easiest to understand It might be either one I like the one below When the transfer function has the general form A032 Boa C 11sz Em F G032 H0 1 jJoa2 Koo L To find the dominant term as n gt 0 look for the lowest power of n in the numerator and the lowest power in the denominator example R joaL Next multiply num and den by joaC R joaL ljoaC joaRC oazLC Next find dominant terms as JO gt0 ijC och 1 joaRC Reduce l joaRC Use this to approximate H003 at JO gt0 To find the dominant term as n gt 00 look for the highest power of n in the denominator and the highest power in the numerator example R Next multiply num and den by joaC R joaL ljoaC Susan Bonner 27 Revised 1162006 Rensselaer Polytechnic Institute Troy New York USA Electronic Instrumentation ENGR43 00 39nRC Next nd dominant terms as Jo gt00 ijC och 1 joaRC Reduce ofLC jR This can be used for H003 at Jo gt00 03L Once you have the dominant term for both the numerator and the denominator you can decide how the function behaves as n approaches the desired limit I made a chart of the different cases as this is easier on the computer comments term in term in 0 oo 0 oo 0 oo 0 oo 0 oo 0 oo 0 oo 0 00 Appendix C Examples of Transfer Functions Resonant or CIRCUIT Transfer Function Corner Frequency 1 R 1 W1 H AwC 1 0 RC 0 R ijC 1 1 1a 1 J C f0 27ch Susan Bonner 28 Revised 1162006 Rensselaer Polytechnic Institute Troy New York USA VSH 34021 MN Kou 900mm spasmael 6Z 9171111574 011411091fZOC JBDZQSSMBH 19141405 MUSHS 0AM 0750 1 H 0AM 071m g 70f 9m H7mf H Darvf137zm DWI a307709f T H H 07209 0wa 07209 H 7MHDmf1 709 H 1 DWI 37lov I 307g7mf H 009 H ngf1 HDm DHmf H 2H 00 EV39HDNEI uoumuewnnsul oguonoelg VSH 34021 MN Kou 900mm spasmael 9171111574 0114110211104 JBDZQSSMBH 5 19141405 MUSHS 07 le I of L 1 qg 0731M 7M owzm H7mf 129 lt lt 1 owzm H7mf 30 a I a o t gt I H 307 x70 0 37am H q H709f 1 Hx7mf I H709f H HXTW AAA WV 0 7M owzm H 7mf 9m 7 W H o Dm 70f H a Dmx7mf 1 071m 03ml 3 H HOLM Om7mfg H Om7mf H 00 EV39HDNEI uoumuewnnsul oguonoelg Electronic Instrumentation ENGR43 00 Appendix D Algebra Review We want our transfer functions to be a ratio of two polynomials In order to achieve this we often have to eliminate fractions from the numerator and denominator of an expression Sometimes it is simply one fraction we need to get rid of In this case we multiply the top and the bottom by the denominator A 1 A 1 D43 l xg f Notel C 5 C For example 1 1 ij ij ij 1 H 1 1 X39Cz39mRC1 R wc R wc 1w Other times we have more than one fraction in the denominator This can be dealt with in one step by multiplying by the product of both denominators BC ABC BC gtlt Note 1 E E BC EBCDC BC or by eliminating one fraction at a time 3y ABC1ABC1 C ABC B B X X EIy E B EBD EBD C EBCDC Forexample ijL ijL ijR ijR XjcoLR ijL ijL ijL ijR ijR jcoC ijR jcoC ijR jcoC ijL XjaC a2RLC ijRijL jaC ijR co2RLC jcoC H ijL Susan Bonner 31 Revised 1162006 Rensselaer Polytechnic Institute Troy New York USA
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