1 00:00:19,410 --> 00:00:31,580 Welcome to this second lecture on concept of image impedance Now I hope you agree with 2 00:00:31,580 --> 00:00:39,870 me that any 2 port network . can always be represented by either a T section or a PIE 3 00:00:39,870 --> 00:00:47,690 section without losing a generality I take that my 2 port network that means that ab 4 00:00:47,690 --> 00:00:58,960 issh transmission matrix characterization is ABCD that is a T section the same analysis 5 00:00:58,960 --> 00:01:16,170 hold for PIE section so now I have a T section this is Z1 this is Z2 this is Z3 all this 6 00:01:16,170 --> 00:01:25,110 are impedances complex impedances so this is many port 2 this is my port 1 this is the 7 00:01:25,110 --> 00:01:41,970 internal description of the network now I want when I will excide the port 1 with that 8 00:01:41,970 --> 00:01:51,940 voltage BS and I have some internal impedance let us call that internal impedance Zi1last 9 00:01:51,940 --> 00:02:01,009 time I called it ZS this time I am calling it ZI1 is simply change of numb…ain this 10 00:02:01,009 --> 00:02:08,640 is my port 1 this is my port 2 now you see all of you are familiar with maximum power 11 00:02:08,640 --> 00:02:16,590 transfer theorem maximum power transfer theorem says that if the load impedance complex is 12 00:02:16,590 --> 00:02:24,400 a complex conjugate of the source impedance then maximum power gets transferred I think 13 00:02:24,400 --> 00:02:30,880 that you have noticed that in low frequency particularly this VLSI people etc when they 14 00:02:30,880 --> 00:02:37,930 work upto gigahertz range they do not give any consideration to these maximum power transfer 15 00:02:37,930 --> 00:02:46,290 theorem because in baseband unless you go upto radio frequency and transmit it you have 16 00:02:46,290 --> 00:02:52,870 plenty of power so you are more concerned with your voltage maximization thats why you 17 00:02:52,870 --> 00:03:03,680 design a good C amplifier with very high voltage gain but voltage gain does not necessarily 18 00:03:03,680 --> 00:03:13,000 mean a maximum power gain but we when we go to radio frequency we know power microwave 19 00:03:13,000 --> 00:03:19,140 power is very precious to produce microwave power lots of complete complicated circuits 20 00:03:19,140 --> 00:03:26,790 are required So also when power is received by a receiver in radio frequency it is very 21 00:03:26,790 --> 00:03:35,700 small amount so its life and death for RF engineer or a RF circuits to maximize power 22 00:03:35,700 --> 00:03:43,900 So maximum power transfer theorem always is the design of RF circuits always aa ba we 23 00:03:43,900 --> 00:03:52,570 try to pay regard to the maximum power transfer theorem So can I have this whole thing suppose 24 00:03:52,570 --> 00:04:00,820 this I will terminate by some load impedance let that load impedance is called ZI2 this 25 00:04:00,820 --> 00:04:08,630 one I called ZI1 source impedance is ZI2 now the idea of image impedance is at this point 26 00:04:08,630 --> 00:04:18,321 obviously if I look at I will get some impedance Now if this impedance is equal to ZI1 then 27 00:04:18,321 --> 00:04:29,330 I can from the source I can have maximum power transfer ok now according to maximum power 28 00:04:29,330 --> 00:04:39,400 transfer I suppose these I am looking at some Zin or something Zin now if ZI is complex 29 00:04:39,400 --> 00:04:48,020 conjugated ZI1 star then I know that maximum power transfer will takes place But as I said 30 00:04:48,020 --> 00:04:58,240 that our consideration now is filter which is lossless network so this Z1 Z2 Z3 there 31 00:04:58,240 --> 00:05:10,410 is no R involved ideally So they are complex but they are generally they will be either 32 00:05:10,410 --> 00:05:18,350 real or not real either all of them will be pure imaginary term there will be pure reactance 33 00:05:18,350 --> 00:05:28,199 So in that case I say that now I say that I will look into here ZI1 then ZI1 is equal 34 00:05:28,199 --> 00:05:36,590 to ZI1 not comp not ness suppose any complex conjugate for pure imaginary things it is 35 00:05:36,590 --> 00:05:46,680 equal as you know that suppose I have two complex number A is equal to B star suppose 36 00:05:46,680 --> 00:05:52,310 A is complex number B is a complex number if A is pure real and B is pure real then 37 00:05:52,310 --> 00:05:59,130 I can say A is equal to B Similarly if A is pure imaginary and B is pure imaginary then 38 00:05:59,130 --> 00:06:07,250 only I can say A is equal to B Since we know that this will be my all these are pure reactances 39 00:06:07,250 --> 00:06:17,030 this may be a pure resistance so that is why I can call that my demand is ZI1 here should 40 00:06:17,030 --> 00:06:24,900 be equal to ZI1 here That means source impedance and these input impedance looking at this 41 00:06:24,900 --> 00:06:32,300 port should be equal but you see this ZI1 is a function of this load resistance ZI2 42 00:06:32,300 --> 00:06:39,490 But now so why it is called this ZI1 is called image impedance if I can find an impedance 43 00:06:39,490 --> 00:06:45,100 load impedance Zi2 so that if I terminate this network this is already given network 44 00:06:45,100 --> 00:06:52,060 and I look at here and see that the input impedance is ZI1 then I know that I can transfer 45 00:06:52,060 --> 00:07:01,300 maximum power from the source to this network this to this port one of this network now 46 00:07:01,300 --> 00:07:08,639 why ZY ZI1 is called an image impedance because if at this plane I look so I looking at this 47 00:07:08,639 --> 00:07:14,720 side I am getting to the right side and getting and ZI1 looking at left side I am seeing the 48 00:07:14,720 --> 00:07:22,280 impedances ZI1 So the this side is image of these that is why this is an image impedance 49 00:07:22,280 --> 00:07:36,400 it so ZI1 is an image impedance Same thing here and here I want that if I look at here 50 00:07:36,400 --> 00:07:48,240 I should look at some output impedance that should be equal to Zi2 So at port 2 if I look 51 00:07:48,240 --> 00:07:56,330 to this side I am getting ZI2 if I look this side I should get ZI2 So if I can find PR 52 00:07:56,330 --> 00:08:06,199 to ZI1 ZI2 so that if I terminate by ZI2 then I get here ZI1 impedance similarly here if 53 00:08:06,199 --> 00:08:14,720 I terminate by ZI1 and excite here I should see here is ZI2 these two pairs are called 54 00:08:14,720 --> 00:08:21,050 image impedance Since this is an a symmetrical network because ZI1 is not equal to ZI2 I 55 00:08:21,050 --> 00:08:39,889 will have two impedance ZI1 and ZI2 Let us see that whether this image impedance can 56 00:08:39,889 --> 00:08:51,070 be represented in terms of this impedance of this network . so the same diagram I can 57 00:08:51,070 --> 00:09:10,009 write that ZI1 or do like this so this ZI1 I can write as what is ZI1 obviously it is 58 00:09:10,009 --> 00:09:38,629 Z1 plus you see Z3 parallel to Z2 plus ZI2 Likewise what is ZI2 it is Z2 plus Z3 parallel 59 00:09:38,629 --> 00:09:53,249 to Z1 plus ZI1 Now these two equations you see ZI1 here I have ZI2 I have ZI2 here I 60 00:09:53,249 --> 00:10:04,570 have ZI1 I have two equations so I can solve for ZI1 and ZI2 in terms of Z1 Z2 Z3 If I 61 00:10:04,570 --> 00:10:23,369 do that upon solving this I get ZI1 is equal to root over Z1 plus Z3 into Z1 Z2 plus Z2 62 00:10:23,369 --> 00:10:56,089 Z3 plus Z3 Z1by Z2 plus Z3 and ZI2 is Z2 plus Z3 into Z1 Z2 plus Z2 Z3 plus Z3 Z1 So you 63 00:10:56,089 --> 00:11:05,319 see that this image impedance can be represented in terms of the component impedances of the 64 00:11:05,319 --> 00:11:14,939 T section Now always we won’t be knowing Z1 Z2 Z3 as I said that let us consider two 65 00:11:14,939 --> 00:11:23,339 port network as a black box But we can do measurements and always find these image impedances 66 00:11:23,339 --> 00:11:33,329 How you know that any measurement requires either an open circuit or short circuit of 67 00:11:33,329 --> 00:11:41,230 the one of the port So for any impedance measurement you need to do this also you have seen that 68 00:11:41,230 --> 00:11:46,879 if you want to find any 2 port parameter you need some port condition either short or open 69 00:11:46,879 --> 00:12:02,050 etc etc . So if we measure measurement of image impedance 70 00:12:02,050 --> 00:12:16,959 let us say that one port one we measure impedance 71 00:12:16,959 --> 00:12:30,910 when port 2 open We call that measurement as Z1 since we are doing at port 1 open circuit 72 00:12:30,910 --> 00:12:42,639 Z1OC means I am measuring the impedance input impedance at port 1 with port 2 open So if 73 00:12:42,639 --> 00:12:52,860 we look at the circuit if I open circuit this what will be Z1OC it will be simply Z1 plus 74 00:12:52,860 --> 00:13:06,860 Z3 Similarly if we measure impedance at port 2 with port 2 with sorry ah imp the imp again 75 00:13:06,860 --> 00:13:21,480 port 1 measure impedance when port 2 is short it now let me short this port So I call Z1SC 76 00:13:21,480 --> 00:13:31,019 second port is shortened you look at the circuit If I short it it will be Z1 plus Z2 Z3 parallel 77 00:13:31,019 --> 00:13:52,689 So Z1 plus Z2 parallel Z3 Ok now what is this this is Z1 Z2 plus Z2 Z3 plus Z3 Z1 by Z2 78 00:13:52,689 --> 00:14:04,629 plus Z3 Now you observe the image impedance terms already I have solved ZI1 can I just 79 00:14:04,629 --> 00:14:20,299 compare can I say that Zi1 is equal to Z1OC into Z1SC So by measurement I can always find 80 00:14:20,299 --> 00:14:29,609 Z1 OC I can find Z1 SC I know what image impedance is immediately I can calculate from these 81 00:14:29,609 --> 00:14:37,529 Similarly instead of port 1 if i measure in port 2 by once open circuiting port 1 find 82 00:14:37,529 --> 00:14:45,690 the input impedance at port 2 and then again you short the port 1 and measure the input 83 00:14:45,690 --> 00:14:54,930 impedance and port 2 . So port 2 things if we do you will see the same thing that ZI2 84 00:14:54,930 --> 00:15:10,329 can be expressed as Z2 OC into Z2 SC So this shows that in image impedances can be always 85 00:15:10,329 --> 00:15:22,079 obtained from short or open circuit measurements on any network So we can easily do this suppose 86 00:15:22,079 --> 00:15:31,499 I am given a network I can always find this Z1 OC Z1 SC Z2 OC Z2 SC and find out tis Zi1 87 00:15:31,499 --> 00:15:40,850 Zi 2 and then I choose a source with that internal impedance that I want and terminate 88 00:15:40,850 --> 00:15:51,919 or choose the load as Zi 2 I know I can achieve maximum power transfer 89 00:15:51,919 --> 00:16:02,389 Already I said so that means I can have maximum power transfer is guaranteed if I use image 90 00:16:02,389 --> 00:16:13,309 impedance as the terminating impedance at both the ports Now 91 00:16:13,309 --> 00:16:21,549 now we know we have said that we will be using the two port network as only lossless components 92 00:16:21,549 --> 00:16:34,249 that means you do not use any hard so their own many internal loss there So by terminating 93 00:16:34,249 --> 00:16:45,990 with image impedances I assume maximum power transfer no loss in the circuit lossless So 94 00:16:45,990 --> 00:16:54,959 image impedance is an important thing performance measure of the power transport power transmission 95 00:16:54,959 --> 00:17:09,310 that is taking place to a network so you see that we can specify something on it later 96 00:17:09,310 --> 00:17:17,380 when we will design a filter So instead of ABCD we can specify image impedances and that 97 00:17:17,380 --> 00:17:28,020 will solve one many of our problems but think one point that I have image impedance here 98 00:17:28,020 --> 00:17:36,810 I have image impedance here also you see with this I require to know that ok by this Zi 99 00:17:36,810 --> 00:17:45,190 1 and Z Zi 2 terminations Zi 1 here and Zi 2 here I have to ensure that I am giving maximum 100 00:17:45,190 --> 00:17:53,390 power am delivering to this load But I am assuming that here there is no loss but the 101 00:17:53,390 --> 00:17:58,790 power is flowing in this direction it may so happen since I am using reactive elements 102 00:17:58,790 --> 00:18:06,950 power may be locally confined that is not flowing there So I need to also see how propagation 103 00:18:06,950 --> 00:18:14,990 is taking place inside this 2 port network So we need to have the transmission of power 104 00:18:14,990 --> 00:18:26,450 to the network also that we will next see that this is called propagation of power . So 105 00:18:26,450 --> 00:18:45,030 what we define that again two port network I have this V1 I have V2 I have I1 I have 106 00:18:45,030 --> 00:18:57,480 I 2 as before now let me define V1 I1 by V2 I2 VI is volt ampere the concept you have 107 00:18:57,480 --> 00:19:07,950 learnt in you electrical circuit class So input volte ampere these are complex quantities 108 00:19:07,950 --> 00:19:20,580 input volte ampere and output volt ampere What is the what is this ratio that will be 109 00:19:20,580 --> 00:19:30,130 something now I want to ensure that that is fully making the power transmission possible 110 00:19:30,130 --> 00:19:35,090 So we call this I can name it any number this will be some number you see input by output 111 00:19:35,090 --> 00:19:43,550 volt ampere but we have certain advantage if we instead of defining any number here 112 00:19:43,550 --> 00:19:52,390 we write it as some exponential factor e to the power two gamma why because you see when 113 00:19:52,390 --> 00:19:59,500 I will cascade many such networks this one will have some this ratio e to the power 2 114 00:19:59,500 --> 00:20:07,340 gamma 1 this will have E to the power 2 gamma 2 another will have E to the power 2 gamma 115 00:20:07,340 --> 00:20:16,610 3 etc Now from this input to this input if I want to find what is this transmission ratio 116 00:20:16,610 --> 00:20:24,060 of all the volte ampire if I express it exponential factor the final thing will E to the power 117 00:20:24,060 --> 00:20:31,710 2 gamma 1 plus 2 gamma 2 plus 2 gamma 3 but if I do not use this exponential factor If 118 00:20:31,710 --> 00:20:39,100 I just write it as gamma suppose then I will have to work out and I will have to work out 119 00:20:39,100 --> 00:20:44,630 and I will have to find out what is the magnitude and phase all these things here But exponential 120 00:20:44,630 --> 00:20:49,530 factor makes simply be an addition in if it is an absolute value it would have been some 121 00:20:49,530 --> 00:21:00,550 multiplication We always prefer addition to multiplication that is why it initially people 122 00:21:00,550 --> 00:21:07,220 did like that they put this as propagation constant But now with after some learning 123 00:21:07,220 --> 00:21:14,290 people understood that if we represent this ratio is it exponential factors and also you 124 00:21:14,290 --> 00:21:19,680 see I have taken a factor two here why because many times will be interested to see what 125 00:21:19,680 --> 00:21:26,130 is the voltage ratio what is the current ratio But this is actually a volt ampere ratio which 126 00:21:26,130 --> 00:21:33,860 is for that it is actually product of voltage and current So I have taken two gamma this 127 00:21:33,860 --> 00:21:44,160 gamma is called propagation constant So what is the definition of propagation constant 128 00:21:44,160 --> 00:21:56,800 you see gamma is equal to half LN V1 I1 by V2I2 a very important definition propagation 129 00:21:56,800 --> 00:22:03,870 constant you see it shows that how input power is propagating through the network inside 130 00:22:03,870 --> 00:22:14,480 the 2 port network . so my job is now to find out what is this e to the power 2 gamma ratio 131 00:22:14,480 --> 00:22:26,610 is equal to V1 I1 by V2 I2 is equal to in terms of ABCD parameters AV2 plus BI2 into 132 00:22:26,610 --> 00:22:41,530 CV2 plus DI2 by V2 I2 also I know V2 is equal to ZI 2I image impedance it is terminated 133 00:22:41,530 --> 00:22:48,020 with image impedance so if you do that finally you can solve that this ratio will turn out 134 00:22:48,020 --> 00:22:57,540 to be this simple manipulation put thus and you know the value of Zi1ZI2 So you will get 135 00:22:57,540 --> 00:23:08,850 this will be simply this or e to the power gamma is equal to root AD plus root BC Now 136 00:23:08,850 --> 00:23:17,890 here you see this propagation constant I have expressed in terms of ABCD parameters One 137 00:23:17,890 --> 00:23:24,310 more thing is remaining I have already said about characteristic impedances is characteristic 138 00:23:24,310 --> 00:23:31,530 impedances also expressible in terms of ABCD parameters . let us see I have the same 2 139 00:23:31,530 --> 00:23:45,610 port network I have I1 here I have V1 here and I want this should be ZI1 and here this 140 00:23:45,610 --> 00:23:58,710 should be terminated by ZI2 and this is V2 this is my I2 So I can write I know this is 141 00:23:58,710 --> 00:24:16,310 ABCD so V1I1 is equal to ABCD the definition of transmission parameters also I have V2 142 00:24:16,310 --> 00:24:30,640 is equal to I2 ZI2 and V1 is equal to I1 ZI1 So put these equations and find out what the 143 00:24:30,640 --> 00:24:43,930 Zi1 you will see you will get AZI2 plus B by CZi2 plus D let me call this for timing 144 00:24:43,930 --> 00:24:59,080 equation 1 . Now reverse the picture that same transmission line this time I am putting 145 00:24:59,080 --> 00:25:11,280 the excitation here So I am looking at it here I will get ZI2 and this is my V2 dashed 146 00:25:11,280 --> 00:25:23,360 as before this is my I2 dashed as before and from here I am taking terminating you to Zi 147 00:25:23,360 --> 00:25:36,690 1 this is my V1dashed this is my I1 dashed so here again I can write that V1 dashed I1 148 00:25:36,690 --> 00:25:51,700 dashed is equal to ABCD V2 dashed minus I2 dashed and what about the ports V1dashed is 149 00:25:51,700 --> 00:26:06,300 equal to ZI1 I1 dashed V2 dashed is equal to ZI2 I2 dashed Then find out that what is 150 00:26:06,300 --> 00:26:18,360 your ZI2 So or you find what is your ZI1 which is nothing but V1 dash by I1 dash that will 151 00:26:18,360 --> 00:26:35,270 turn out to be AZi2 minus B by minus C ZI2 plus D So this let me call equation 2 you 152 00:26:35,270 --> 00:26:40,290 have equation 1 you have equation 2 to solve for Zi1. 153 00:26:40,290 --> 00:27:01,460 If you solve even so from 1 & 2 you can solve for ZI1 and that will be equal to 154 00:27:01,460 --> 00:27:27,190 or ZI1 will be equal to AB by CD and ZI2 will be equal to root over BD by AC Now I am happy 155 00:27:27,190 --> 00:27:38,520 because I know that ZI1 one of the image impedance can be expressed in terms of four ABCD parameters 156 00:27:38,520 --> 00:27:46,840 ZI2 also I can express in terms of ABCD parameters and I have already seen that propagation constant 157 00:27:46,840 --> 00:28:00,750 gamma you see this propagation constant that also I can express in terms of ABCD parameters 158 00:28:00,750 --> 00:28:08,400 ABCD parameters completely characterization 2 port network I say equivalently I can say 159 00:28:08,400 --> 00:28:21,580 two image impedance ZI1 ZI2 and e to the and gamma these three also characterizes a network 160 00:28:21,580 --> 00:28:32,110 But what is the beauty if I have ZI ZI 2 I know what is i impedance level of the excitations 161 00:28:32,110 --> 00:28:36,590 of the network that means what is the source impedance what is the load impedance they 162 00:28:36,590 --> 00:28:42,350 are according to the power matching So that no maximum power sorry they are according 163 00:28:42,350 --> 00:28:49,490 to the maximum power will flow and by putting conditions and gamma I will be able to say 164 00:28:49,490 --> 00:28:55,510 whether these frequency will pass or not So instead of ABCD parameters this is a better 165 00:28:55,510 --> 00:29:04,140 description of a 2 port network if I want to design a filter And already I have seen 166 00:29:04,140 --> 00:29:21,800 that I can that I can do the yes I can do the measurement of image impedances that time 167 00:29:21,800 --> 00:29:30,559 I said in terms of the by opening and shorting the port I will also have to prove that I 168 00:29:30,559 --> 00:29:37,630 can do this for propagation constant also because this is a new thing that time I didn’t 169 00:29:37,630 --> 00:29:43,920 say these So that I will do now that measurement of image impedance and propagation constant 170 00:29:43,920 --> 00:29:56,300 . So what we will do the same network this is port 2 this is open circuit and I am looking 171 00:29:56,300 --> 00:30:13,070 here at let me call this Z01 So I know V1equal to AV2 plus BI2 I1 is equal to CV2 plus Di2 172 00:30:13,070 --> 00:30:22,380 etc and open circuit means I2 is equal to 0 So if you enforce that is ZO1 that will 173 00:30:22,380 --> 00:30:34,470 be A by C then you short circuit so or open circuit this port purpose you open circuit 174 00:30:34,470 --> 00:30:48,620 and measure here ZO2 so ZO2 that will be turn out to be D by C then you do that short circuit 175 00:30:48,620 --> 00:31:00,120 port one and measure the from this port you measure ZS2 you will see ZS2 will turn out 176 00:31:00,120 --> 00:31:16,809 to be B by A And which one I missed ZS2 ZS1 this so you short 177 00:31:16,809 --> 00:31:31,290 circuit this port and measure here ZS1 ZS1 will be B by D . Once you have that you can 178 00:31:31,290 --> 00:31:41,460 immediately write because already we have seen ZI1 is equal to the ZO1 into ZO2 etc 179 00:31:41,460 --> 00:31:53,430 So you will get that is equal to AB by CD and that is nothing but ZO1 ZSL similary Zi2 180 00:31:53,430 --> 00:32:05,550 is equal to root over BD by AC and that is D by C into B by A that is nothing but ZO2 181 00:32:05,550 --> 00:32:16,310 ZS2 And you see what is TAN gamma TAN hyperbolic gamma all of you are familiar with hyperbolic 182 00:32:16,310 --> 00:32:25,680 functions So this is E to the power gamma minus E to the power minus gamma by E to the 183 00:32:25,680 --> 00:32:41,080 power gamma plus and that is nothing but BC by AD that is ZS1 by ZO1 and or ZS2 by ZO2 184 00:32:41,080 --> 00:32:51,840 So you see that gamma can be expressed completely in terms of short circuit and open circuit 185 00:32:51,840 --> 00:33:02,170 measurement So I can measure image impedance by open circuit short circuit measurements 186 00:33:02,170 --> 00:33:08,990 I can also measure gamma by open circuit short circuit measurements Previously I showed that 187 00:33:08,990 --> 00:33:15,860 Zi1 Zi2 and gamma the completely characterizes the network reciprocal 2 port network Now 188 00:33:15,860 --> 00:33:20,410 I have now I have shown that they also can be measured so you do not have a difficulty 189 00:33:20,410 --> 00:33:29,860 any 2 port reciprocal network lossless network you can represent like this So an alternate 190 00:33:29,860 --> 00:33:38,232 description for characterization of 2 port network is in terms of ZI1 and ZI2 and gamma 191 00:33:38,232 --> 00:33:49,340 I think in the next class will introduce another criteria all symmetrical network and we will 192 00:33:49,340 --> 00:33:51,669 simplify this procedure Thank you