| Transcriber: | OOPS SJTU: 郑晔鑫，宋仕仲, 朱晨, 汪冠春 | | Brief Bio: | Shanghai Jiao Tong University, China, bush@sjtu.edu.cn, jduith0605@hotmail.com， arvid@sjtu.edu.cn | | Timecode: | OOPS SJTU: 季雯 | | Brief Bio: | Shanghai Jiao Tong University, China, claragee@sjtu.edu.cn | | Date finished: | 25, Sep, 2005 | | Proofreader: | OOPS SJTU: 郑晔鑫、季雯 , 邹扬文, 汪冠春 | | Brief Bio: | Shanghai Jiao Tong University, China, zhenyexin@citiz.net, claragee@sjtu.edu.cn | | Date finished: | 25, Sep, 2005 |
1
00:00:01,450 --> 00:00:05 早上好，OK，让我们开始吧， 2
00:00:05,940 --> 00:00:10,050
We have one handout today. That's your lecture notes. 我们有一份材料，那就是我们今天的讲课材料 3
00:00:10,050 --> 00:00:15,880
The same copy is still outside for those who haven't picked up. 4
00:00:15,880 --> 00:00:26,200
In general,what I will do is that in the lecture notes I leave out large parts of large amounts of material. 5
00:00:26,200 --> 00:00:34,730
So this will enable you to give your hands busy while on lecture and take down some notes and so on. 6
00:00:34,730 --> 00:00:42,660
So know the assume that everything that I talk about on here,please follow along. OK. 7
00:00:48,160 --> 00:00:58,560
So as is my usual practice,let me start with the brief review of what we covered so far. 8
00:00:58,560 --> 00:01:12,010
So what we did primarily was look at this discipline that we called the LMD,which was very similar, 9
00:01:12,010 --> 00:01:16,380
very...... of the point mass simplification in physics. 10
00:01:16,380 --> 00:01:21,590
And this discipline,this set of constrains we import to ourselves 11
00:01:21,590 --> 00:01:30,020
allowed us to move from Maxwell equations to a very very simple form of algebraic equations. 12
00:01:30,020 --> 00:01:41,820
And specifically discipline do two forms. One is be set that we will leave the elements 13
00:01:41,820 --> 00:01:50,720
for whom the rate of change of make that flux is zero outside the elements. 14
00:01:50,720 --> 00:02:02,870
And for whom the rate of change of charge,I want the charge inside the element was zero. 15
00:02:02,870 --> 00:02:06,730
So if I took any element,any lumps,any element 16
00:02:06,730 --> 00:02:12,550
that I called a lumped circuit element like resistor or a voltage source,and put a black box around it. 17
00:02:12,550 --> 00:02:20,580
Then what I am saying is that the net charge inside that is going to be zero. And this is not true in general. 18
00:02:20,580 --> 00:02:26,130
OK. We'll see examples where if you choose some pieces of element for example. 19
00:02:26,130 --> 00:02:31,990
They might be charge built up. ...... but net inside the,if I put the box on the entire element, 20
00:02:31,990 --> 00:02:37,770
I'm gonna assume that the rate of change of charge is gonna be zero. 21
00:02:37,770 --> 00:02:52,430
So what this did was enabling us to create Lumped Circuit Abstraction. Well,I could give elements. 22
00:02:52,430 --> 00:02:56,780
Some element of the sort just could be a resistor,a voltage source,whatever, 23
00:02:56,780 --> 00:03:05,380
and I could now describe a voltage,some voltage across the element, 24
00:03:05,380 --> 00:03:10,300
and also some current that was going into the element. 25
00:03:10,300 --> 00:03:20,550
And as it goes forward,when I label the voltage and currents,across and through elements,and would be following in convention. 26
00:03:20,550 --> 00:03:26,250
OK. The convention is that if I label V in the following manner, 27
00:03:26,250 --> 00:03:33,840
then I would label i for that element as a current flowing into the positive terminal. 28
00:03:33,840 --> 00:03:47,970
By doing this,just convention. By doing this,it turns out that the power consumed by the element is Vi,is positive. OK. 29
00:03:47,970 --> 00:03:56,370
So by choosing i going in this way,into the positive terminal the power consumed by the element is gonna be positive. 30
00:03:56,370 --> 00:04:01,550
OK. So in general,we simply follow the convention when I label voltage and currents, 31
00:04:01,550 --> 00:04:08,330
I'm labeling the current into an element,entering into and through the plus terminal. 32
00:04:08,330 --> 00:04:13,700
Remember of course if the current is going this way. Let's have one amp flowing this way, 33
00:04:13,700 --> 00:04:19,000
then when I compute the current,I would come out to be negative. 34
00:04:19,000 --> 00:04:23,490
OK. So by making these assumptions, 35
00:04:23,490 --> 00:04:29,080
of the assumptions of the lumped matter discipline I said I was able to simplify my life tremendously 36
00:04:29,080 --> 00:04:37,910
And in particular what it did was allowed me to take Maxwell equations. 37
00:04:37,910 --> 00:04:42,760
OK. And simplify them into a very simple algebra form 38
00:04:42,760 --> 00:04:52,520
which is the voltage law and the current law that I called Kirchhoff Voltage Law and Kirchhoff Current Law. 39
00:04:52,520 --> 00:05:01,710
KVL simply states that if I have some circuits and if I measure the voltages in any loop in the circuit. 40
00:05:01,710 --> 00:05:11,650
So if I look at that voltages in any loop,then the voltages in the loop were summed to zero. 41
00:05:11,650 --> 00:05:15,470
OK. So measure the voltages in the loop and they were summed to zero. 42
00:05:15,470 --> 00:05:24,730
Similarly,for the current,if I take a node of a circuit,the very circuit, 43
00:05:24,730 --> 00:05:29,360
a node is a point in a circuit where multiple edges connect. 44
00:05:29,360 --> 00:05:40,470
If I take a node,then the current coming into that node,the net current coming into a node is gonna be zero. 45
00:05:40,470 --> 00:05:49,790
OK. So if I take any node of the circuit and sum up all the currents coming into that node,they will all net sum to zero. 46
00:05:49,790 --> 00:05:59,030
So notice that what have done is by this discipline,by this constraint I import to myself, 47
00:05:59,030 --> 00:06:07,890
I was able to make this incredible leap for Maxwell equations to these really really simple algebraic equations KVL,KCL. 48
00:06:07,890 --> 00:06:11,920
And I promise you going forward to the rest of 6002, 49
00:06:11,920 --> 00:06:18,600
if this is all you know,you can pretty much solve any circuits using these two very simple relations. 50
00:06:18,600 --> 00:06:25,430
Exactly really really simple,all very simple algebra. OK. 51
00:06:25,430 --> 00:06:32,510
So just to show you an example. Let me do a demonstration,let me bring you a small circuit. 52
00:06:32,510 --> 00:06:41,190
And measure some voltages for you,and show you that the voltages indeed add up to zero. 53
00:06:41,190 --> 00:06:45,040
So here's my little circuit. 54
00:07:25,600 --> 00:07:32,490
So I would show you some simple circuits that looks like this and let's go ahead and measure some voltages and currents. 55
00:07:32,490 --> 00:07:39,350
In terms of terminology,member,this is called a loop,so if I take 56
00:07:39,350 --> 00:07:46,260
so if I start from the point C and I travel to the voltage source,come to the node A,down to R1 57
00:07:46,260 --> 00:07:50,360
and all the way down to R2,backed to C. That's a loop. 58
00:07:50,360 --> 00:07:59,860
Similarly this point A is a node,where this is to R1,the voltage source V0,and R4 are connected. 59
00:07:59,860 --> 00:08:03,700
OK. Just make sure your terminology is correct. 60
00:08:03,700 --> 00:08:12,750
So what I do is let me make some quick measurements for you and show you that these KVL and KCL are in deed true. 61
00:08:12,750 --> 00:08:20,590
So the circuits up there. Could I have a volunteer? 62
00:08:20,590 --> 00:08:28,820
Any volunteer? All you have to do is write things on the board. Come on over. 63
00:08:28,820 --> 00:08:35,560
OK. So let me take some measurements and volunteer write down what I measure on the board. 64
00:08:35,560 --> 00:08:40,150
What I do is (let me use a chalk here) 65
00:08:40,150 --> 00:08:49,600
What I do is focus on this loop here,and focus on this node and make some measurements. 66
00:08:59,290 --> 00:09:03,780
All right. So you see the circuit up there. 67
00:09:08,830 --> 00:09:16,410
OK. So I get the 3 volts for the voltage from C to A,so would you write down 3 volts. 68
00:09:30,600 --> 00:09:36,400
OK. So the next one is minus 1.6. 69
00:09:42,440 --> 00:09:47,150
And so that be on doing ab,Vab. 70
00:09:49,410 --> 00:10:00,870
OK. And then let me do the last one and it is a minus 1.37. 71
00:10:06,990 --> 00:10:14,560
I get the measurements I guess have been this way,so it's what certainly is V. It's OK for now go on everybody. 72
00:10:14,560 --> 00:10:26,080
So thank you. I appreciate your help here. 73
00:10:26,080 --> 00:10:37,380
Within the bounds of experimental error,the voltages satisfy....add up these three voltages they nicely sum up to zero. 74
00:10:37,380 --> 00:10:40,170
Next let me focus on the node here. 75
00:10:40,170 --> 00:10:44,320
And at this node,let me go ahead and measure some currents. 76
00:10:46,310 --> 00:10:52,220
What I do now is a change to an AC voltage. 77
00:10:52,220 --> 00:10:58,390
So that I could go ahead with measuring currents without breaking my circuit. 78
00:10:58,530 --> 00:11:04,350
As it settles down,you will get to see the measurements that I am taking as well. 79
00:11:16,460 --> 00:11:18,050
So what I have here. 80
00:11:21,020 --> 00:11:22,880
I guess you can see in this way. 81
00:11:22,880 --> 00:11:26,990
What I have here is three wires that have pulled out from D. 82
00:11:26,990 --> 00:11:32,170
And This is the node D. So three wires coming into Node D. 83
00:11:32,170 --> 00:11:35,170
Just make it a little bit easier for me to measure stuff. 84
00:11:35,170 --> 00:11:37,750
So everybody keeps your fingers crossed 85
00:11:37,750 --> 00:11:41,520
So don't look like a fool here. Hope this works out. 86
00:11:41,520 --> 00:11:55,140
So you roughly get,what's that,so it's a little more than 10 mV,peak to peak out there. 87
00:11:55,140 --> 00:11:59,970
And let's say that if the waveform rises on the left hand side,it's positve. 88
00:11:59,970 --> 00:12:03,080
So it's +10 mV. 89
00:12:05,640 --> 00:12:11,000
And another +10mV,so that's 20mV. 90
00:12:11,000 --> 00:12:13,880
And this time let's make it negative. 91
00:12:13,880 --> 00:12:20,010
Roughly 20,I guess. -20. 92
00:12:20,010 --> 00:12:29,820
I am getting,in terms of currents,I have +10,+10 and -20 that add up to zero. 93
00:12:29,820 --> 00:12:36,000
But more interestingly,I can show you the same thing by holding the current measuring probe 94
00:12:36,000 --> 00:12:45,130
directly across that node and notice that the net current that is entering into this node here is 0. 95
00:12:45,130 --> 00:12:54,080
That just shows you that KCL does indeed hold in practice,and does not just stay in the figment of our imaginations. 96
00:13:01,670 --> 00:13:08,500
So,the following go on,i want to point one other thing out 97
00:13:08,500 --> 00:13:13,610
Notice that I've written and I'll do it in the assumption of the lumped method discipline,O.K.? 98
00:13:14,210 --> 00:13:16,650
There is the third assumption of the lumped method discipline 99
00:13:16,650 --> 00:13:20,650
and that assumption is inspirited at least 100
00:13:20,650 --> 00:13:24,150
shared by the point mass simplification and physics as well 101
00:13:24,150 --> 00:13:26,910
Can somebody told me what the assumption is？ 102
00:13:26,910 --> 00:13:30,030
as all the assumptions which I did not mention which you can 103
00:13:30,030 --> 00:13:38,800
read in your notes in section A point 2 in the appendix 104
00:13:38,800 --> 00:13:42,280
What's told the assumption that is shared in spirit 105
00:13:42,280 --> 00:13:48,700
with the point mass simplification, anybody? 106
00:13:48,700 --> 00:13:52,570
I've told the assumption be made here is that in all the signals 107
00:13:52,570 --> 00:13:55,260
that you are going to study in this courses 108
00:13:55,260 --> 00:13:59,060
It made the assumption that the signal speeds of interest 109
00:13:59,060 --> 00:14:00,940
eg .transition speed and so on 110
00:14:00,940 --> 00:14:04,350
are much slower than the speed of light 111
00:14:04,350 --> 00:14:11,030
My signal's transition speeds of interest are much slower than the speed of light 112
00:14:11,030 --> 00:14:16,160
Remember, the three laws of motion break down 113
00:14:16,160 --> 00:14:19,870
if your objects begin moving at the speed of light 114
00:14:19,870 --> 00:14:26,030
The same token here are lumped circuit abstractions breaks down if we approach the speed of light 115
00:14:26,030 --> 00:14:29,750
And the following on courses let's talk about the wave guides 116
00:14:29,750 --> 00:14:32,110
and other distributed analysis techiques 117
00:14:32,110 --> 00:14:37,400
that deal with the signals that travels at the speed of light 118
00:14:37,400 --> 00:14:46,480
O.K.,so,with that,let me go on to talking about method 1 119
00:14:46,480 --> 00:14:48,600
of circuit analysis 120
00:14:48,600 --> 00:14:56,750
This is called the basic KVL and KCL method 121
00:14:56,750 --> 00:14:58,950
So just based on those simple logic...... 122
00:14:58,950 --> 00:15:04,060
I can analyse very interesting and complicated circuits 123
00:15:04,060 --> 00:15:08,200
The method goes as follows: 124
00:15:08,200 --> 00:15:11,330
So let's say our goal is given the circuit like this 125
00:15:11,330 --> 00:15:13,840
our goal is to solve it 126
00:15:13,840 --> 00:15:15,610
In this course,you'd be doing two kinds of things 127
00:15:15,610 --> 00:15:16,740
analysis and synthesize 128
00:15:16,740 --> 00:15:20,600
And analysis says given the circuit 129
00:15:20,600 --> 00:15:23,170
O.K.,what can you tell me about the circuit 130
00:15:23,170 --> 00:15:25,660
So it solves the existing circuits 131
00:15:25,660 --> 00:15:27,640
for all of the voltages and currents circuits 132
00:15:27,640 --> 00:15:31,010
voltages across elements and currents through those elements 133
00:15:31,010 --> 00:15:34,840
Synthesis says given a function 134
00:15:34,840 --> 00:15:37,200
I mean ask to go and build the circuits 135
00:15:37,200 --> 00:15:40,600
So from analysis here 136
00:15:40,600 --> 00:15:42,810
we can apply this method that i want to show you 137
00:15:42,810 --> 00:15:46,080
And idea here is that given a circuit like this 138
00:15:46,080 --> 00:15:51,760
Let us figure out all the voltages and currents that are the function 139
00:15:51,760 --> 00:15:54,290
of the way these elements are connected 140
00:15:54,290 --> 00:15:58,910
So,the basic KVL and KCL method has the following steps 141
00:15:58,910 --> 00:16:07,040
The 1st step is to write down the element vi relationships 142
00:16:07,040 --> 00:16:11,020
Writh down the element vi relationships through the elements 143
00:16:11,020 --> 00:16:16,590
2nd step is write KCL for all of the notes 144
00:16:16,590 --> 00:16:22,140
And third step is write KVL for all the loops in the circuit 145
00:16:22,140 --> 00:16:24,000
that's it 146
00:16:24,000 --> 00:16:26,100
just go ahead and write down the element rules 147
00:16:26,100 --> 00:16:30,750
and KVL and KCL and go ahead and solve circuit 148
00:16:30,750 --> 00:16:34,410
so we will do an example of course 149
00:16:34,410 --> 00:16:39,210
but just for your,as a referential 150
00:16:39,210 --> 00:16:42,700
Look at a bunch of elements so far 151
00:16:42,700 --> 00:16:45,960
And for the resister 152
00:16:45,960 --> 00:16:51,090
The element relation says that v is i R 153
00:16:51,090 --> 00:16:56,990
R is the resistance of the element here 154
00:16:56,990 --> 00:17:03,610
For a volt source,v is equal to Vo 155
00:17:03,610 --> 00:17:05,490
that's the element relationship 156
00:17:05,490 --> 00:17:08,400
And for a current source 157
00:17:08,400 --> 00:17:21,130
the relation is i is something the current going through the element 158
00:17:21,130 --> 00:17:26,680
so,these are some simple elements rules for the devices that, 159
00:17:28,030 --> 00:17:30,850
the current source,the volt source and the resister 160
00:17:30,850 --> 00:17:34,570
so,let's go ahead and solve a simple circuit 161
00:17:34,570 --> 00:17:38,220
and what we will do is go ahead and solve the circuit for you 162
00:17:38,220 --> 00:17:46,760
ok,turn to the page 5 of your notes. 163
00:17:46,760 --> 00:17:51,120
And we would go ahead and edit this circuit here 164
00:17:51,120 --> 00:17:58,980
You can the scribble values on your notes on page 5 165
00:17:58,980 --> 00:18:01,630
so at the first step 166
00:18:01,630 --> 00:18:05,270
of my KVL and KCL method 167
00:18:05,270 --> 00:18:07,390
I need to write down all of my elements vi relationships 168
00:18:07,390 --> 00:18:08,740
So before I do that 169
00:18:08,740 --> 00:18:12,170
Let me go ahead and label all the voltages and currents 170
00:18:12,170 --> 00:18:14,800
that are unknowns in the circuit 171
00:18:14,800 --> 00:18:20,920
So let me label the voltages and currents associated with the voltage source 172
00:18:20,920 --> 00:18:24,460
i 0 173
00:18:24,460 --> 00:18:26,720
Notice i continue to follow this convention 174
00:18:26,720 --> 00:18:32,380
whenever i label voltages and currents for an element 175
00:18:32,380 --> 00:18:40,800
I will show the current going into the positive terminal of the element valuable voltage 176
00:18:40,800 --> 00:18:44,650
so you would have Vo and Io 177
00:18:44,650 --> 00:18:47,410
let me bother you for five seconds 178
00:18:47,410 --> 00:18:51,330
and show you a point of confusion that happen sometimes 179
00:18:51,330 --> 00:18:55,770
often people confused between 180
00:18:55,770 --> 00:18:58,330
what is called the valuable associated with the element 181
00:18:58,330 --> 00:18:59,950
what is the element value 182
00:18:59,950 --> 00:19:06,350
Notice that here capital Vo is the voltage that this voltage source porvides 183
00:19:06,350 --> 00:19:12,030
While this name here Vo is simply a variable that we used 184
00:19:12,030 --> 00:19:15,930
to lable the voltage accross the element 185
00:19:15,930 --> 00:19:20,910
So certainly,I can label V1 as the voltage across the resister 186
00:19:20,910 --> 00:19:23,620
and i1 as the current flowing through the resister 187
00:19:23,620 --> 00:19:31,050
so this method of labeling where I follow the convention 188
00:19:31,050 --> 00:19:33,720
as the current flows into the positive terminal 189
00:19:33,720 --> 00:19:44,950
is called the associated variable discipline 190
00:19:44,950 --> 00:19:50,510
I try to use the word discipline in situations where you have a choice 191
00:19:50,510 --> 00:19:54,020
of a variety of possible choices,you pick one as the convention 192
00:19:56,510 --> 00:20:01,140
so,here,as the convention be used as associated variable discipline 193
00:20:01,140 --> 00:20:04,140
and use that method to consistently label the unknown 194
00:20:04,140 --> 00:20:06,690
voltage and currents in our circuits 195
00:20:06,690 --> 00:20:09,970
O.K. Let me continue to labeling here 196
00:20:09,970 --> 00:20:14,440
v4,i4 197
00:20:14,440 --> 00:20:17,670
i3,v3 here 198
00:20:17,670 --> 00:20:25,600
and v2,i2 199
00:20:25,600 --> 00:20:29,000
v5,and i5 200
00:20:29,000 --> 00:20:30,640
i think that's it 201
00:20:30,640 --> 00:20:33,520
so go ahead and label all my unknowns 202
00:20:33,520 --> 00:20:36,000
see each of these voltages and currents are 203
00:20:36,000 --> 00:20:39,330
the voltages and currents are associated with each of the elements 204
00:20:39,330 --> 00:20:44,550
and my goal is to solve for these 205
00:20:44,550 --> 00:20:51,750
so,in terms of the five solution here 206
00:20:51,750 --> 00:20:54,460
Let's follow the method that I outline for you 207
00:20:54,460 --> 00:20:56,360
so as the first step 208
00:20:56,360 --> 00:21:02,000
i'm simply going to go ahead and write down all the element vi relationships 209
00:21:02,000 --> 00:21:06,130
So the first step,I'm to go ahead and write down all the vi relationships 210
00:21:06,130 --> 00:21:16,780
Can someone yell out for me the vi relationship for the voltage source? 211
00:21:16,780 --> 00:21:27,240
so V0 is Vo,that is the variable Vo is simply equals to the voltage V0 212
00:21:29,010 --> 00:21:29,160
Similarly i can write the others 213
00:21:30,400 --> 00:21:32,310
v1 is i1 r1 214
00:21:32,540 --> 00:21:36,800
v2 is i2 r2 and so on 215
00:21:36,740 --> 00:21:45,630
and i have one two ...six elements,so i will get six such equations 216
00:21:45,630 --> 00:21:53,710
step2,i would go ahead and write KCL for the nodes in my system 217
00:21:53,710 --> 00:22:02,260
let me start with node a 218
00:22:02,260 --> 00:22:06,080
so for node a,let me take as possitive the currents went out of the node. 219
00:22:06,080 --> 00:22:20,270
as i get (see the picture) and they must sum to zero. okey for node a. 220
00:22:20,270 --> 00:22:25,150
then i can go ahead and do the other nodes let's say for example 221
00:22:25,150 --> 00:22:38,600
i do node b,for node b,i have (see the picture) 222
00:22:38,600 --> 00:22:48,380
so i have four nodes,so i will get four equations. 223
00:22:48,380 --> 00:22:55,540
so it turns out four equation is not independent,you can derived from the others 224
00:22:55,540 --> 00:23:00,540
I get three independent equations out of these 225
00:23:00,540 --> 00:23:03,730
i can then write KVL 226
00:23:03,730 --> 00:23:08,030
and for KVL,i just go down my loops here 227
00:23:08,030 --> 00:23:16,160
let me go through this first loop here in this manner 228
00:23:16,160 --> 00:23:22,580
the simply trick that i use,get to be incredible careful when you go through this 229
00:23:22,580 --> 00:23:25,010
in keeping your minuses and pluses correct 230
00:23:25,010 --> 00:23:28,140
otherwise you can get hopelessly muddled 231
00:23:28,140 --> 00:23:29,820
once you label it 232
00:23:29,820 --> 00:23:33,090
you need to be sure that you get all the minuses and pluses correct 233
00:23:33,090 --> 00:23:39,860
so for KVL,what i like to do is let's say I start at c 234
00:23:39,860 --> 00:23:45,450
and from c i want to go to a,form a go to b,and from b I would come back to c 235
00:23:45,450 --> 00:23:49,060
That's all I travel my loops 236
00:23:49,060 --> 00:23:55,680
the trick that I would follow is as my finger walks through that loop, 237
00:23:55,680 --> 00:24:00,560
i would label the voltage as the first signing that i see for that voltage3 238
00:24:00,560 --> 00:24:03,290
so i get started from c and I go up 239
00:24:03,290 --> 00:24:07,500
I start my punching into the voltage source element 240
00:24:07,500 --> 00:24:13,510
I punching into it,I add the minus sign for the Vo 241
00:24:11,540 --> 00:24:14,970
I just gonna write down (see the picture) 242
00:24:14,970 --> 00:24:19,130
then I go through and as I come up to a and go down to b 243
00:24:19,130 --> 00:24:24,030
I punch to the plus sign of the V1 so that (see the picture) 244
00:24:24,030 --> 00:24:28,440
and I punch into the plus sign of the V2 and so I get(see the picture) 245
00:24:28,440 --> 00:24:30,750
and add to zero 246
00:24:30,750 --> 00:24:35,300
That matches the what you have in your notes as well 247
00:24:35,300 --> 00:24:40,270
I can go through my other loops and write down equations for each of the loops 248
00:24:40,270 --> 00:24:43,880
And the convention that I like to follow is go through the loop 249
00:24:43,880 --> 00:24:47,080
write down as the sign of the voltage the first sign 250
00:24:47,440 --> 00:24:50,350
that I encountered for the element 251
00:24:50,350 --> 00:24:53,130
you can write the exact opposite if you want just to be different 252
00:24:53,130 --> 00:24:57,400
but as long as the taken insistence you will be okay. 253
00:24:57,400 --> 00:25:00,300
all right,so the same manner here 254
00:25:00,300 --> 00:25:03,590
there are four loops that I can have so four equations 255
00:25:03,590 --> 00:25:09,400
Again one of them turns out to be dependent on the others. 256
00:25:09,400 --> 00:25:11,470
So end up getting three independent equations. 257
00:25:11,470 --> 00:25:23,030
So I get 12 equations. 6 elements,a voltage source and 5 resisters. 258
00:25:23,030 --> 00:25:27,870
So there are 6 unknown voltages and 6 unknown currents,few. 259
00:25:27,870 --> 00:25:33,160
So 12 equations and 12 unknowns. 260
00:25:33,160 --> 00:25:41,990
Take all these equations and put them through a big crank and ...... 261
00:25:41,990 --> 00:25:46,640
And if I was really cruel,I'd give this as homework problem and have you drained 262
00:25:46,640 --> 00:25:52,360
drained and drained you get your 6 voltages and 6 currents. 263
00:25:52,360 --> 00:26:05,560
O.k. It works. And get all the equations and this method just works. 264
00:26:05,560 --> 00:26:12,270
However,notice that this is quite a grubby method,quite grunge. 265
00:26:12,270 --> 00:26:20,260
I get all the equations and it's quite famed for the simple circuits like this. 266
00:26:20,260 --> 00:26:22,340
However,if I suppose to say that this fundamental method 267
00:26:22,340 --> 00:26:27,990
that is just one step away from Maxwell equations simply works. O.K? 268
00:26:27,990 --> 00:26:31,690
So what we going to do with the rest of this lecture is 269
00:26:31,690 --> 00:26:33,230
I introduce a couple of methods. 270
00:26:33,230 --> 00:26:38,130
What is an intuitive method and another one called node method is a little bit formal 271
00:26:38,130 --> 00:26:53,930
but is much more... I guess much more terse than the KVL, KCL methods. 272
00:26:53,930 --> 00:27:02,370
Method 2,so the relevant section to read in the course notes is section 2.4. 273
00:27:02,370 --> 00:27:06,290
What kind of things I am discussing in this semester is intuition. 274
00:27:06,290 --> 00:27:13,770
o.k. What you will find that as you become a EECS major in sort and go on. 275
00:27:13,770 --> 00:27:21,670
Or if you talk to your TAs or prefessors in sort, you will find that very rarely that 276
00:27:21,670 --> 00:27:22,830
they actually go ahead and apply the formal methods of analysis. 277
00:27:22,830 --> 00:27:30,920
By large, engineers are able to look at the circuit and simply by observation write down the answer. 278
00:27:30,920 --> 00:27:36,280
Usually the path what you try to do is gonna be know that process 279
00:27:36,280 --> 00:27:39,440
and don't like students look for the teaching of formal methods 280
00:27:39,440 --> 00:27:43,540
and you know you will develop your own intuition and be able to do it. 281
00:27:43,540 --> 00:27:47,710
What I try to do this term is try to stress the intuitional methods 282
00:27:47,710 --> 00:27:50,360
and try to show you how the intuitional possess goes 283
00:27:50,360 --> 00:27:55,460
so you can very quickly solve many of the circuits simply by inspections. 284
00:27:55,460 --> 00:28:00,950
So this method I will show you here is one such intuitional method 285
00:28:00,950 --> 00:28:08,900
called the element combination rules. 286
00:28:08,900 --> 00:28:14,590
For many simple circuits you can solve them very quickly by applying this method. 287
00:28:14,590 --> 00:28:20,380
The components of this method are these. 288
00:28:20,380 --> 00:28:30,180
I learn about how to compose a branch of elements,so let's have an example I have a set of resistors. 289
00:28:30,180 --> 00:28:48,680
R1 to Rn,in series,you can use KVL and KCL to show that this is equivalence to a single resistor 290
00:28:48,680 --> 00:28:56,300
whose value is given by the sum of the resistances. 291
00:28:56,300 --> 00:29:01,940
So I have resistors in series, and effectively its as same as a single resistor 292
00:29:01,940 --> 00:29:04,920
whose value is the sum of the resistances. 293
00:29:04,920 --> 00:29:11,840
You can look at the course notes for the proof of verification of this fact. 294
00:29:11,840 --> 00:29:21,730
Similarly if I have resistances in parellel,so let me call them conductances. 295
00:29:21,730 --> 00:29:28,770
Conductance is the reciprocal of a resistance. 296
00:29:28,770 --> 00:29:37,850
If resistances are measured in Ohms,conductances are measured in mhos. 297
00:29:37,850 --> 00:29:45,870
So if the conductances are G1 G2 G3,then effectively this is the same as having a single conductance 298
00:29:45,870 --> 00:29:53,320
whose effective value given by the sum of the conductances. 299
00:29:53,320 --> 00:29:57,820
Conductances in parellel add and resistances in series add. 300
00:29:57,820 --> 00:30:11,450
Similarly for voltage sources,if I put voltage sources in series, 301
00:30:11,450 --> 00:30:14,040
then they are tend to amount to the sum of the voltages. 302
00:30:14,040 --> 00:30:20,830
And similarly,for currents if I have currents in parellel they can be viewed as a single current source 303
00:30:20,830 --> 00:30:24,100
whose currents are the sum of the individual power currents. 304
00:30:24,100 --> 00:30:32,050
So let's do a quick example, so let's do this example. 305
00:30:32,050 --> 00:30:50,520
So let's say I have a circuit that looks like this and three resistances 306
00:30:50,520 --> 00:30:57,300
and the only care about is the current I that flows through this wire. 307
00:30:57,300 --> 00:30:59,760
Only care about that current. 308
00:30:59,760 --> 00:31:05,300
Of course you can go ahead and write KVL and KCL, you will get four equations 309
00:31:05,300 --> 00:31:08,460
and four unknowns and you can solve it. 310
00:31:08,460 --> 00:31:18,310
But I can apply my element combination rules and very quickly figure out what the current i is. 311
00:31:18,310 --> 00:31:22,730
It is the following technique. So I can do is I can first forget this circuit and I can compose these two resistances 312
00:31:22,730 --> 00:31:32,580
and show that the circuit is equivalent as far as the current i is concerned to the following circuit. 313
00:31:32,580 --> 00:31:47,190
R1 and I take the sum of the conductances and the conductances turn out to be (see the picture). 314
00:31:47,190 --> 00:32:05,110
And then I can further simplify and I get a single resistance whose value is (see the picture). 315
00:32:05,110 --> 00:32:12,350
I'm just simplifying the circuit. Now from this circuit I can get the answer I need i is 316
00:32:12,350 --> 00:32:23,910
simply the voltage V divided by (see the picture) 317
00:32:23,910 --> 00:32:27,220
So in situation like this, when I am looking for a single current I can very quickly 318
00:32:27,220 --> 00:32:30,360
get to the answer by applying this element combination rules 319
00:32:30,360 --> 00:32:35,640
and I can get rid of having to go to a set of formal steps. 320
00:32:35,640 --> 00:32:39,800
So in general, whenever you encounter to a circuit, 321
00:32:39,800 --> 00:32:43,970
by large, I tempt to use a total of method so as to solve it. 322
00:32:43,970 --> 00:32:48,150
And go to formal method only if some intuitive method fails. 323
00:32:48,150 --> 00:32:54,620
Even in your homeworks,by large,the homeworks are not meant to be grunge. 324
00:32:54,620 --> 00:32:58,850
If you set a lot of grunge in your homework,just remember, 325
00:32:58,850 --> 00:33:03,750
you are probably not using some intuitional method. 326
00:33:03,750 --> 00:33:13,730
Just be cautious. All right,so let me go on to the third method of circuit analysis. 327
00:33:13,730 --> 00:33:27,570
The third method is called the node method. 328
00:33:27,570 --> 00:33:33,270
The node method is simply a specific application of the KVL and KCL method 329
00:33:33,270 --> 00:33:40,670
and results in a much much more compact form of the final equations. 330
00:33:40,670 --> 00:33:48,010
If there is one method that you have to remember for life then I would say just remember this method. 331
00:33:48,010 --> 00:33:53,990
The node method is the would course of the EECS industry. 332
00:33:53,990 --> 00:33:58,850
There is one method that you will consistently apply then this is the one to remember. 333
00:33:58,850 --> 00:34:05,290
So let me quickly outline for you the method and then work out an example for you. 334
00:34:05,290 --> 00:34:17,240
The first step of the node method will be the select a reference of the ground node. 335
00:34:17,240 --> 00:34:18,930
This is the symbol for the ground node. 336
00:34:18,930 --> 00:34:24,670
The ground node simply says that I would denote the voltage at that point to be zero 337
00:34:24,670 --> 00:34:28,600
and measure all my other voltage in reference to that point. 338
00:34:28,600 --> 00:34:31,590
So select the ground node in my circuit. 339
00:34:31,590 --> 00:34:44,690
Second I would label the remaining voltages with respect to that ground node. 340
00:34:44,690 --> 00:34:47,700
So label voltages for all the other nodes with respect to the ground node. 341
00:34:47,700 --> 00:34:58,870
Next write KCL for each of the nodes, write KCL. 342
00:34:58,870 --> 00:35:07,060
Don't try the KCL for the ground node ,remember if you have n nodes,the node equations would give you n-1 independent equations. 343
00:35:07,060 --> 00:35:12,480
So write KCL for the nodes but don't do so for the ground node. 344
00:35:12,480 --> 00:35:17,850
Then solve for the node voltages. 345
00:35:17,850 --> 00:35:23,290
So let's,if you let me label node voltages,I would be labeling them as e something go the other. 346
00:35:23,290 --> 00:35:28,000
So solve for the other unknown node voltages 347
00:35:28,000 --> 00:35:33,580
and then once you know all the voltages associated with nodes, 348
00:35:33,580 --> 00:35:43,010
I can then back solve for all the branch voltages and currents. 349
00:35:43,010 --> 00:35:46,190
Once you know the node voltages I can then go ahead and figure out 350
00:35:46,190 --> 00:35:49,220
all the branch voltages and branch currents. 351
00:35:49,220 --> 00:36:06,770
So let's go ahead and apply this method and work out an example. 352
00:36:06,770 --> 00:36:10,530
Again remember,if there is one method that you should remember,its node method. 353
00:36:10,530 --> 00:36:15,060
And when in doubt,consistantly apply the node method and it could work, 354
00:36:15,060 --> 00:36:22,930
whether the circuit is linear or nonlinear,the resistors built in U.S or ...... it doesn't matter. 355
00:36:22,930 --> 00:36:28,220
The node method will simply work,linear or nonlinear. 356
00:36:28,220 --> 00:36:34,770
So what I'm going to do is I would build the circuit that's my old faithful 357
00:36:34,770 --> 00:36:40,240
plus I make it a little bit more complicated by adding in a current source. 358
00:36:40,240 --> 00:36:45,120
So let's go and have some fun. Let's do this. 359
00:36:45,120 --> 00:36:59,830
So is a voltage source as before. 360
00:36:59,830 --> 00:37:06,810
What I'll do is for fun. 361
00:37:09,540 --> 00:37:11,290
Add a current source over there. 362
00:37:11,290 --> 00:37:21,070
And you can convice yourselves that if you go and apply KVL or KCL method,it can really be a mass of equations. 363
00:37:21,070 --> 00:37:35,380
R1,R3,R4,R2,R5,so let's follow our method, just ------ 364
00:37:35,380 --> 00:37:38,640
So let's apply the first step,I select the ground node. 365
00:37:38,640 --> 00:37:43,750
It's a reference node from which I measure all the other voltages. 366
00:37:43,750 --> 00:37:52,060
Without knowing anything about node method,the very useful intuition as which node you should choose as the ground node. 367
00:37:52,060 --> 00:37:59,360
You will label the ground node as the voltage zero,and measure all the other voltages with respect to that node. 368
00:37:59,360 --> 00:38:08,630
A usual trick is to pick a node which has the largest number of elements connected to it as the ground node. 369
00:38:08,630 --> 00:38:13,370
And in particular you will find out later the use of the pick of node 370
00:38:13,370 --> 00:38:17,980
to which the maxium number of voltage sources are also connected. 371
00:38:17,980 --> 00:38:23,960
In this instance,I will choose this as my ground node. 372
00:38:31,300 --> 00:38:33,780
Okey,that's my first step. 373
00:38:33,780 --> 00:38:40,060
I chose that as my ground node and I will lable that as having a voltage zero. 374
00:38:40,060 --> 00:38:47,510
Second step,I label the voltages of the other branches with respect to the ground node. 375
00:38:47,510 --> 00:38:59,810
So I will do is as this node here,I will label that voltage e1,this is an unknown. 376
00:38:59,810 --> 00:39:03,570
So remember,node method,because my node voltages are my unknowns. 377
00:39:03,570 --> 00:39:16,420
So I label this as e1,I label this one as my unknown voltage e2. 378
00:39:20,450 --> 00:39:24,750
What about this one here,is that voltage unknown? 379
00:39:24,750 --> 00:39:35,280
No,I know the voltage because I know this node is add the voltage V0 higher than the ground node. 380
00:39:35,280 --> 00:39:40,510
Notice that the go from here to here as exactly the go through a voltage source. 381
00:39:40,510 --> 00:39:45,680
So this node has voltage V0,and I simply write down V0. 382
00:39:45,680 --> 00:39:51,180
If you want,you know,simplify the number of steps that you have to go through, 383
00:39:51,180 --> 00:39:56,000
so directly go and write down the voltage V0 for that node. 384
00:39:56,000 --> 00:40:06,300
What I've lost to do is for convenience,I will use conductance, 385
00:40:06,300 --> 00:40:14,250
I will use Gi in place of Ri. I would write down the multiple equations. 386
00:40:14,250 --> 00:40:25,010
Thus,step 1 I'v chosen my ground node. Step 2 I've labeled my node voltages e. I've done that 2 steps. 387
00:40:25,010 --> 00:40:27,840
Now let me go ahead 388
00:40:41,060 --> 00:40:44,660
and so let me go and apply step 3. 389
00:40:44,660 --> 00:40:59,060
And step 3 said go and apply KCL for each of the nodes I've chosen as unknown voltages and then that will fill your equations. 390
00:40:59,060 --> 00:41:10,600
So let me start my apply of KCL at e1. 391
00:41:10,600 --> 00:41:18,090
I do one more thing. Notice that we don't have any current there. So how do I get the KCL. 392
00:41:18,090 --> 00:41:22,450
KCL discipline says that the sum of currents into a node is zero. 393
00:41:22,450 --> 00:41:27,000
And remember by the LMD,so there might have a current there. 394
00:41:27,000 --> 00:41:34,140
So the trick I adopt is that to write KCL use node voltages 395
00:41:34,140 --> 00:41:41,730
and implicitely substitute ...... voltages divided by the element resistance for instance. 396
00:41:41,730 --> 00:41:45,210
So take the node voltages and divide by the resistors and get the current. 397
00:41:45,210 --> 00:41:49,450
So I implicitely apply element relationships to get the node current. 398
00:41:49,450 --> 00:41:50,980
So I get the example to make it clear. 399
00:41:50,980 --> 00:41:58,800
So I take the node e1 and again put the currents screen out I would assume to have...... 400
00:41:58,800 --> 00:42:13,390
So the current going up is (see the picture). That's the current going up. 401
00:42:13,390 --> 00:42:27,550
Plus the current going down is(see the picture). 402
00:42:27,550 --> 00:42:44,030
Plus the current that is going through resistor 3 which is simply (see the picture) is equal to zero. 403
00:42:44,030 --> 00:42:52,390
See all I got this. The simply KCL,to get my currents I simply get the difference of voltages across the elements 404
00:42:52,390 --> 00:42:58,380
and divided by the element resistance and I get the currents. 405
00:42:58,380 --> 00:43:06,980
So I can similarly write KCL at e2. The KCL at e2,again,let me go on works. 406
00:43:06,980 --> 00:43:26,120
The current going up is (see the picture). The current going left is (see the picture). 407
00:43:26,120 --> 00:43:44,430
The current going down is (see the picture). The current going down is (see the picture). 408
00:43:44,430 --> 00:43:49,330
Gonna be careful with ...... zero. So all the currents going out sum to zero. 409
00:43:49,330 --> 00:43:54,340
So here are the currents going out at this point. 410
00:43:54,340 --> 00:44:04,150
So I'll do next is I can move the constant terms to left hand side and collect my unknowns. 411
00:44:04,150 --> 00:44:23,380
So let me turn out here. So let's say I get e1 here,and from this equation I have Vo,G1 which comes out here. 412
00:44:23,380 --> 00:44:32,750
So minus Vo,G1 comes over the other side and let collect all the values that multiply e1, 413
00:44:32,750 --> 00:44:38,250
so I get,G1 is one example. 414
00:44:38,250 --> 00:44:51,850
have G2 and have G3 and then for e2 I have minus G3. 415
00:44:51,850 --> 00:44:59,130
So I simply express this as the element voltages multiply by some terms in ...... 416
00:44:59,130 --> 00:45:03,030
and I put my external sources on the right hand side. 417
00:45:03,030 --> 00:45:12,310
Similar I go ahead to the same thing here. In this instance,let me move my sources to the right. 418
00:45:12,310 --> 00:45:22,480
So I get I1 coming out there and I get Vo,G4 coming out there. 419
00:45:22,480 --> 00:45:27,700
By the way,I just want to mention to you that if you looking to fall asleep this is good time to do so 420
00:45:27,700 --> 00:45:33,680
because as soon as I write down this two equations from now on it's,you know,a nap time. 421
00:45:33,680 --> 00:45:36,340
There is nothing you can learn from here now. 422
00:45:36,340 --> 00:45:42,640
It's just a ....... having fun at the blackboard pushing symbols on them. 423
00:45:42,640 --> 00:45:47,590
So the one to write down the two equations the rest of us just go to nap. 424
00:45:47,590 --> 00:45:50,820
So let me just have some fun and let me go and do that. 425
00:45:50,820 --> 00:45:57,120
So move my voltages and currents to the other side and let me collect all the equations for e1 here. 426
00:45:57,120 --> 00:46:06,370
So e1 is minus G3,and that's it,yes. 427
00:46:06,370 --> 00:46:18,630
Then I do the same for e2. So I get G4 and I get G3 and I get G5. 428
00:46:18,630 --> 00:46:26,270
So notice here I have two equations and two unknowns. 429
00:46:26,270 --> 00:46:36,280
The two equations are on the right hand side I have some voltages and currents which are my drive voltages and drive currents. 430
00:46:50,300 --> 00:46:55,070
I guess this get quite boring and ...... and I will pause you and talk about something else. 431
00:46:55,070 --> 00:46:59,310
So you can take this and you can put this into a matrix form so that can help you understand. 432
00:46:59,310 --> 00:47:04,800
Get all matrix form,I know that. You can use anything you need to solve it. 433
00:47:04,800 --> 00:47:10,610
Use algebra techniques,use linear algebra methods to solve it or use a computer,whatever you want. 434
00:47:10,610 --> 00:47:18,520
And when computers analyze circuits to write down these equations in and it will solving matrixes. 435
00:47:18,520 --> 00:47:23,310
So when you took a linear algebra class. How many of you have taken linear algebra class? 436
00:47:25,750 --> 00:47:31,140
How many people here have heard the Gaussian Elimination? 437
00:47:31,140 --> 00:47:36,210
How could more people know Gaussian Elimination than take a linear algebra class? 438
00:47:36,210 --> 00:47:48,740
So now you know when you took a linear algebra classes and if I just collect these two matrix form. 439
00:48:06,470 --> 00:48:10,680
So I just simply express these two equations in linear algebra form. 440
00:48:10,680 --> 00:48:13,490
And here is my column vector of unknowns 441
00:48:13,490 --> 00:48:23,220
and you can apply any of the techniques you've learnt in linear algebra to solve this Gaussian Elimination work. 442
00:48:23,220 --> 00:48:28,600
If you do a searching in computer techniques to solve this equations simply give a huge equations like this. 443
00:48:28,600 --> 00:48:35,110
Putting in computer programs and giving equations like this can go ahead and solve them. 444
00:48:35,110 --> 00:48:42,710
Let me stop here and reemphasize that what you've done is made a huge leap from maths equations 445
00:48:42,710 --> 00:48:50,670
to using the LMD to KCL KVL which end up giving this simply linear algebra equations to solve 446
00:48:50,670 --> 00:48:58,080
and not to have to worry about ......equations that will,the form of Maxwell equations. 447
00:00:00,000 --> 00:00:00,000
448
00:00:00,000 --> 00:00:00,000
449
00:00:00,000 --> 00:00:00,000
Last Modified 5/19/06 8:53 AM
|
