a. Determine ICQ, VC EQ, and IB Q for the network of Problem 15 (Fig. 125 ) using the approximat

A. Determine ICQ, VC EQ, and IB Q for the network of Problem...

Key Concepts

Quiescent Operating Point

The quiescent operating point, or Q-point, is the DC bias point of a transistor circuit where the currents and voltages are set when no signal is applied. Determining the Q-point is essential for ensuring that the transistor functions in its intended operating region, typically the active region, and for establishing the linearity and stability of the amplifier circuit.

DC Biasing Analysis

DC biasing analysis involves calculating steady-state voltages and currents in a transistor circuit to ensure that the device is appropriately biased. This analysis is crucial in understanding how various circuit elements interact under DC conditions and forms the basis for evaluating a transistor's performance under small-signal conditions.

Approximate Analysis Methods

Approximate analysis methods simplify the problem by making assumptions, such as neglecting small variations or linearizing non-linear components, to reduce complex circuit equations into more manageable forms. These methods are useful for quick evaluations and provide reasonable estimates when the assumptions are justified, although they may lead to discrepancies if the conditions validating the approximations are not met.

Exact Analysis Methods

Exact analysis methods involve solving the full set of nonlinear equations that describe the transistor circuit without making simplifying assumptions about the operating conditions. This approach ensures that all factors, including non-ideal behaviors or deviations from linearity, are taken into account, thereby providing a more accurate determination of the circuit's DC operating points.

Error Estimation and Validity of Approaches

Comparing the results of approximate and exact analysis methods provides insight into the magnitude of errors introduced by simplifying assumptions. This error estimation is important for deciding whether an approximate method is sufficiently accurate for practical purposes or if the complexities of the circuit require a more detailed, exact approach. Evaluating this discrepancy helps in understanding the limitations of the approximation techniques and guides the selection of the appropriate analysis method.

Transcript

00:01 Given the water flow through a network of pipes in thousands of cubic meters per hour, shown in the following figure, and considering that the total flow into each junction is equal to the total flow out of each junction, we are going to solve the associated system of linear equations using matrices.

00:27 In this case, the linear system represents the water flow, using the variables x1 through x7.

00:36 In part p we are going to find the flow pattern when x6 and x7 are equal to 0.

00:44 And in part c, we will do the same thing when x5 equals 400 and x6 equals 500.

00:54 So the idea here is to write the system that arises when we apply the condition given here that the total flow into each junction is equal to the total flow out of each each junction so it's a kind of conservation law here so we are going to write for each junction we are going to apply that criterion and with that we are going to find a linear system then we solve this linear system using matrices that will give us the variables x1 x2 up to x7 and then we are going to find specific configurations of that solution in part a in parts b and c so we start with the calculation of the general solution of the linear system so we first write down the linear system arising from the property given here so for example when we are going to write the equation associated with this junction here, this one.

02:26 So here we have only one flow in, that is this one here, whose value is 600, and we have two flows out, corresponding to x3 and x1.

02:47 So the law given here reads 600, 600 equal xx plus x3 so that's our first equation x1 plus x3 equals 600 now we do the same thing for the junction to the right this one here and for that junction we have one flow in x1 and 2 out x2 and x4 so x1 must be equal to x2 plus x4 now we go to the next junction here and we have that x the flow in is coming from x2 and x5 we see here x5 is going up to this junction and x2 is going from the left to the same junction and the flow out corresponds to 500 here so we will have that x2 plus x5 is 500.

04:45 Now we go to this junction here, and we have two flow ins, x6 and x3, and one out 600.

04:57 So x3 plus x6 is 600.

05:14 Now we write the equation corresponding to this junction here.

05:20 And we have two in, two flows in, x6, and x7 and one out x6 so x6 equal x4 plus x7 and we have one junction left this one and for this one we have two flows out x5 and x7 and one in is 500 so x5 plus x7 is 500 so we have this linear system here we have six equations and we have seven variables.

06:26 So we are likely to have infinite solutions and finally many infinitely many solutions.

06:33 So we are going to calculate that and for that we rewrite the system this way.

06:49 Remember we got to put the variables on the left of these of each equation.

06:59 So the first equation is the same because we have already all the variables to the left of the equal sign and the independent variable to the right independent value sorry the second equation this one here can be written this way x1 x1 minus x2 minus x4 equals zero and this is equivalent to negative x1 plus x2 plus x4 equals 0 we are going to use this form here now this equation here we write it as is because all the variables are to the left x2 plus x5 equals 500 and this equation here the same thing x3 plus x6 equals 600 then this equation can be written this way x4 minus x6 which is passed to the left plus x7 equals 0 and finally this equation here is written the same way and now we have this linear system and we can write the augmented matrix which is our first in order to solve the linear system.

08:58 So now the augmented matrix will be the following.

09:03 For the first row we have 1 which is coefficient of x1 x2 has coefficient 0 because it doesn't appear on the equation, the first equation x3 has coefficient 1 and x4c 4567 x coefficient 0 so we have 4 0 and the right side is 600.

09:31 Now for the second equation we have x1 has coefficient negative 1 x2 coefficient 1 x3 doesn't appear so coefficient 0 x4 coefficient 1 and the last 3 variables x 5 6 and 7 has coefficient 0 so we have 3 zeros that in this second row and the last coefficient will be also 0 because we have the right hand side equal 0.

10:07 Now for the third equation we have 0 x1 doesn't appear 1 x2, 0 x3, 0 x4, 1x5, 0 x6 and 0 x7, and the right hand side is 500.

10:26 Now the fourth equation is 0 0 0 1, 0 0 0 0 0, 0, 0, 0, 0, 0, 0, 0, 0, because x 6, 0, 0, 0, 0, because x 0, 0, 0, 0, 0, 0, 0, because x 0, 0, 0, 0, 0, 0, 0, 0, 6 has coefficient 1 and the right hand side is 6 .1.

10:49 And the fifth equation is 0 0 .1 because we have x4, 8 5 is 0.

11:01 It doesn't appear.

11:02 X6 is negative 1 and x7 has coefficient 1 and the right side is 0.

11:10 And the last equation is 0 001 500 so this is our augmented matrix which now we reduce to row echelon form so we apply some row operations in order to find a row echelon form of this matrix so so the first operation we're going to apply is we can see that we have already a 1 in this position here.

12:12 So it's good.

12:13 And we have 0, 0 all these terms, so we had to nullify only this term for this column.

12:20 And that can be done adding the first and two rows and assigning the result to second row.

12:29 So the row operation will be row 2 becomes row 2 plus row 1.

12:43 And we get 1 0 ,100, that is the first row remains the same.

12:59 And the second row becomes 0 -1 -1 -3 times 0 and 600.

13:19 The other rows are the same, 0 -1 -00 -100, then we have 0 -0 -1 -0 -0 -0 -0 -60 -600, then 0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 and 0 -0 -0 -0 and 1 -0 -0 -0 -0 and 1 -0 -0 -0.

14:09 So this is our first transformation of the matrix of the automatic matrix.

14:23 Now we are going to nullify.

14:30 Let's see, we're going to notify this one because we have already one here that it's very good.

14:36 So we get to notify this value here.

14:40 And that we do by or through the following operation.

14:47 Row 3 is row 3 minus row 2 so we have first two rows are the same 1 -0 -1 -0 -0 -600 then 0 -1 -11 -0 -0 -0 -600 and the third row becomes row 3 minus row 2 0 minus 0 is 0 -0 -1 is negative 1 0 minus 1 is negative 1, 1 minus 0 is 1, then 0 minus 0 and 0 minus 0 0 0 and 500 is negative 100.

15:51 Okay, so the last 3 rows are the same, so we write them again, 001, 0600, 600, 600, 0 ,000, 0 ,000, 0 ,000, 0 ,000, negative 1 1 0 and 0 0 0 0 0100.

16:36 So this is the second matrix and now we have this row here get to be a transforming to 1 so we can do that by changing the sign of the third row so we do row 3 becomes row 3 times negative 1 that is negative row 3 and so we have the first two rows are the same so it's 1 -0 -1 -0 -0 -0 -0 -600 the second row is the same 0 -1 -0 -0 -1 -6 and the third row is 0 -0 -1 -1 negative 1 -0 -0 and 100 okay the last three rows are the same so 0 -0 -1 -0 -0 -600 then 0 -0 -0 -0 -0 -0 -0 -0 and 0 -0 -0 -0 and 0 -0 -0 -0 and 0 -0 -0 -0 -0 and 0 -0 -0 -0 -0 -0 and 0 -0 -0 -0 -0 -0.

18:25 101 500 okay so now we have to nullify this one here and we do that by the operation row 4 will be row 4 minus row 3 and we get the following metrics again we get first three rows are the same 0 -0 -0 -0 -0 -1 -11 -0 -0 -0 -0 -0 -600 -0 -0 -0 -100 -1 -0 -0 -100 let's see correct and the fourth row becomes 0 -0 negative 1 -0 negative 1 1 1 one zero and five hundred.

20:00 The last two row are the same zero zero zero one one and zero and the last row is zero zero zero zero zero zero one five hundred.

20:33 Okay now in this matrix here we get to change the sign of this coefficient to have one so which is the sign of that row of the fourth row and we get that fourth row is negative fourth row so the three first equations are the same one one zero one zero zero zero six hundred zero one one zero zero six hundred and zero zero and zero six hundred and zero and zero 011 negative 1 000 100 and the fourth row is 0 001 negative 1 0500 and the last 2 rows are the same 0 0 090 1 0000 negative 500 and the last 2 rows are the same 0 0 0 0 0191 1 1 0 0 0 2 the hand side and 0 0001 0500 and we see in this matrix we get to nullify this one and we do that by the row operation row 5 is row 5 minus row 4 this give us the first four equations are the same so i write them down again 1 01 600 then 01111 600 600 011 -0 -0 -0 -1 -negative 1 0 -0 -100 0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -500 and the 4 the 5th row becomes 0 -0 -0 -0 -101 -500 the last row is the same 0 -0 -0 -0 -0 -1 -500 and we have found the last two rows are the same so we are going to have a row of zeros at the end of the matrix and that is a sign of a sign of having infinitely many solutions...

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