00:01
In this question, a circuit is given.
00:03
A circuit diagram is given.
00:09
So this is the circuit diagram.
00:11
Let us assume it is figure and if we draw this figure in terms of s domain, it can be redrawn as in s domain.
00:27
So it becomes something like this.
00:37
Wait a minute.
00:38
Something like this.
01:00
And then this is joined by, here we have this and here we have this.
01:11
So this is 4 by s, this is 2 by s, this is s and this is 0 .5 ohm.
01:24
And here the current, amount of current, it is i1s and here the current becomes i2s.
01:36
Here s, what is s? s is equals to j omega where omega is frequency of sinusoidal input.
02:01
So moving further, we have to find the transmission parameters for this question.
02:09
So the transmission parameters, we have like this.
02:35
This is a two -part system and this is by omega.
02:48
Here the input is current i1, voltage is v1 and here the current is i2 and the voltage is v2.
03:03
This is positive and this is negative.
03:05
This is again positive and this is negative.
03:10
This is figure 2 and this is figure.
03:14
So therefore, the transmission parameters are, let us assume a, b, c and d because this is a four -part system.
03:40
So we have assumed a, b, c and d.
03:44
This implies we have v1 is equals to a, v2 plus b, i2.
03:53
Similarly, we have i1 is equals to c, v2 plus d, i2.
04:01
Moving further, if we represent it in matrix form, we will have in matrix form, this will be 2 cross 1 matrix v1, i1 is equals to a, b, c, d that is a 2 cross 2 matrix multiplied by 2 cross 1 matrix v2 and i2.
04:43
Let us assume this is equation 1.
04:45
So now we know that when two networks of parameters like a1, b1, c1, d1 and a2, b2, c2 and d2 are cascaded, then the resultant transmission parameters are resultant transmission parameters are given as a, b, c, d in matrix form will be equal to a1, b1, c1, d1 multiplied by another 2 cross 2 matrix whose entries are a2, b2, c2 and d2.
06:18
This is equation 2...