00:01
High, so we are given that the transfer function tf is equal to the 200 of s plus sub 20 divided by the s twice of s plus 1 into the s plus 40.
00:20
So we just have to draw the bode plot magnitude and phase angle plot for this transfer function.
00:28
As you can see on comparing k value will be goes to 100 and it's starting from 0 at s is equal to the minus of 20 here.
00:41
So basically the pulse is at 0, 0 point, okay let me show you the pulse is at 0, negative of 0 .5.
00:55
So in accordance to that add 20 log of k that will be equals to 20 log of k by 2.
01:08
So it will be 50 here.
01:12
So this will comes to near about 33 .98 and to above the plot.
01:20
So the bode plot, bode plot this will become, so let me draw you an arrows.
01:30
So let's suppose these are the vertical and horizontal axis.
01:35
Here i am considering 11, this is as 20, this as 40, over there we are taking the frequency of omega.
01:47
So the plot will be something like in this way that you will get is from the given of bode plot here.
01:58
So this is 20, not this one.
02:00
Now what are the readings right over here? minus negative 24 .08, here it will be negative of 30 .1.
02:09
How did we draw? by using the scepter, by using the given transfer function.
02:16
Now in the same way we can draw the phase plot.
02:19
For that we have to consider again the vertical axis and let's say the horizontal axis or let it be above something like right here.
02:29
Now i can take the measurements like 1, 20, here it is 40...