00:01
In this question, we need to determine the poles and zeros.
00:18
So we can write down the equation for open loop transfer function which is denoted by gs and it is equals to k multiplied with s plus 2 divided by s plus 1 plus j multiplied with s plus 1 minus j.
00:52
Similarly, we can write down the equation for closed loop transfer function and it is denoted by ts which is equals to gs multiplied with 1 divided by 1 plus gs.
01:26
So on simplifying, we get the value of ts equals to k multiplied with s plus 2 divided by s square plus 2s plus 2 minus k plus j square is minus kj minus 1.
01:50
Now we have to convert it into the characteristic equation which is given by s square plus 2s plus 2 minus k plus j square is minus kj minus 1 equals to 0 and this is for a closed loop.
02:31
We know that for real part, the equation will be equal to 0.
02:41
So we can write down further that is plus 2s plus 1 minus k equals to 0.
02:54
Solving open this, we can get the final roots of the quadratic equation that is given by s equals to minus 1 plus minus k.
03:20
Now we have to find the number of asymptotes...