‫Welcome back.
‫I hope you gave it a thought with regards to the exercise and now let's discuss it, in the first case,
‫both lambdas must be less than zero, so no one must be less than zero.
‫And Lunda, too, must be less than zero.
‫Why?
‫Well, if you look at the general solution here, then regardless of your A and B constants, negative
‫lambdas would mean that your Oilor numbers would have negative powers.
‫So if you have an oil number to the power of minus T or to the power of minus two times T or to the
‫power of minus two point five times T, whatever lambda you have here for as long as it's negative,
‫then eight times the oil and number to the power of minus eight times T and let's say that your lambda
‫one equals minus A and then your B times oil.
‫And number two, the power of minus B times T.
‫So your lambda two now equals minus B for as long as your powers are negative.
‫These things here and here, they will approach to zero as time goes to infinity and they have to the
‫negative powers will drive the entire thing to zero.
‫How about if one lambda is less than zero but the other lambda is greater than zero?
‫Let's say lambda one is less than zero and then lambda two is bigger than zero.
‫How about that?
‫What's going to happen then?
‫Well then this would be false in order to achieve this behavior here in the first graph, because both
‫numbers must be less than zero if no one is less than zero.
‫But Lambda two is greater than zero than the first term here eight times.
‫And then the oilor number to the power of lambda one time stae, this term will approach to zero.
‫And it is logical because lambda one is less than zero.
‫So here you will have a negative power.
‫So this term will approach zero.
‫Over time, however, the second term B times the oil number to the power of lambda two times T now
‫this term it has a positive power and therefore this term will simply shoot up.
‫So this term will approach infinity.
‫So mathematically, you could write it down like this limit as time goes to infinity, and then here
‫you have your total error, which is here, and then that equals limit as time goes to infinity, the
‫first term, which is right here, plus limit as time goes to infinity and the second term, which is
‫right here, the first term has a negative power.
‫The second term has a positive power.
‫So the first term will approach zero.
‫But the second term will approach infinity, so you will have zero plus infinity equals infinity.
‫So even if you have one lambda that is negative.
‫If the other lambda is not negative, then your total error will still go to infinity.
‫Your first term is not going to be able to hold the second term back because your first term is not
‫going to become negative.
‫It will only approach zero.
‫And since the other one can grow without any boundaries, then this entire error will go to infinity.
‫As time goes to infinity, even if you're error as a function of time looks like this eight times the
‫oil and number to the power of minus 1000 times T plus B times the oil number to the power of zero point
‫zero zero zero one time T.
‫Then still, one day as time goes to infinity, the first term goes to zero and then the second term
‫will go to infinity over time.
‫And so eventually your error as a function of time will still approach zero by the first term and infinity
‫by the second term.
‫And so in total, the error will approach infinity over time.
‫So in the first graph, both lambda one must be less than zero and Lambda two must be less than zero.
‫And that is true for this graph.
‫Or even if your graph is like this is the same thing is just here in this graph in the negative region,
‫you're A and B are negative, or maybe one of them is negative and the other one is positive.
‫But the one that is negative is more dominant.
‫But still, if both of your numbers are less than zero, then this entire thing will approach zero no
‫matter what you're A and B constants are.
‫So if you want your errors to approach zero, then lambdas are the ones that matter, not A and B constants.
‫But lambdas in the second graph is the same thing.
‫But now Lambda one and number two, they must be even more negative to get a stronger response.
‫So let's say that the numbers in this graph, they have an asterisk here in order to differentiate them
‫from the lambda in the first graph.
‫So they have to be more negative than the lambda in the first graph.
‫And of course, all of them must be less than zero.
‫And it makes sense, right, if you have an oil no to the power of minus five thousand times time and
‫then you have another oil and no to the power of minus five times time, then which one will get you
‫closer to zero faster?
‫Well, of course, this one here will get you closer to zero faster because its power here is way more
‫negative than here, minus five times T in the third graph lambda one must be greater than zero or Lunda
‫two must be greater than zero or both.
‫So it's in or here or both.
‫As we discussed earlier, even if one lambda is negative and the other one is positive, then you will
‫still have this kind of response.
‫So if you're lambda, one is less than zero and Lambda two is greater than zero, then you will still
‫have this kind of response.
‫And the same thing with the fourth graph here, you also need to have at least one lambda that is greater
‫than zero.
‫Let's say that your Londa one is greater than zero, but now, of course, your positive lambda must
‫be even stronger in order to get that stronger rise in the fifth graph you have oscillation and oscillation
‫means that you have to end up with a cosine or sine terms or both of them.
‫They need to be present in your solution and that can only happen if your numbers are complex numbers.
‫So your line does need to be.
‫Complex.
‫All right, but since you also wanted to damp out, in other words, you want your error to approach
‫zero as time approaches infinity, then that means that your lambdas also need to have a negative real
‫number.
‫So they look like this number one equals X plus eye times Y, and then number two equals X minus items
‫Y.
‫Both of them have the same X, the same real number.
‫However, if you want this thing to dump out, then that means that your X needs to be less than zero.
‫So the imaginary part gives you oscillation and then the negative X part gives you error.
‫That is approaching to zero in the sixth graph.
‫You also need to have complex Londis.
‫However, here your real number X must be zero.
‫Then you are only left with imaginary lambdas, two of them like this.
‫And then you will only have oscillation, the oscillation will not die out.
‫And finally in the seventh graph, you also need to have complex lambdas.
‫However, now your real number must be greater than zero.
‫Then you will have oscillation, but then your real number will make the error grow in magnitude as
‫well.