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Sketch Bode plots for the following systems: St2 tf = (Also write the sinusoidal steady state solution of this system t0 the 82+20 $ input 10 sin (2 t).) s+1 tf = (...

Question

Sketch Bode plots for the following systems: St2 tf = (Also write the sinusoidal steady state solution of this system t0 the 82+20 $ input 10 sin (2 t).) s+1 tf = (s+10)(s2+s+100)

Sketch Bode plots for the following systems: St2 tf = (Also write the sinusoidal steady state solution of this system t0 the 82+20 $ input 10 sin (2 t).) s+1 tf = (s+10)(s2+s+100)



Answers

Determine the equation of motion for an undamped system at resonance governed by
$$ \begin{array}{l}{\frac{d^{2} y}{d t^{2}}+y=5 \cos t} \\ {y(0)=0, \quad y^{\prime}(0)=1}\end{array} $$
Sketch the solution.

It was the 80 PICO Ju Julia Thames depressed Science three on 20 and we find a steady state. You coaching a limit on the 80. It's digger to infinity. Therefore again, a limit on the two attempts. Depressed side day over 30 years to infinity, you compute this limit the first time I want to do with the any by humanity. And then again, the limit becomes two times on a time again and one less inside the Darmody give anybody one. And now we're seeing a stigma to infinity. This one goes to zero and then we have left. Wondered, too, Tim's one on the one and equal to the to.

So the equation is why is equal to 3.5 sunny hi over to Times Team? Best is, you know, plank five. So first thing is, was the amp acute. The amplitude is 3.5 was the opinion period is to pipe divided by pi or two, and that comes out to be called to four and then face shift. So there's a face shift off negative 40.5, and the graph is right here. You can take a look at the amplitude is 3.5 and that the period is four and also that the face shift is negative 40.5.

What happens to this function? A As T becomes large, this a function of time. So what happens to this value as time goes by as T approaches infinity in order to evaluate that we're going to rearrange this just a little bit. First thing we're gonna do is separate the fraction. So in a t over tea plus sine of t over tea in this tea over T here, that's equal to one. We're going to distribute the two so two times one is two. Never lost to have two times the sine of t over tea. Two is going to remain, too, no matter how big T is. Two stays too. But this part right here, we're gonna have to figure out what is. The LTD's t approaches infinity, and that's gonna be a job for the squeeze. Their, um so we're gonna start off making the observation that the sine of T is gonna oscillated between positive one and negative one. It's always gonna be greater than negative one, and it's always gonna be less center equal to positive one. So now if we multiply everything by two, we get that negative two is less than or equal to two times the sine of T, which is less than or equal to two. And if we divide every single side of our inequality by T, it looks like this. And now we've got what we were going for. So here we have a function to scientist over T, which is always going to be between negative to over tea and positive two over tea. So the Squeeze serum tells us if we take a limit of the left side and the right side, and they come out to be the same than for sure, the center guy will have that same limits. So let's look at what happens as T approaches positive infinity, one over tea approaches zero. As this denominator gets larger and larger and larger, the value of the fraction approaches zero. And it doesn't matter if the numerator is a constant like negative to our positive two. It's still approaches zero. So on the left side Here we have the limit as T approaches infinity of negative two over T that's going to approach zero. So the limited zero on the right side. Here we get that the limit as T approaches infinity of two over t also equal zero. These two things are equal. Therefore, according to the squeeze serum, the limit s t approaches infinity for the center guy, which is what we were actually trying to determine two times the sine of t over t. That limit has toe also equal zero. So that tells us that this right here approaches zero as T approaches infinity so F is going to equal or be about equal to two plus zero the limit as T approaches infinity of our function a F T is going to equal to. So that tells us that indeed we will reach some equilibrium value and that value is to

All right this time were given the equation y equals negative 10 plus 20 sign of one half Times Data -120. So our amplitude Again, we start with this multiplier, so it is 20 and then the period take this one half. So we're gonna do 3 60/1 half. So it's really 360 times too. So we get 720 for the period Phase displacement we have fate of -120 equals zero. So we add the 120, It failed to equals 120, So it's 120°. Then sonia soil is we have a -10 right here, so it's gonna be y equals negative 10. So now for our graph This time we're shifting down by 10 And our amplitude was 20, so that means we're gonna be at 10 And then negative 10 and negative 30. So now our sign graph are period with 720, we're shifting to the right by 120. So we're gonna start at 120 so right there and then again and we start in the middle of this time because this is a sign graph instead of a cosign graph. So now we take that 7 20 we divide by four so that's 1 80 so we're gonna go up to 300 and that's going up here And then for 80, so it comes back to here And then 660 which comes down to here and then 800 and 40 it's going to be back here and there's our curve. And so now we can go in and check this on our graph. So go to Y equals We have negative 10 was 20 sign of five x minus 60. Check our window. So for X0 Let's see. We only went up to 800. So we called 900 but then we went down to negative 30 so say -35. We're up at 10 so we'll save 15. So there we go. And that is what this graph looks like as we have grafted.


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