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Remark from [[User:Bell|Steve Bell]]: | Remark from [[User:Bell|Steve Bell]]: | ||
| − | That's life for engineers. The solution corresponds to a hammer hit on a spring-mass system (with damping) at time pi/2 followed by turning on a vibration force at time pi. You will note that the solution you get, although piecewise defined, is continuous. The velocity jumps at the hammer hit. After you experience trying to graph it with your bare hands, I will show you how to use maple to graph these things.<br> | + | That's life for engineers. The solution corresponds to a hammer hit on a spring-mass system (with damping) at time pi/2 followed by turning on a vibration force at time pi. You will note that the solution you get, although piecewise defined, is continuous. The velocity jumps at the hammer hit. After you experience trying to graph it with your bare hands, I will show you how to use maple to graph these things.<br> |
| − | Question from Mickey Rhoades [[User:Mrhoade|Mrhoade]] | + | Question from Mickey Rhoades [[User:Mrhoade|Mrhoade]] |
| − | If I remember correctly from my undergrad years ago, in | + | If I remember correctly from my undergrad years ago, in systems these portions of the solution are zero-state and zero-input? I know all through this course we talk about homogeneous and non-homogeneous, but when dealing with Laplace transforms and tranfer functions is the portion that pops out from the initial conditions and no forcing function called the zero-input? Then we set the initial conditions to zero and obtain the zero-state? In this particular problem, it seems the the two portions of the general solution seperate themselves naturally. 1/(s+2)(s+3) is the zero input and the other two sections are zero-state. |
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Revision as of 11:45, 14 October 2013
Homework 6 collaboration area
From Mnestero:
On 6.4 prob 10 I am getting a pretty intense solution that is difficult to graph. Am I on the right track? How in depth are we supposed to graph the solution?
Response from Mickey Rhoades Mrhoade
I thought the same thing. My solution is pretty intense as well. It seems like there is the portion from the initial conditions which is e-2t - e-3t and then there is the portion from the impulse function which is added beginning at pi/2, epie-2t - e3pi/2e-3t and then there is the portion of the output due to the cosine input beginning at pi. This section looks like a sin/cos wave inside an exponential envelope. Did anyone else come up with something different? -Mick
Remark from Steve Bell:
That's life for engineers. The solution corresponds to a hammer hit on a spring-mass system (with damping) at time pi/2 followed by turning on a vibration force at time pi. You will note that the solution you get, although piecewise defined, is continuous. The velocity jumps at the hammer hit. After you experience trying to graph it with your bare hands, I will show you how to use maple to graph these things.
Question from Mickey Rhoades Mrhoade
If I remember correctly from my undergrad years ago, in systems these portions of the solution are zero-state and zero-input? I know all through this course we talk about homogeneous and non-homogeneous, but when dealing with Laplace transforms and tranfer functions is the portion that pops out from the initial conditions and no forcing function called the zero-input? Then we set the initial conditions to zero and obtain the zero-state? In this particular problem, it seems the the two portions of the general solution seperate themselves naturally. 1/(s+2)(s+3) is the zero input and the other two sections are zero-state.
Remark from Steve Bell:
I got a question about p. 231: 14a in class today. Split the integral up like the book suggests. For the integral
$ \int_{np}^{(n+1)p} e^{-st}f(t)\ dt, $
make the change of variables
τ = t + n'p
and take it from there. In part b, you just need to apply the formula derived in part (a) to the function they give you.
Question from Luo Shibo;
On 6.4 problem 14(b): half wave rectifier
I can get the f=sin(wt) for 2n*pi<t<(2n+1)*pi ,and f=0 for (2n+1)*pi<t<(2n+2)*pi. But I have no idea about how to combine them together, I mean combine them into one equation: f=sin(wt-xx)*u(t-xx) such kind of form. Could someone give me a hint?
Response from Mickey Rhoades Mrhoade
Luo,
write out L[f(t)] = 1 / (1-e^(-2*pi*s/w)) from part (a) and multiply it times the laplace transform of the half-wave from 0 to p. i.e. the wave is u(t)*sin(wt) - u(t-pi/w) * sin(wt). Your period (p) is 2pi/w. You proved this theory in part (a) so no need to run though it all again. The big trick here is to use the identity that sin(wt) = -sin(wt-pi) You then recognize that you need to divide out the w because it is the t portion that needs to be shifted. You have the required shift of t-pi/w in the sin function. you should be taking the Laplace transform from 0 to p of sin(wt) + u(t-pi/w) * sin(w(t-pi/w)). Take your transform and the solution pops out.
