Answer I compute the invL of each term separately, starting with the left. The e^-s factor tells me that there will be a shift of 1 in the t domain and that I must multiply by the heaviside function. Looking at the table, I see that invL(s/(s^2-1))=cosh(t), but since hyperbolic trig is meaningless to me, I rewrite cosh(t) as (e^t-e^-t)/2. Now remembering the shift, I decide that invL(s*e^-s/(s^2-1))=H(t-1)*[e^(t- 1)-e^-(t-1)]/2.
Next, in order to find invL(s^2/(s+2)^2), I rewrite s^2 as (s+2-2)^2= (s+2)^2-4*(s+2)+4. Then I divide each term by (s+2)^2 and get 1-4/(s+2)+4/(s+2)^2, which i can easily inverse laplace transform with the help of a table or perhaps even without one. invL(s^2/(s+2)^2) =invL(1 - 4/(s+2) + 4/(s+2)^2)=dirac(t)-4*e^(-2*t)+4*t*e^(-2*t).
Putting together the two pieces, I find that invL(s*e^-s/(s^2-1)-s^2/ (s+2)^2)=H(t-1)*[e^(t-1) - e^-(t-1)]/2 - dirac(t)+4*e^(-2*t)-4*t*e^(- 2*t).
Sent by N. Wilson at Fri, 20 Feb 2004 18:34:08
Answer f(t)=(H(t)t^2) F(s)=2/s^3
g(t)=(H(t)cos(t)) G(s)=s/(s^2+1)
L((f*g)(t))=F(s)G(s) =2s/(s^3(s^2+1)) =2/(s^2(s^2+1)) =2/s^2 - 2/(s^2+1)
(f*g)(t)= 2t - 2 sin(t)
Sent by C. Graziadio at 20 Feb 2004 00:23:33
Answer First: I Laplace transformed the equation Second: I solved for Y(s) Third: I manipulated the equation for Y(s) using partial fractions Fourth: I used Lerch's(???) Theorem to get y(t)
Comment by the Management Mr. Woodrow tried several times to get every darn number correct. Here is what Maple reports as the solution, following Mr. Woodrow's outline above: (-1/9 exp(-3 t + 6) + 7/9 - t/3) Heaviside(t - 2) + exp(-t) cosh(2 t) Of course, the exp(-t)cosh(2t) term could be rewritten as (1/2)et+(1/2)e-3t since cosh(x)=(ex+e-x)/2.
Sent by M. Woodrow at Thu, 19 Feb 2004 22:35:42
Answer b) t*H(t-1)+(1-t)*H(t-3) c) e^(-s)*(1/s^2 + 1/s) + e^(-3s)*(-2/s - 1/s^2)
Comment by the Management Because Management can't receive pictures, the picture was created locally. Also, to see why, say, the second part of the answer is correct, consider (1-t)*H(t-3)=(-2+3-t)*H(t-3)=(-2-(t-3))*H(t-3).
Sent by T. Obbayi at Fri, 20 Feb 2004 13:45:56
Answer This solution is by the Management, and is copied from an e-mail I just wrote.
I copied this problem from a 421 exam given last year. We discussed two methods of solution at the review session yesterday evening. Neither of them is both easy and obvious.
1. Write s/(s^2+36)^2 as the product of s/(s^2+36) and 1/(s^2+36). Each of these functions is the Laplace transform of a "known" function (its in the table). The first comes from (1/6)cos(6t) and the scond, fron (1/6)sin(6t). Since Laplace transform changes convolution to product, you could (!) compute the convolution of these functions directly:the integral from 0 to of (1/6)cos(6(t-tau))(1/6)sin(6tau)dtau. This can be done. Outline: "expand" the cosine using the addition formula for cosine. Then
2. This is easier, but calls for an inspiration: s/(s^2+36)^2 is quite close to the derivative of something ... yes, it IS the derivative of 1/(s^2+36) after we divide by 2 (because of the 2 coming in from the derivative of s^2). So d/ds[(1/2)(1/(s^2+36))] is s/(s^2+36)^2. But the Laplace transform of t^n f(t) is (-1)^n d^n/ds^n (F(s)). That means (adjusting for the - sign) that what we have is the Laplace transform of -t times a function whose Laplace transform is (1/2)(1/(s^2+36)). Therefore (!!) the answer is -t(1/2)(1/6)sin(6t).
#2 is fast, but it certainly requires cleverness.
Sent by no one, at about 9:30 AM Tuesday, Feb 24
Sent by
Answer Int(delta(t-e)*ln(t) + delta(t-pi)*H(t-2))dt Filtering property pg148 substitute e for t and the pi for t new equation is ln(e) + H(pi-2) = 1 + 1 = 2
Sent by J. Marchitello at Fri, 20 Feb 2004 13:52:16
Answer Laplace transform of any periodic function, if exists, can be represented L[f(t)](s) = 1/[1-exp(-sT)] * int[exp(-s*t)*f(t)*dt,t=0..T] , where t goes from 0 to T, where T is the period of the function, that is, f(t+T) = f(t). On interval t = 0..T, or t = 0..2, our function is represented by f(t) = (1/2)*t, therefore L[f(t)](s) = 1/[1-exp(-sT)] * int[exp(-s*t)*f(t)*dt,t=0..T] = = 1/[1-exp(-s*T)] * int[exp(-s*t)*(1/2)*t*dt,t=0..T] = = 1 / [1-(exp(-s*2)](-2exp(-2s)/s-exp(-2s)/s^2+1/s^2) (integration by parts) Substituting T=2, we get L[f(t)](s) = 1/[1-(exp(-2s)](-2exp(-2s)/s-exp(-2s)/s^2+1/s^2)
Sent by D. Ivanov at Mon, 23 Feb 2004 20:07:56
Answer I believe the answer to number 10 could be t^4(e^(t^4)). e^(t^4) grows faster then any simple exponential as t approaches infinity and also t^4 makes it grow even faster. So therefore you can't take it's laplace transform.
Sent by V. Inserra at Sun, 22 Feb 2004 20:53:28
Answer laplace of -t is -1/s^2 so at s=5 the laplace of -t is -1/25 and at s=10 the laplace of -t is -1/100. being that -1/100>-1/25 then the laplace of -t is greater at 10 then at five.
Comment by the Management Mr. Cohen's solution is certainly a correct response to the question. However, it ispossible to find a Laplace transform which is positive and whose value at 10 is larger than its value at 5. For example ...
Sent by B. Cohen at Thu, 19 Feb 2004 22:45:04
Answer I) a(0,1,3,4) + b(1,2,4,-1) + c(-2,1,1,0) = 0 II) 0a + b - 2c = 0 => b = 2c => c = 1/2b 1a + 2b + c = 0 3a + 4b + c = 0 4a - b + 0c = 0 => b = 4a => a = 1/4b III) plug c and a back into the second equation => 1/4b + 2b + 1/2b =0, therefore, b=0, a=0, and c=0. The vectors are linearly independent in R^4.
Sent by A. Lu at Thu, 19 Feb 2004 21:49:40
Answer Yes it is in the span. This can be shown by setting up a system of linear equations and solving for a=0, b=0, c=1/2. Or, simply through inspection it can be seen that (1,1,1,1,1) is a scalar multiple of (2,2,2,2,2), and is in the linear combination of 0(0,2,0,2,1)+0(1,0,1,0,1)+1/2(2,2,2,2,2).
Comment by the Management Mr. Hunt certainly has solved this problem as stated. The Management sincerely regrets that the vector (2,2,2,2,2) was given instead of the vector (2,2,2,2,-2). With that replacement, the problem would have been a bit more challenging.
Sent by C. Hunt at Thu, 19 Feb 2004 22:47:25
Answer S={x,y,z,t,0,0,0} with a basis{[1000000] [0100000] [0010000] [0001000]}.
Sent by M. Mudrak at Tue, 24 Feb 2004 08:20:19
A=(3 0 -1) and B=(3 2 0) (1 2 2) (2 -1 1) (4 2 1)
Answer A^2=
[5 -2 -4 13 8 5 18 6 1]
[11 4 1 9 0 -3]
[14 -2 -7 16 14 11 30 12 4]
Sent by M. Mudrak at Thu, 19 Feb 2004 15:19:44
Answer F(s)=int_0^infty e^(-st)t dt Its an improper integral so take the limit as a or some constant goes to infinity so you can evaluate it.
e^(-st)*t has to be integrated by parts
-d/dt | int ------------------ t e^(-st) -1 -1/s*e^(-st) 0 1/s^2*e^(-st)
Sent by M.McNair at Sat, 21 Feb 2004 14:55:10
means no answer is available yet.
#6
S. DiVento
4. Consider the function f(t)= 0 if 0<t<1 and t if 1<t<3
and 1 if 3<t.
0inftydelta(t-e)ln(t)+delta(t-Pi)H(t-2) dt
8. What is the Laplace transform of the function shown? (The function
is piecewise linear, and extends periodically for all positive t
with period 2. Try to find a compact representation of its Laplace
transform.)
means no answer is available yet.