This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 1. Suppose , and let Starting with the definition of the Laplace transform, compute that the Laplace transform of and the region of convergence. DO NOT use the tables, and DO show all your work. Solution There is no restriction on required to evaluate the integral, so the region of convergence is the entire complex plane. 2. An impulse is applied to a system with zero initial conditions, and the corresponding output is measured and found to be . Find the transfer function for this system. Solution . Since , this is the transfer function. 3. Using the Laplace transform, compute the solution to the ordinary differential equation where and Solution Implies that Thus Partial fraction expansion yields and the solution is 4. Find the single transfer function that represents the block diagram below where , , and are transfer functions. + X +  G + F Y H Solution + +  G + W F H Label the node shown above as . Then and Noting that , 5. Find the transfer function for the block diagram below. x(t) 1 s y(t) 8 1 s 18 1 s 12 SOLUTION 6. Exercise 4.312 (page 482, i, ii, iii, v) SOLUTIONS (i) roots are . BIBO stable (ii) roots are , not BIBO stable (iii) Careful, this one has a higher order numerator than denominator. We say that the transfer function is not proper. To see that his can cause problems, let us compute the corresponding impulse response (inverse Laplace transform of ). To do so, we must pull out an and a constant. Multiple both sides by then, Thus to clear the denominator, and . The impulse response is For BIBO stability, we need to be bounded. The integral (antderivative) of is , which is not bounded. (v) roots are . Not BIBO stable 7. Exercise 4.411(a) Solution all have negative real parts, and thus is stable and the initial Note that the poles of and final value theorems can be applied. (i) , (ii) 8. Exercise 4.53 (page 485), (a) and (b). For part (b), do (i), (ii), and (iii). By "discuss," you are to determine if the system is stable. You are welcome to compute the poles of a transfer function numerically. For example, you can find the roots of the polynomial in Matlab using the command roots([1 2 10 25 50]) Solution (a) The closedloop transfer function is , which has a pole at and is therefore stable. (b) The closedloop transfer function is (i) For , the poles are 4.7608 0.6196 + 1.3102i 0.6196 1.3102i which have negative real parts. Thus the system is stable. (ii) For , the poles are 6.0449 0.0225 + 2.8759i 0.0225 2.8759i two of which have positive real parts. Thus the system is not stable. , the poles are (iii) For 6.0000 0.0000 + 2.8284i 0.0000 2.8284i two of are on the imaginary axis and do not have negative real parts. Thus the system is not stable. ...
View Full
Document
 Spring '08
 stilwell

Click to edit the document details