Question

2y"(t) + 3 y' (t) + y(t)=x"(t) +x'(t) - x(t), y(0) = -2, y'(0) = 0, u(t) is the step function. 1. Write an expression for Y(s); at first leave U(s) symbolic. Identify which part is the zero-state and which part is the zero-input frequency-domain solution. Identify which part is the transfer function and which part is the initial condition polynomial. You will need to use the following transform pairs or properties, noting that they apply to the input as well except that the initial conditions for inputs are 0. C{y()} = Y() C{y' (t)} = Y()- (0) L{x"()} = 2Y() - *(0) - (0) After you identify the solution parts, substitute for U(s) = 1/s (the pair u(t) + 1/s and the pair 8(t) 1 should be memorized). 2. How many zeros does the system or H(s) have? How do you know? What are they? 3. How many modes and how many poles does the system have and what are they? 4. For each pole you found, write if it is real-valued or complex-valued Also indicate the number of times any pole repeats (if any repeats). True or false? : The number of poles equals the order of the denominator (which is the highest power of s in the denominator) and also equals the order of the system and the order of the corresponding differential equation.

Answer 1

24"4) + 3 + 4 +44) = x^2+x4) X(4) 5- domain 2[5* Y29)- syro)?yko)] +3CsYess - yra] +Y44)= ($75 -1)x() (25 +3541) 968) + 4 = (s”+5-1)XC9 Yes) = (5 73-1)X(9) estost) Estos - step input; XCS) = US) Yes)- 575-1 Ves) - 25 +35+1 4 . 25 +35+1 Transfer function ' Zaro-state (5+1.618 )C6 -0.618) (5+1)(5+0.5) Transfer function Yes) = 5 +5-1 ves) 25'+35+1 H(S) – (5+1-618)(5-0.618) (5+)( 5 +0.5) Zeros' 01618; -1618 No of zeros = 2 po s-plan -4.618 0.618 poles & -0.5;-| No. of poles=2 - po s-plan ajw -05 a s-plan so -1.618 -| -0-5 0.618 all goles are real only. No-repeated poles & no complexe poles