EECE 290 Exam with Solution, Spring 2011

EECE 290 Quiz 3, May 7, 2011 10 H 1 1’ 2 2’ 5 V 10 H 1 F 1 F + – 1. Determine which of the following statements is or are true of the two-port circuit shown. (If a false statement is marked true, the question is considered incorrect). A. The circuit is a symmetric two-port circuit. B. The circuit is symmetric but not reciprocal. C. The circuit is not a valid two-port circuit. D. The circuit is valid at some frequencies and invalid at other frequencies. E. The circuit can be described in terms of a and b parameters. Answer. The only true statement is C. 2. Specify the two-port parameter equations that are directly represented by the circuit shown. Solution: The circuit directly represents the g-parameter quations. 3. ssuming A = 1. The Laplace transform should apply of the triangle, which A/2. The Laplace Method 2: + – V1 + – V2 I2 I1 6V1 4I2 – + 2 Ω 3 S e e the Laplace transform of the time integral of f(t), ∫ − f t dt 0 ( ) , a Determin t f (t) t, s 1 A for t ≥ 2 s. Solution. Method 1: ∫ − f t dt 0 ( ) is the area 2 -A transform is therefore A/2s. ⎥ = ⎤ ⎥⎦ ⎢ ⎡ ⎟⎞ ⎜⎛− − ⎟⎞ ⎜ ⎝ − ⎛− ⎟⎞ ⎜⎛− = − = − − − − ∫ 1 1 1 1 1 1 1 ) (1 ) ( st st st st e s e t e A dt t e A F s ⎢⎣ ⎠ ⎝ − − 0 0 s ⎠ ⎝ ⎠ − − 0 0 s s ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + − ⎥ = ⎦ ⎤ ⎢ ⎣ ⎡ − + + + − − − − − 2 2 2 2 1 1 1 1 s e s A s s s e s e s s e A s s s s . L {f(-1)(t)} ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + − = − 3 3 2 1 1 s e s s A s ; f(-1)(t) = ( )⎥ ⎦ ⎣ 2 2 ⎤ ⎢ ⎡ − − + − = 1 )1 ( 2 2 u t t t A t . For t ≥ 2 s, f(-1)(t) = A/2, and the Laplace transform is A/2s. Version 1: A = 1, A/2s = 0.5/s Version 2: A = 2, A/2s = 1/s 1 Version 3: A = 3, A/2s = 1.5/s Version 4: A = 4, A/2s = 2/s Version 5: A = 5, A/2s = 2.5/s 4. Given that the Laplace transform of ( ) ( ) / t u t A e −t is + 1 s A π , determine the Laplace transform of ( ) t u(t) Ae t − , assuming A = 1. Solution. The function whose Laplace transform is required is times the given function. Hence, its Laplace transform is t ds dF s ( ) − , where 1 ( ) + = s A F s π . Thus, ( ) + 1 3/ 2 2 ( ) = − s A ds dF s π . Version 1: A = 1, ( ) ( ) 3/2 3/ 2 1 1 2 + + s s = 0.5 A π π Version 2: A = 2, ( ) ( ) 3/2 3/ 2 1 1 2 + + s s = A π π Version 3: A = 3, ( ) ( ) 3/2 3/ 2 1 1 2 + + s s = 1.5 A π π Version 4: A = 4, ( ) ( ) 3/2 3/ 2 1 1 2 + + s s 2 = A π π Version 5: A = 5, ( ) ( ) 3/2 3/ 2 1 1 2 + + s s 5. e true. (If a false statement is A. e same circuit can be bounded for a certain input and unbounded in complex conjugate pairs. E. e transfer function are in the right half of the s is. 6. f(t)∗g(t), with f(t) = sint, g(t) = 2δ(t) + δ(2)(t), and δ(2)(t) being the second derivative of δ(t). = 2.5 A π π Determine which of the following statements is or ar marked true, the question is considered incorrect). The response of th for another input. B. The poles of a first-order circuit must always occur C. A second-order circuit cannot have a double pole. D. A pole of a voltage source at s = -2 corresponds to dc voltage of -2 V. A circuit is unstable if the poles of th plane including the imaginary ax Answer. The only true statement is A. Determine the value of the convolution integral y(t) at t = 0.5 s, where y(t) = 2 Solution. L { } s2 +1 , and L 1 sint = { } 2 (2) 2 2 ( ) s t + = + δ δ ; 1 1 1 1 2 ) ( 2 2 2 + + = + + = s s s Y s ; t t y t sin ( ) ( ) + = δ . At t = T s, y(T) = sinT. 48 00 ersion 5: T = 2.5, sin(2.5) = 0.60 7. n for the Laplace transform of f(t – b)u(t – a), where a and b are ution. L ∫ ∫ − = − − = − − − a dt b e f t dt a e b u t f t a b u t f t ) ( ) ) ( ( ) ) ( ( 0 . Substituting, t’ = t – ∫ − ′ = − ′ + = − − 0 ) ( ) ) ( ( e dt b e a f t a b u t f t L Version 1: T = 0.5, sin(0.5) = 0. Version 2: T = 1, sin(1) = 0.84 Version 3: T = 1.5, sin(1.5) = 1. Version 4: T = 2, sin(2) = 0.91 V Derive an expressio positive Sol constants. { } ∞ − ∞ − st st a, { } ∞ − ′+ − ) ( as a s t { }) ( b a f t + − L . 8. Determine f(t) if 20 4 ( ) = Ks F s 2 + + s s , assuming K = 1. Solution. ( ) ( ) 16 2 2 2) ( 16 2 ) ( 2 2 + + − + = + + = = s K K s Ks s s Ks F s 20 4 2 + + s t e sin 4 − 2 K t Ke f t t t cos 4 ( ) 2 2 − − = t ersion 5: K = 5, t e t e f t t t sin 4 2.5 cos 4 5 ( ) 2 2 − − − = 9. If Version 1: K = 1, 4 e t e f t t t sin 0.5 cos 4 ( ) 2 2 − − − = Version 2: K = 2, t e t e f t t t sin4 cos 4 2 ( ) 2 2 − − − = Version 3: K = 3, t e t e f t t t sin 4 1.5 cos 4 3 ( ) 2 2 − − − = Version 4: K = 4, t e t e f t t t sin 4 2 cos 4 4 ( ) 2 2 − − − = V 1 1 ) ( 2 + + = s s K F s 2 3 + + + s s s , determine f(t) for large values of t Solution. , assuming K = 1. ⎥ ⎤ ⎢⎣ ⎡ + + + + = + + + + = + + + + + 1 1 1 2 )1 1)( ( 1 1 1 2 2 2 2 3 2 s s s K s s s s K s s s K 1⎦ s s . [ ]t t K e f t t sin cos 2 ( ) + + = − . For large values of t, [ ] / 4) cos( 2 sin cos ( ) − π = + → t K t t K f t . The final-value theorem does 2 3 not apply because of the poles at s = ±j. Version 1: K = 1, / 4) 1 cos( / 4) cos( π π − = − t t K 2 2 Version 2: K = 2, / 4) 2 cos( / 4) cos( 2 π π − = − t t K Version 3: K = 3, / 4) cos( 2 / 4) cos( 2 π π − = − t t 3 K Version 4: K = 4, / 4) 2 2 cos( / 4) cos( 2 π π − = − t t K Version 5: K = 5, / 4) cos( 2 5 / 4) cos( 2 π π − = − t t K 10. The figure shows two identical, consecutive pulses, each of duration a, the second pulse being inverted with respect to the first. If F(s) is the Laplace transform of the two pulses shown, determine, in terms of F(s the Laplace tran ), sform of f(t) shifted to the left by a. t = a. Then, Assume a = 1. Solutio single pulse that extends from t = 0 to n. Let G(s) be the Laplace tr G s e as G s F s − − = ( ) ( ) ( ) , ansform of a or e as F s G s − − = 1 ( ) ( ) . f (t) t a 2a When shifted to the left, the Laplace transform of f(t + a) is as G s − − = − 1 ( ) Version 1: a = 1, e − F s ) ( e s G s − − = − 1 ( ) − F s ) ( Version 2: a = 2, s e G s 2 1 ) ( − − = − F s ( ) − Version 3: a = 3, s e G s 3 1 ) ( − − = − F s ( ) − Version 4: a = 4, s e G s 4 1 ) ( − − = − F s ( ) − Version 5: a = 5, s e F s G s 5 1 ( ) ) ( − − = − − 11. ng ω = 1 rad/s (5% Determine the h parameters for the circuit shown, assumi for each parameter). 2vx 1 F + vx – 1 Ω 1 Ω 1 F 4 Solution. With the output short circuited, the impedance of the paralleled capacitors and resistor is 2 1 1 1 1/ 2 1/ 2 j j j + = + ; h11 = = 1 1 I V 5 2 1 2 1 j j + + x ) 2(3 ) 2(1 1 1 j j − = + = + Ω; V = 2 1+ 1 I j 1, and 2Vx = 2 x x I + j V , or V = − j 2 1 I2. Equating Vx, 2 1 1 + j I1 = 2 − j I2; h21 = 1 = 1 2 I I j . j j = − + − 2 1 2 Vπ = V1 = With the input open circuited, ) 1/(1 j j + + 2 = 1/ 1 V ; hence, h ) /(1 j + 12 = 2 1 V V 5 2 2 1 j j j + = + . I2 = ) 1/(1 1/ 1 j j + + V2 + 2Vx = 2 1 + j 2 ) (1 j j + V + 2 1 + j 2 j2 V , so h22 = = 2I V2 ) 1+ j2 1 S. ( 3 1 + = − + j j 12. Determine the Laplace transform of: 10% (a) f(t). 10% (b) df(t)/dt. Solution. (a) Method 1: f(t) = − − − − )1 )1 ( 2( u t t ( )1 ( )t( 2tu 2) 2) ( ( )1 )1 ( − − + − − u t t u t t ; hence, − t − u s s e s e s s s 2 Method 2: F F s 2 1 3 1 ) ( − ⎟⎞ − + ⎜⎛ + − = . − 2 st 2 2 ⎠ ⎝ = ∫ − − 1 0 2 ( ) dt te s st 2 ∫ = − + 1 ) (2 dt t e − ⎥⎦ ⎤ ⎢⎣ ⎡− + ⎥⎦ ⎤ ⎢⎣ ⎡− + ⎥⎦ ⎤ ⎢⎣ ⎡− − − − − − 2 1 1 0 1 0 1 2 1 2 2 st st st s e s e s s e t 2 1⎥⎦ ⎤ ⎢⎣ ⎡− s e−st t = ⎥⎦ ⎤ ⎢⎣ ⎡− − − 2 1 1 1 s e st s − − − − s s s e s s e s e s 2 2 2 2 2 1 1 2 + − + + − + − − − s s s e e e 2 2 2 2 − 2 − e−s 2 s s s s 1 e−s = s2 s s e e 2 1 3 1 2 − ⎟⎞ − + ⎜⎛ + − . s s s s 2 2 2 ⎠ ⎝ (b) L {df/dt)} = sF(s) – f(0-) = s 1 3 1 2 1 ⎟ − + ⎠ ⎞ ⎜ ⎝ − ⎛ + + . s s e s e s 2 − + – Vx 1 Ω 1 Ω – V2 – V1 1/j Ω + + I1= 0 I2 1/j Ω 2Vx t δ(t) 2 -δ(t) -1 f (1)(t) 2 f (t) 1 1 -1 t, s 2 2 + – Vx I1 j Vx I2 1 Ω 1 Ω – V1 + 1/j Ω 2Vx 5 As a check, f(1)(t) = δ(t) + 2u(t) – 3u(t – 1) – δ(t – 1) + u(t – 2). It follows that L {df/dt)} = s s s s e e s e s 2 1 3 2 1 − − − + − − + . Note that in order to obtain the Laplace transfor has an initial value of -1 at t = 0-, the m of f(t) from f(1)(t), then because f(t) correct expression is: F(s) = 6 s 1 L {df/dt)} + s f (0 ) − . Thus, s e s e s s s s s 1 1 3 1 2 1 2 2 2 2 − + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − + − − . s 13. The d ened at t = 0 osed for a long time. s domain for s shown. The hown f + mutual 1/4s)I2 ouble switch is op after having been cl 8% (a) Draw the circuit in the t ≥ 0+. 6% (b) Derive VO(s). 6% (c) Derive vO(t) for t ≥ 0+. Solution. (a) At t = 0-, there will be an initial current of 2 A in each winding and an initial voltage of 4 V across the capacitor a circuit in the s domain is s secondary inductor is L2 – M = 2 is (2 H)×(-2 A) = -4. The current in the 4 H or t ≥ 0 . The H and the source inductor is 2 + 2 = 4 A directed downwards, and the source is (4 H)×(4 A) = 16. (b) From KVL: (6s + 2 + +16 + 4 + 4/s = 0, which gives: 0.25 2 6 2 + s + s )1 4(5 2 + = − s I ; 0.25 2 0.25 2 6 3 24 1 25 1 5 1 4 4 4 1 4 4 2 2 2 2 2 2 + + ⎥ = ⎦ ⎤ ⎢ ⎣ ⎡ + + + = ⎥⎦ ⎤ ⎢ ⎡ + + + − ⎥ = ⎤ ⎢ ⎡ + = + = s s s s s s s s s I s s s I VO . (c) 6 )1 3(8 + s 0. 2 6 ⎣ ⎦ ⎣ s s ( ) ( ) ( ) ( ) 2 2 2 2 2 1/ 6 2 / 6 1 1/ 6 2 2 1/ 6 2 1/ 6 1/ 24 / 3 + + − + + = + + = s s s s VO . It follows that 1/ 6) 4( 0.5 4 + + s s ( ) ( 2) sin 6 / 2 2 cos 6 / 4 ( ) /6 /6 t e t e t v t t O − − − = V. 2 Ω 4 H 5 H 6 H 4 F 1 Ω 3 Ω t = 0 + – + 6 V – vO 2 Ω + – + – + – + – 2s Ω s Ω 4s Ω 1/4s Ω 2 4 I2 16 4/s + VO – 2 Ω 4 V 3 Ω + + – – 2 A 2 A 1 Ω