Chap4Solutions - M. J. Roberts - 10/7/06 Chapter 4 -...

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Unformatted text preview: M. J. Roberts - 10/7/06 Chapter 4 - Continuous-Time System Properties Solutions Modeling CT Systems 1. In a chemical molecule the atoms are mechanically coupled by interatomic binding forces. A salt molecule consists of one sodium atom bound to one chlorine atom. The atomic mass of sodium is 22.99 and the atomic mass of chlorine is 35.45 and one atomic mass unit is 1.6604 10 27 kg. Model the molecule as two masses coupled by a spring whose spring constant is K s = 1.2 1059 N/m . In a system of this type the two atoms can accelerate relative to each other but (in the absence of external forces) the center of mass of the system does not accelerate. It is convenient to let the center of mass be the origin of the coordinate system describing the atom’s positions. Let the unstretched length of the spring be 0 and let the position of the sodium atom be ys t and let the position of the chlorine atom be yc t . Write two coupled differential equations of motion for this mechanical system, combine them into one differential equation in terms of the amount of spring stretch y t = ys t yc t a nd show that the resonant radian frequency is 0 () () () () () ms + mc where ms is the mass of the sodium atom and mc is the mass of the ms mc chlorine atom and find the resonant frequency for a salt molecule. (This model is unrealistic because salt molecules rarely form in isolation. Salt occurs in crystals and the other molecules in the crystal also exert forces on the molecule making the realistic analysis much more complicated.) Ks ms ys t + K s ys t c c () m y (t ) Ks () y (t ) s yc t c () y (t ) 0 =0 =0 0 Let y t = ys t () () yc t () 0 . Then ms ys t + K s y t = 0 c c () m y (t ) () Ks () y (t ) = 0 () Adding equations, ms ys t + mc yc t = 0 . Solutions 4-1 M. J. Roberts - 10/7/06 Define the total mass as M = ms + mc and the center of mass as ycm t = () ms ys t + mc yc t ms + mc () ( ) . Then M y cm (t ) = 0 , confirming that the center of mass does () K s y t = 0 by mc and not accelerate. Divide ms ys t + k y t = 0 by ms and mc yc t take the difference of the resulting equations. ys t + K s / ms y t = 0 c s c () () () () ( ) () y (t ) ( K / m ) y (t ) = 0 y ( t ) y ( t ) + K (1 / m + 1 / m ) y ( t ) = 0 s c s s c yt () Therefore y t + Ks () ms + mc y t = 0. ms mc () The eigenvalues are 1,2 = ± Ks ms + mc . In this exercise ms mc 27 ms + mc 22.99 + 35.45 = 1.6604 10 22.99 35.45 ms mc kg = 1.19 10 28 kg . Therefore the eigenvalues are ±3.78 1015 rad/s which corresponds to a resonant frequency of 6 1014 Hz. 2. Write the differential equation for the voltage v C t in the circuit in Figure E-0 for () time t > 0 then find an expression for the current i ( t ) for time t > 0 . R1 = 2 Ω i s(t) C=3F iC (t) -+ i(t) vC(t) t=0 R2 = 6 Ω Vs = 10 V Figure E-0 i t = i s t + iC t () () () , is t = () Vs R1 , iC t = C () d vt dt C ( ( )) Solutions 4-2 M. J. Roberts - 10/7/06 v C t + iC t R2 = 0 v C t = Ke () () t , v C t + R2C , () d v t =0 dt C v C t = Ke ( ( )) () () , t R2 C = 1 / R2C =K K = 10 t / R2 C v C 0 = 10 = Ke () , t / R2 C v C t = 10e = 10 e R2 t 18 t / R2 C () t / R2 C iC t = C () d 10 vC t = Ce dt R2C Vs t / R2 C ( ( )) 10 it = + e R1 R2 3. () 5 = 5+ e 3 The water tank in Figure E-3 is filled by an inflow x t and is emptied by an outflow y t . The outflow is controlled by a valve which offers resistance R to the flow of water out of the tank. The water depth in the tank is d t and the surface area of the water is A, independent of depth (cylindrical tank). The outflow is related to the water depth (head) by dt yt = . R () () () () () The tank is 1.5 m high with a 1m diameter and the valve resistance is 10 s . m2 (a) Write the differential equation for the water depth in terms of the tank dimensions and valve resistance. (b) If the inflow is 0.05m 3 /s , at what water depth will the inflow and outflow rates be equal, making the water depth constant? (c) Find an expression for the depth of water versus time after 1 m 3 of water is dumped into an empty tank. m3 (d) If the tank is initially empty at time, t = 0 , and the inflow is a constant 0.2 s after time, t = 0 , at what time will the tank start to overflow? Solutions 4-3 M. J. Roberts - 10/7/06 Inflow, x(t) Surface area, A d(t) R Valve Outflow, y(t) Figure E-3 Water tank with inflow and outflow (a) yt = () dt R () d Ad t dt () =x t ( ) y (t ) dt R volume Ad t = x t () () () () Ad t + (b) dt ( )=x t () R For the water height to be constant, d t = 0 . Then d t = R x t = 10 () () () s m2 0.05 m3 = 0.5m s (c) Dumping 1 m 3 of water into an empty tank is exciting this system with a unit impulse of water inflow. The impulse response, h t , of the system is the solution of Ah t + The solution is () () ht R ( )= t / AR (t ) h t = Ke () ut . () Solutions 4-4 M. J. Roberts - 10/7/06 We can find K by finding the initial water depth in response to 1 m 3 being suddenly dumped in. The surface area is 0.7854 m 2 . Therefore the initial depth is 1.273 m and h t = 1.273e (d) () t / AR u t = 1.273e () t / 7.854 ut () . The response to a step of flow is the convolution of the impulse response with the step excitation. d t = h t x t = 1.273e t / AR u t 0.2 u t () () () / AR () () d t = 0.2546 e () u ( ) u (t ) d = 0.2546 e 0 / AR ut ( )d For t < 0 , d t = 0 For t > 0 , d t = 0.2546 e 0 () t () / AR d = 0.2546 AR e d t =2 1 e / AR t 0 = 0.2546 AR 1 e ( t / AR ) For all time, () ( t / 7 /854 ) u (t ) . Solving for a depth of 1.5 m, 1.5 = 2 1 e 4. ( t / 7.854 ) 0.25 = e t / 7.854 1.386 = t / 7.854 t = 10.886 s As derived in the text, a simple pendulum is approximately described for small angles, , by the differential equation, mL (t ) + m g (t ) xt () where m is the mass of the pendulum, L is the length of the massless rigid rod supporting the mass and is the angular deviation of the pendulum from vertical. (a) Find the general form of the impulse response of this system. mL h t + m gh t The form of the homogeneous solution is h () ej ( ) (t ) j g/ Lt (t ) = ( K K h1 cos g/ Lt h1 + K h 2e ) u (t ) or, more conveniently, h (t ) = ( g/ Lt + K h 2 sin ) ( g/ Lt u t . ) () Solutions 4-5 M. J. Roberts - 10/7/06 There can be no discontinuity or impulse in the impulse response therefore this is also the impulse response. Integrate the differential equation once through zero. mL ( g/ L K h 2 ) 1 Kh 2 = L / g / mL = 1 / g L / m Now integrate again through zero. K h1 = 0 . Therefore ht = () 1 1 / g L sin m ( g/ Lt u t ) () (b) If the mass is 2 kg and the rod length is 0.5 m, at what cyclic frequency will the pendulum oscillate? The cyclic frequency is 5. 1 2 9.8 = 0.704 Hz . The mass is irrelevant. 1/ 2 A block of aluminum is heated to a temperature of 100 °C. It is then dropped into a flowing stream of water which is held at a constant temperature of 10°C. After 10 seconds the temperature of the ball is 60°C. (Aluminum is such a good heat conductor that its temperature is essentially uniform throughout its volume during the cooling process.) The rate of cooling is proportional to the temperature difference between the ball and the water. (a) Write a differential equation for this system with the temperature of the water as the excitation and the temperature of the block as the response. (b) (c) Compute the time constant of the system. Find the impulse response of the system and, from it, the step response. (d) If the same block is cooled to 0 °C and dropped into a flowing stream of water at 80 °C, at time, t = 0 , at what time will the temperature of the block reach 75°C? The controlling differential equation is d T t = K Tw Ta t dt a or 1d T t + Ta t = Tw K dt a where Ta is the temperature of the aluminum ball and Tw is the temperature of the water. The solution is Solutions 4-6 () ( ( )) () () M. J. Roberts - 10/7/06 Ta t = 90e () Kt + 10 We can find the constant, K, by using the temperature after 10 seconds, Ta 10 = 60 = 90e () 10 K + 10 K = 0.0588 . The impulse response is the solution of the equation, 1d h t +h t = K dt The form of the solution is h t = Khe ( ) ( ) (t ) . Kt () ut () Kh = K 0.0588 t Integrating both sides of the differential equation through t = 0 , K 1 h 0+ = 1 = h K K Therefore h t = Ke () Kt () u t = 0.0588e () ut () and the time constant is 1 / 0.0588 = 17 s . The unit step response is the integral of the impulse response, h 1 (t ) = (1 e 0.0588 t ) u (t ) . ) u (t ) . The response is the response to a step of 80 °C . Ta t = 80 h () 1 (t ) = 80 (1 e 0.0588 t To find the time at which the temperature is 75 °C, t75 , solve Ta t75 = 75 = 80 1 e Solving, t75 = 47.153 . Properties of CT Systems 6. Show that a system with excitation x t and response y t described by y t =u x t () ( 0.0588 t75 ) u (t ) . 75 () () ( ) ( ( )) Solutions 4-7 M. J. Roberts - 10/7/06 is non-linear, time invariant, stable and non-invertible. Homogeneity: Let x1 t = g t . Then y1 t = u g t . 2 2 () () ( ) ( ( )) Let x ( t ) = K g ( t ) . Then y ( t ) = u ( K g ( t )) K y1 t = K u g t . () ( ( )) Not homogeneous () () ( ) ( ( )) Let x ( t ) = h ( t ) . Then y ( t ) = u ( h ( t )) . Let x ( t ) = g ( t ) + h ( t ) . Then y ( t ) = u ( g ( t ) + h ( t )) y ( t ) + y ( t ) = u ( g ( t )) + u ( h ( t )) 2 2 3 3 1 2 Additivity: Let x1 t = g t . Then y1 t = u g t . Not additive Since it is not homogeneous and not additive, it is not linear. It is also not incrementally linear because incremental changes in the excitation do not produce proportional incremental changes in the response. It is statically non-linear because it is non-linear without memory (lack of memory proven below). () () ( ) ( ( )) Let x ( t ) = g ( t t ) . Then y ( t ) = u ( g ( t t )) = y ( t t ) . 2 0 2 0 1 0 Time Invariance: Let x1 t = g t . Then y1 t = u g t . Time Invariant Stability: The unit step function can only have the values, zero or one, therefore any bounded (or unbounded) excitation produces a bounded response. Stable Causality: The response at any time, t = t0 , depends only on the excitation at time, t = t0 and not on any future values. Causal Memory: The response at any time, t = t0 , depends only on the excitation at time, t = t0 and not on any past values. System has no memory. Solutions 4-8 M. J. Roberts - 10/7/06 Invertibility: There are many value of the excitation that all cause a response of zero and there are many values of the excitation that all cause a response of one. Therefore the system is not invertible. 7. Show that a system with excitation x t and response y t described by y t =x t 5 () () () ( ) x (3 t ) is linear but not causal and not invertible. Homogeneity: Let x1 t = g t . Then y1 t = g t 5 2 2 () () ( ) ( ) g (3 t ) . Let x ( t ) = K g ( t ) . Then y ( t ) = K g ( t 5) K g ( 3 t ) = K y ( t ) . 1 Homogeneous () () ( ) ( ) g (3 t ) . Let x ( t ) = h ( t ) . Then y ( t ) = h ( t 5) h ( 3 t ) . Let x ( t ) = g ( t ) + h ( t ) . Then y ( t ) = g ( t 5) + h ( t 5) g ( 3 t ) h ( 3 t ) = y ( t ) + y ( t ) 2 2 3 3 1 2 Additivity: Let x1 t = g t . Then y1 t = g t 5 Additive Since it is both homogeneous and additive, it is also linear. It is also incrementally linear, since any linear system is incrementally linear. It is not statically non-linear because it is linear. Time Invariance: Let x1 t = g t . Then y1 t = g t 5 2 0 2 () () ( ) ( ) g (3 t ) . Let x ( t ) = g ( t t ) . Then y ( t ) = g ( t t 5) g ( 3 0 t + t0 = y1 t t0 . ) ( ) Time Invariant Stability: If x t i s bounded then x t 5 and x 3 t y t =x t 5 Stable () () ( ) x (3 t ) . () ( ) ( ) a re bounded and so is Causality: At time, t = 0 , y 0 = x 5 x 3 . Therefore the response at time, t = 0 , depends on the excitation at a later time, t = 3 . Not Causal ( ) () Solutions 4-9 M. J. Roberts - 10/7/06 Memory: At time, t = 0 , y 0 = x 5 x 3 . Therefore the response at time, t = 0 , depends on the excitation at a previous time, t = 5 . System has memory. () ( ) () Invertibility: A counterexample will demonstrate that the system is not invertible. Let the excitation be a constant, K . Then the response is y t = K K = 0 . This is the response, no matter what K is. Therefore when the output is a constant zero, the input cannot be determined. Not Invertible. () 8. Show that a system with excitation x t and response y t described by y t =x t/2 is linear, time variant and non-causal. () () () () () () () ( ) Let x ( t ) = K g ( t ) . Then y ( t ) = K g ( t / 2 ) = K y ( t ) . 2 2 1 Homogeneity: Let x1 t = g t . Then y1 t = g t / 2 . Homogeneous Additivity: Let x1 t = g t . Then y1 t = g t / 2 . 2 2 () () () ( ) Let x ( t ) = h ( t ) . Then y ( t ) = h ( t / 2 ) . Let x ( t ) = g ( t ) + h ( t ) . Then y ( t ) = g ( t / 2 ) + h ( t / 2 ) = y ( t ) + y ( t ) 3 3 1 2 Additive Since it is both homogeneous and additive, it is also linear. It is also incrementally linear, since any linear system is incrementally linear. It is not statically non-linear because it is linear. Time Invariance: Let x1 t = g t . Then y1 t = g t / 2 . () () () ( () ( ) Let x 2 t = g t t0 . Then y2 t = g t / 2 t0 Time Variant Stability: ) () ( ) y1 t t0 = g ( ) t t0 2 . Solutions 4-10 M. J. Roberts - 10/7/06 If x t is bounded then y t is bounded. Stable Causality: At time, t = 2 , y 2 = x 1 . Therefore the response at time, t = 2 , depends on the excitation at a later time, t = 1 . Not Causal () () () () Memory: At time, t = 2 , y 2 = x 1 . Therefore the response at time, t = 2 , depends on the excitation at a previous time, t = 1 . System has memory. () () Invertibility: The system excitation at any arbitrary time, t = t0 , is uniquely determined by the system response at time, t = 2t0 . Invertible. 9. Show that a system with excitation x t and response y t described by y t = cos 2 t x t () () () ( ) () is time variant, BIBO stable, static and non-invertible. Homogeneity: Let x1 t = g t . Then y1 t = cos 2 t g t . 2 2 () () () ( ) () Let x ( t ) = K g ( t ) . Then y ( t ) = cos ( 2 t ) K g ( t ) = K y ( t ) . 1 Homogeneous () () () ( ) () Let x ( t ) = h ( t ) . Then y ( t ) = cos ( 2 t ) h ( t ) . Let x ( t ) = g ( t ) + h ( t ) . Then y ( t ) = cos ( 2 t ) g ( t ) + h ( t ) 2 2 3 3 Additivity: Let x1 t = g t . Then y1 t = cos 2 t g t . = y1 t + y2 t () () Additive Since it is both homogeneous and additive, it is also linear. It is also incrementally linear, since any linear system is incrementally linear. It is not statically non-linear because it is linear. Time Invariance: Let x1 t = g t . Then y1 t = cos 2 t g t . () () () ( ) () Solutions 4-11 M. J. Roberts - 10/7/06 () ( ) Then y ( t ) = cos ( 2 t ) g ( t Let x 2 t = g t t0 . 2 t0 ) y1 t t0 = cos 2 ( ) ( ( t t )) g ( t t ) . 0 0 Time Variant Stability: If x t is bounded then y t is bounded because it is multiplied by a cosine which is bounded. Stable () () Causality: The response at any time, t = t0 , depends only on the excitation at that same time and not on the excitation at any later time. Causal Memory: The response at any time, t = t0 , depends only on the excitation at that same time and not on the excitation at any earlier time. System has no memory (static). Invertibility: This system is not invertible because when the cosine function is zero the unique relationship between x and y is lost; any x produces the same y, zero. Not Invertible. 10. Show that a system whose response is the magnitude of its excitation is non-linear, BIBO stable, causal and non-invertible. yt =xt Homogeneity: Let x1 t = g t . Then y1 t = g t . 2 2 () () K y1 t . () () () () Let x ( t ) = K g ( t ) . Then y ( t ) = K g ( t ) = If K is negative, K y ( t ) K y ( t ) . 1 1 () Not Homogeneous. () () () () Let x ( t ) = h ( t ) . Then y ( t ) = h ( t ) . Let x ( t ) = g ( t ) + h ( t ) . Then y ( t ) = g ( t ) + h ( t ) 2 2 3 3 Additivity: Let x1 t = g t . Then y1 t = g t . y1 t + y2 t = g t + h t () () () () Not Additive Since it is neither homogeneous nor additive, it is also nonlinear. Solutions 4-12 M. J. Roberts - 10/7/06 It is not incrementally linear because it is not linear except for the addition of a constant. It is statically non-linear because it is non-linear without memory. (No memory is proven below.) Time Invariance: Let x1 t = g t . Then y1 t = g t . 2 0 2 () () () () Let x ( t ) = g ( t t ) . Then y ( t ) = g ( t Time Invariant t0 = y1 t t0 . ) ( ) Stability: If x t is bounded then y t is bounded. Stable () () Causality: The response at any time, t = t0 , depends only on the excitation at that same time and not on the excitation at any later time. Causal Memory: The response at any time, t = t0 , depends only on the excitation at that same time and not on the excitation at any earlier time. System has no memory (static). Invertibility: Any response, y, can be caused by either x or –x. Not Invertible. 11. Show that the system in Figure E-11 is linear, time invariant, BIBO unstable, and dynamic. 0.1 x(t) + - + + + + 14 ∫ ∫ ∫ -7 y(t) Figure E-11 A CT system The 10 y 25 () differential t 14 y t + 7 y t () () equation of 25 y t = x t . () () the system is Solutions 4-13 M. J. Roberts - 10/7/06 Homogeneity: Let x1 t = g t . Then 10 y1 t 2 2 () () ( ) 14 y (t ) + 7 y (t ) 25 y (t ) = g (t ) . Let x ( t ) = K g ( t ) . Then 10 y ( t ) 14 y ( t ) + 7 y ( t ) 25 y ( t ) = K g ( t ) . 1 1 1 2 2 2 If we multiply the first equation by K, we get 10 K y1 t 14 K y1 t + 7 K y1 t Therefore () () () 25 K y1 t = K g t () () () 25 y2 t 10 K y1 t () 14 K y1 t + 7 K y1 t () () 25 K y1 t = 10 y2 t () () 14 y2 t + 7 y2 t () () This can only be true for all time for an arbitrary excitation if y2 t = K y1 t . Homogeneous () () () () ( ) 14 y (t ) + 7 y (t ) 25 y (t ) = g (t ) . Let x ( t ) = h ( t ) . Then 10 y ( t ) 14 y ( t ) + 7 y ( t ) 25 y ( t ) = h ( t ) . Let x ( t ) = g ( t ) + h ( t ) . Then 10 y ( t ) 14 y ( t ) + 7 y ( t ) 25 y ( t ) = g ( t ) + h ( t ) 1 1 1 2 2 2 2 2 3 3 3 3 3 Additivity: Let x1 t = g t . Then 10 y1 t Adding the first two equations, 10 y1 t + y2 t Therefore () () () (t ) 14 y1 t + y2 t () () + 7 y1 t + y2 t () () 25 y1 t | + y2 t () () =g t +h t () () 10 y1 t + y2 t = 10 y3 10 y1 t | + y2 t 3 () 14 y1 t + y2 t 14 y3 3 ( ) + 7 y (t ) + y (t ) (t ) + 7 y (t ) 25 y (t ) 1 2 3 () 25 y1 t | + y2 t () () () () () = 10 y ( t ) 14 y1 t | + y2 t 14 y3 t + 7 y3 t () () () () + 7 y1 t | + y2 t 25 y3 t () () () 25 y1 t | + y2 t () This can only be true for all time for an arbitrary excitation if y3 t = y1 t + y2 t . Additive Since it is homogeneous and additive, it is also linear. It is also incrementally linear. It is not statically non-linear because it is linear. Time Invariance: Solutions 4-14 () () () M. J. Roberts - 10/7/06 () () ( ) 14 y (t ) + 7 y (t ) 25 y (t ) = g (t ) . Let x ( t ) = g ( t t ) . Then 10 y ( t ) 14 y ( t ) + 7 y ( t ) 25 y ( t ) = g ( t t ) . Let x1 t = g t . Then 10 y1 t 2 0 1 1 1 2 2 2 2 0 The first equation can be written as 10 y1 t t0 Therefore ( ) 14 y1 t t0 + 7 y1 t t0 ( ) ( ) 25 y1 t t0 = g t t0 ( )( t0 ) 10 y1 t t0 ( ) ( ) 25 y (t = 10 y ( t ) 14 y ( t ) + 7 y ( t ) 25 y ( t ) 14 y1 t t0 + 7 y1 t t0 2 2 1 2 2 ) ( ) This can only be true for all time if y2 t = y1 t t0 . Time Invariant Stability: The characteristic equation is eigenvalues are 1 2 3 () ( ) 10 3 14 2 +7 25 = 0 . The =1.7895 =-0.1948 + j1.1658 =-0.1948 - j1.1658 So the homogeneous solution is of the form, y t = K1e1.7895t + K 2 e( () -0.1948 + j1.1658 t ) + K 3e( -0.1948 - j1.1658 t ) . If there is no excitation, but the zero-excitation response is not zero, the response will grow without bound as time increases. Unstable Causality: The system equation can be rewritten as t 3 2 1 yt = 10 () x () 1 t t 3 2 d 1d 2 d 2 + 25 3 t y ( )d 1 1 d 2d 3 7 y () 1 d 1d + 14 y 2 ( )d 1 1 Solutions 4-15 M. J. Roberts - 10/7/06 So the response at any time, t = t0 , depends on the excitation at times, t < t0 and not on any future values. Causal Memory: The response at any time, t = t0 , depends on the excitation at times, t < t0 . System has memory. Invertibility: The system equation, 10 y (t ) 14 y t + 7 y t () () 25 y t = x t () () expresses the excitation in terms of the response and its derivatives. Therefore the excitation is uniquely determined by the response. Invertible. System Models 12. The suspension of a car can be modeled by the mass-spring-dashpot system of Figure E-12 Let the mass, m, of the car be 1500 kg, let the spring constant, K s , be 75000 N/m and let the shock absorber (dashpot) viscosity coefficient, K d , be 20000 N s/m . At a certain length, d0 , of the spring, it is unstretched and uncompressed and exerts no force. Let that length be 0.6 m. (a) (b) What is the distance, y t ( ) x (t ) , when the car is at rest? ( ) ( ) x (t ) constant such that, when the Define a new variable z t = y t system is at rest, z t = 0 and write a describing equation in z and x which describes an LTI system. Then find the impulse response. (c) The effect of the car striking a curb can be modeled by letting the road surface height change discontinuously by the height of the curb, h c . Let h c = 0.15 m . Graph z t versus time after the car strikes a curb. () () Solutions 4-16 M. J. Roberts - 10/7/06 Automobile Chassis Spring Shock Absorber y(t) x(t) Figure E-12 Car suspension model Using the basic principle, F = ma , we can write Ks y t or ( ) x (t ) d0 + K d d yt dt ( ) x (t ) + mg = m y t () m y t + K d y t + K s y t = K d x t + K s x t + K s d0 At rest all the derivatives are zero and Ks y t Solving, yt () () () () () mg . ( ( ) x (t ) mg = d0 + m g = 0 . ) ( ) x (t ) = K s d0 Ks 75000 0.6 1500 9.8 = 0.404 m 75000 The describing equation is m y t + K d y t + K s y t = K d x t + K s x t + K s d0 which can be rewritten as m y t + Kd y t or m y t + Kd y t Let z t = y t () () () () () mg . () () xt xt () + Ks y t ( ) x (t ) K s d0 + m g = 0 () () () + Ks y t ( ) x (t ) d0 + m g/ K s = 0 ( ) ( ) x (t ) d0 + m g/ K s . Then y t = z t + x t and () () () m z t +x t or () () + Kd z t + Ks z t = 0 () () m z t + Kd z t + Ks z t = m x t Solutions 4-17 () () () () M. J. Roberts - 10/7/06 This equation is in a form which describes an LTI system. We can find its impulse response. After time, t = 0 , the impulse response is the homogenous solution. The eigenvalues are = 2 Kd ± Kd 4 mK s 1,2 2m = Kd 2m ± 2 Kd Ks m 4m 2 = 6.667 ± j 2.357 . The homogeneous solution is h t = K h1e () 1t + K h 2e 2t = K h1e( 6.667 + j 2.357 t ) + K h 2 e( 6.667 j 2.357 t ) . Since the system is underdamped another (equivalent) form of homogeneous solution will be more convenient, h t = e 6.667 t K h1 cos 2.357t + K h 2 sin 2.357t . () ( ) ( ) The impulse response can have a discontinuity at t = 0 and an impulse but no higher-order singularity there. Therefore the general form of the impulse response is h t =K () (t ) + e + 6.667 t K h1 cos 2.357t + K h 2 sin 2.357t u t 0+ ( ) ( ) () Integrating both sides of the describing equation, mh 0 ( ( ) h (0 )) + K ( h (0 ) h (0 )) + K h (t ) dt = 0 . + d s 0 0+ (The integral of the doublet, which is the derivative of the impulse excitation, is zero.) Since the impulse response and all its derivatives are zero before time, t = 0 , it follows then that m h 0 + Kd h 0 and () + ( ) + K h (t ) dt = 0 + s 0 m 6.667 K h1 + 2.357 K h 2 + K d K h1 + K s K = 0 . Integrating the describing equation a second time, mh 0 or mK h1 + K d K = 0 . Integrating the describing equation a third time, ( ) ( )+ K + 0+ d 0 h t dt = 0 () Solutions 4-18 M. J. Roberts - 10/7/06 0+ m h t dt = m 0 () or mK = m K= 1. Solving for the other two constants, K h1 = K d / m and m or Kh 2 = Kd m Therefore ht = Ks / m 2 K d / m2 + 6.667 K d / m 2.357 6.667 Kd m + 2.357 K h 2 + K d Kd m Ks = 0 () (t ) + e 6.667 t 13.333cos 2.357t ( ) 16.497 sin 2.357t u t ( ) () The response to a step of size 0.15 is then the convolution, z t = 0.15u t or z t = 0.15 () ( ) h (t ) () () { ( )+ e { ( )+ e () 6.667 13.333cos 2.357 ( ) ) 16.497 sin 2.357 ( ) u ( )} u (t ) d ) } u (t ) d z t = 0.15 0 6.667 13.333cos 2.357 ( 16.497 sin 2.357 ( For t < 0 , z t = 0 . For t > 0 , using ea x sin b x d x = () ea x a sin b x a 2 + b2 () () b cos b x () () ea x e cos b x d x = 2 a cos b x + b sin b x a + b2 ax () Solutions 4-19 M. J. Roberts - 10/7/06 we get z t = 0.15u t + 0.15 () () 13.333 e 6.667 50 e 6.667 16.497 50 6.667 t 6.667 cos 2.357 6.667 sin 2.357 ( ) + 2.357 sin ( 2.357 ) ) 2.357 cos 2.357 t ( ( ) 0 or 13.333 z t = 0.15u t + 0.15 6.667 cos 2.357t + 2.357 sin 2.357t 50 e 6.667 t 6.667 sin 2.357t 2.357 cos 2.357t 16.497 50 2.357 6.667 + 16.497 13.333 50 50 e ( ) ( ) () () ( ) ( ) z t = 0.15u t + 0.15 e or z t = 0.15e z(t) 0.1 () () { 3.333t 2.812sin 2.357t ( ) cos 2.357t ( ) +1 u t } () () 3.333t 2.812sin 2.357t ( ) cos 2.357t u t ( ) () t 2 -0.2 13. Pharmacokinetics is the study of how drugs are absorbed into, distributed through, metabolized by and excreted from the human body. Some drug processes can be approximately modeled by a “one compartment” model of the body in which V is the volume of the compartment, C t is the drug concentration in that compartment, () ke is a rate constant for excretion of the drug from the compartment and k0 is the infusion rate at which the drug enters the compartment. (a) Write a differential equation in which the infusion rate is the excitation and the drug concentration is the response. (b) Let the parameter values be ke = 0.4 hr 1 , V = 20 l and k0 = 200mg/hr (where “l” is the symbol for “liter”). If the initial drug concentration is C 0 = 10mg/l , plot the drug concentration as a function of time (in hours) for the first 10 hours of infusion. Find the solution as the sum of the zero-excitation response and the zero-state response. () The differential equation is Solutions 4-20 M. J. Roberts - 10/7/06 V or d C t = k0 Vke C t dt k d C t + ke C t = 0 V dt ( ( )) () ( ( )) () The eigenvalue is –0.4 and the zero-excitation response is C t = 10e () 0.4 t mg/l (“t” in hours). The impulse response (to a unit impulse of “infusion rate”) is e 0.4t ht = u t mg/l . The step response to an infusion rate of k0 = 200mg/hr is 20 then C t = 25 1 e 0.4t mg/l . The sum of the two responses is () () () C t = 10e C (t) 25 () ( ( ) 0.4 t + 25 1 e ( 0.4 t )) mg/l = ( 25 15e 0.4 t ) mg/l . 10 t (hours) 14. A well-stirred vat has been fed for a long time by two streams of liquid, fresh water at 0.2 cubic meters per second and concentrated blue dye at 0.1 cubic meters per second. The vat contains 10 cubic meters of this mixture and the mixture is being drawn from the vat at a rate of 0.3 cubic meters per second to maintain a constant volume. The blue dye is suddenly changed to red dye at the same flow rate. At what time after the switch does the mixture drawn from the vat contain a ratio of red to blue dye of 99:1? Let the concentration of red dye be denoted by Cr t and the concentration of blue dye be denoted by Cb t . The concentration of water is constant throughout at 2 / 3 . The rates of change of the dye concentrations are governed by d VCb t = Cb t f draw dt d VCr t = f r dt () () ( ( )) () ( ( )) Cr t f draw () where V is the constant volume, 10 cubic meters, f draw is the flow rate of the draw from the vat and f r is the flow rate of red dye into the tank. Solving the two differential equations, Solutions 4-21 M. J. Roberts - 10/7/06 f /V t Cb t = 1 / 3 e ( draw ) () ( ) () and Cr t = 1 f /V t 1 e ( draw ) . 3 ( ) Then the ratio of red to blue dye concentration is ( ) = 3 (1 e 1( C (t ) e 1 Cr t b ( fdraw / V )t f draw / V t ) )=1 f /V t e ( draw ) f /V t = e( draw ) ( fdraw / V )t e 1. 3 Setting that ratio to 99 and solving for t99 , 99 = e( 15. 0.3/10 t99 ) 1 t99 = 153.5 seconds Some large auditoriums have a noticeable echo or reverberation. While a little reverberation is desirable, too much is undesirable. Let the response of an auditorium to an acoustic impulse of sound be ht = n=0 () e n (t n/5 . ) We would like to design a signal processing system that will remove the effects of reverberation. In later chapters on transform theory we will be able to show that the compensating system that can remove the reverberations has an impulse response of the form, hc t = Find the function, g n . Removal of the reverberation is equivalent to making the overall impulse response, h 0 t , an impulse. That means that () gn n=0 (t n/5 . ) () ho t = h t () () hc t = () e n=0 n (t ) n/5 ) gm m= 0 (t m/5 = K ) (t ) e n= 0 m= 0 n (t n/5 gm (t m/5 = K ) (t ) Solutions 4-22 M. J. Roberts - 10/7/06 e ng m n= 0 m= 0 t n+m =K 5 n+m =K 5 (t ) (t ) gm m= 0 n=0 e n t g0 +g 1 +g 2 (t ) + e (t 1 / 5) + e (t 2 / 5) + (t 1 / 5) + e (t 2 / 5) + e (t 3 / 5) + (t 2 / 5) + e (t 3 / 5) + e (t 4 / 5) + 1 2 1 2 1 2 =K (t ) g 0 =K g 1 + e 1g 0 = 0 g 1 = Ke 1 2 g 2 + e 1g 1 + e 2 g 0 = 0 g 2 = Ke Ke 2 = 0 3 g 3 + e 1g 2 + e 2g 1 + e 3g 0 = 0 g 3 = Ke Ke 3 = 0 So the compensating impulse response is h c t = K t Ke and the function, g, is () () n 1 (t 1/ 5 ) . g n =K 16. Ke 1 n1 A car rolling on a hill can be modeled as shown in Figure E-16. The excitation is the force f t for which a positive value represents accelerating the car forward with the motor and a negative value represents slowing the car by braking action. As it rolls, the car experiences drag due to various frictional phenomena which can be approximately modeled by a coefficient k f which multiplies the car’s velocity to produce a force which tends to slow the car when it moves in either direction. The mass of the car is m and gravity acts on it at all times tending to make it roll down the hill in the absence of other forces. Let the mass m of the car be 1000 kg, let the Ns and let the angle be . friction coefficient k f be 5 m 12 () (a) Write a differential equation for this system with the force f t excitation and the position of the car, y t as the response. () () as the Solutions 4-23 M. J. Roberts - 10/7/06 (b) If the nose of the car is initially at position y 0 = 0 with an initial () m and no applied acceleration or braking force, graph the t =0 s velocity of the car y t for positive time. velocity, y t = 10 () () (c) If a constant force f t of 200 N is applied to the car what is its terminal velocity ? () f(t) y(t) mg sin (θ) θ Figure E-16 Car on an inclined plane Summing forces, ft or () m g sin () kf y t = my t () () () m y t + k f y t + m g sin () () ( ) = f (t ) () The zero-excitation response can be found by setting the force, f t , to zero yielding m y t + k f y t = m g sin () () The homogeneous solution is y h t = K h1 + K h 2 e m . The particular solution must be in the form of a linear function of t, to satisfy the differential equation. Choosing the form, () kf t and solving, K p = ( m g/ k ) sin ( ) . Then the total zero-excitation response is f y p t = K pt () y t = K h1 + K h 2 e m m g/ k f sin Using the initial conditions, y 0 = 0 = K h1 + K h 2 and () kf t ( ) ( )t () Solutions 4-24 M. J. Roberts - 10/7/06 y 0 = 10 = Solving, Kh 2 = 2 () (k f f / m Kh 2 ) ( m g/ k ) sin ( ) . f ( m / k ) g sin ( ) ( m / k )10 = f 1.0146 105 2000 = 1.0346 105 and K h1 = 1.0346 105 y t = 1.0346 105 1 e 1.0346 105 e yt= 200 () ( t / 200 ) 507.28t 507.28 = 517.28 e () ( t / 200 ) 507.28 = 517.28e t / 200 ( t / 200 1 + 10 ) y’(t) 1000 t -550 The differential equation is m y t + k f y t + m g sin We can re-write the equation as my t + kf y t = f t () () ( ) = f (t ) m g sin () () () () treating the force due to gravity as part of the excitation. Then the impulse response is the solution of mh t + kf h t = t () () () kf m t which is of the form, h t = K h1 + K h 2 e () ut () . Integrating both sides of the differential equation through t = 0 we get m h 0+ + k f h 0+ = 1 = m Integrating a second time yields, m h 0+ = 0 = m K h1 + K h 2 Solving, Solutions 4-25 () () ( ( k / m) K ) + k ( K f h2 f h1 + Kh 2 ) () ( ) . M. J. Roberts - 10/7/06 kf 1 we get 0 1 K h1 Kh 2 = 1 0 K h1 = 1 / k f , K h 2 = 1 / k f . So the impulse response is 1e ht = kf () ( k / m) t f ut () . Now, if we say that the force, f t , is a step of size, 200 N, the excitation of the system is x t = 200 u t () () () m g sin () . But this is going to cause a problem. The problem is that the term, m g sin , is a constant, therefore presumed to have acted on the system for all time before time, t = 0 . The implication from that is that the position at time, t = 0 , is at infinity. Since we are only interested in the final velocity, not position, we can assume that the car was held in place at y t = 0 until the force was applied and gravity was allowed to act on the car. That makes the excitation, x t = 200 m g sin ut and the response is () () () ( ) () y t =x t or () ( ) h (t ) = yt = 200 m g sin ( ) u (t ) 1e f 1e kf ( k / m) t f ut () () 200 m g sin kf () t t ( k / m) d 0 or yt = () 200 m g sin kf () m ( k / m) + ef kf = 0 200 m g sin kf () t+ m ( k f / m) t e kf m kf The terminal velocity is the derivative of position as time approaches infinity which, in this case is 200 m g sin 200 2536.43 = = 467.3 m/s . y+ = 5 kf () () Obviously a force of 200 N is insufficient to move the car forward and its terminal velocity is negative indicating it is rolling backward down the hill. Solutions 4-26 M. J. Roberts - 10/7/06 System Properties 17. A CT system is described by the block diagram in Figure E-17. x(t) 0.25 ++ + 1 ∫ ∫ 3 y(t) Figure E-17 A CT system Classify the system as to homogeneity, additivity, linearity, time-invariance, stability, causality, memory, and invertibility. 4 y t + y t + 3y t = x t () () () () () () () () () () Let x ( t ) = K g ( t ) . Then 4 y ( t ) + y ( t ) + 3y ( t ) = K g ( t ) . 2 2 2 2 Homogeneity: Let x1 t = g t . Then 4 y1 t + y1 t + 3y1 t = g t . If we multiply the first equation by K, we get 4 K y1 t + K y1 t + 3K y1 t = K g t Therefore 4 K y1 t + K y1 t + 3K y1 t = 4 y2 t + y2 t + 3y2 t This can only be true for all time if y2 t = K y1 t . Homogeneous () () () () () () () () () () () () () () () () () () Let x ( t ) = h ( t ) . Then 4 y ( t ) + y ( t ) + 3y ( t ) = h ( t ) . Let x ( t ) = g ( t ) + h ( t ) . Then 4 y ( t ) + y ( t ) + 3y ( t ) = g ( t ) + h ( t ) 2 2 2 2 3 3 3 3 Additivity: Let x1 t = g t . Then 4 y1 t + y1 t + 3y1 t = g t . Adding the first two equations, 4 y1 t + y2 t Therefore () () + y1 t + y2 t () () + 3 y1 t + y2 t () () =g t +h t () () Solutions 4-27 M. J. Roberts - 10/7/06 4 y3 t + y3 t + 3y3 t = 4 y1 t + y2 t 4 y3 t + y3 t + 3y3 t = 4 y1 t + y2 t () () () () () + y1 t + y2 t () () + 3 y1 t + y2 t () () () () () () () + y1 t + y2 t () () + 3 y1 t + y2 t () () This can only be true for all time if y3 t = y1 t + y2 t . Additive Since it is homogeneous and additive, it is also linear. It is also incrementally linear. It is not statically non-linear because it is linear. () () () () () () () () () Let x ( t ) = g ( t t ) . Then 4 y ( t ) + y ( t ) + 3y ( t ) = g ( t t ) . 2 0 2 2 2 0 Time Invariance: Let x1 t = g t . Then 4 y1 t + y1 t + 3y1 t = g t . The first equation can be written as 4 y1 t t0 + y1 t t0 + 3y1 t t0 = g t t0 Therefore ( ) ( ) ( )( ) () 4 y1 t t0 + y1 t t0 + 3y1 t t0 = 4 y2 t + y2 t + 3y2 t This can only be true for all time if y2 t = y1 t t0 . Time Invariant Stability: The eigenvalues are 1 2 ( ) ( ) ( ) () () () ( ) =-0.1250 + j0.4841 =-0.1250 - j0.4841 Therefore the homogeneous solution is of the form, y t = K1e( () -0.1250 + j 0.4841 t ) + K 2 e( -0.1250 - j 0.4841 t ) . If there is no excitation, but the zero-excitation response is not zero, the response will decay to zero as time increases. Since the particular solution has the same form as the excitation plus all its unique derivatives, the response to any bounded input will be a bounded output. Stable Solutions 4-28 M. J. Roberts - 10/7/06 Causality: The system equation can be rewritten as 1 yt = 4 () t 2 x ( )d 1 t d 1 2 y ( )d 1 t 1 2 3 y ( )d 1 1 d 2 So the response at any time, t = t0 , depends on the excitation at times, t < t0 and not on any future values. Causal Memory: The response at any time, t = t0 , depends on the excitation at times, t < t0 . System has memory. Invertibility: The system equation, 4 y t + y t + 3y t = x t () () () () expresses the excitation in terms of the response and its derivatives. Therefore the excitation is uniquely determined by the response. Invertible. 18. A system has a response that is the cube of its excitation. Classify the system as to homogeneity, additivity, linearity, time-invariance, stability, causality, memory, and invertibility. y t = x3 t () () Homogeneity: Let x1 t = g t . Then y1 t = g 3 t . 2 2 () () () () Let x ( t ) = K g ( t ) . Then y ( t ) = K g ( t ) Not homogeneous Additivity: Let x1 t = g t . Then y1 t = g 3 t . 3 2 2 3 = K 3 g3 t () K y1 t = K g 3 t . () () () () () () Let x ( t ) = h ( t ) . Then y ( t ) = h ( t ) . Let x ( t ) = g ( t ) + h ( t ) . 3 Then y3 t = g t + h t Not additive () () () 3 = g 3 t + h 3 t + 3g 2 t h t + 3g t h 2 t () () () () () () y1 t + y2 t () () Solutions 4-29 M. J. Roberts - 10/7/06 Since it is not homogeneous and not additive, it is also linear. It is also not incrementally linear. It is statically non-linear because it is non-linear without memory (lack of memory proven below). () () () Let x ( t ) = g ( t t ) . Then y ( t ) = g ( t t ) = y ( t 2 0 3 2 0 1 Time Invariance: Let x1 t = g t . Then y1 t = g 3 t . () t0 . ) Time Invariant Stability: If x t is bounded then y t = x 3 t is also bounded. Stable () () () Causality: The response at any time, t = t0 , depends only on the excitation at time, t = t0 and not on any future values. Causal Memory: The response at any time, t = t0 , depends only on the excitation at time, t = t0 and not on any past values. System has no memory. Invertibility: Solve y t = x 3 t for x t . x t = y 3 t . The cube root operation is multiple valued. Therefore the system is not invertible, unless we assume that the excitation must be real-valued. In that case, the response does determine the excitation because for any real y there is only one real cube root. 19. A CT system is described by the differential equation, ty t () () () () 1 () () 8y t = x t . () () Classify the system as to linearity, time-invariance and stability. Homogeneity: Let x1 t = g t . Then t y1 t 2 () () ( ) 8 y (t ) = g (t ) . Let x ( t ) = K g ( t ) . Then t y ( t ) 8 y ( t ) = K g ( t ) . 1 2 2 If we multiply the first equation by K, we get Solutions 4-30 M. J. Roberts - 10/7/06 tK y1 t Therefore tK y1 t () 8 K y1 t = K g t () () 8 y2 t () 8 K y1 t = t y2 t () () () This can only be true for all time if y2 t = K y1 t . Homogeneous Additivity: Let x1 t = g t . Then t y1 t 2 2 () () () () ( ) 8 y (t ) = g (t ) . Let x ( t ) = h ( t ) . Then t y ( t ) 8 y ( t ) = h ( t ) . Let x ( t ) = g ( t ) + h ( t ) . Then t y ( t ) 8 y ( t ) = g ( t ) + h ( t ) 1 2 3 3 3 Adding the first two equations, t y1 t + y2 t Therefore t y1 t + y2 t t y1 t + y2 t () () 8 y1 t + y2 t 8 y1 t + y2 t () () =g t +h t () () 8 y3 t () () () () = t y3 t () () () () 8 y1 t + y2 t () () = t y3 t () 8 y3 t () This can only be true for all time if y3 t = y1 t + y2 t . Additive Since it is homogeneous and additive, it is also linear. It is also incrementally linear. It is not statically non-linear because it is linear. () () () () () ( ) 8 y (t ) = g (t ) . Let x ( t ) = g ( t t ) . Then t y ( t ) 8 y ( t ) = g ( t t ) . 1 2 0 2 2 0 Time Invariance: Let x1 t = g t . Then t y1 t The first equation can be written as (t Therefore t0 y1 t t0 )( ) 8 y1 t t0 = g t t0 ( )( ) () (t t0 y1 t t0 )( ) 8 y1 t t0 = t y2 t Solutions 4-31 ( ) () 8 y2 t M. J. Roberts - 10/7/06 This equation is not satisfied if y2 t = y1 t t0 therefore y2 t Time Variant Stability: () ( ) () y1 t t0 . ( ) The homogeneous solution to the differential equation is of the form, t y t = 8y t () () To satisfy this equation the derivative of “y” times “t” must be of the same functional form as “y” itself. This is satisfied by a homogeneous solution of the form, y t = Kt 8 If there is no excitation, but the zero-excitation response is not zero, the response will increase without bound as time increases. Unstable Causality: The system equation can be rewritten as yt = () () t x ( )d t +8 y ( )d So the response at any time, t = t0 , depends on the excitation at times, t < t0 and not on any future values. Causal Memory: The response at any time, t = t0 , depends on the excitation at times, t < t0 . System has memory. Invertibility: The system equation, ty t () 8y t = x t () () expresses the excitation in terms of the response and its derivatives. Therefore the excitation is uniquely determined by the response. Invertible. 20. A CT system is described by the equation, Solutions 4-32 M. J. Roberts - 10/7/06 yt = () t 3 x ( )d . Classify the system as to time-invariance, stability and invertibility. Homogeneity: Let x1 t = g t . Then y1 t = t 3 () () () () g ( )d Kg . Let x 2 t = K g t . Then y2 t = Homogeneous Additivity: Let x1 t = g t . Then y1 t = t 3 () () t 3 ( )d =K g t 3 ( )d = K y1 t . () () () () g t 3 ( )d ( )d t 3 . Let x 2 t = h t . Then y2 t = Let x 3 t = g t + h t . Then y3 t = Additive () () () t 3 () h . () () () g ( ) + h( ) d= g ( )d + t 3 h ( )d = y1 t + y2 t () () Since it is both homogeneous and additive, it is also linear. It is also incrementally linear, since any linear system is incrementally linear. It is not statically non-linear because it is linear. Time Invariance: Let x1 t = g t . Then y1 t = Let x 2 t = g t t0 . Then y2 t = Time Variant Solutions 4-33 t 3 () () () ( () t 3 () g ( )d () . ) g ( t0 d = ) t t 30 t t0 g u du y1 t t0 = ( ) 3 g ( )d . M. J. Roberts - 10/7/06 Stability: If x t is a constant, K, then y t = without bound. Unstable Causality: The response at time, t = 3 , depends partially on the excitation at time t = 1 which is in the future. Not causal Memory: The response at any time, t = t0 , depends partially on excitations in the past, t < t0 . System has memory. Invertibility: Differentiate both sides of y t = it follows that x t = y 3t . Invertible. t 3 () () t 3 Kd = K t 3 d and, as t , y t increases () () x ( )d w.r.t. t yielding y t = x t / 3 . Then () () () () 21. A CT system is described by the equation y t = () t +3 x ( )d . Classify the system as to linearity, causality and invertibility. Homogeneity: Let x1 t = g t . Then y1 t = Let x 2 t = K g t . Then y2 t = Homogeneous Additivity: Let x1 t = g t . Then y1 t = Let x 2 t = h t . Then y2 t = Let x 3 t = g t + h t . Solutions 4-34 () () () () t +3 g ( )d Kg . () () t +3 () t +3 d =K g ( )d = K y1 t . () () () () t +3 g t +3 ( )d ( )d . . () () () h () () () M. J. Roberts - 10/7/06 Then y3 t = Additive () t +3 g () () +h t +3 d= g () t +3 d+ h ( )d = y1 t + y2 t () () Since it is both homogeneous and additive, it is also linear. It is also incrementally linear, since any linear system is incrementally linear. It is not statically non-linear because it is linear. Time Invariance: Let x1 t = g t . Then y1 t = Let x 2 t = g t t0 . Then y2 t = Time Invariant Stability: If x t () () () ( () () t +3 g ( )d () . ) t +3 g ( t0 d = ) t t0 + 3 g u du = y1 t t0 . ( ) () is a constant, K , then y t = () t +3 t +3 Kd = K d and, as t , yt () increases without bound. Unstable Causality: The response at any time, t = t0 , depends partially on the excitation at times t0 < t < t0 + 3 which are in the future. Not causal Memory: The response at any time, t = t0 , depends partially on excitations in the past, t < t0 . System has memory. Invertibility: Differentiate both sides of y t = it follows that x t = y t 3 . Invertible. 22. i s additive but not S how that the system described by y t = Re x t homogeneous. (Remember, if the excitation is multiplied by any complex constant and the system is homogeneous, the response must be multiplied by that same complex constant.) Solutions 4-35 () ) t +3 x ( )d w.r.t. t yielding y t = x t + 3 . Then () ( ) () ( () ( ( )) M. J. Roberts - 10/7/06 Homogeneity: () () () () () Then y ( t ) = Re ( g ( t ) + j h ( t )) = g ( t ) . Let x ( t ) = ( K + jK ) g ( t ) + j h ( t ) , where K 1 2 r i Let x1 t = g t + j h t , where g t and h t are both real-valued functions. r and K i are both real constants. Then y2 t = Re K r + jK i y2 t = Re K r g t () (( ) g (t ) + j h (t ) ) ( )) () Ki h t () ( () K i h t + jK i g t + jK r h t = K r g t () () () If we multiply the first equation by K r + jK i , we get (K Therefore r + jK i y1 t = K r + jK i Re g t + j h t ) () ( ) ( () ( )) (K unless K i = 0 . Not homogeneous Additivity: r + jK i y1 t = K r + jK i g t y2 t ) () ( r ) () ( ) (K + jK i y1 t ) () () () () () () Then y ( t ) = Re ( g ( t ) + j h ( t )) = g ( t ) . Let x ( t ) = g ( t ) + j h ( t ) , where g ( t ) and h ( t ) are both real-valued functions. Then y ( t ) = Re ( g ( t ) + j h ( t )) = g ( t ) . Let x ( t ) = g ( t ) + j h ( t ) + g ( t ) + j h ( t ) . Then y ( t ) = Re ( g ( t ) + j h ( t ) + g ( t ) + j h ( t )) = g ( t ) + g ( t ) = y ( t ) + y ( t ) . Let x1 t = g1 t + j h1 t , where g1 t and h1 t are both real-valued functions. 1 1 1 1 2 2 2 2 2 2 2 2 2 3 1 1 2 2 3 1 1 2 2 1 2 1 2 Additive Solutions 4-36 ...
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