6.003 Homework #7 Solutions

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1 6.003 Homework #7 Solutions Problems. Secon-orer systems The impulse response of a secon-orer CT system has the form h(t) = e σt cos(ω t + φ)u(t) where the parameters σ, ω, an φ are relate to the parameters of the characteristic polynomial for the system: s 2 + Bs + C. a. Determine expressions for σ an ω (not φ) in terms of B an C. Express the impulse response in terms of complex exponentials: h(t) = 2 e σt e jω t+jφ + e jω tjφ u(t) = 2 ejφ e (σ+jω )t u(t)+ 2 e jφ e (σjω )t u(t) The impulse response is a weighte sum of moes of the form e s 0t an e s t where s 0 an s are the poles. Thus the poles of the system are at s = σ ± jω. The characteristic polynomial has the form s 2 + Bs + C = (s + σ + jω )(s + σ jω ) = (s + σ) 2 + ω 2. Thus B = 2σ an C = σ 2 + ω 2. Solving, we fin that σ = B 2 ω = C 4 B2. b. Determine the time require for the envelope e σt of h(t) to iminish by a factor of e, the perio of the oscillations in h(t), an the number of perios of oscillation before h(t) iminishes by a factor of e. Express your results as functions of B an C only. The time to ecay by a factor of e is 2 =. σ B The perio is =. ω C 4 B2 The number of perios before iminishing a factor of e is 2 C B 4 = B2. πb C 4 B2 ω Notice that this last answer is equivalent to Q/π where Q =. 2σ

2 6.003 Homework #7 Solutions / Fall 20 c. Estimate the parameters in part b for a CT system with the following poles: 00 s-plane 2 0 From the plot σ = 0 an ω = 00. The time to ecay by a factor of e is 0.. The perio is = = ω The unit-sample response of a secon-orer DT system has the form n h[n] = r 0 cos(ω 0 n + Φ)u[n] 00 0 The number of cycles before ecaying by e is.6 where the parameters r 0, Ω 0, an Φ are relate to the parameters of the characteristic polynomial for the system: z 2 + Dz + E.. Determine expressions for r 0 an Ω 0 (not Φ) in terms of D an E. Express the unit-sample response in terms of complex exponentials: h[n] = r n 0 2 e jω 0n+jΦ + 2 e jω 0njΦ u[n] = 2 e jφ r n 0 e jω 0n u[n]+ 2 e jφ r n 0 e jω 0n u[n] The poles have the form z = r 0 e jω 0 an z = r 0 e jω 0. The characteristic equation is z 2 + Dz + E = (z r 0 e jω 0 )(z r 0 e jω 0 ) = z 2 2r 0 cos Ω 0 + r0 2. Thus D = 2r 0 cos Ω 0 an E = r0 2. Solving, we fin that r 0 = E D Ω 0 = cos = cos D 2r 0 2 E e. Determine the length of time require for the envelope rn 0 of h[n] to iminish by a factor of e. the perio of the oscillations (i.e., Ω ) in h[n], an 0 the number of perios of oscillation in h[n] before it iminishes by a factor of e. Express your results as functions of D an E only.

3 6.003 Homework #7 Solutions / Fall 20 3 The time to iminish by a factor of e is r0 n = e. Taking the log of both sies yiels n ln r 0 = so that the time is = ln r 0 ln E The perio is Ω 0 which is cos D 2 E The number of perios before the response iminishes by e is ln E = cos D 2 E ln r 0 cos D 2 E f. Estimate the parameters in part e for a DT system with the following poles: z-plane From the plot Ω 0 = tan raians an r 0 = The time to ecay by a factor of e is ln The perio is = Ω The number of cycles before ecaying by e is Matches The following plots show pole-zero iagrams, impulse responses, Boe magnitue plots, an Boe angle plots for six causal CT LTI systems. Determine which correspons to which an fill in the following table. Pole-zero iagram has a single pole at zero. The impulse response of a system with a single pole at zero is a unit step function (3). We evaluate the frequency response by consiering frequencies along the jω axis. As we move away from the pole at the origin the log-magnitue ecays linearly (5). The phase is constant since the angle between the pole an any point along positive sie of the jω axis remains constant at π/2. The angle of the frequency response is therefore π/2 (4). Pole-zero iagram 4 has a single pole at at s =. The impulse response has the form e st u(t) = e t u(t) (2). As we move along the jω axis, we move away from the pole at the origin, an the log-magnitue will eventually ecay linearly. Because the pole is not

4 6.003 Homework #7 Solutions / Fall 20 4 exactly at the origin, this ecay is not significant until ω = (6). The phase starts at 0, an eventually moves to π/2. Note that as we move farther up the jω axis, this system behaves like the system of iagram (2). Pole-zero iagram 3 as a zero at the origin. A zero at the origin correspons to taking the erivative, so we take the impulse response of pole-zero iagram 4 (2) an take its erivative (4). When ω is small, the zero is ominant. As we move away from ω = 0, the effect of the zero iminishes an the log-magnitue increases linearly. For sufficiently large ω we are far enough that the zero an pole appear to cancel each other, an the magnitue becomes a constant (3). A zero at the origin means that we take the phase response of pole-zero iagram 4 (2) an a π/2 to it (6). Pole-zero iagram 2 contains complex conjugate poles H(s) = (s + σ + jω )(s + σ jω ) = ja ja. s + σ + jω s + σ jω The impulse response has the form h(t) e σt (e jω t e jω t ) e σt sin ω t which is response (). The magnitue response will eventually ecay twice as fast as that of pole-zero iagram 4 (6). Since there are two poles, there will be a bump at aroun ω = (2). At the origin, the angular contributions of the two poles cancel each other out, hence the angle is zero. As we move up the jω axis, the angles a up to π, with each pole contributing π/2 (3). Pole-zero iagram 6 as a zero at the origin, meaning that we take the erivative of the impulse response of pole-zero iagram 2 (). The erivative ens up being the combination of a ecaying cos(t) term minus a ecaying sin(t) term (5). The zero at the origin as a linearly increasing component to the magnitue function (4). It also as π/2 to the phase response everywhere (5). Pole-zero iagram 5 has complex conjugate poles an zeros at the same frequency ω. The system function has the form H(s) = s2 ω 0 Q s + ω2 0 s 2 + ω 0 Q s +. ω2 0 This enominator has the same form as pole-zero iagrams 2 an 6, but has an aitional power of s (corresponing to ifferentiation) in the numerator. This leas to a response of the form in (6). The symmetry of the poles an zeros means they cancel each other s effect on magnitue (). The phase response at ω = 0 is zero, as the contributions cancel each other out. As we move past ω = where the conjugates are locate, the phase moves in the negative irection faster, but eventually settles back at 0 as we move farther an the contributions again cancel each other out (). h(t) Magnitue Angle PZ iagram : PZ iagram 2: 2 3 PZ iagram 3: PZ iagram 4: PZ iagram 5: 6 PZ iagram 6: 5 4 5

5 6.003 Homework #7 Solutions / Fall 20 Engineering Design Problems 3. Desire oscillations The following feeback circuit was the basis of Hewlett an Packar s founing patent. V in + R R R C C C V out 5 a. With R = kω an C = µf, sketch the pole locations as the gain varies from 0 to, showing the scale for the real an imaginary axes. Fin the for which the system is barely stable an label your sketch with that information. What is the system s oscillation perio for this? The close-loop gain is H(s) = (+s) 3 + (+s) 3 = The enominator is zero if ( + s) 3 = ( + s) = 3 s = + 3 ( + s) 3 + There are three cube roots of : 3, 3 e jπ/3, an 3 e jπ/3 an three corresponing poles: s = 3, + 3 e jπ/3, an + 3 e jπ/3 3 3 The point of marginal stability is where the root locus crosses the jω axis. This occurs when the real part of + 3 e jπ/3 equals zero: 3 = 2 so that = 8. The frequency of oscillation is ω = 3 T = ω =. 3 so the perio of oscillation is

6 6.003 Homework #7 Solutions / Fall 20 For = ms (as given), the perio T = 3.63 ms. 6 b. How o your results change if R is increase to 0 kω? Increasing R by a factor of 0 increases the perio T by a factor of 0, to T = 36.3 ms. It has no effect of the critcal value of = Robotic steering Design a steering controller for a car that is moving forwar with constant velocity V. θ p You can control the steering-wheel angle w(t), which causes the angle θ(t) of the car to change accoring to θ(t) t V = w(t) where is a constant with imensions of length. As the car moves, the transverse position p(t) of the car changes accoring to p(t) = V sin θ(t) V θ(t). t Consier three control schemes: a. w(t) = e(t) b. w(t) = v ė(t) c. w(t) = e(t) + v ė(t) where e(t) represents the ifference between the esire transverse position x(t) = 0 an the current transverse position p(t). Describe the behaviors that result for each control scheme when the car starts with a non-zero angle (θ(0) = θ 0 an p(0) = 0). Determine the most acceptable value(s) of an/or v for each control scheme or explain why none are acceptable. Part a. This system can be represente by the following block iagram: W + V Θ X A V A P We are given a set of initial conitions p(0) = 0 an θ(0) = θ 0 an we are aske to characterize the response p(t). Initial conitions are easy to take into account when a system is escribe by ifferential equations. However, feeback is easiest to analyze for systems expresse as operators or (equivalently) Laplace transforms. Therefore we first calculate the close-loop system function, H(s) Y (s) X(s) = V V s 2 + V V s 2 = V 2 s 2 + V 2

7 6.003 Homework #7 Solutions / Fall 20 which has two poles: ±jω 0 where ω 0 = V. We can convert the system function to a ifferential equation: p(t) + V 2 p(t) = V 2 x(t) an then fin the solution when x(t) = 0, p(t) + V 2 p(t) = 0 so that p(t) = C sin ω 0 t since p(0) = 0. From p(t) we can calculate θ(t) = ṗ(t)/v = C V ω 0 cos ω 0 t. θ(0) = θ 0, it follows that C = V θ 0 /ω 0 an p(t) = V θ 0 sin ω 0 t = θ 0 ω 0 sin V t for t > 0. From the initial conition If is small, then the oscillations are slow, but they have a large amplitue. If is large, then the oscillations are fast (an therefore uncomfortable for passengers), but the amplitue is small. While none of these behaviors are esireable, it woul probably be best to increase so that the amplitue of the oscillation is small enough so that the car stays in its lane. Part b. The system can be represente by the following block iagram: X The close-loop system function is H(s) = vs V 2 s 2 + v s V 2 W + v s V Θ V s s = s 2 v s V 2 s(s + v V 2 ). The close-loop poles are at s = 0 an s = v V 2. Since p(0) = 0, the form of p(t) is given by ( ) p(0) = C e v V 2 t for t > 0. We can fin C by relating C to the initial value of θ(t) = ṗ(t)/v. Since θ(0) = θ 0, ṗ(0) = V θ 0. Therefore C = V, so that v p(t) = θ 0 v V for t > 0 as shown below. ( e v V 2 t ) θ 0 V v V 2 v We woul like to make v large because large v leas to fast convergence. Large values of v also lea to smaller steay-state errors in p(t). There are no oscillations in p(t) with the velocity sensor, which is an avantage over results with the position sensor in part a. However, there is now a steay-state error in p(t), which is worse. Fortunately the steay-state error can be mae small with large v. t P

8 6.003 Homework #7 Solutions / Fall 20 Part c. The system can be represente by the following block iagram: 8 X W + v s+ V Θ V s s P The close-loop system function is H(s) = ( vs + ) V 2 s 2 + ( v s + ) V 2 = ( V 2 vs + ) s s 2 + ( 2 v s + ) V 2 This secon-orer system has a resonant frequency ω 0 = = ( vs + ) V 2 s 2 + Q sω 0 + ω0 2 V 2 an a quality factor Q = v ω 0. There is an enormous variety of acceptable solutions to this problem, since there are many values of an v that can work. Here, we focus on one line of reasoning base on our normalization of secon-orer system in terms of Q an ω 0. To avoi excessive oscillations, we woul like Q to be small. Try Q =. Then H(s) = ( vs + ) V 2 s 2 + ω 0 s + ω0 2. Then p(t) has the form p(t) = Ce ω 0t/2 sin ( ) 3 2 ω 0t. As before, we can use the intial conition of θ(0) = θ 0 to etermine C. θ(t) = ṗ(t)/v so ṗ(0) = θ 0 V = C 3ω 0 /2. Therefore p(t) = 2V θ ( ) 0 3 e ω0t/2 sin 3ω0 2 ω 0t. In general, V θ 0 3ω0 ω 0 t Increasing Q woul reuce the overshoot but slow the response. We coul compensate for the slowing of the response by increasing ω 0. Performance can be ajuste to be better than either part a or part b. By ajusting Q an ω 0 we can get convergence of p(t) to zero with minimum oscillation. Although the steay-state value of the error is zero an the oscillation is minimize, there is still a transient behavior, which coul momentarily move the car into the other lane!

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