6.003 Homework #14 Solutions

6.003 Homework #14 Solutions Problems 1. Neural signals The following figure illustrates the measurement of an action potential, which is an electrical pulse that travels along a neuron. Assume that this pulse travels in the positive z direction with constant speed ν = 10 m/s (which is a reasonable assumption for the large unmyelinated fibers found in the squid, where such potentials were first studied). Let Vm (z, t) represent the potential that is measured at position z and time t, where time is measured in milliseconds and distance is measured in millimeters. The right panel illustrates f (t) = Vm (30, t) which is the potential measured as a function of time t at position z = 30 mm. f(t) = Vm(30,t) f(t) = Vm(30,t) Axon 0 10 20 30 z [mm] 40 50 +50 0 -50 -100 0 2 4 6 Time [ms] 8 10 Part a. Sketch the dependence of Vm on t at position z = 40 mm (i.e., Vm (40, t)). It will take the action potention 1 ms to travel from the reference position at z = 30 mm to its new position at z = 40 mm. Thus, the new waveform Vm (40, t) is a version of f (t) that is shifted by 1 ms to the right. Millivolts +50 Vm(30,t) 0 Vm(40,t) -50 -100 0 2 4 6 Time [ms] 8 10 Part b. Sketch the dependence of Vm on z at time t = 0 ms (i.e., Vm (z, 0)). The action potential peaks at z = 30 mm when t = 2 ms. Since it is traveling to the right at speed ν = 10 mm/ms, it must also peak at z = 10 mm when t = 0. Thus f (2) must map to z = 10 mm in the new figure. Similarly, the following function locations map to new positions: f (0) maps to 30 6.003 Homework #14 Solutions / Fall 2011 f (1) f (2) f (3) f (4) maps maps maps maps to to to to 2 20 10 0 −10 Millivolts +50 Vm(z,0) 0 -50 -100 -20 -10 0 10 Distance [mm] 20 30 Part c. Determine an expression for Vm (z, t) in terms of f (·) and ν. Explain the relations between this expression and your results from parts a and b. f t− Vm (z, t) = z − 30 ν The definition of f (t) provides a starting point: Vm (30, t) = f (t). In part a, we found that Vm (40, t) = f (t − 1). This result generalizes: shifting to a more positive location (i.e., adding z0 to z) adds a time delay of z0 /ν. Expressed as an equation, Vm (30 + z0 , t) = f (t − zν0 ). Substituting z = 30 + z0 , we get the general relation Vm (z, t) = f t − z − 30 ν . To understand our result from part b, substitute t = 0 to obtain Vm (z, 0) = f (0 − z−30 ν ). Thus we must scale the x-axis by ν (to convert the time axis to a space axis) then shift the space axis by 30 mm (so that the peak is now at z = −10 mm) and finally, flip the plot about the x-axis (bringing the peak to z = 10 mm). 6.003 Homework #14 Solutions / Fall 2011 3 2. Characterizing block diagrams Consider the system defined by the following block diagram: + X + A − Y − 3 2 1 2 a. Determine the system functional H = A A Y X. Let W represent the output of the topmost integrator. Then 1 1 W = A(X − AY ) = AX − A2 Y 2 2 and 3 Y = W − AY . 2 Substituting the former into the latter we find that 1 3 Y = AX − A2 Y − AY . 2 2 Y yields the answer, X A Y = . 3 X 1 + 2 A + 12 A2 Solving for b. Determine the poles of the system. Substituting A → Y = X 1+ 1 in the system functional yields s 1 s 31 1 1 2 s + 2 s2 = s s2 + 32 s + 1 2 = s . (s + 21 )(s + 1) The poles are then the roots of the denominator: − 12 , and −1. c. Determine the impulse response of the system. Expand the system functional using partial fractions: A βA A Y αA 2A = + − = = X 1 + A 1 + 12 A 1 + A 1 + 12 A 1 + 32 A + 12 A2 Each term in the partial fraction expansion contributes one fundamental mode to h, h(t) = (2e−t − e−t/2 ) u(t) 6.003 Homework #14 Solutions / Fall 2011 4 3. Bode Plots Our goal is to design a stable CT LTI system H by cascading two causal CT LTI systems: H1 and H2 . The magnitudes of H(jω) and H1 (jω) are specified by the following straightline approximations. We are free to choose other aspects of the systems. |H(jω)| [dB] −20 −40 ω [log scale] 1 10 100 1000 1 10 100 1000 |H1 (jω)| [dB] 24 6 ω [log scale] H1 and H2 have to be stable as well as causal because we’re talking about their frequency responses, and H has to be causal because H1 and H2 are. This implies that all poles must be in the left half-plane. a. Determine all system functions H1 (s) that are consistent with these design specifi­ cations, and plot the straight-line approximation to the phase angle of each (as a function of ω). The frequency response of H1 breaks up at ω = 1 and then down at ω = 8 and 40. The two breaks downward require poles at s = −8 and s = −40 respectively, The break upward can be achieved with a zero at s = 1 (blue) or at s = −1 (red). ∠H1 (jω) [rad] π 2 ω [log scale] 1 10 100 1000 − π2 H1 could also be multiplied by −1 without having any effect on the magnitude function. Multiplying by −1 would shift the phase curves up or down by π. b. Determine all system functions H2 (s) that are consistent with these design specifi­ cations, and plot the straight-line approximation to the phase angle of each (as a function of ω). 6.003 Homework #14 Solutions / Fall 2011 5 To compensate for H1 , the frequency response of H2 must break downward at ω = 1 and upward at ω = 40. In addition, H2 must break downward at ω = 8 so that the slope of H changes from 0 to −40 dB/decade at ω = 10. H2 can be achieved with poles at s = −1 and −8 and a zero at s = 40 (blue) or at s = −40 (red). ∠H2 (jω) [rad] π 2 ω [log scale] 1 10 100 1000 − π2 H2 could also be multiplied by −1 without having any effect on the magnitude function. Multiplying by −1 would shift the phase curves up or down by π. 6.003 Homework #14 Solutions / Fall 2011 6 4. Controlling Systems Use a proportional controller (gain K) to control a plant whose input and output are related by R2 F = 1 + R − 2R2 as shown below. + X − K F Y a. Determine the range of K for which the unit-sample response of the closed-loop system converges to zero. Using Black’s equation, we can write KR2 KR2 Y 2 = 1+R−2R = 2 KR X 1 + R − (2 − K)R2 1 + 1+R−2R 2 The closed-loop poles can be found by substituting R → z1 : K Y = 2 X z + z − (2 − K) and solving for the roots of the denominator: 1 z=− ± 2 1 +2−K 4 The unit-sample response will converge to zero iff the poles are inside the unit circle. When K = 0, the poles are at z = −2 and z = 1 (not convergent). As K increases, the poles move toward each other, creating a double pole at z = − 12 when K = 94 . The response will converge when the pole that started at z = −2 reaches z = −1, i.e., at K = 2. The poles will split away from z = − 12 for K > 94 and will stay inside the unit circle if 14 + 2 − K > − 43 , i.e., if K < 3. These results are shown in the following graphical representation. K =3 K = 9/4 K =0 K =2 K =2 K =0 K =3 Thus, the unit-sample response will converge if 2 < K < 3. 6.003 Homework #14 Solutions / Fall 2011 7 b. Determine the range of K for which the closed-loop poles are real-valued numbers with magnitudes less than 1. From the plot in the previous part, it follows that the closed-loop poles are on the real axis and have magnitudes less than one when 2 < K < 94 . 8 6.003 Homework #14 Solutions / Fall 2011 5. CT responses We are given that the impulse response of a CT LTI system is of the form A t T where A and T are unknown. When the system is subjected to the input x1 (t) 3 t 1 −2 the output y1 (t) is zero at t = 5. When the input is   πt u(t), x2 (t) = sin 3 the output y2 (t) is equal to 9 at t = 9. Determine A and T . Also determine y2 (t) for all t. The first fact implies that Z ∞ Z y1 (5) = x1 (τ )h(5 − τ )dτ = A −∞ 5 x1 (τ )dτ = 0. 5−T If the lower limit is 1, the area of the triangle between τ = 1 and τ = 3 is 2 and cancels the area of the rectangle between τ = 4 and τ = 5. Therefore T = 4. From the second fact, we have Z 9 9 = y2 (9) = A x2 (τ )dτ 5 Z 9  πτ  dτ =A sin 3 5  πτ  9 A =− cos π/3 3 5 9A = , 2π so A = 2π. There are three ranges to consider in computing y2 (t). For t < 0, there is no overlap between x2 (τ ) and h(t − τ ) and hence y2 (t) = 0. For 0 ≤ t < 4, there is partial overlap and y2 (t) is given by    Z t  πτ   πτ  t πt 2π y2 (t) = 2π = 6 1 − cos sin dτ = − cos . 3 π/3 3 0 3 0 For t ≥ 4, the overlap is total and we have      Z t  πτ  π(t − 4) πt y2 (t) = 2π sin dτ = 6 cos − cos . 3 3 3 t−4 6.003 Homework #14 Solutions / Fall 2011 9 Hence  0, 

 πt 6 1 − cos 3  , y2 (t) =   −4) π(t   6 cos − cos 3 t < 0, 0 ≤ t < 4, πt 3
 , t ≥ 4. 6.003 Homework #14 Solutions / Fall 2011 10 6. DT approximation of a CT system Let HC1 represent a causal CT system that is described by ẏC (t) + 3yC (t) = xC (t) where xC (t) represents the input signal and yC (t) represents the output signal. xC (t) HC1 yC (t) a. Determine the pole(s) of HC1 . The the Laplace transform of the differential equation to get sYC (s) + 3YC (s) = XC (s) and solve for YC (s)/XC (s) = 1/(s + 3). The pole is at s = −3. Your task is to design a causal DT system HD1 to approximate the behavior of HC1 . xD [n] HD1 yD [n] Let xD [n] = xC (nT ) and yD [n] = yC (nT ) where T is a constant that represents the time between samples. Then approximate the derivative as dyC (t) yC (t + T ) − yC (t) = . dt T b. Determine an expression for the pole(s) of HD1 . Take the Z transform of the difference equation yD [n + 1] − yD [n] + 3yD [n] = xD [n] T to obtain zYD (z) − YD + 3YD (z) = XD (z) T Solving (z − 1 + 3T )YD (z) = T XD (z) so that HD (z) = T YD (z) = . XD (z) z − 1 + 3T There is a pole at z = 1 − 3T . 6.003 Homework #14 Solutions / Fall 2011 11 c. Determine the range of values of T for which HD1 is stable. Stability requires that the pole be inside the unit circle −1 < 1 − 3T < 1 or −2 < −3T < 0 so that 0 |p0 |. This will be true if α > |p0 | − 2 or if α < −|p0 | − 2. 6.003 Homework #14 Solutions / Fall 2011 13 8. Finding a system a. Determine the difference equation and block diagram representations for a system whose output is 10, 1, 1, 1, 1, . . . when the input is 1, 1, 1, 1, 1, . . .. Notice that Y = 10X − 9RX. This relation suggests the following difference equation y[n] = 10x[n] − 9x[n − 1] and block diagram + 10 X Delay Y −9 b. Determine the difference equation and block diagram representations for a system whose output is 1, 1, 1, 1, 1, . . . when the input is 10, 1, 1, 1, 1, . . .. The difference equation for the inverse relation can be obtained by interchanging y and x in the previous difference equation to get x[n] = 10y[n] − 9y[n − 1]. So y[n] = 9y[n − 1] + x[n] , 10 which has this block diagram X + 0.1 0.9 Y Delay c. Compare the difference equations in parts a and b. Compare the block diagrams in parts a and b. The difference equations for parts a and b have exactly the same structure. The only difference is that the roles of x and y are reversed. The block diagrams have similar parts (1 delay, 1 adder, 2 gains), but the topologies are completely different. The first is acyclic and the second is cyclic. 14 6.003 Homework #14 Solutions / Fall 2011 9. Lots of poles All of the poles of a system fall on the unit circle, as shown in the following plot, where the ‘2’ and ‘3’ means that the adjacent pole, marked with parentheses, is a repeated pole of order 2 or 3 respectively. Im z × (×)2 (×)3 Re z × Which of the following choices represents the order of growth of this system’s unit-sample response for large n? Give the letter of your choice plus the information requested. a. y[n] is periodic. If you choose this option, determine the period. b. y[n] ∼ Ank (where A is a constant). If you choose this option, determine k. c. y[n] ∼ Az n (where A is a constant). If you choose this option, determine z. d. None of the above. If you choose this option, determine a closed-form asymptotic expression for y[n]. A partial fraction expansion of the system functional will have terms of the following forms:  2 3  2  1 1 1 1 1 1 , , , and , , . 1−R 1−R 1−R 1 + R2 1+R 1+R The third one will have the fastest growth for large n. Its expansion has the form (1 + R + R2 + R3 + · · ·) × (1 + R + R2 + R3 + · · ·) × (1 + R + R2 + R3 + · · ·) . Multiplying the first two: 1 R R2 R3 ··· 1 R R2 R3 ··· 1 R R2 R3 ··· R R2 R3 R4 ··· R2 R3 R4 R5 ··· R3 R4 R5 R6 ··· ··· ··· ··· ··· ··· Group same powers of R by following reverse diagonals: 1 + 2R + 3R2 + 4R3 + · · · Multiplying this by the last term: 1 2R 3R2 4R3 ··· 1 R R2 R3 ··· 1 2R 3R2 4R3 ··· R 2R2 3R3 4R4 ··· R2 2R3 3R4 4R5 ··· R3 2R4 3R5 4R6 ··· ··· ··· ··· ··· ··· 6.003 Homework #14 Solutions / Fall 2011 15 Group same powers of R by following reverse diagonals: 1 + 3R + 6R2 + 10R3 + · · · This expression grows with (n + 1)(n + 2)/2 which is on the order of n2 . Thus b is the correct solution with k = 2. 6.003 Homework #14 Solutions / Fall 2011 16 10.Relation between time and frequency responses The impulse response of an LTI system is shown below. h(t) 1 5 t −1 If the input to the system is an eternal cosine, i.e., x(t) = cos(ωt), then the output will have the form y(t) = C cos(ωt + φ) The impulse response has the form of a decaying sinusoid. The time constant of decay is approximately 2, so the exponential part has the form e−t/2 . The sinusoid has approx­ imately 8 periods in 5 time units so 8 ω2πd = 5. Solving this, we find that ωd ≈ 10. The impulse response therefore has the form h(t) = e−t/2 sin(10t)u(t) . There are two poles associated with such a response and no zeros. The poles have real parts of −σ = − 21 and imaginary parts of ±j10. The characteristic equation is (s − p0 )(s − p1 ) = (s + 12 + j10)(s + 12 − j10) = s2 + s + 100.25 = s2 + ωQ0 s + ω02 . Thus ω0 ≈ 10 and Q ≈ 10. The system function is the Laplace transform of the impulse response, 10 ωd ≈ 2 H(s) = 2 ω0 s + s + 100 s + Q s + ω02 a. Determine ωm , the frequency ω for which the constant C is greatest. What is the value of C when ω = ωm ? The gain of the system is largest at a frequency ωm = ω02 − 2σ 2 ≈ 10. The gain is 1 . Thus C ≈ 1. then approximately Q ≈ 10 times the DC gain, which is ≈ 10 b. Determine ωp , the frequency ω for which the phase angle φ is − π4 . What is the value of C when ω = ωp ? The phase angle varies from 0 when ω = 0 to −π as ω → ∞. The phase angle is equal to − π2 when ω = ω0 [notice that when ω = ω0 the ω02 term in the denominator of the system function is cancelled by s2 = (jω0 )2 ]. The phase angle will be − π4 when √ at ωp than ω = ωp = ω0 − σ (so that the vector from the upper pole is 2 times longer √ at ω0 . At ωp , the gain is reduced from its maximum by 3 dB (a factor of 2). Thus C ≈ √12 .