NPTEL Online Course: Control Engineering Dr. Ramkrishna Pasumarthy and Dr.Viswanath Assignment - 0 : s. A passive band pass filter with is one which: (a) Attenuates signals between the two cut-off frequencies and amplifies the signal outside the cutoff frequencies (b) Amplifies signals between the two cut-off frequencies and attenuates the signal outside the cutoff frequencies (c) Passes the signal between the two cut-off frequencies and attenuates the signal outside the cut-off frequencies [Correct] (d) Passes the signals at all frequencies and attenuates the signal exactly at the cut-off frequencies Standard definition of a band pass filter is that it passes the signal (with minor attenuation) in the pass band and attenuates the signals outside the pass band. Hence [c] is correct answer. 2. Consider a 20Hz signal which is sampled at 00Hz. Assume that the signal is corrupted with high frequency noise (of about 50Hz). To remove this noise the signal is to be passed through a low pass filter. A good choice for the cut off frequency for the low pass filter would be: (a) 30Hz [Correct] (b) 20Hz (c) 0Hz (d) 70Hz We want to pass the signal and attenuate the noise. Signal is at 20Hz and noise is at 50Hz. So a 30Hz cut off low pass is enough. The sampling has been deliberately included to confuse students a bit and it does NOT have a role in the choice of the cut off frequency. Hence [a] is the correct answer. 3. Which of the following Bode magnitude plots correspond to a second order low pass filter (note: frequency axis is in Hertz and not rad/s)
(a) (b) (c) Correct A simple second order low pass filter will have a 40dB slope. By looking at the Y-axis only [d] is the correct solution. 2
4. The noise present in sensor measurements can be easily eliminated [Ans: c] (a) Usually True: Because the noise is constant (w.r.t. time) and a simple calibration step can remove it (b) Usually True: because noise is negligible compared to signal magnitude (c) Usually Not true: Requires complex filtering techniques [Correct] (d) All of the above Noise comes in various types not just a simple bias. And the removal of this noise often requires advanced filtering techniques. [c] is correct. 5. Consider a gyroscope strapped to a vehicle to measure its rotational velocity. The gyroscope has a very low frequency noise component (around Hz). We want to design a simple high pass filter to attenuate the noise signal. For a RC high pass filter, if the cut-off frequency is 5Hz and the value of R is MΩ, the value of C in nf should be (rounded to one decimal point)? Given the cut off frequency F c = 5Hz and R = MΩ F c = 2πRC C = 2πF c R C = 2 3.4 5 0 6 C = 3.8nF (Any answer between 3.5 and 32.5 will be marked right) 6. Consider an accelerometer strapped to a vehicle and is measuring the vehicles linear acceleration. The accelerometer has a high frequency noise component (around 70Hz). We want to design a simple low pass filter to attenuate the noise signal. For a simple RC low pass filter, if the cut-off frequency is 0Hz and the value of C is µf, the value of R is: (rounded to one decimal point) 3
Given the cut off frequency F c = 0Hz and C = µf F c = 2πRC R = 2πF c C R = 2 3.4 5 0 6 R = 5.92kΩ (Any answer between 5.5 and 6.5 will be marked right) 7. The attitudes (roll or pitch angles) computed using an accelerometer s measurements are erroneous. This is because of: (Hint: carefully study the formulae to compute roll and pitch angles from accelerometer measurements) (a) Drift in the angles computed from accelerometer measurements (b) Noise in accelerometer measurements [Correct] (c) Influence of earth s gravity on the accelerometer s measurements (d) All of the above Drift occurs only when numerically integrating a sensor measurement since the integration process integrates not just the correct measurement but also the noise. To compute attitudes using accelerometer measurements we do NOT perform any integration. It is purely a trigonometric relationship. Thus drift is not the answer to why the solution may be erroneous. Earth s gravity is always present (on all 3 axis) and does not affect accuracy. It is only the noise which is different along each axis that affects the accuracy. Hence [b] is correct answer. 8. A 3-axis gyroscope fixed to a vehicle is measuring the rotational velocity (of that vehicle). It is known that one axis has a constant bias, the remaining two axes have a noise component of 50Hz. It is also known that the vehicle s rotational motion is around 20Hz. All three gyroscope axes data are passed through the same single filter. Which filter would you use in this situation (a) A low pass filter with cut-off frequency 20Hz (b) A band pass filter with lower cut-off frequency 20Hz and upper cut off frequency 60Hz (c) A high pass filter with cut-off frequency 60Hz (d) A band pass filter with lower cut-off frequency 5Hz and upper cut off frequency 30Hz [Correct 4
The 3 axis gyroscope has a constant noise (bias) in axis and a 50Hz noise component in axis 2 and axis 3. We have only one filter which we can use. And the signal of interest has a 20Hz component. So we want to pass the 20Hz an we want to eliminate the constant bias and the 50Hz noise. Hence, only [d] is correct solution. 9. You are given a low pass filter, designed with a cut off frequency of 30Hz, which can filter an accelerometer s signals. Suppose that an accelerometer is strapped to a vehicle. Select all the following examples where such a filter would be practically useful. (a) A vehicle whose dynamics have dominant frequency component at 20Hz [Correct] (b) A vehicle as in option (a), and where the accelerometer s measurement has a time varying noise component with frequency around 40Hz [Correct] (c) A vehicle as in option (a), and where the accelerometer s measurement has a time varying bias noise component with frequency at 0Hz (d) A vehicle as in option (a), and where the accelerometer s measurement has a constant bias noise component. The low pass filter has a cut off frequency at 30Hz. Option [c] is not correct, since the 0Hz noise component would also be passed through the filter. For the same reason option [d] is not correct as the bias component (0Hz) would also get passed. Thus [a] and [b] are correct. 0. If we design a first order passive RC circuit to filter noise in a sensor, what will be the transfer function of the circuit? (a) G(s) = +RCs [Correct] (b) G(s) = R (c) G(s) = +RCs +RCs C (d) G(s) = R+Cs The figure of an RC circuit to filter noise is as shown below: Figure : First order low pass filter 5
Transfer function is obtained as follows: V in = RI + jωci V out = jωci Applying Laplace transforms we get, V out (s) = Cs V in (s) = R + Cs G(s) = V out(s) V in (s) G(s) = + RCs. Consider the filter circuit shown in the figure below. What is the transfer function of the filter shown in the circuit? Figure 2: Filter circuit (a) V out(s) V in (s) = R R 2 C C 2 s 2 +(R C +R C 2 +R 2 C 2 )s+ [Correct] (b) V out(s) V in (s) = R R 2 C C 2 s 2 +(R C +R C 2 +R 2 C 2 )s+ (c) V out(s) V in (s) = s 2 +(R C +R C 2 +R 2 C 2 )s+r R 2 C C 2 (d) V out(s) V in (s) = R R 2 C C 2 s 2 +(R C +R C 2 +R 2 C 2 )s+r R 2 C C 2 For Loop : R I + (I I 2 )dt = V in C 6
For Loop 2: R 2 I 2 + C 2 I 2 dt + (I 2 I )dt = 0 C On applying Laplace transforms we get: I (s) = V in(s)c s + I 2 (s) R C s + sc 2 V in (s) I 2 (s) = R R 2 C C 2 s 2 + (R C + R C 2 + R 2 C 2 )s + V out (t) = I 2 dt C 2 V out (s) = I 2(s) sc 2 Therefore, on solving above equations we get: Substituting I 2 (s) in the above equation we get: ( R2 + C 2 s + ) I2 (s) = I (s) C s C s ( R2 + C 2 s + ) I2 (S) = V in(s)c s + I 2 (s) C s C 2s2 R +C s ( R2 + C 2 s + C s )( Vout c 2 (s)c 2 s ) = V in(s)c s s2 R +C s C 2s2 R +C s V out (s) V in (s) = C 2 s 2 (R 2 C 3C 2R ) + s(r 2 C 2C 2 + R C 2C 2 + R C 3 + R C 2C 2) +C 2 V out (s) V in (s) = R R 2 C C 2 s 2 + (R C + R C 2 + R 2 C 2 )s + 2. In Problem No., given that R = 2kΩ,R 2 = kω,c = µf,c 2 = 2µF and if you draw the bode plot, what would be the DC gain in db? DC gain, K = G(0) = DC gain in db = 20log(K) = 0 This can also be observed from the bode plot that the magnitude at zero frequency is 0. 3. In Problem No 2, in the bode plot, what is the corner frequency in rad/sec? 7
You can draw the bode plot and observe the corner frequency. This can also be calculated without drawing the bode plot: ω n = R R 2 C C 2 ω n = 500rad/s 4. In Problem No. 2, in the bode plot, what is the slope of the plot in db/dec after the corner frequency? There is a quadratic pole at the corner frequency and the slope before corner frequency is 0db/dec. Therefore the slope after the corner frequency is 40db/dec. 8