Designing Information Devices and Systems II Spring 2019 A. Sahai, J. Roychowdhury, K. Pister Midterm 1: Practice

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1 EES 16B Designing Information Devices an Systems II Spring 019 A. Sahai, J. Roychowhury, K. Pister Miterm 1: Practice 1. Speaker System Your job is to construct a speaker system that operates in the range from 0 Hz to 0,000 Hz. Such a system woul ieally have a transfer function H esire ( f ), where the magnitue plot of H esire ( f ) is as follows: 10 1 Transfer function H esire ( f ) 10 0 Hesire( f ) f Within the operating range from 0 Hz to 0,000 Hz, the transfer function has magnitue equal to 1, an otherwise it is H esire ( f ) = { 1 if f [0,0000], 10 3 else. Say you alreay have the parts to make a speaker system that operates in the mirange section of the auio frequency spectrum, f between 50 Hz to 4,000 Hz. Mathematically, this speaker has a transfer function H 1 where { 1 if f [50,4000], H 1 ( f ) = 10 3 else. (a) Draw the boe plot of H 1 ( f ). Miterm 1: Practice UB EES 16B, Spring 019. All Rights Reserve. This may not be publicly share without explicit permission. 1

2 10 1 Transfer function H H1( f ) f (b) You want to buil upon what you currently have to buil the esire speaker system. You alreay have a system that implements the transfer function H 1 ( f ). Your system for H 1 ( f ) can be cascae with another system in series. In between these two systems, you place a unity gain buffer to ecouple these systems. In other wors, your secon system can be compose with your first system. This means that the resulting transfer function of the composition is the multiplication of the two systems. H composition ( f ) = H 1 ( f ) H ( f ), where H ( f ) is the transfer function of the secon system. The goal of the composition is to achieve H esire ( f ). Draw the Boe plot of H ( f ) such that H composition ( f ) = H esire ( f ) Transfer function H 10 3 H( f ) f. Phun with Phasors Miterm 1: Practice UB EES 16B, Spring 019. All Rights Reserve. This may not be publicly share without explicit permission.

3 (a) Suppose that we have a voltage source v(t) = 4sin(t) V connecte in parallel to a resistor with R = Ω. Draw this circuit. ω = 1 Note that the given input signal has amplitue A = 4, phase φ = 0 an angular frequency v ( t) Ω i(t) v out (t)) (b) For the previous part, compute the phasor Ṽ that represents the voltage across the voltage source. Draw this phasor Ṽ as a vector on the complex plane. (HINT: Remember that the phasor is the coefficient of the complex exponential corresponing to the positive frequency.) First, we shoul convert the sine to a cosine, ( v(t) = 4sin(t) = 4cos t π ) Next, as per the hint, we want to first write the voltage as a sum of exponentials. v(t) = 4 e j(t π ) e j(t π ) Ṽ = e jπ = j (3) (c) ontinuing, what is the phasor Ĩ R for i R (t), the current into the resistor? Draw this phasor Ĩ R on the complex plane. Using Ohm s aw, we have i R (t) = 4sin(t) = sin(t) = cos ( t π ). onverting this to a phasor, we have: i R (t) = e j(t π ) e j(t π ) (4) Ĩ R = j (5) (1) () () Suppose that instea of a resistor, we ha an inuctor with inuctance = 4 H. What is the phasor Ĩ for i (t), the current into the inuctor? Draw this phasor Ĩ on the complex plane. Rather than working in the time omain, let s convert the circuit first into the phasor omain. We know Z = jω = 4 jω. Hence, Ĩ = Ṽ Z (6) = j 4 jω (7) = 1 ω (8) = 1 (9) Miterm 1: Practice UB EES 16B, Spring 019. All Rights Reserve. This may not be publicly share without explicit permission. 3

4 In the last line, we have substitue for ω = 1. (e) Suppose that instea of an inuctor or resistor, we have a capacitor = 1 F connecte in parallel with this voltage source. What is the phasor Ĩ for i (t), the current into the capacitor? Draw this phasor Ĩ on the complex plane. Again working in the phasor omain, we have Z = 1 jω = jω. Hence, Ĩ = Ṽ Z (10) = j (11) jω = ω (1) = 1 (13) (f) onsier a capacitor ( = 1) an an inuctor ( = 1) connecte in series. Give expressions for the impeances of Z,Z for each of these elements as a function of the angular frequency ω. As before, we have Z = jω = jω an Z = 1 jω = 1 jω. (g) For the following values of ω, raw the iniviual impeances of the inuctor an capacitor as vectors on a single complex plane for each value of ω. Also raw the combine impeance Z total of their series combination. Give the magnitue an phase of Z total for each. A logically soun graphical argument is sufficient justification. i. ω = 1 ra/sec We have Z = j 1 an Z = j. Z total = Z Z = 3 j. ii. ω = 1 ra/sec We have Z = j an Z = j. Z total = Z Z = 0. iii. ω = ra/sec We have Z = j an Z = 1 j. Z total = Z Z = 3 j. (h) Derive the expression for the impeance Z in terms of the angular frequency ω an the inuctance from first principles. You can assume: The efinition of the phasor representation is the coefficient of e jωt in any real sinusoial quantity at frequency ω. The efinition of impeance as the ratio of the phasor representation of the voltage across an element to the phasor representation of the current through the element. The basic ifferential equation V (t) = t I (t) that governs an inuctor s current-voltage relationship. i v Figure 1: A simple inuctor circuit Miterm 1: Practice UB EES 16B, Spring 019. All Rights Reserve. This may not be publicly share without explicit permission. 4

5 onsier a simple inuctor circuit as in Figure 1, with current being Then, we can write the phasor as, i(t) = I 0 cos(ωt φ) By the inuctor equation, Ĩ = I 0 e jφ (14) v(t) = i t (t) = I 0 ω sin(ωt φ) ( = I 0 ω cos ωt φ π ) ( = (ω)i 0 cos ωt φ π ) Hence, Hence, from Equations (14) an (16), Ṽ = (ω)i 0 e j(φ π ) (15) = jω I 0 e jφ (16) Ṽ = jωĩ The impeance of an inuctor is an abstraction to moel the inuctor as a resistor in the phasor omain. This is enote Z. Z = Ṽ Ĩ = jω (i) If the time-omain current waveforms i 1 (t),i (t),i 3 (t) entering a noe satisfy K (i.e. sum to zero) at that noe for all time, show that their phasor representations Ĩ 1,Ĩ,Ĩ 3 also must satisfy K at that noe. From K, we have t, Base on our efinitions of phasors, we can write i 1 (t), i (t), an i 3 (t) as follwing: i 1 (t) = Ĩ 1 e jωt Ĩ 1 e ( jωt) (17) i (t) = Ĩ e jωt Ĩ e ( jωt) (18) i 3 (t) = Ĩ 3 e jω1 Ĩ 3 e ( jωt) (19) i 1 i i 3 = e jωt (Ĩ 1 Ĩ Ĩ 3 ) e ( jωt) (Ĩ 1 Ĩ Ĩ 3 ) (0) = e jωt (Ĩ 1 Ĩ Ĩ 3 ) e ( jωt) (Ĩ 1 Ĩ Ĩ 3 ) (1) = 0 () We know that the expoenetial e jωt is always non-zero. Hence, we must have Ĩ 1 Ĩ Ĩ 3 = 0 for the last line to be true. Miterm 1: Practice UB EES 16B, Spring 019. All Rights Reserve. This may not be publicly share without explicit permission. 5

6 3. Analog-to-Digital onverter (AD) In this question, we ll be examining common pitfalls of riving successive approximation ADs, which are the kin of ADs foun on many evices, incluing the MSP430s you use in lab. We can moel the AD internally as having a switch connecte to a sampling capacitor with capacitance. In orer to rea an analog voltage, the AD closes the switch for a sampling perio τ. harge collects on the capacitor uring this time. At the en of the sampling perio, the switch is opene an the voltage across the capacitor V c is rea an converte into a igital value. (a) Our AD can only rea voltages from 0V to 3V. We have a voltage source V s we want to measure that we know outputs from 0V to 9V. Your lab partner suggests you use a resistor ivier with resistors R1 an R so you can rea the entire range of the voltage source. Fin the equation for the voltage on the capacitor V c (t) for 0 t τ (when the switch is close) assuming V c (0) = 0. AD R 1 t = 0 Vc (t) V s R Figure : AD circuit charge via voltage ivier i R1 AD V s R 1 R i R i t = 0 Vc (t) Figure 3: AD circuit charge via voltage ivier First using KV we can fin that V c (t) = V s i R1 R 1 Now we can apply K to fin that i R1 = i R i Miterm 1: Practice UB EES 16B, Spring 019. All Rights Reserve. This may not be publicly share without explicit permission. 6

7 Using Ohm s law we solve i R = V c(t) R an using the capacitor equations we solve i = t V c(t) Substituting in to the original equation we get Rearranging the equation gets us V c (t) = V s R 1 ( V c(t) R t V c(t)) V c (t) = R R 1 R V s R 1R R 1 R t V c(t) By using a variable substitution an solving for the first orer ifferential equation we get R V c (t) = V s (1 e R 1 R R 1 R R 1 R t ) (b) The nominal voltage we want to rea is V n = R R R 1 V s. Suppose R = 100kΩ, R 1 = 00kΩ, an = 10pF. What is the minimum sampling time T min such that our measure V c (T min ) is within 5% of V nom? (That is, V nom V c (T min ) 0.05V nom.) (Note: Some useful approximations are e 1 = 0.37, e = 0.14, e 3 = 0.05, e 4 = 0.0.) First let s solve for the time constant τ for V c (t): τ = R 1R (00)(100) 106 = R 1 R τ = τ = Using the hint, we can see that: T min 3τ For us to get within 5% of V nom we nee 3τ time to pass. For a charging capacitor we have: V nom (1 e t τ ). So for us to get within 5% of V nom we must charge to V nom (1 0.05) which is when t 3τ. This means T min 10 6 secons. (c) Your lab partner thinks this time is too slow an that "we gotta go faster" to be able to sample inputs that are quickly changing. Using a single op-amp, how coul you moify the non-ad part of the circuit to look like an ieal voltage source with output voltage V n? Use a unity gain, non inverting amp as a buffer. The ieal buffer has infinite input resistance an very low (0) output resistance. This ecouples the voltage ivier from the latter part of the circuit an prevents R 1 an R from affecting the latter part of the circuit. The output of the buffer just looks like a perfect voltage source to the latter part of the circuit. Miterm 1: Practice UB EES 16B, Spring 019. All Rights Reserve. This may not be publicly share without explicit permission. 7

8 () Now that the voltage input to the AD is ieal, we nee to consier the effects of the parasitic inuctance an resistance R 3 in the circuit. Set up a system of ifferential equations for V c (t) an i(t) in matrix form an solve for the eigenvalues of the system. R 3 V c (t) V s i(t) Figure 4: AD circuit with perfect voltage source inuctor, capacitor an resistor Using our capacitor equations we know that t V c(t) = i(t) Applying our inuctor equations we can fin that V S = V c (t) V V R3 V = V S V c (t) V R3 t i(t) = V S V c (t) i(t)r 3 t i(t) = V S V c (t) i(t)r 3 To get ri of the V s term we use the variable substitution Ṽ (t) = V c (t) V s Our final system is [ t i(t) = Ṽ (t) i(t)r 3 i(t) Ṽ (t) = t ] tṽ (t) t i(t) = [ ] 1 0 ][Ṽ (t) 1 R 3 i(t) To solve for eigenvalues we solve for [ ] 1 0 et( 1 R λi) = 0 (3) 3 This leas to the characteristic equation: λ R 3 λ 1 = 0 (4) Miterm 1: Practice UB EES 16B, Spring 019. All Rights Reserve. This may not be publicly share without explicit permission. 8

9 Using the quaratic equation we get that λ = R 3 ± 1 R 3 4 (5) (e) is still 10pF an = 1nH. We can increase R 3 by aing series resistance to our voltage source. What is the minimum value R 3 shoul be to eliminate all oscillation in V c (t)? In orer for their to be no oscillation, the eigenvalues of the system must all be real. Thus we want R min 4 R min (6) Plugging in values we get that R min 0Ω. 4. D Motor Driver R V sup (t) t = 0 R motor I sup V s motor Figure 5: omplete motor river circuit In this question, we ll examine the operation of a D motor river similar to the one you ll make in lab. Our circuit moel for a motor is a resistor with resistance R motor an inuctor with inuctance motor. Our motor has an ieal ioe in parallel with the inuctor. Remember that an ieal ioe is an open circuit when there is negative voltage across it, but is a close circuit when there is positive voltage across it. (a) To begin with, we ll moel the supply as an ieal voltage source V sup. At time t = 0 we close the switch to provie current to the motor. If we leave the switch close long enough that the circuit is in steay state, what will the current I max through the motor resistor R motor be? This is the maximum possible current through the motor. t = 0 R motor V sup motor Figure 6: Motor river circuit with ieal voltage source We know at steay state, an inuctor looks like a wire. In this case, the ioe is reversebiase an no current flows through it. Miterm 1: Practice UB EES 16B, Spring 019. All Rights Reserve. This may not be publicly share without explicit permission. 9

10 Thus the only circuit element that we nee to worry about is R motor. Using Ohm s law we can fin that I max = V s R motor. (b) et s now assume that we open the switch again before the circuit reaches steay state, because we on t want the current to actually reach this maximum value. We want to control the torque of the motor, which means we want I motor (t) to have a time-average value of I avg = I max 4. Assuming I motor(0) = I avg, at what time t = T 1 shoul we open the switch so that I motor (T 1 ) = 1.1 I avg? Answer symbolically, without plugging in for particular component values. We know that: From KV an Ohm s law we can fin: V = motor t I motor (7) V = V sup I motor R motor (8) We substitute an get: t I motor = V sup I motorr motor (9) motor motor Doing a variable substitution an solving for the first orer iffeq we get (This is like the ifferential equations that you have seen before in the context of R charging/ischarging circuits): Simplifying out the I max = V sup R motor By substituting in I avg = I max 4 we get I motor (t) = V sup R motor (1 (1 I avgr motor V sup )e R motor t ) (30) term we get that. Fining where the current is equal to 1.1 I avg gives: I motor (t) = I max (1 (1 I avg )e Rmotor motor t ) (31) I max I motor (t) = I max (1 3 4 e R motor motor t ) (3) I motor (T 1 ) = I max (1 3 4 e R motor motor T 1 ) = 1.1 I avg = 1.1 Imax 4 (33) T 1 = t = motor ln( (1 )) (34) R motor 3 4 (c) Now that I motor = 1.1 I avg we ll open the switch an reset our timer so t = 0. In this new time frame, I motor (0) = 1.1 I avg. At what time T will I motor (T ) = I avg? eave your answer symbolically. In typical ioes, there will be some amount of voltage roppe across the ioe when forwar biase, which we call V f w. But in our case we our assuming an ieal ioe, an ignoring this rop in voltage, so V f w = 0 V. Using KV, we know that V f w = V I motor R motor (35) Miterm 1: Practice UB EES 16B, Spring 019. All Rights Reserve. This may not be publicly share without explicit permission. 10

11 this gets us Rearranging gets us V f w motor t I motor I motor R motor = 0 (36) t I motor = V f w I motorr motor (37) motor Doing the variable substitution gets an plugging in initial conitions gets us: I motor (t) = (1.1I avg V f w )e Rmotor motor t V f w (38) R motor R motor Plugging in for V f w = 0 V, I motor (t) = 1.1I avg e R motor motor t (39) I motor (T ) = I avg = 1.1I avg e R motor motor T (40) T = R ln = ln1.1 (41) R Note that this equation is only vali for t when I motor (t) 0. () Base on the previous sections, we will control the average current through the motor by alternating between opening the switch for T 1 time an closing the switch after T time. We can now moel the motor an switch as a current sink attache to our supply that alternates between sinking I avg current for T 1 time (when the switch is open) an sinking 0 current for T time (when the switch is close), seen in the circuit below. Assume for now that = 0F. Sketch a graph for V sup (t) for from time 0 t T 1 T. Make sure to label the value for V sup (t) at time t = 0, t = T 1, an t = T 1 T. et I sup (t) in Figure 8 be the I sup (t) in the waveform shown below. I avg Isup(t) 0 0 T 1 T 1 T T 1 T t Figure 7: urrent sink over time Miterm 1: Practice UB EES 16B, Spring 019. All Rights Reserve. This may not be publicly share without explicit permission. 11

12 R V sup (t) I sup V s I sup Figure 8: omplete motor river circuit If = 0F then the capacitor shoul look like an open circuit. Thus the voltage shoul look like a square wave with V sup = V s I avg R for 0 t T 1 an V sup = V s for T 1 t T 1 T (e) Given V s = 9V, R = 3Ω, I avg = 3A, an V sup (0) = V s, for what value of will V sup (T 1 ) = 0.9 V s? You may leave T 1 as a variable. (Note: You may use the approximation e ) Setting up the KV we get that An substituting in I c = t V sup we get V sup = V s R(I avg I c ) (4) V sup = V s R(I avg t V sup) (43) Rearranging the terms results in t V sup = V s RI avg V sup R By plugging in values we can see that V s = RI avg So we get t V sup = V sup R Solving the iffeq an plugging in initial conitions we get (44) (45) At time T 1, Thus using the hint we see that an by plugging in values we get that V sup (t) = V s e t R (46) V sup (T 1 ) = V s e T 1 R = 0.9 Vs (47) T 1 = 0.1 (48) R = T 1 0.1R = T (49) Miterm 1: Practice UB EES 16B, Spring 019. All Rights Reserve. This may not be publicly share without explicit permission. 1

13 5. High-pass Filter You have a 1 kω resistor an a 1 mh inuctor wire up as a high-pass filter. (a) Draw the filter, labeling the input noe, output noe, an groun. R V out V in Figure 9: A simple inuctor circuit (b) Write own the transfer function of the filter, H(ω). First, we convert everything into the phasor omain. We have, Z R = R = Ω (50) Z = jω = jω 10 3 H (51) In phasor omain, we can treat these impeances essentially like we treat resistors. Hence, Hence, the corner frequency ω = R = 106. Ṽ out = Ṽ out Ṽ in = H(ω) = jω R jω Z Z Z R Ṽ in (5) (53) = jω/( R ) 1 jω/( R ) (54) (c) Draw a straight-line approximation to the Boe plot (both magnitue an phase) of the filter on the attache graph paper. abel the horizontal an vertical axes clearly. (55) Miterm 1: Practice UB EES 16B, Spring 019. All Rights Reserve. This may not be publicly share without explicit permission. 13

14 H(ω) og-log plot of transfer function magnitue ω H(ω) Semi-log plot of transfer function phase ω Miterm 1: Practice UB EES 16B, Spring 019. All Rights Reserve. This may not be publicly share without explicit permission. 14

15 H(ω) og-log plot of transfer function magnitue ω H(ω) Semi-log plot of transfer function phase ω () Annotate your Boe plot with three ots, each representing where the straight line approximation has its worst errors. One ot shoul be on the magnitue plot, an two shoul be on the phase plot. abel each ot with the error at that point. Miterm 1: Practice UB EES 16B, Spring 019. All Rights Reserve. This may not be publicly share without explicit permission. 15

16 H(ω) og-log plot of transfer function magnitue Point ω H(ω) Errors for these points are: Semi-log plot of transfer function phase Point Point ω Point 1: Magnitue greatest error at ω = ω c = 10 6 ra/sec The actual value of H(ω = 10 6 ) = 1. The Boe plot approximation is H Boe (ω = 10 6 ) 1. The error at this point is: error 1 = 1 1 Point : Phase greatest error at ω = 0.1 ω c = 10 5 ra/sec The actual value of H(ω = 10 5 ) 84. The Boe plot approximation is H Boe (ω = 10 5 ) 90. The error at this point is: error 6 Point 3: Phase greatest error at ω = 10 ω c = 10 7 ra/sec The actual value of H(ω = 10 7 ) 6. The Boe plot approximation is H Boe (ω = 10 7 ) 0. The error at this point is: error 3 6 Miterm 1: Practice UB EES 16B, Spring 019. All Rights Reserve. This may not be publicly share without explicit permission. 16

17 ( ) (e) Write own the steay state response of this filter to an input voltage V (t) = 10cos 10 6 t π 3 V. In the given input, our ω = 10 6, an at steay state response, the magnitue of our transfer function is 1 an the phase is 45 (or π 4 raians), hence the output will be scale by this magnitue an shifte by π 4 : V out (t) = 10 ( cos 10 6 t π 3 π ) (56) 4 = 5 ( cos 10 6 t 7π ) (57) 1 NOTE: Above, we are not using the boe approximation of the magnitue at the corner frequency. 6. R ircuit: State-Space representation onsier the R circuit you saw in HW. R V R (t) V s (t) V out (t) i R (t) i (t) i (t) (a) et x 1 (t) = V (t) be[ the voltage ] on the capacitor, an let x (t) = i (t) be the current through the inuctor. et x(t) = x 1 (t) x (t). Write out the system of ifferential equations governing this circuit to express it as: t x(t) = A x(t) bv S (t) (58) Recall from class the equations: i c (t) = t V c(t) (59) V (t) = t i (t) (60) Using KV we can see that V s (t) = V R (t) V (t). This an the above equations allow us to write the following equations: V c (t) = V s (t) i R (T )R (61) i R (t) = i (t) i (t) = t V c(t) i (t) (6) V c (t) = V s (t) R( t V c(t) i (t)) (63) Miterm 1: Practice UB EES 16B, Spring 019. All Rights Reserve. This may not be publicly share without explicit permission. 17

18 Using the fact that the capacitor an inuctor are in parallel we know that: V (t) = V c (t) = t i (t). Plugging in V c (t) = x 1 (t) an i (t) = x (t), we get: Rearranging the equations to the esire format we get: x 1 (t) = t x (t) (64) x 1 (t) = V s (t) R( t x 1(t) x (t)) (65) t x 1(t) = x 1(t) R x (t) A = [ 1 b = V s(t) R t x (t) = x 1(t) R [ 1 R 0 ] ] (66) (67) (68) (69) (b) What are the eigenvalues of A in terms of R,,? A = haracteristic equation to fin the eigenvalues: [ 1 R λ 1 1 λ ] (70) (λ)( 1 R λ) 1 = 0 (71) λ λ 1 R 1 = 0 (7) (73) Using the quaratic formula we get: λ = 1 R ± ( 1 R ) 4 (74) ontributors: Sirej Dua. Miterm 1: Practice UB EES 16B, Spring 019. All Rights Reserve. This may not be publicly share without explicit permission. 18

19 Jaijeet Roychowhury. Anant Sahai. Harrison Zheng. Nikhil Shine. Regina Eckert. Kris Pister. Geoffrey Négiar. Miterm 1: Practice UB EES 16B, Spring 019. All Rights Reserve. This may not be publicly share without explicit permission. 19