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HW6_Soln

HW6_Soln - BENG 1863 Winter 2010 Homework 6 Solutions...

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Unformatted text preview: BENG 1863 Winter 2010 Homework 6 Solutions Problem 1 [30 points]: The electrical activity of the heart can be modeled, to first order, as a current dipole varying over time. This time-varying current dipole generates the ECG signal in space due to volume conduction. Assume a current dipole with two current sources I and —/, separated by a vector d (pointing from the center of —l to the center of l). The volume conductivity in the body is c. We place a reference electrode at a distance r from the center of the heart (dipole), and a signal electrode a short distance away from the reference electrode. We define vector r (with magnitude r) pointing from the dipole center to the location of the reference electrode, and vector a pointing from the reference electrode to the signal electrode. The voltage measured between the two electrodes is Va. a. Assuming that both d and a are much shorter than r, show that the voltage Va can be approximately written as in Eqn. (6.1): V6] = M - a. Find an expression for the cardiac vector M in terms of l, G, d, and r. b. Now we repeat the voltage measurement for three perpendicular lead vectors a1 (to the right), a2 (to the front) and a3 (to the top), giving Va1, V82, and V33 respectively. Show that these three lead measurements give estimates for the components of M. How would you use these three components, varying over time, to find estimates for the conventional ECG lead signals I, ll, Ill, aVR, aVL, aVF, V1, V2, V3, V4, V5, and V6 (Section 6.2)? 0. Where would you place electrodes to make these three lead measurements? Why don’t clinical practitioners use a simple and compact three-lead device, and instead use 10 electrodes across the body for 12 leads of the ECG signal? I i ( i I 2 ”It. \ 2L ’ oL‘ ) LHTO‘ I "‘ 2(7ng 1 +2”‘%9 (x i 3- ((l+ 4649) ’(I~ 339950) my FL 1’“ - T ' LN“ a} LfﬂF/Lg’ m :: at: ’L «.9 0" V (£0.00 WM» M > ._.) - w) l —‘J 7") ‘9 :. o D ' /L ’11 "‘ 0L 0L. , ﬂ ‘7 ~7 “7 , ‘9 33 .3; Va»: W m2) -W‘°) Mm >. my— mm. a , w» . v _ a f) .. 21 ~ ~ N :Laaf (If ’1)» )ob l<< } ’L’ W...) 7,4. ’7. _> m (flax) rv MOE: Lﬁmrt’ “7 ~» ‘3 “‘7 —> “ j: I ,> \{Ob ~ m ,(m w ~ M , w M Wm; HWoLCoWWWLCVme Magma; RYAN“ 10k (Lamina CMXU» wok L Lav/(X 05 ‘ \on M Ume. Cymbgwh .M/PL )9 oQquu/o mi 1?, (we to , k“ Mada (5“wa WV amt Z20» rnm bu WXUQMUV‘K "new owiauzﬁccb a; WW~MM§FW mm mm” ‘ Problem 2 [20 points]: Consider the driven-right-leg (DRL) circuit in Figure 1, connected as shown to the LA, RA, LL, and RL terminals of the body. a. Show that this circuit drives the Wilson central terminal of the body towards the ground potential. Where is the Wilson central terminal in this circuit? b. Find the effective resistance between the RL terminal and ground, assuming that all terminals LA, RA, LL, and RL of the body are at the same common-mode potential. c. If R0 = 1 M9, RRL = 100 kg), and the op amp power supplies are +1.2V and -1.2V, what is the maximum displacement current for which the DRL circuit will operate? RL LL l—x/W—/ Auxiliary op amp 3 Wm ”M + m \fwcT ’-‘ 3 :3 WW : \S’W CT Will/)9“ (,9va \WM WW3 Rm»? \l/oV 2’ 0 w cue/Maw _. “I O 3 N, 3 V) f0 : "' ”6—? d” W CT Kw / K. ' ’1 \W’Lb WNCT : R), («at r1, U, .. K KL. ~. D NUT ’ bat Md: K 3 "” 37B 7 % % [+3.9 Km 6. Turn WNW“ L'OL 59¢ WW ”3%” ow M MM: ) c.e. may“? ”KM fox/Jr i» M‘W -/»2‘( Mat “'2‘, ; W3 '25 WT” 458(30sz \$3»ka («r ”0) I RF»; JWCT :30 T 4—~—- 4 \$35 (ML (“L WW (1320 RO*RF1_ ’ 1M4“ EKAX myzwv (Wk/1W): #3 >' I MJL M FM: 7 3OLJZ. ' Rita). K0 , r 6m : - (Km 2+ f»: ()+K§)>LCL — - 12m. 00L Problem 3 [20 points]: A biopotential amplifier has a differential gain of 60dB. Each of the inputs of the amplifier has an input resistance of 100 M0. The amplifier is connected through electrodes to the body to measure a differential signal. The impedance of the electrode on the non-inverting input is 115kQ, and the impedance of the electrode on the inverting input is 130kQ. a. Find the common mode rejection ratio (CMRR) of the biopotential amplifier, due to the mismatch in impedance of the electrodes. You may assume that the amplifier circuit by itself has zero common mode gain. b. Now account for the effect of amplifier input capacitance, 10pF from each of the amplifier inputs to ground. Find the CMRR as a function of frequency, assuming the differential gain is constant. What is the corner frequency? CO ﬁnd hm CMW‘ (<3,ij) / - ' ‘ \lp I :~é—lﬂ——— Vt W\ 4;: \L‘u’ﬁ» + 33 VP ' V117 ; 2: iv“; ,. Z lv" M a“? «“21. ah/t , 4 _~ - -———\ a“; l ‘w -‘> “A V93. A ,, x ,, w a ‘ (-9 :3 3 \f‘ x "m. L; “he L M I\ R om ilk 'ﬁ/M_, @1th <3 é" it‘)"t’<.imL~«/2 w] , , D 9% :2 x m ' . T] th : IUU Mil I #. , km“ i m ~—-.._... ‘: "3 N z? i ” }'A:)(,.U"\r2 - {'(JUK it)” I «”4; x» H. 3 !+ m2 {jdz‘ xx mé‘y'w} ”7:3 WWW—“Mm” I g" C7 m . ‘ , ‘ "“" iL,r‘/H\R u [w ULﬁLxg—7 ‘ erLi>m\$w{ Problem 4 [30 points]: Consider the non-contact biopotential electrode, buffer, and amplifier shown in Figure 2. The electrode is a metal plate separated from the skin by a distance d = 1 mm, and with surface area A = 1 cm2. The shield is a second metal plate, at a distance t= 0.8 mm from the electrode, andwith the same surface area. The medium in between the skin and the electrode is air (with relative dielectric permittivity s, = 1), and the medium in between the two plates of the electrode and shield is FR4 printed circuit board (with a, = 4). a. From these parameters, estimate the coupling capacitance CC between the skin and electrode, and the shield capacitance Cs between the shield and electrode. For small signals (such as EEG and ECG), the two diodes can be modeled as a linear resistor, R, = 1T9. What is the function of the diodes other than a high resistance? Show that the first stage of the circuit is a high-pass filter with unity gain. Find the transfer function of voltage gain from the skin biopotential V8 to the buffer output and shield potential vb. Does it depend on the shield capacitance Cs? Why is the shield needed? Show that the second stage of the circuit is a low-pass filter with non-inverting gain. Find the transfer function of the voltage gain from the amplifier input vb to the amplifier output v0, for R6] = 1kQ, Rf: 1M9, and Cf: 160pF. Sketch the overall transfer function, from the skin biopotential V3 to the amplifier output v0, as a function of frequency. Indicate the midband gain and the corner frequencies. Now consider the effect of diode capacitance and buffer input capacitance, modeled as a capacitance C,- from v,- to ground. How does this affect the voltage gain of the buffer? How would you modify the circuit to cancel this effect? Hint: use the principle of the negative— input-capacitance amplifier in Section 6.6. You already have a capacitor from the buffer output to the electrode input, in the form of the shield capacitor Cs. Skin Electrode Shield R Highpass Lowpass d buffer amplifier Figure 2: Non-contact electrode, buffer, and amplifier in Problem 4. . ~ ’11. 7. Cc: 206/1213 Xilﬁ’uf 1 Lo M ; g3 ~13,“ ‘3 at.” M 10‘3”" L9 r “08ng C : (E E A '8 L041]? ' loﬂ‘fmz ' "1 5 Off: ‘3 m k=tr+rw fzwrF ”f “\‘O 5W" OXIOJ’W‘ ovm\ 09 (5.3003 Avila/WK W54)? 0W2) WE M0 139» ohm/«AV, W aka/3 ml” {Mk ( 9 ﬂak \ﬂr I I I ’ + ’mﬁokcé ’ + r W A d , \ WM» “HF ” "" NM’MKVD WWW «9»: PW- km wk Aw KC), o’h \ij maﬁa: O awcc (69/ 66) + bwcs (\fﬁ'6i) " [3“) CL: (L: ‘ \ .AW’ W W ...
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HW6_Soln - BENG 1863 Winter 2010 Homework 6 Solutions...

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