8pm, 28th November, 2012 Marking scheme and brief answers: http://www.ee.iitb.ac.in/~belur/ee225/endsem-soln.txt Any mistake purely calculational: at most 0.5 marks are deducted (out of 100). ---------------------------------------------- Answer to Q1 of end sem: Foster I: (2 marks) Z(s) = s + 1/((2/3)s) + (s/2)/(s^2 + 2) L1 = 1, C1 = 2/3, L2 = 1/4, C2 = 2 Foster II: (2 marks) Y(s) = (s/2)/(s^2 + 3) + (s/2)/(s^2 + 1) L1 = 2, C1 = 1/6, L2 = 2, C2 = 1/2 Cauer I: (3 marks) Z(s) = s + 1/[0.5s + 1/{4s + 1/((1/6)s)}] L1 = 1, C2 = 0.5, L3 = 4, C4 = 1/6 Cauer II: (3 marks) Z(s) = 3/(2s) + 1/[4/(5s) + 1/{25/(2s) + 1/(1/(5s))}] C1 = 2/3, L2 = 5/4, C3 = 2/25, L4 = 5 ---------------------------------------------- Answer to question 2: A: A transfer function G(s) is called positive real if - all its poles are in CLHP - Re[G(s)]>=0 for all s in ORHP (This is *one* of the definitions. Any other (equivalent) definition is ok too.) (It turns out that such conditions are equivalent to G(s) being the driving point immittance of some RLCT network.) B: No, because degree relative degree of G(s) is 2. For a positive real G(s), its Nyquist plot is in the CRHP. If relative degree is 2, then the Nyquist plot will have to come into the left half complex plane while it is approaching the origin. (This is due to angle becoming 180 degrees asymptotically as omega tends to infinity.) C: Poles in ORHP imply the existence of sources, as some state variables blow up in time. Also, the impulse response keeps growing, suggesting sources. Since positive real is same as no sources, all poles have to be in CLHP. C: No zeros in ORHP because - inverse of G(s) is also positive real, whose poles are the zeros of G(s). Hence, zeros of G(s) can't be in ORHP either. - Suppose G(s) has a zero s=q in ORHP, meaning G(q)=0. As G(s) is a continuous function of s, there exists a neighborhood around q where Re[G(s)] is both negative and positive. So, Re[G(s)]<0 for some s in ORHP, which contradicts the positive realness of G(s). (There can be other correct reasons too.) (Marks for Q2: A:2, B:2, C:2+2+2 (For C, 2 marks per reason.) ---------------------------------------------- Answer to Q3: (Tellegen's theorem and proof): Statement: 4 marks, proof 6 marks. In proof, rank counts will fetch 3 marks. (Rank count should aim to prove *complementarity*, in addition to orthogonality (below).) (Rank count is also same as showing v = A^T v_N (node potentials) and i = B^T i_l (loop currents). Showing A*B^T = 0 : 3 marks. ---------------------------------------------- Answer to Q4: (a) Question is similar to the one that was given in the Mid-semester exam. Basic idea is that if there needs to be at least one loop around which resistances add up to zero. (This is a sufficient, but not necessary, it appears.) 5 marks for this question. (We will think of partial marks, but no guarantee.) (b) If all resistances to be chosen are positive, then singular is ruled out. 2 marks for stating this, and 3 marks for reason. One reason: if all resistances are positive, and that matrix is singular, then, at least one resistance is absorbing nonzero power, but all resistances being positive also means no device is *supplying* power. Hence by conservation of power the contradiction tells us matrix had better be nonsingular. (3 marks for this reason). A more elaborate reason, by using Tellegen's strong from, we eventually arrive at a loop impedance matrix which was shown to be positive definite (and hence nonsingular) in the class. That reasoning is the more rigorous one. (3 marks for this reason). ---------------------------------------------- Answer to Q5 Key concept behind this question is: For DC, capacitor becomes open ckt and inductor becomes short ckt. For High Freq., capacitor becomes short ckt. and inductor becomes open A) For given ckt, if we apply DC then capacitor becomes open and equivalent ckt. will remain with 2 resistors of resistance R, so Z(0)=2R. If we apply high frequency then capacitor becomes shorted and inductor will get opened. So equivalent ckt. will remain with 3 resistors in which 2 are in parallel, so Z(infinity)=3R/2 B) For given ckt, if we apply DC then capacitor becomes open and equivalent ckt. will become open ckt, so Z(0) = infinity. If we apply high frequency then capacitor becomes shorted and inductor which is in series with capacitor will get opened. So equivalent ckt. will again remain open ckt, so Z(infinity) = infinity. (5 marks for A, 5 marks for B.) ---------------------------------------------- Answer to Q6: A: 1 mark, B: 5 marks, C:2, D:1+1 For A, with different definition (about sign of I_2) also you get 1 mark. For C, if you choose sign of I_2 differnetly and get AD + BC = 1 as equivalent to reciprocity, then you still get 2 marks.) For D: "cascade" actually refers to "chain", when used in the context of *2-port* interconnection. (But cascade also (ambigously) refers to series/parallel cascading, etc. Unfortunately, I missed this ambiguity. Any confusion about "cascade" will get 1 mark. If series is written wrongly, then zero mark, instead of 1. ---------------------------------------------- Answer to Q7. A: Just RC is not possible to have non-monotonic magnitude plot. (Magnitude will be always monotonically decreasing or increasing depending on impedance or admittance.) But if somebody has successfully realized using *negative* values of R and C, then you can get full marks. (Continuous fraction expansion procedure is systematic enough to get full marks, though values were not found fully.) Any other structure *assumed* may not work (unless numerical values have been calculated.) Please note: unspecified means just assume something or whatever specific value you get is fine. In case you are required to solve *in generality* of omega_1 and omega_2, I would ask realize the circuit and obtain parameters "in terms of omega_1 and omega_2". (Problem is too hard like that. No extra marks for trying so tough things. But let's see (slightly) liberally. ) B: Can realize. Not a unique answer. Method/approach will fetch surely. A+B : 5 + 5 marks. ---------------------------------------------- Answer to Question 8: A+B: 5+5 A: 2+3 for magnitude+phase (Overall shape: 1 for magnitude, 1 for phase. cut-off values of frequencies: 1 for phase. Initial and final values of magnitude+phase: 1 + 1) ---------------------------------------------- Answer to Q9. A +B : 5 + 5 A: correct answer: high pass. Reason as follows: A: The delay (or advance, if T is < 0) does not change high or low pass characteristics, since the same signal comes later or before. Hence assume T = 0 for the purpose of discussion of A. The impulse response is delta + another function, call that h_1(t). Convolution with just h_1(t) is clearly low-pass. (It is negative of another function, which just averages. Multiplying by -1 does not change high/low pass characteristics. Hence convolving with h_1 is low pass. (meaning, reduces high-pass). The delta(t) function's Laplace transform is 1. Let the Laplace transform of the h_1(t) be H_1(s). Let a function f(t) (with Laplace F(s), say) be convolved with the function h(t) given in the figure. Thus their Laplace transform is F(s) - F(s)H_1(s). As H_1(s) is a low pass system thus here we are extracting low frequencies from the function F(s). Thus the resulting system is high pass. (This is a little vague explanation, the DC gains sure play a role. But if the values are positive, etc, this should work.) B: If we take T greater than zero then we will have to introduce delays in the system which is not possible to introduce with the help of an inductor or a capacitor (or a *finite* number of memory elements). If T is less than zero, then it is a non-causal system, i.e. response to the input at zero time comes out *before* t = 0. Since some realization has been sought, causality is required. Hence only T=0 is allowed. So to get a RC, RL or a RLC circuit we need to take T = 0. Just RC should be possible. (If multiple memory elements are used, then non-uniqueness.) If just one exponent is assumed for h_1(t), then RC should be possible. This turned out to be relatively hard. If high-pass conclusion has been done after realizing RC or RL and using transfer function, then 9 out of 10 (since causality argument and finite-memory etc: 1 mark.) ---------------------------------------------- Answer to Question 10: (A + B = 5 + 5 ) The given T says: v1 = v2, and i1 = -i2. A: So, neither v1 and v2 can together be chosen independently, nor can i1 and i2 together be chosen independently. Hence (respectively) neither Y nor Z parameters exist. B: Take a network in which ports 1 and 2 are just connected straight almost like a T network, but without the vertical resistor and no resistor for the horizontal ones. In this network, two independent voltage sources would form a loop, and two independent current sources would form a cut-set. Hence two independent voltage sources cannot be put together (hence Y does not exist), etc. (Construction of this network was sought in the pre-endsem problem sheet.) --------------