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y'' - y' = 0 Consider the following differential equation to be solved using a power series. Using the substitution y = n = 0 cn x^n, find an expression for ck+2 in terms of ck+1 for k = 0, 1, 2,

y'' - y' = 0 Consider the following differential equation to be solved using a power series. Using the substitution y = n = 0 cn x^n, find an expression for ck+2 in terms of ck+1 for k = 0, 1, 2,

y'' - y' = 0 Consider the following differential equation to be solved using a power series. Using the substitution y = n = 0 cn x^n, find an expression for ck+2 in terms of ck+1 for k = 0, 1, 2,

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y'' - y' = 0 Consider the following differential equation to be solved using a power series. Using the substitution y = n = 0 cn x^n, find an expression for ck+2 in terms of ck+1 for k = 0, 1, 2,

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y'' + xy = 0 Consider the following differential equation to be solved using a power series. Using the substitution y = n = 0 cn x^n, find an expression for ck+2 in terms of ck-1 for k = 1, 2, 3,
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For the amplifier below, following design information is provided: VDD = 3.3 V, R = 2.4 kΩ, ISS = 1.8 mA, k1 = k2 = 5 mA/V2, VTH,1 = VTH,2 = 0.5 V, λ = 0.1 V−1, γ = 0, ki′ = μ0Cox, ki = μ0 Cox(W L)i. Assume that both transistors are identical and in the saturation region. Finally, NMOS and PMOS bulk terminals are connected to the ground and VDD, respectively. a) What is the DC bias voltage for M1 and M2 given that the voltage drop across ISS is 0.4 V ? You can neglect the channel length modulation, here. Check that the transistors are in the correct operation-mode. b) The resistors will be realized using identical PMOS transistors M3 and M4. If the transistors to be used will have k3 = k4 = 1.5 mA/V2, what will be their overdrive voltages? Check that the transistors are in the correct operation-mode. c) What is the differential voltage gain AV,diff = Vout/Vid in dB ? d) ISS will be realized as a cascode current source. Assume that the voltage drop of 0.4 V is equally shared by the two transistors, M5 and M6 having k5′ = k6′ = 0.5 mA/V2 and VTH, 5 = VTH, 6 = 0.5 V. Assuming the gate voltages of both transistors are 1 V, find the W/L ratios of each device. Once again, you can neglect the channel length modulation here. e) Considering (d), how would the W/L ratios qualitatively change (increase/decrease/stays the same), if γ > 0 ? Explain your reasoning analytically, i. e. , using appropriate equations. f) Using the cascode current source, calculate the common-mode gain of the circuit AV, CM = Vout /Vin in dB, where Vin . s applied to the gates of both M1 and M2. Assume λ = 0.1 V−1 for M5 and M6. g) Calculate the CMRR of the circuit in dB.
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This is a short reading exercise on reindexing. There is no standard name to this operation. Some people call it shifting the summation. Consider the notation for a series P∞ n=1 an. The symbol n is called the index of summation. The lower bound of the series is 1 . If we write P∞ n=u bn, then the lower bound of the series is u. Reindexing means to change lower bound of the series without changing the series. For example, the series 1 + 1 4 + 1 9 + 1 16 + 1 25 + • • • can be written as P∞ n=1 1 n2 or P∞ n=3 1 (n−2)2 . Please verify this by writing down the first few terms of both series. Therefore, we have an equality between series X∞ n=1 1 n2 = X∞ n=3 1 (n − 2)2 Notice we DO NOT simply replace n = 1 by n = 3. The expression for the terms of the series also have to change for the equality to hold. To reindex by symbolic manipulation, we introduce a change of variable. In this example, we start with the series P∞ n=1 1 n2 . We wish to reindex from n = 1 to n = 3, meaning the lower bound of the series changes from n = 1 to n = 3 without changing the series. We introduce m = n + 2 (because m = 3 when n = 1 ). Isolate for n to get n = m − 2, and we replace every occurrence of n in the series by the expression on the right-hand side (which is m − 2 in this example). The series becomes X∞ n=1 1 n2 = X∞ m−2=1 1 (m − 2)2 ( replace n by m − 2) = X∞ m=3 1 (m − 2)2 ( reorganize the lower bound of summation) = X∞ n=3 1 (n − 2)2 (change m back to n ) In general, if we wish to reindex a series P∞ n=u an from n = u to n = v, we introduce a change of variable m = n+(v−u) and perform similar manipulations as shown above. In the example above, u = 1 and v = 3. Let r be a non-zero constant. You are asked to reindex the series P∞ n=1 1 rn−1 from n = 1 to n = 0. What is the result of reindexing? Select one or more: a. P∞ n=0 1 rn+1 b. P∞ n=0 1 rn−1 c. P∞ n=0 1 rn

This is a short reading exercise on reindexing. There is no standard name to this operation. Some people call it shifting the summation. Consider the notation for a series P∞ n=1 an. The symbol n is called the index of summation. The lower bound of the series is 1 . If we write P∞ n=u bn, then the lower bound of the series is u. Reindexing means to change lower bound of the series without changing the series. For example, the series 1 + 1 4 + 1 9 + 1 16 + 1 25 + • • • can be written as P∞ n=1 1 n2 or P∞ n=3 1 (n−2)2 . Please verify this by writing down the first few terms of both series. Therefore, we have an equality between series X∞ n=1 1 n2 = X∞ n=3 1 (n − 2)2 Notice we DO NOT simply replace n = 1 by n = 3. The expression for the terms of the series also have to change for the equality to hold. To reindex by symbolic manipulation, we introduce a change of variable. In this example, we start with the series P∞ n=1 1 n2 . We wish to reindex from n = 1 to n = 3, meaning the lower bound of the series changes from n = 1 to n = 3 without changing the series. We introduce m = n + 2 (because m = 3 when n = 1 ). Isolate for n to get n = m − 2, and we replace every occurrence of n in the series by the expression on the right-hand side (which is m − 2 in this example). The series becomes X∞ n=1 1 n2 = X∞ m−2=1 1 (m − 2)2 ( replace n by m − 2) = X∞ m=3 1 (m − 2)2 ( reorganize the lower bound of summation) = X∞ n=3 1 (n − 2)2 (change m back to n ) In general, if we wish to reindex a series P∞ n=u an from n = u to n = v, we introduce a change of variable m = n+(v−u) and perform similar manipulations as shown above. In the example above, u = 1 and v = 3. Let r be a non-zero constant. You are asked to reindex the series P∞ n=1 1 rn−1 from n = 1 to n = 0. What is the result of reindexing? Select one or more: a. P∞ n=0 1 rn+1 b. P∞ n=0 1 rn−1 c. P∞ n=0 1 rn
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