15 n=0. zz = re jθ re jθ = r 2. (b) For division and multiplication, it is handy to use the polar representation: z = rejθ. = z 1z 2.

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1 Professor Fearing EECS0/Problem Set v.0 Fall 06 Due at 4 pm, Fri. Sep. in HW box under stairs (st floor Cory) Reading: EE6AB notes. This problem set should be review of material from EE6AB. (Please note, vector notation here is x = x, u is a scalar, and A is a matrix.). (8 pts) Complex review. Given z = x + jy = re jθ. Derive the following relations: a. zz = r z b. z = e jθ c. (z z ) = z z d. ( z z ) = z z e. Show that (ejωt + e jωt ) = cos ωt f. Find all integer k {0,,...5} for which 5 ejπnk/6 = 0. Solution: (a) Since z = x jy = re jθ, we have zz = re jθ re jθ = r (b) For division and multiplication, it is handy to use the polar representation: z z = rejθ re jθ = ejθ (c) Let z = x + jy = r e jθ, and z = x + jy = r e jθ, then we can show that: (z z ) = (r r e j(θ +θ ) ) = r r e j(θ +θ ) = r e jθ r e jθ = z z (d) Similar to (c), we can derive that: ( ) ( ) z r = e j(θ θ ) = r e j(θ θ ) r r z = r e jθ r e jθ = z z By the way, according to the Euler s formula, for z = x + jy = re jθ, we have x = r cos θ, y = r sin θ. (e) We apply the Euler s formula to expand the left hand side (LHS): (ejωt + e jωt ) = ejωt + e jωt = cos ωt (f) If k = 0, e jπnk/6 =, and the sum 5 ejπnk/6 is equal to 6, which clearly doesn t satisfy the condition. For k 0, the sum of geometric series is given by: 5 e jπnk/6 = ejπk e jπk/6 = 0 since e jπk = for any integer k. Therefore k {,...5} will satisfy the condition.

2 . (4 pts) Phasors and Operational Amplifiers Consider the circuit in Fig.. Use the golden rules (ideal) ideal op amp assumptions. a. Using phasors, determine the transfer function H = Vo V i. b. Sketch the magnitude Bode plot (log-log scale) for H(ω). c. What filter function is performed (low pass, high pass, etc.)? Solution: (a) Recap that for any sinusoidal time-varying function x(t), it can be represented in the form x(t) = RXe jωt, where X is a time-independent function called the phasor counterpart of x(t). x(t) is defined in the time domain, while phasor X is defined in the phasor domain. Step : To use the phasors, first denote the phasors of v i (t), v o (t) as V i, V o. Step : Transform the circuits components to phasor domain, i.e., impedance, using the formula: Z R = R for resistors, Z C = for capacitors, and Z L = jωl for inductors, where R, C, L are the resistance, capacitance, and inductance, and ω is the frequency, as is shown in Fig. sol. Step 3: Use KCL and/or KVL equation in the phasor domain, where we apply the golden rule of the ideal op amp to obtain the phasor V = 0 for v (t) = v (t) = v + (t) = 0 (ground). By KCL, and let R = 00kΩ, C = µf, we have: V i R + = Step 4: Obtain the transfer function H = Vo V i : H = V R + R o R+ = V i R + = jω/ω c + (jω/ω c ) (jω/ω c + ) where ω c = RC = 5Hz. (b) The Bode plot is shown below. V o R + R R+ R R + R+ = R + = R ω C R (R + ) (c) The filter is performing high pass. 3. (8 pts) State space For the circuit shown in Fig., let the state variables x be v c (t), and x be the inductor current.

3 The input u = v i (t) and output y(t) is the voltage across the MΩ resistor. Let C = 0 6 F and R = 0MΩ. a. Write the differential equation for the circuit in state space form (find A,B,C,D): ẋ = Ax + Bu y = Cx + Du b. Determine the eigenvalues for A. Is the system stable? Solution: (a) First, using the physical properties of the inductor and capacitors, and let x = v c (t), x = i L (t), we have i c (t) = C vc(t) = C dx, and v L(t) = L di L(t) = L dx, where i c (t), v L (t) are the current and voltage across the capacitor and inductor, respectively. ow, use the KCL and KVL, and u = v i (t), we have: u = L dx + R C dx + x x = C dx + R C dx + x where R = MΩ. Rearranging the above equations for dx and dx, we have: Therefore, using x = x x T, we have: Also, we have dx = R C + R C x R + R C + R C x dx = L x + L u + R R L + R L x R R R L + R L x ẋ = R C+R C R R R C+R C L + R R L+R L R R R L+R L 0 x + u L y(t) = R C dx + x = R x + R R x + x R + R R + R = R R R R +R R +R x R Therefore, we have A = R C+R C R C+R C 0 6 L + R R L+R L R R = R L+R L, B = 07 R R R R +R R +R = 0 7, and D = 0. (b) To obtain the eigenvalues, we use the characteristic equations: ( ) λ 0 6 det(a λi) = det 07 λ = λ L + 06 = 0, C = Therefore, the eigenvalues are λ = 0., λ = Since both of them are negative values, the system is stable. 3

4 Fig. Fig. 4

5 4. (0 pts) LDE solutions Consider ẋ = Ax where initial condition x o = 5 0 and A = 0 0 a. Show that the general solution is c e λ t v + c e λ t v where λ, λ are eigenvalues of A with corresponding eigenvectors v, v, and find eigenvalues and eigenvectors. b. Plot each component of the solution and the phase portrait. The phase portrait is a dimensional plot of x (t) vs x (t). (Hand sketch is fine.) Solution: (a) To show that c e λt v + c e λt v is the general solution to the LDE, simply substitute the solution to the differential equation: LHS = d ) (c e λt v + c e λt v = c λ e λt v + c λ e λt v ) RHS = A (c e λt v + c e λt v = c e λt Av + c e λt Av = c λ e λ t v + c λ e λ t v = LHS where we use the fact that v and v are eigenvectors of A. To find the eigenvalues of A, we use the characteristic equation: ( ) λ det(a λi) = det = λ + λ + 0 = (λ + )(λ + 0) 0 λ Therefore the eigenvalues are λ =, λ = 0. To find the eigenvectors, let v =, then we v v have Av = v =, therefore, v 0v v v = T w. Similarly, let v =, we w w have Av = w = 0, so v 0w w w = T. (b) First, let s find the solution to the LDE given the initial condition. c + c = 5 0 c 0 0 c = 0 v where we find c = 0 3, c = To plot the phase portrait, we start from the initial condition 5 x o = and plot the x 0 (t) = 0 3 e t 5 3 e 0t vs x (t) = 0 3 e t e 0t components. ote that as t, both components decacy to 0. Here we show one way of ploting it, by first sketch the evolution of x (t) (bottom) and x (t) (top right), and then find corresponding points on the phase portrait figure. 5

6 5. (0 pts) DFT basics The Discrete Time Fourier Series (DTFS) is defined as ak = xne jπnk = xnw nk where W e jπ/, and k {0,,..., }. a. Show that the DTFS can be written as a = Ux for = 4, and write U in terms of W 4. b. What special property or properties do the columns of U have? c. For = 4, and xn = cos πn, find a in terms of W 4 (simplify). d. For = 8, and x = T, find a in terms of W 8 (simplify). W W W Solution: (a) By the definition of DTFS, U = W W 4 W ( ).... W For a = a 0 a a T and x = x 0 x x T, we have a = Ux. For = 4, we have: U = W 4 W4 W W4 W4 4 W4 6 W4 3 W4 6 W4 9 W ( ) W ( )( ) (b) Since it can be verified that U is orthogonal matrix, the columns of U are orthogonal among each other. The orthogonality can be seen by taking the dot product of column k and p: U :,k U:,p = W nk W np = 0, k p. 6

7 which follows by the reasoning of prob (f). (c) By Euler s formula, xn = W n 4 + W n 4, therefore, ak is given by: ak = 3 ( 4 W 4 n + ) W 4 n W4 nk { 0 k, 3 = k =, 3 where we use the fact that W 4n 4 =. Therefore, a = 0 0. (d) By the definition of DTFS: ak = 8 7 xnw nk 8 = 8 W 4k 8 = 8 ( )k 7