20095ee110_1_hw1_solutions - Problem 1(a x(t =10...

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Unformatted text preview: Problem 1 (a) x(t) =10 cos(1000t‐60o) X = 10 exp{‐j60o} (b) x(t) =0.1 sin(1000t‐60o)= 0.1 cos(1000t‐90o ‐60o) X = 0.1 exp{‐j150o} (c) x(t) = 3 sin(400πt‐270o)= 3 cos(400πt‐90o‐270o)= 3 cos(400πt) X = 3 exp{j0o} (d) x(t) = 3 cos(50πt‐60o)+4 sin(50πt‐60o)= 3 cos(50πt‐60o)+4 cos (50πt‐90o‐60o) = 3 cos(50πt‐60o)+4 cos (50πt‐150o) X = 3 exp{‐j60o} + 4 exp{‐j150o} = (1.5‐2.598j)+(‐3.464‐2j) =‐1.9641‐4.598j =5 exp{‐j113.13o} (e) x(t) consists of two sinusoidal waveforms at different frequencies, and so no phasor representation exists. Problem 2 (a) X x(t) = 3+4j = 5 exp{j53.13o} = 5 cos(ωt + 53.13o) (b) X x(t) = ‐j = 1 exp{‐j90o} = cos(ωt‐90o) (c) X x(t) = 10 exp{‐j(π/6)} = 10 cos(ωt+ π/6) = 10 cos(ωt+30o) (d) X x(t) = 4 exp{j(5π/6)} = 4 cos(ωt+ 5π/6) = 10 cos(ωt+150o) (d) X x(t) = 4 exp{‐j(5π/6)} = 4 cos(ωt‐5π/6) = 10 cos(ωt‐150o) = 10 cos(ωt+210o) Problem 3 (a) (b) We can simplify the circuit as follows: If and , the transfer function from Vin to Vc is: A zero phase‐shift requires that: To satisfy this equation: (c) If L =1mH, the parallel inductance and capacitor present an infinite impedance. And so the resistor current evaluates to: Since no current flows through the resistor, there is no voltage drop across the resistor. This means that . Problem 4 Applying Mesh analysis to the above circuit, we get the following simultaneous equations: Solving for each current: Therefore: Problem 5 (a) (b) Rewriting the impedance in rectangular form, we notice that the imaginary component of the impedance is negative. A negative imaginary component implies a capacitance, and so the “black‐box” must be a RC series circuit. (c) Problem 6 (a) (b) Applying Mesh analysis to the above circuit, we get the following simultaneous equations: Solving for and (c) To obtain the Norton equivalent current, short output and calculate the output current, i.e. applying mesh analysis to the above circuit gives And solving for the output current gives To get the Norton equivalent impedance, short the voltage source and remove the current source: ...
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