ECE 45 Discussion 2 Notes

Transcription

1 UC San Diego J. Connelly Frequency Response ECE 45 Discussion Notes The inputs and outputs of RLC circuits are generally either voltages or currents. The output of the circuit depends on the frequency of the input. An input/output RLC circuit is defined by its frequency response (also called the transfer function): H(ω) Out(ω), where Out(ω) and In(ω) are the phasor transforms of out(t) and in(t), In(ω) for an arbitrary frequency ω. The inputs and outputs of RLC circuits are generally either voltages or currents. In general, out(t) in(t)h(ω) (a common mistake in this course). H(ω) is a complex number, which is a function of ω, since input/output relationships change with different frequencies. H(ω) ( ) Im{H(ω)} Re{H(ω)} +Im{H(ω)} and H(ω) tan. Re{H(ω)} Suppose x(t) is sinusoidal, x(t) A cos(ω 0 t + φ), so its phasor representation is Ae jφ. If x(t) is the input to an RLC circuit with frequency response H(ω), then the output phasor is Y A H(ω 0 ) e j(φ+ H(ω 0)), so y(t) A H(ω 0 ) cos(ω 0 t+φ+ H(ω 0 )) The frequency response describes how the system responds to a sinusoid of a particular frequency. When x(t) A i cos(ω i +φ i ), by super position, the output is y(t) A i H(ω i ) cos(ω i +φ i + H(ω i )) This allows us to analyze RLC circuits whose inputs are voltages and currents consisting of multiple frequencies. Please report any typos/errors to jconnelly@ucsd.edu

2 Example Find the frequencyh(ω) of an RLC circuit, with inputx(t) and outputy(t), given by d y(t) dt +x(t) dx(t) dt Determine the output when the inputx(t) cos(t π/4). +y(t). Assumex(t) is sinusoidal, then we can represent the differential equation using phasors: (jω) Y +X jωx +Y H(ω) Y X jω (jω) jω +. When x(t) cos(t), the output is given by y(t) H() cos(t π/4+ H()) where H() 5 and H() tan (). Example (a) Find the frequency response,h(ω), of the circuit below with inputi in (t) and outputi o (t). (b) Find the valueω 0 for which H(ω) is maximized. (c) Find the valueω > 0 such that H(ω ) H(ω 0 ) / (d) With the value ofrfixed, what value shouldltake so that we have H(0 4 ) H(ω 0 ) / i in (t) i o (t) L R R 00kΩ, L mh (a) We can use a current divider to find the current through the inductor: I o I i Z R + Z R + H(ω) I o I i +jω(l/r)

3 (b) To maximize H(ω) with respect to ω, note that H(ω) +(ωl/r) H max H(0) (c) H(ω ) +(ω L/R) +(ω L/R) ω R/L 0 5 / ω R/L 0 5 /L L 0H Example 3 R R v (t) C v o (t) v (t) R Ω, R Ω, andc /F. (a) Whenv (t) 0 andv (t) is the input to the circuit, find the frequency responseh (ω) V o /V. (b) Whenv (t) 0 andv (t) is the input to the circuit, find the frequency responseh (ω) V o /V. (c) Find the outputv o (t) when v (t) cos(3t+π/3) andv (t) 3sin(t) (a) When v (t) 0, by using a voltage divider on the phasor-transformed circuit, we have (R // V o V ) jωc R +(R // ) V R R jωc +R +jωcr R H (ω) V o V 3+jω 9+ω e jtan (ω/3) (b) Similarly, whenv (t) 0, by using a voltage divider on the phasor-transformed circuit, we have (R // V o V ) jωc R +(R // ) V R R jωc +R +jωcr R H (ω) V o V 3+jω 9+ω e jtan (ω/3)

4 (c) We know that when v (t) 0 andv (t) cos(3t+π/3), the output is v o, (t) H (3) cos(3t+π/3+ H (3)) and whenv (t) 0 and v (t) 3sin(t), the output is v o, (t) 3 H () sin(t+ H ()). So by super-position, the output is v o (t) v o, (t)+v o, (t) cos(3t+π/3 tan ()+ 6 sin(t tan (/3)) cos(3t+π/)+ 6 3 sin(t tan (/3)) Example 4 Suppose the output of an RLC circuit is 6+ω cos(ωt tan (ω/4) when the input iscos(ωt), whereω 0. (a) What is the output when the input is Acos(ωt+θ) for some real numbersa,θ? (b) What is the output when the input is +sin(4t)? (a) When the input to an RLC circuit iscos(ωt), the output is H(ω) cos(ωt+ H(ω)). Hence H(ω) e jtan (ω/4) 6+ω and the output when Acos(ωt+θ) is A 6+ω cos(ωt+θ tan (ω/4)). This is a useful trick to use in general. That is, we can find the output when cos(ωt) is the input, then apply any scaling and shifting after the fact. This can simplify some of the phasor analysis of circuits. This utilizes the linearity and time invariance of an RLC circuit.

5 (b) Let x (t) and x (t) sin(4t), and suppose y (t) and y (t) are the outputs when x (t) and x (t), respectively, are the inputs. Then by super position, when x (t) +x (t) is the input, y (t)+y (t) is the output. x (t) is the case in (a), wherea,ω 0, and θ 0, soy (t) 4. x (t) is the case in (b), wherea, ω 4, and θ π/, so y (t) 3 cos(4t π/ tan ()). Thus the desired output is + 4 sin(4t π/4) Example 5 For an RLC circuit with frequency response { jω ω / H(ω) 0 otherwise find the outputy(t) when the input is x(t) k0 +k cos(kt/3). For k 0,,,..., let x k (t) +k cos(kt/3), and let y k(t) be the output of the system when x k (t) is the input. Then, sinceh(ω) is the frequency response of an RLC circuit, y k (t) H(k/3) +k cos(kt/3+ H(k/3)) and we have H(k/3) 0 fork, andh(0), andh(/3) j/3 +4/9e jtan ( /3), so 7 y 0 (t) and y (t) 6 cos(t/3 tan (/3)) By super-position/linearity, we have y(t) y k (t) + k0 7 6 cos(t/3 tan (/3)).