Causas de Extinción de la Concesión

Exercise 3.11.1 (Solution on p. 116.)

Suppose the complex amplitudes of the voltage and current have xed magnitudes. What phase relationship between voltage and current maximizes the average power? In other words, how are φ and θ related for maximum power dissipation?

Because the complex amplitudes of the voltage and current are related by the equivalent impedance, average power can also be written as

P_ave= 1 2Re (Z) (|I|) 2 =1 2Re  1 Z  (|V |)2

These expressions generalize the results (3.3) we obtained for resistor circuits. We have derived a fundamental result: Only the real part of impedance contributes to long-term power dissipation. Of the circuit elements, only the resistor dissipates power. Capacitors and inductors dissipate no power in the long term. It is important to realize that these statements apply only for sinusoidal sources. If you turn on a constant voltage source in an RC-circuit, charging the capacitor does consume power.

Exercise 3.11.2 (Solution on p. 116.)

In an earlier problem (Section 1.5.1: RMS Values), we found that the rms value of a sinusoid was its amplitude divided by√2. What is average power expressed in terms of the rms values of the voltage and current (Vrms and Irms respectively)?

3.12 Equivalent Circuits: Impedances and Sources

24

When we have circuits with capacitors and/or inductors as well as resistors and sources, Thévenin and Mayer- Norton equivalent circuits can still be dened by using impedances and complex amplitudes for voltage and currents. For any circuit containing sources, resistors, capacitors, and inductors, the input-output relation for the complex amplitudes of the terminal voltage and current is

V = ZeqI + Veq

I = V Zeq

− Ieq

with Veq= ZeqIeq. Thus, we have Thévenin and Mayer-Norton equivalent circuits as shown in Figure 3.27

(Equivalent Circuits).

Equivalent Circuits veq + – Req + – v i Sources and Resistors + – v i + – v i Req ieq

Thévenin Equivalent Mayer-Norton Equivalent

(a) Equivalent circuits with resistors.

+ – V I Zeq Ieq Mayer-Norton Equivalent Veq + – Zeq + – V I Thévenin Equivalent Sources, Resistors, Capacitors, Inductors + – V I

(b) Equivalent circuits with impedances.

Figure 3.27: Comparing the rst, simpler, gure with the slightly more complicated second gure, we see two dierences. First of all, more circuits (all those containing linear elements in fact) have equivalent circuits that contain equivalents. Secondly, the terminal and source variables are now complex amplitudes, which carries the implicit assumption that the voltages and currents are single complex exponentials, all having the same frequency.

Simple RC Circuit Vin + – R C + – V I Figure 3.28

Let's nd the Thévenin and Mayer-Norton equivalent circuits for Figure 3.28 (Simple RC Cir- cuit). The open-circuit voltage and short-circuit current techniques still work, except we use impedances and complex amplitudes. The open-circuit voltage corresponds to the transfer function we have already found. When we short the terminals, the capacitor no longer has any eect on the circuit, and the short-circuit current Isc equals Vout_R . The equivalent impedance can be found by

setting the source to zero, and nding the impedance using series and parallel combination rules. In our case, the resistor and capacitor are in parallel once the voltage source is removed (setting it to zero amounts to replacing it with a short-circuit). Thus, Zeq = R k _{j2πf C}1 = _{1+j2πf RC}R .

Consequently, we have Veq= 1 1 + j2πf RCVin Ieq= 1 RVin Zeq= R 1 + j2πf RC

Again, we should check the units of our answer. Note in particular that j2πfRC must be dimen- sionless. Is it?

3.13 Transfer Functions

25

The ratio of the output and input amplitudes for Figure 3.29 (Simple Circuit), known as the transfer function or the frequency response, is given by

Vout

Vin = H (f )

= _{j2πf RC+1}1 (3.15)

Implicit in using the transfer function is that the input is a complex exponential, and the output is also a complex exponential having the same frequency. The transfer function reveals how the circuit modies the input amplitude in creating the output amplitude. Thus, the transfer function completely describes how the circuit processes the input complex exponential to produce the output complex exponential. The circuit's function is thus summarized by the transfer function. In fact, circuits are often designed to meet transfer

function specications. Because transfer functions are complex-valued, frequency-dependent quantities, we can better appreciate a circuit's function by examining the magnitude and phase of its transfer function (Figure 3.30 (Magnitude and phase of the transfer function)).

Simple Circuit vin + – R C vout + –

Figure 3.29: A simple RC circuit.

Magnitude and phase of the transfer function

-1 0 1 1 1 /√2 |H(f)| 1 2πRC 1 2πRC f (a) -1 1 π/ 4 –π/ 4 –π/ 2 π/ 2 ∠H(f) f 0 1 2πRC 1 2πRC (b)

Figure 3.30: Magnitude and phase of the transfer function of the RC circuit shown in Figure 3.29 (Simple Circuit) when RC = 1. (a) |H (f) | = _√ 1

(2πf RC)2₊₁ (b) ∠ (H (f)) = −arctan (2πfRC)

This transfer function has many important properties and provides all the insights needed to determine how the circuit functions. First of all, note that we can compute the frequency response for both positive and negative frequencies. Recall that sinusoids consist of the sum of two complex exponentials, one having the negative frequency of the other. We will consider how the circuit acts on a sinusoid soon. Do note that the magnitude has even symmetry: The negative frequency portion is a mirror image of the positive

frequency portion: |H (−f) | = |H (f) |. The phase has odd symmetry: ∠ (H (−f)) = −∠ (H (f)). These properties of this specic example apply for all transfer functions associated with circuits. Consequently, we don't need to plot the negative frequency component; we know what it is from the positive frequency part.

The magnitude equals _√1

2 of its maximum gain (1 at f = 0) when 2πfRC = 1 (the two terms in the

denominator of the magnitude are equal). The frequency fc = _2πRC1 denes the boundary between two

operating ranges.

• For frequencies below this frequency, the circuit does not much alter the amplitude of the complex exponential source.

• For frequencies greater than fc, the circuit strongly attenuates the amplitude. Thus, when the source

frequency is in this range, the circuit's output has a much smaller amplitude than that of the source. For these reasons, this frequency is known as the cuto frequency. In this circuit the cuto frequency depends only on the product of the resistance and the capacitance. Thus, a cuto frequency of 1 kHz occurs when 1

2πRC = 10

3 _{or RC =} 10−3

2π = 1.59 × 10

−4_{. Thus resistance-capacitance combinations of 1.59 kΩ and}

100 nF or 10 Ω and 1.59 µF result in the same cuto frequency.

The phase shift caused by the circuit at the cuto frequency precisely equals −π

4. Thus, below the cuto

frequency, phase is little aected, but at higher frequencies, the phase shift caused by the circuit becomes −π

2. This phase shift corresponds to the dierence between a cosine and a sine.

We can use the transfer function to nd the output when the input voltage is a sinusoid for two reasons. First of all, a sinusoid is the sum of two complex exponentials, each having a frequency equal to the negative of the other. Secondly, because the circuit is linear, superposition applies. If the source is a sine wave, we know that

vin(t) = Asin (2πf t)

= _2jA ej2πf t_{− e}−(j2πf t)
(3.16)

Since the input is the sum of two complex exponentials, we know that the output is also a sum of two similar complex exponentials, the only dierence being that the complex amplitude of each is multiplied by the transfer function evaluated at each exponential's frequency.

vout(t) = A 2jH (f ) e j2πf t₋ A 2jH (−f ) e −(j2πf t) _(3.17)

As noted earlier, the transfer function is most conveniently expressed in polar form: H (f) = |H (f ) |ej∠(H(f ))_{. Furthermore, |H (−f) | = |H (f) | (even symmetry of the magnitude) and ∠ (H (−f)) =}

−∠ (H (f)) (odd symmetry of the phase). The output voltage expression simplies to vout(t) = _2jA|H (f ) |ej2πf t+∠(H(f ))−_2jA|H (f ) |e(−(j2πf t))−∠(H(f ))

= _{A|H (f ) |sin (2πf t + ∠ (H (f )))} (3.18)

The circuit's output to a sinusoidal input is also a sinusoid, having a gain equal to the magnitude of the circuit's transfer function evaluated at the source frequency and a phase equal to the phase of the transfer function at the source frequency. It will turn out that this input-output relation description applies to any linear circuit having a sinusoidal source.

Exercise 3.13.1 (Solution on p. 117.)

This input-output property is a special case of a more general result. Show that if the source can be written as the imaginary part of a complex exponential vin(t) = Im V ej2πf t
 the output

is given by vout(t) = Im V H (f ) ej2πf t

. Show that a similar result also holds for the real part. The notion of impedance arises when we assume the sources are complex exponentials. This assumption may seem restrictive; what would we do if the source were a unit step? When we use impedances to nd the transfer function between the source and the output variable, we can derive from it the dierential equation that relates input and output. The dierential equation applies no matter what the source may be. As

we have argued, it is far simpler to use impedances to nd the dierential equation (because we can use series and parallel combination rules) than any other method. In this sense, we have not lost anything by temporarily pretending the source is a complex exponential.

In fact we can also solve the dierential equation using impedances! Thus, despite the apparent restric- tiveness of impedances, assuming complex exponential sources is actually quite general.

3.14 Designing Transfer Functions

26

If the source consists of two (or more) signals, we know from linear system theory that the output voltage equals the sum of the outputs produced by each signal alone. In short, linear circuits are a special case of linear systems, and therefore superposition applies. In particular, suppose these component signals are complex exponentials, each of which has a frequency dierent from the others. The transfer function portrays how the circuit aects the amplitude and phase of each component, allowing us to understand how the circuit works on a complicated signal. Those components having a frequency less than the cuto frequency pass through the circuit with little modication while those having higher frequencies are suppressed. The circuit is said to act as a lter, ltering the source signal based on the frequency of each component complex exponential. Because low frequencies pass through the lter, we call it a lowpass lter to express more precisely its function.

We have also found the ease of calculating the output for sinusoidal inputs through the use of the transfer function. Once we nd the transfer function, we can write the output directly as indicated by the output of a circuit for a sinusoidal input (3.18).

Example 3.5 RL circuit vin + – R + – v i L iout Figure 3.31

Let's apply these results to a nal example, in which the input is a voltage source and the output is the inductor current. The source voltage equals Vin = 2cos (2π60t) + 3. We want the

circuit to pass constant (oset) voltage essentially unaltered (save for the fact that the output is a current rather than a voltage) and remove the 60 Hz term. Because the input is the sum of two sinusoidsa constant is a zero-frequency cosineour approach is

1. nd the transfer function using impedances;

2. use it to nd the output due to each input component; 3. add the results;

4. nd element values that accomplish our design criteria.

Because the circuit is a series combination of elements, let's use voltage divider to nd the transfer function between Vin and V , then use the v-i relation of the inductor to nd its current.

Iout Vin = j2πf L R+j2πf L 1 j2πf L = 1 j2πf L+R = H (f ) (3.19) where voltage divider = j2πf L R + j2πf L and inductor admittance = 1 j2πf L

[Do the units check?] The form of this transfer function should be familiar; it is a lowpass lter, and it will perform our desired function once we choose element values properly.

The constant term is easiest to handle. The output is given by 3|H (0) | = 3

R. Thus, the value we

choose for the resistance will determine the scaling factor of how voltage is converted into current. For the 60 Hz component signal, the output current is 2|H (60) |cos (2π60t + ∠ (H (60))). The total output due to our source is

iout= 2|H (60) |cos (2π60t + ∠ (H (60))) + 3 × H (0) (3.20)

The cuto frequency for this lter occurs when the real and imaginary parts of the transfer function's denominator equal each other. Thus, 2πfcL = R, which gives fc = _2πLR . We want this

cuto frequency to be much less than 60 Hz. Suppose we place it at, say, 10 Hz. This specication would require the component values to be related by R

L = 20π = 62.8. The transfer function at 60

Hz would be | 1 j2π60L + R| = 1 R| 1 6j + 1| = 1 R 1 √ 37 ' 0.16 × 1 R (3.21)

which yields an attenuation (relative to the gain at zero frequency) of about 1/6, and result in an output amplitude of 0.3

R relative to the constant term's amplitude of 3

R. A factor of 10 relative

size between the two components seems reasonable. Having a 100 mH inductor would require a 6.28 Ω resistor. An easily available resistor value is 6.8 Ω; thus, this choice results in cheaply and easily purchased parts. To make the resistance bigger would require a proportionally larger inductor. Unfortunately, even a 1 H inductor is physically large; consequently low cuto frequencies require small-valued resistors and large-valued inductors. The choice made here represents only one compromise.

The phase of the 60 Hz component will very nearly be −π

2, leaving it to be 0.3 Rcos 2π60t − π 2
= 0.3

Waveforms

0

0.1

0

1

2

3

4

5

Time (s)

Voltage (v) or Current (A)

input

voltage

output

current

Figure 3.32: Input and output waveforms for the example RL circuit when the element values are R = 6.28Ωand L = 100mH.

Note that the sinusoid's phase has indeed shifted; the lowpass lter not only reduced the 60 Hz signal's amplitude, but also shifted its phase by 90◦_.

In document BASES DE LICITACIÓN CONCESIÓN RUTA 5. TRAMO PUERTO MONTT PARGUA (página 114-118)