BIOEN 302, Section 3: AC electronicscourses.washington.edu/bioen303/Lab/BIOEN302_3.pdf · BIOEN...

BIOEN 302, Section 3: AC electronics

BIOEN 302, 3: AC electronics 1 Instructor: Albert Folch

For this section, you will need to have very present the basics of complex number calculus (see Section 1 for a brief overview) and EE215’s section on phasors. 1. Representation of sinusoids in the complex plane For a magnitude (say, a voltage) W(t) varying sinusoidally with time, we can write: W(t) = M cos(ωt + θ) hence W(t) = Re [M ej(ωt + θ)] where M ej(ωt + θ) = W(t) is the complex-plane representation of W(t). W(t) can be thought of as a rotating vector the real part of which represents a sinusoid. 2. Representation of sinusoids as phasors The phasor representation of a time-varying sinusoid v(t) = Vm cos(ωt + θ) is exactly the same as its complex representation, most often used in its exponential or polar form (for convenience). The nomenclature is slightly different:

where Vm is the maximum value of the sinusoid, ωt is suppressed and θ is the phase angle of the sinusoid. Thus, the phasor is a “transformed” version of a sinusoidal voltage or current waveform and consists of the magnitude and phase angle information of the sinusoid. Equivalently, v(t) is then referred to as the representation of the phasor in the time domain: Phase lead is expressed by a positive angle θ in the phasor notation, while phase lag is expressed by a negative angle θ in the phasor notation. See the diagrams below.

Important: As we shall see, the phasor concept may be used ONLY when: 1) the circuit is linear, 2) the steady-state response is sought 3) all independent sources are sinusoidal and have the same frequency.

( )( )[ ]eVRe)t(vtcosV)t(vtj

m

mθ+ω=

θ+ω=

θVeV

m

θ)tj(m

∠≡≡ +

VV ω



3. Operations with phasors a) Addition of phasors: just like addition of vectors! (Please do NOT remember by ear!)

b) Multiplication of phasors: just like exponentials (NOT vectors)!

c) Division of phasors: just like exponentials (NOT vectors)!

Exercise 1: Given V0=120∠30° and V1=120∠120°, give V0 + V1 and V0/V1 Answer: V0 + V1=169.7∠75° V0/V1=1∠-90° Example 2:

Reduce the equation

v(t) = 10 cos(ωt) + 5 sin(ωt + 60o) + 5 cos(ωt + 90o)

to an equation of the form

v(t) = Vm cos(ωt + θ)

This operation is greatly simplified by using complex (phasor) notation. Converting all sinusoids to cosines, we get

v(t) = 10 cos(ωt) + 5 cos(ωt + 60o – 90o) + 5 cos(ωt + 90o)

All cosines are in turn the real part of complex numbers:

v(t) = 10 Re[ejωt] + 5 Re[ej(ωt–30o)] + 5 Re[ej(ωt+90o)]

( )θθθθ

1010

1100

+∠⋅=⋅

∠=∠=

VVVV

VVVV

10

10

( )θθ 101

0 −∠=VV

VV

1

0

[ ] [ ] [ ] [ ] ( )

( ) ( )[ ]

+

+=

∠+++=+

+++=+

+++=+

∠=∠=

θθθθθ

θθθθθ

θθθθ

θθ

1100

11002

221

2

1100

2

1100

11001100

1100

coscossinsin

arctan

sinsincoscos

sinsincoscos

VVVV

VVVV

VVVV

VV

j

j

VV

VVVVVVVV

VV

10

10

101010

10

ImImReRe



v(t) = Re[(10 + 5 e–j30o + 5 ej90o) ejωt]

In polar form, we have

v(t) = Re[(10∠0o + 5∠– 30o + 5∠90o) ejωt]

Now, we can combine the complex numbers as

10∠0o + 5∠– 30o + 5∠90o = 10 + 4.33 – 2.5j + 5j = 14.33 + 2.5j

= 14.54∠9.9o = 14.54 ej9.90o

Thus we obtain

v(t) = Re[(14.54 ej9.90o) ejωt] = Re [14.54 ej(ωt+9.90o)] = 14.54 cos(ωt + 9.90o)

Answer: Vm = 14.54 and θ = 9.90o Example 3:

Exercise 4: Reduce the following expressions using phasors.



Exercise 5.

Exercise 6.

Exercise 7.

Summarizing, when performing operations with phasors, rather than remembering the formulas, think of phasors as if they were arrows described by a length and an angle when you are multiplying or dividing, and think of them as arrows with two rectangular coordinates when you are adding or subtracting – it’s as easy as that! 4. Complex impedance

a) Inductance

Suppose the sinusoidal current going through an inductance is given by

iL(t) = Imax sin(ωt+θ) ; in other words, iL(t) = Imax cos(ωt+θ-π/2)

In complex notation:

iL(t) = Re[Imax ej(ωt+θ-π/2)]

Answer:



The voltage across an inductance is given by vL(t) = L diL(t)/dt, hence

vL(t) = Re[jωLImax ej(ωt+θ-π/2)] = Re[ωLImax ejπ/2 ej(ωt+θ-π/2)] = ωLImax Re[ej(ωt+θ)]

or

vL(t) = ωLImax cos(ωt+θ)

In phasor notation, we have that the phasors for current and voltage are

IL = Im ∠(θ – 90o)

VL = ωLIm ∠θ

[Discuss: is this equivalent to writing θ+90o for VL and θ for IL?] It is the current, and not the voltage, that carries an extra 90o when put into phasor notation, because otherwise it does not satisfy iL(t) = Re [IL]. We subtract 90o to keep the sign of the real part constant.

VL = (ωL∠90o) × (Im ∠(θ – 90o)) = (ωL∠90o) × IL

VL = jωL IL Remember this !!!!

We refer to the term (ωL∠90o) = jωL ≡ ZL as the impedance of the inductance.

Summarizing,

Note: Current lags voltage by 90o in a pure inductance, as illustrated below



b) Capacitance

Suppose the voltage across a capacitor is given by

vC(t) = Vmax sin(ωt+θ) ; in other words, vC(t) = Vmax sin(ωt+θ-90o)

In complex notation:

vC(t) = Re[Vmax ej(ωt+θ-π/2)]

The current charging this capacitor is given by iC(t) = C dvC(t)/dt, hence

iC(t) = Re[jωCVmax ej(ωt+θ-π/2)] = ωCVmax Re [ejπ/2ej(ωt+θ-π/2)] = ωCVmax Re[ej(ωt+θ)]

or

iC(t) = ωCVmax cos(ωt+θ)

In phasor notation, we have that the phasors for current and voltage are

VC = Vmax ∠(θ – 90o)

IC = ωCVmax ∠θ

We refer to the term (1/(ωC))∠-90o = 1/jωC ≡ ZL as the impedance of the capacitance. VC = IC / jωC IC = jωC VC Remember this !!!! ZC = 1/(jωC) = 1/(ωC) ∠ -90o Summarizing:

Note: Current leads voltage by 90o in a pure capacitance:



c) Resistance v(t) = R i(t) VR = R IR IR = VR / R You already knew this … ZR = R ∠ 0o Summarizing:

Note: VR and IR are in phase in a pure resistance:



Overall summary:

1) Imprint in your memory that ZC = 1/jωC and ZL = jωL because you will be using it a lot.

2) Appreciate the simplification provided by the phasor notation:

Element Differential equation Phasor notation Impedance

L v = L di/dt V = jωL I jωL

C i = Cdv/dt I = jωC V 1/jωC

In other words, except for the fact that we use complex arithmetic, sinusoidal steady-state analysis of RLC circuits is virtually the same as the analysis of resistive circuits – no RLC differential equations! (We pay a price: the phasor notation applies only to steady-state sinusoidal signals.)

3) Rules of conversion from one notation to another: W = A+jB W = W∠θ where W = (A2+B2)½ and θ = arctan (B/A) W = W∠θ W = A+jB, where A = Wcosθ and B = Wsinθ

Exercise 8:

Exercise 9:



Exercise 10:

Exercise 11:



5. Circuits with multiple impedances We now know:

1) The expressions for the impedances (i.e. phasor notations) of R, L, and C

2) The rules for adding R, L, and C in series vs. parallel in non-phasor notation

What about the rules for adding impedances?

Importantly, impedances add just like resistors (albeit in the complex plane): in series, they sum to yield the total impedance; in parallel, their inverses (also called admittances, denoted by Y) add to yield the inverse of the total impedance.

Note: In Z ≡ R + jX, R is called resistance and X reactance (both in ohms) In Y ≡ 1/Z = G + jB, G is called conductance and B susceptance

(both in siemens) Important: G ≠ 1/R !!! It is straightforward to see that both Kirchhoff’s laws hold for phasors. Example 12: A series RLC circuit

The exercise will consist of finding the currents and voltages given the following circuit:

The given expression for the source voltage vs(t) tells us that the peak voltage is 100 V, the angular frequency is 500, and the phase angle is 30o. The phasor for the voltage source is thus

Important: a disadvantage of the polar notation is that the angular frequency is implicit – it’s up to you to remember its value!

The complex impedances of the inductance and the capacitance are, respectively,

The circuit in phasor/complex notation is shown below:



Note that all three elements are in series, therefore we can find the equivalent impedance of the circuit by adding the impedances of the elements: Zeq = R + ZL + Zc. Substituting values, we obtain Zeq = 100 + 150j – 50j = 100 + 100j. Converting to polar form,

Now we can find the phasor current by dividing the phasor voltage by the equivalent impedance:

As a function of time, the current is:

Therefore, we can express the phasors for the voltage drops across R, L and C as:

As a function of time, the voltages are:

vR(t) = 70.7 cos (500t – 15o)

vL(t) = 106.1 cos (500t + 75o)

vC(t) = 35.4 cos (500t – 105o)

Exercise 13: Consider the circuit below:



Find the values and units of Vs, ZR, ZC, ZR, ZL1, and ZL2. Find an expression for the total impedance, for vo(t), for the currents through each branch of the circuit (call them i1(t), i2(t), and i3(t)), and for the voltages across each circuit element (call them vR(t), vC(t), vL1(t), and vL2(t)). Exercise 14. Suppose that the input signal for the circuit shown below is

Exercise 15.

i1(t)

i2(t)

i3(t)



Exercise 16.

Exercise 17.



Exercise 18.

Exercise 19.



6. Low-pass and high-pass filters Until now we have only talked generically about how we can analyze circuits with a pure (single frequency) sinusoidal input – so we didn’t have to worry about the response of the system to other frequencies. Let’s take a closer look at the equations for impedance and try to think about what will happen when the frequency of the input is varied. Consider the following simple circuit: Note that the current I and the voltage V1 are related as follows:

Importantly, the function of the circuit depends on what part of the circuit we consider the “output”:

• If the output Vo ≡ VR = IR, then Vo varies with ω in the same way as I does.

( )[ ]

−∠

+=

RωLarctan

ωLR

V

21

22

RRV

If you plot the magnitude of the voltage gain, Gv ≡ Vo / V, you will see that this is a “low-pass” filter:

As you can see, the response is far from “ideal”.

( )

( )[ ]

−∠

+=

+−

=+

==

RωLarctan

ωLR

VωLR

LjRLjR

21

22

22

I

VV

ZVI ω

ω

V

I

R

L



An approximate idea of what the filter response is like can be obtained by looking at the limits when ω 0 and ∞:

What happens when ω = R/L?

°−∠= 452R

1VI

The frequency ω = R/L is therefore referred to as the “half-power frequency” (or also “break frequency” for reasons we will see later).

• If the output Vo ≡ VL = IjωR, then I lags Vo by 90o.

( )[ ]

∠

ω

+

=

ω−

=

−∠

+

ω=

ω+ω

=

ωLRarctan

LR1

V

LRj1

1alsoor

RωLarctan90

ωLR

VLLjR

Lj

2L

o

21

22

22

L

VV

VV

If you plot the magnitude, you will see that this is a “high-pass” filter:

Exercise 20. Repeat the previous exercise but with an RC circuit instead of RL. Consider also both cases, Vo ≡ VC (left) and Vo ≡ VR (left).

°−∠=°∠=∞

900lim0R

lim 10

IVIωω



What type of filter are these? Sketch a plot for the voltage gain Gv ≡ Vo / V1 in both cases. 7. The Transfer Function We have started talking about voltage gain, Gv ≡ Voutput / Vinput. The voltage gain is just one of several “transfer functions”, which are defined as the ratio of the output phasor (of interest) to the input phasor (defined by the source). They are expressed usually as a function of frequency, H(f), where f = ω/2π is measured in Hertz. Let’s find the transfer function of the following circuit:

The phasor current is the input voltage Vin divided by the complex impedance ZT of the circuit:

Since the output is defined across the capacitor, the phasor for the output voltage is the product of the phasor current and the impedance of the capacitance, 1/jωC:

Substituting for I, we have

The transfer function is thus:

If is useful to define the parameter fB,

Then, the transfer function can be written as:

We see that the transfer function is a complex quantity having a magnitude and a phase:

CjRZT ω

1+=



Plots of the magnitude and phase are shown below:

This is a low-pass filter. Also, fB is the half-power frequency and the frequency at which the output lag is half of 90o. It is important to note that the low-pass filter built with R and L (VL as the output, see above) would have the same transfer function, even though the definition of fB would be different. This makes the concept of fB all the more useful – it characterizes the filter, regardless of how it is physically built. The same applies to the two high-pass filters: 1) series RL circuit (output taken across L), and 2) series RC circuit (output taken across R). 8. Bode Plots Most of the times, it is easier to visualize the ranges of H(f) and of f in a logarithmic scale, in what is called a Bode plot. In a Bode plot, we choose to represent 20 log H(f) as a function of log ω instead of G(ω) as a function of ω. In this representation, a change in 20 dB (decibels) represents a change in an order of magnitude in H(f). A change in √2 represents 3 dB. Let’s see an example, taking again the transfer function of the low-pass filter:

In decibels,

Thus,

Using the properties of the logarithm, we obtain



Since the logarithm of 1 is zero and the logarithm of x½ is ½ log(x),

The magnitude Bode plot for the first-order low-pass filter thus looks like:

Note the following:

• For very small f, f << fB , H(f)dB ≈ 0 (a horizontal line)

• For very large f, f >> fB , H(f)dB ≈ – 20 log (f / fB) (a straight line in the log(f) scale that intersects the frequency axis at fB)

• The high- and low-frequency asymptotes appear to intersect (or “break”) at fB, hence the name “break frequency”.

A table of useful values for the f >> fB approximation is shown below:

Often in the context of decibels and Bode plots, you will hear the terms “decade” (a frequency range spanning one order of magnitude, i.e. a factor of 10) and “octave” (spanning a factor of 2). In the logarithmic scale:



An important feature of the Bode plots and their decibel representation is that a product of two transfer functions “appears” as a sum of the two:

This is important when one tries to figure out the transfer function corresponding to a given, more complex Bode plot with multiple breaks. For example, a band-pass filter can be straightforwardly seen as the addition of the Bode plots of the high-pass and the low-pass filters, but in reality the two filters can be assembled in two stages (e.g. the input of the high-pass stage is the output of the low-pass stage, so the total transfer function is the product of the transfer functions of each stage).

Exercise 21.

Exercise 22.

Exercise 23.

Exercise 24.



Example 25.

Example 26.



Example 27.



9. Resonance in RLC circuits Suppose we connect a series RLC circuit to a voltage supply as indicated. The impedance of the circuit is:



ω−ω∠

ω−ω+=

−+=

CR1

RLarctan

C1LR

Cω1LωjR

22Z

Note we can also write which has a minimum in magnitude when ωL = 1/ωC, i.e. at ω0 = (LC)–½. Most important, at this frequency the phase becomes zero. By definition, the frequency at which the output (either current or voltage, or both) is in phase (“in resonance”) with the input (a condition termed “resonance”) is called the resonant frequency, which is often (but not always) characterized by yielding a maximal output. The circuits themselves are said to be in resonance at this frequency. Remember this value because it will come back again and again! For a parallel RLC circuit we would write the total admittance (i.e. the inverse of the total impedance) of the circuit as:

Lj1Cj

R1)j(Y

ω+ω+=ω

where Y(jω) ≡ 1/Z(jω). It is straightforward to see that a plot of the magnitude of Y(jω),Y(jω), for the parallel RLC circuit is essentially the same as the series-RLC Z(jω) impedance plot, with a minimum at ω = ω0 = (LC)–½.

Summarizing for the series as well as parallel RLC circuits:

• ω0 = (LC)–½

• At resonance (by definition) Z(ω0) = R∠0o (no imaginary component, i.e. no phase either)

• At low frequencies,Zis dominated by the capacitive term for the series circuit and by the inductive term for the parallel circuit

• At high frequencies,Zis dominated by the inductive term for the series circuit and by the capacitive term for the parallel circuit

As a general rule (for any circuit),



• Resonance conditions are defined as those that satisfy Im(Z) or Im(Y) = 0, not necessarily ω0 = (LC)–½

• At resonance current and voltage are by definition in phase because the impedance won’t change their phase. (Recall the phase is introduced by the imaginary component.)

• Example of a more general case: consider a circuit of the type RL//C (R in series with L, C in parallel with them), then

ω+ω

−ω+ω+

=ω+

+ω= 222222 LRLCj

LRR

LjR1CjY

The frequency that satisfies Im(Y)=0 is

2

2

0 LR

LC1

−=ω

… which, incidentally, is not a maximum or minimum of Y. In other words, in general the output is not necessarily maximum at resonance. “Resonance” is a term more related with the phase of the output than with its magnitude.

Example 28. Consider the voltage drop across the resistor in a series RLC circuit:

In phasor notation, the input voltage Vs divided by the total impedance Z gives us the current going through the circuit, I = Vs/Z. The voltage drop across the resistor R will thus be Vo = I × R = (R/Z) × Vs. Hence we obtain:

[Discuss: what is the expression for the voltage across the inductance? And across the capacitance?] Rearranging,

and substituting values,

The plots of the magnitude and phase of Vo look as follows. Note the resonance peak in Vo at low frequencies.



Because of their steep slopes, magnitude and phase plots are usually shown with frequency on a log axis:

Let’s take a closer look at the phases for each particular phasor in the circuit. In the case of the series RLC circuit, we consider and plot the phasors Vc, VL and VR versus I because I is what is conserved along the circuit. (Here V1 represents Vs in the above circuit.) [Discuss the following plots.]

Note what happens at resonance: the voltage across the resistor and the current through the resistor are in phase. That is precisely because Z(ω0) = R, hence there is no change in phase when we compute I = Vs/Z, and both ZL = ZC = 0 at resonance. [Discuss the implications on energy storage.]



10. Filters, Bandwith, and Quality factor Let’s take a closer look at the expression for the gain in the series RLC circuit as defined above (i.e. output voltage is defined as the voltage across the resistor):

Thus,

RC1

RL

Qwith,jQ1

1

0

0

0

0

s

o

ω=

ω≡

ωω

−ωω

+=≡

VVG (recall ω0 = (LC)-½ )

Writing the phasor G in terms of magnitude and phase, G = M(ω) ejφ(ω), we have:

2

0

0

2Q1

1)(M

ωω

−ωω

+

=ω

and

ωω

−ωω

−=ωφ − 0

0

1 Qtan)(

An approximate log graph of these two functions is shown here: We define bandwidth (BW in the graph) as the difference between the two half-power frequencies or, equivalently (since power is the square of magnitude), as the range of frequencies that satisfy M(ω) > 1/√2. Looking at the above relationship for M(ω), that condition is satisfied when

2Q12

0

0

2 <

ωω

−ωω

+

which in turn is satisfied when

ωω

−ωω

+=

ωωω

−ωωω

+=

ω−ω+

=0

00

0

0

0s

o

jQ1

1

CL

jR

R

C1LjR

RVV



1Qwhenor1Q 0

0

0

0

−<

ωω

−ωω

>

ωω

−ωω

i.e. ω2 – ω0ω/Q – ω02 > 0 or else ω2 + ω0ω/Q – ω0

2 < 0

Each equation has 2 roots, but only one is positive; discarding these “negative frequencies” (no physical meaning), we only have one value per equation. We call the lower value ωLO and the larger one ωHI. In other words, M(ω) > 1/√2 in the interval ωLO < ω < ωHI, where

+

+−ω≡ω 1

Q21

Q21

2

0LO and

+

+ω≡ω 1

Q21

Q21

2

0HI

The difference between ωHI and ωLO gives us (by definition) the bandwidth:

BW ≡ ωHI – ωLO = ω0/Q

i.e. BW = R/L for the series RLC circuit.

The latter is generally used as the general definition of Q, i.e. Q ≡ ω0/BW.

Hence circuits with high Q have narrow bandwidths. In addition, ω02 = ωLO × ωHI, i.e. the

resonance frequency is the geometric mean of the two half-power frequencies.

Bode plots of the transfer function vs. frequency for various values of Q are shown below. As we can see, the series RLC circuit (with output taken across the resistor) is a (second-order) bandpass filter.

Exercise 29.

Show that for a series RLC circuit the ratio of impedance at resonance, Z0 ≡ Z(ω0), to impedance at any radian frequency, Z, is:

RL

Qwhere

jQ

0

0

0

0

0

1

1

ωωω

ωω

≡

−+

==YY

ZZ



Show also that for a parallel RLC circuit the expression is the same provided we invert the definition of Q:

Summarizing:

• The expression of Q for a series RLC circuit is the inverse of the expression of Q for a parallel RLC circuit. For example, Q decreases with increasing R in the series circuit, but it increases with increasing R in the parallel circuit.

10. Other filters using RLC series circuits We have only been considering the series RLC circuit with the output defined as the voltage across the resistor. What would happen if we chose the voltage across the capacitor VC (or across the inductance VL) as the output of the circuit, such as depicted below? Output ≡ Capacitor

Vs R L C

LRQwhere

jQ

0

0

0

0

0

1

1

ω

ωω

ωω

≡

−+

==YY

ZZ

Exercise 30.

Exercise 31.

Exercise 32.



We have:

I = Vs/Z , VL = jωLI and VC = I /(jωC)

where Z = R + jωL + 1/(jωC)

NOTE that the outputs VC and VL can NEVER be in phase with the input Vs, but the current at the output (L, C or R) CAN be in phase with Vs. In such cases, we still speak of resonance when the current at the output is in phase with the input voltage/current. In L or C, we can’t have current and voltage in phase!

Thus, CRjLC1

Cj1LjR

Cj1

2s

sC ω+ω−=

ω+ω+

ω=

VVV

Note that I = IC is in phase with Vs for ω0 = (LC)-½ , but note also that at this frequency the phase of VC is -90o (with respect to Vs).

The transfer function H(f) = Vout/Vin = Vc / Vs, can be written as:

with Qs = ω0L/R = 1/(ω0RC) and f0 = ω0/(2π) = (2π)-1 (LC)-½

The magnitude of the transfer function is plotted below. This is a second-order low-pass filter circuit, similar to the RC (“first-order”) one, compared here side-by-side:



Output ≡ Inductor

When the output of a series RLC circuit is defined as the voltage across the inductor, we have:

LRj

LC11

Cj1LjR

Lj

2

ssL

ω−

ω−

=

ω+ω+

ω=

VVV

The magnitude of the transfer function is plotted below:

where with Qs = ω0L/R = 1/(ω0RC) and f0 = ω0/(2π) = (2π)-1 (LC)-½

Therefore, this is a (second-order) high-pass filter.

Exercise 33.

Suppose we need a filter that passes components higher in frequency than 1 kHz and rejects components lower than 1 kHz. As we know, the series-RCL second-order circuit configuration constitutes a high-pass filter when the output is taken across the inductor. Determine the values or R and C, keeping in mind that we are interested in a filter that has a transfer function that is approximately constant in the pass band (see plot above).

Solution: C = 0.507 µF, R = 314.1 Ω.

Now, let’s see what happens at resonance at both L and C.

At resonance, since Z(ω0) = R, we have

I(ω0) = Vs/R and VR = Vs Does that mean there is no voltage drop across L or C? No!! We have

VL(ω0) = jω0L×I(ω0) and V(ω0) = I(ω0)/(jω0C)



Substituting I(ω0), we get

VL(ω0) = jω0L×Vs/R = j Q Vs

and VC(ω0) = Vs /(jω0CR) = – j Q Vs

In other words, at resonance the voltage across the capacitor and across the inductor have the same magnitude, the gain is Q, and they are out of phase by 180o, i.e. they add to 0.

However, QVs is not the maximum value of the magnitudes of VL and VC. For example,

CRjLC1Cj

1LjR

Cj1

2s

sC ω+ω−=

ω+ω+

ω=

VVV

It is straightforward to show that the frequency that satisfies 0d

d C =ω

V is

0202

202

0

2

max 211

2211 ωω

ωωω ≠−=−=

−=

QQLR

LC

At this frequency, we have

2s

c

Q411

QVV

−=

max

Although ωmax ≠ ω0, for large Q (Q>>1), ωmax ≈ ω0 and the gain at ωmax is Vc/Vs ≈ Q.

Exercise 34.

Derive the transfer function for a series RLC circuit whose output is taken across L and C combined (a “band-reject” filter):



Vs R L C

I

IR IL IC

Note, once more, that our choice of “output” determines what type of filter we have:

11. Resonance in the parallel RLC circuit From what you know of phasors and resonance in RLC series circuit (and some examples with parallel circuits above), you should be able to comfortably analyze the parallel RLC circuit. Let’s take a quick look at the expression for the gain G = Vout/Iin in the parallel RLC circuit:

It is straightforward to show that the magnitude of the gain is:

Exercise 35.

Find the total impedance Z = Vs/I, the resonance frequency ω0, and the currents through each element, i.e. IR, IC, and IL in the parallel RLC circuit below. Compare (in arrow plots) the magnitudes and phases of the resonance values IR(ω0), IC(ω0), and IL(ω0).

ω−ω+

=

C1LjR

RvG Gv = (see Exercise 35 just above)



Hint: We can now plot the phasors Ic, IL and IG versus Vs (≡ V1 below) because Vs is what is conserved along the circuit. Similarly to what we did for the series RLC circuit, we have:

Looking at the magnitude of the gain:

We note first that the magnitude of the gain has a maximum at the resonance frequency, ω0 = (LC) –½ , and that at this frequency Gmax = R. (Note that the same conclusion is reached by observing that by definition Z at resonance must have Im(Z) = 0, and when Z is maximum, Vout and G are maximum.) If we plotted this curve, we would see that it also has one peak (it looks very similar to the one for the series RLC circuit), and that therefore the parallel RLC circuit acts as a bandpass filter.

To find the bandwidth of the circuit, we find the ωHI and ωLO that satisfy that the magnitude of the gain is reduced by √2 (i.e. the power is reduced by 2):

Solving the equation and taking only the positive values of ω we obtain:

Thus the bandwidth (recall that in the series RLC circuit BW = R/L) is:

BW ≡ ωHI – ωLO = 1/RC

and the quality factor Q is:

Q ≡ ω0/BW = RC/(LC)½ = R (L/C) ½



which we already knew from our previous study of the parallel RLC circuit. [To visualize the meaning of the Q factor, recall from your homework the values for the currents through the capacitor and inductance.]

Using this value of Q, we can re-write the expressions for ωHI and ωLO as:

+

+−ω≡ω 1

Q21

Q21

2

0LO and

+

+ω≡ω 1

Q21

Q21

2

0HI

… which are the same expressions as the ones for the series RLC circuit! Hence the usefulness of the Q factor concept.

Exercise 36.

Consider the parallel RLC circuit, with L = 10 mH, C = 100 µF, and R = 1 kΩ. Determine a) ω0, b) ωHI, c) ωLO, d) BW, and e) Q.

Answer: a) ω0 = 1000 rad/s, b) ωHI = 1005 rad/s, c) ωLO = 995 rad/s, d) BW = 10 rad/s, and e) Q = 100.



Exercise 37.

Exercise 38.

BIOEN 302, Section 3: AC electronicscourses.washington.edu/bioen303/Lab/BIOEN302_3.pdf · BIOEN...

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