Ee122 Freq Resp Revnotes

EE122: Frequency Response of Passive Circuits

Review Notes on the Frequency Responseof Passive Circuits

Prepared by Rizwan AhmedStanford University

January 7, 2003

1 Impedance of Resistors, Capacitors, and Inductors

We write the impedance of a component as Z and the frequency as ω. Then the impedance of a resistor isZR = R, The impedance of a capacitor is ZC = 1/(jωC), and the impedance of an inductor is ZL = jωL.Thus a capacitor is a short circuit (ZC = 0) at high frequencies and is an open circuit (ZC = ∞) at DC.An inductor is an open circuit (ZL = ∞) at high frequencies and is a short circuit (ZL = 0) at DC. Tounderstand how we arrive at these equations, recall that inductors store energy with magnetic fields whereas capacitors store energy with electric fields. Using these ”impedances” we can do frequency analysis of ourcircuit using normal methods (KVL, KCL, Ohm’s Law, etc . . .). The input/output relationship is referredto as the Transfer function, and is often denoted as H(jω). At any given frequency, this is just a complexnumber. The magnitude and phase of this number give the magnitude and phase of the output in relationto the input.

2 Identifying the Frequency Response Visually

Good circuit designers can just look at a complex circuit and start identifying components (amplifiers, filters,etc . . . ). Identify the frequency response (Lowpass, Highpass, or Bandpass) of the circuits in the figure below.

R1

R2 L

C

R

R1

C L R2Vout Vout

Vout

hint: Think about how the circuits would behave at DC and very high frequencies.

3 Defining the Cutoff Frequency

The cutoff frequency is defined where

Pout(ωc) =Pmax

2(1)

Plugging in the definition of power across a resistor, we obtain the equation

V 2(ωc)

R=

V 2max

2R(2)

which leads toV (ωc)

Vmax

=1√2

(3)

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So the ratio of the transfer function is equal to 1/√

2 at the cutoff frequency.

The decibel is a convenient unit when measuring values over a very wide range and is often used in en-gineering. If the measurement is voltage, the value in decibels is defined as

VdB = 20 log(|V |) (4)

plugging in equation (3) we get

VdB(ωc) = 20 log

( |Vmax|√2

)

= 20 log(|Vmax|) − 20 log(√

2)

VdB(ωc) = VdBmax − 3dB (5)

So the voltage is 3dB down at the cutoff frequency! In the case of power measurements, the value indecibels is defined as

PdB = 10 log(P ) (6)

plugging in equation (1) we get

PdB(ωc) = 10 log(Pmax

2) = 10 log(Pmax) − 10 log(2)

PdB(ωc) = PdBmax − 3dB (7)

So the power is 3dB down at the cutoff frequency as well.

4 Bode Plots

Before digital computers were a commonplace, a quick way of sketching the frequency response of a systemwas the Bode plot. However, these plots still provide valuable intuition and insight and thus remain an im-portant concept. The Bode plot is a piecewise linear approximation of the transfer function (both magnitudeand phase) that can be constructed using a set of simple rules. To construct a Bode plot, we must writethe transfer function in the following form (known as the standard form). The z and p terms are called thezeros and poles respectively.

H(jω) = k ·(jω)±`(1 − jω

z1

)(1 − jωz2

) · · · (1 − jωzm

)

(1 − jωp1

)(1 − jωp2

) · · · (1 − jωpn

)(8)

Now if we measure the transfer function in decibels and use the properties of logarithms (multiplication →addition, division → subtraction, exponents → coefficients), we obtain

H(jω)dB = 20 log(|k|) ± ` · 20 log(ω) + 20 log

(∣

∣

∣

∣

1 − jω

z1

∣

∣

∣

∣

)

· · · + 20 log

(∣

∣

∣

∣

1 − jω

zm

∣

∣

∣

∣

)

(9)

−20 log

(∣

∣

∣

∣

1 − jω

p1

∣

∣

∣

∣

)

· · · − 20 log

(∣

∣

∣

∣

1 − jω

pn

∣

∣

∣

∣

)

The first term is a constant and the second is linear on a log scale. The other terms are approximated in thefollowing way. If ω z than 1− ω

z≈ 1 and 20 log(1) = 0. On the other hand, if ω z than |1− ω

z| ≈ ω

zand

20 log(ω/z) is a linear term on a log scale. We approximate these terms on the Bode plot by letting themequal zero until ω = z and linear afterwards. Of course, this reasoning also holds for the poles. So here arethe rules to making a Bode plot.

1. Construct a log-log plot with ω on the x-axis and |H(jω)|dB on the y-axis. Let the x-axis start at 1.

2. Calculate 20 log(|k|), this is the starting point on the y-axis at ω = 1.

3. If you have a linear term, then start rising by ` ·20dB/decade. If not, then stay constant. Do this untilyou encounter a pole or zero.

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4. If you encounter a pole, start dropping by 20dB/decade. If you encounter a zero, start rising by20dB/decade.

5. The rises and drops are cumulative. So for example, if you have a linear term with ` = 2 and two polesat ω = 1000Hz, you would rise at 40dB/decade and flatten out after ω = 1000Hz (This is a secondorder high-pass filter).

Now if we look at the phase, we use two very convenient properties of the phase of complex numbers; namelythat multiplication → addition and division → subtraction. So the phase of the transfer function is given by

6 H(jω) = 6 k + 6 (jω)` + 6

(

1 − jω

z1

)

· · · + 6

(

1 − jω

zm

)

− 6

(

1 − jω

p1

)

· · · − 6

(

1 − jω

pn

)

(10)

The phase of the first term is 0 if k ≥ 0 and 180 otherwise. The second term has a phase of ` · 90. Thephase of the pole and zero terms is approximated in the following way. If ω z than 1− jω

z≈ 1 and 6 1 = 0.

If ω = z than 6(

1 − jωz

)

= −45. Finally if ω z than 1 − jωz

≈ − jωz

and 6(

− jωz

)

= −90 (all of thisassumes a positive z). Here are the rules for constructing a phase Bode plot.

1. Construct a semilog plot with ω on the x-axis and 6 H(jω) on the y-axis. Let the x-axis start at 1.

2. Calculate 6 k + 6 (jω)`. This is the starting point of your phase plot.

3. If you encounter a pole, construct a linear drop of 90 from 0.1z to 10z. If you encounter a negativezero (<z < 0), construct a linear rise of 90 from 0.1z to 10z. If you encounter a positive zero(<z > 0), construct a linear drop of 90 from 0.1z to 10z.

If the steps sound a bit confusing, it is only because it is difficult to put them into words. The actual stepsare very straightforward. Constructing Bode plots becomes much clearer after doing a few examples.

5 Some Common Filters

You will all use filters in your projects. The forms for some common transfer functions is given below.

• A lowpass filter with a given ωc and a rolloff of n · 20dB/decade. H(0) is the DC gain and n is theorder of the filter.

H(jω) =1

(1 + jωωc

)nH(0) (11)

• A highpass filter with a given ωc and a rolloff of n ·20dB/decade. H(∞) is the gain at infinite frequencyand n is the order of the filter.

H(jω) =

(

jωωc

)n

(

1 + jωωc

)n H(∞) (12)

• A bandpass filter with 20dB/decade rolloff in both directions. β is the bandwidth (the amount ofspectrum between the 3dB points) and H(ωo) is the transfer function evaluated at the center frequencyof the passband. While we did not put this in standard form it is easy to see that it has a linear termand two poles. Thus it will rise and then eventually fall at 20dB/decade. With some algebraic effort,you can show that the β and ωo are indeed the bandwidth and center frequency respectively.

H(jω) =jβω

(jω)2 + β(jω) + ω2o

H(ωo) (13)

It is easy to realize these transfer functions with circuits. The importance of writing the transfer functionin these forms is that they give an easy way of identifying the important parameters of the filter (cutofffrequency in the case of LP and HP filters, and center frequency and bandwidth in the case of the BP filter).

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6 Examples

Example 1

R

L

Type of FilterVoutVin

First, we see that this circuit is a voltage divider. So we write

H(jω) =R

R + jωL(14)

or in standard form

H(jω) =1

1 + jω/(R/L)(15)

So we can immediately see that the cutoff frequency is ωc = R/L. The magnitude and phase of the transferfunction is given by

|H(jω)| =1

√

1 + (ωLR

)2(16)

6 H(jω) = 6 1 − 6

(

1 +jω

R/L

)

(17)

We can simplify the equation for the phase to get

6 H(jω) = − tan−1

(

ωL

R

)

(18)

The figure below plots the true magnitude and phase response (blue) and the Bode approximations (red).

Example 2

R2 Type of FilterVin

R1

Vout

L

We see that the impedance of R2 ‖ L is jωLR2

R2+jωL. So we write

H(jω) =

jωLR2

R2+jωL

R1 + jωLR2

R2+jωL

(19)

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Doing some algebra, we get

H(jω) =R2

R1 + R2· (jω)/(R1R2/L(R1 + R2))

1 + (jω)/(R1R2/L(R1 + R2))(20)

Comparing with the standard form of the highpass filter we see that, H(∞) = R2

R1+R2

and ωc = R1R2

L(R1+R2).

Now suppose R2 R1 and R1 = 100Ω, and you need a ωc = 1kHz. What value would you choose forthe inductor?

Since R2 R1, we can ignore R1 in the denominator of the expression for the cutoff frequency. ThenR2’s cancel and we are left with ωc = R1/L. Thus L = R1/ωc = R1/(2πfc). Then,

L =100Ω

2π · 1000Hz=

1

20πH ≈ 15.9 mH (21)

The figure below plots the true magnitude and phase response (blue) and the Bode approximations (red).

Example 3

R Type of FilterVin Vout

CL

We write the transfer function

H(ω) =R

R + jωL + 1/(jωC)(22)

or equivalently as

H(ω) =jω(R/L)

j2ω2 + jω(R/L) + 1/(LC)(23)

so ωo = 1/√

LC and β = R/L. We also calculate the magnitude and phase of the transfer function.

|H(jω)| =ω(R/L)

√

(ωRL

)2 + ( 1LC

− ω2)2(24)

6 H(jω) = 6 (jω(R/L)) − 6

(

1

LC− ω2 + jω(R/L)

)

(25)

This simplifies down to

6 H(jω) = 90 − tan−1

(

ω(R/L)

1/(LC) − ω2

)

(26)

The next figure plots the true magnitude and phase response.

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Ee122 Freq Resp Revnotes