II. PASSIVE FILTERS

Frequency-selective or filter circuits pass to the output only those input signals that are in a

desired range of frequencies (called pass band). The amplitude of signals outside this range

of frequencies (called stop band) is reduced (ideally reduced to zero). Typically in these

circuits, the input and output currents are kept to a small value and as such, the current

transfer function is not an important parameter. The main parameter is the voltage transfer

function in the frequency domain, Hv(jω) = Vo/Vi. As Hv(jω) is complex number, it has

both a magnitude and a phase, filters in general introduce a phase difference between input

and output signals.

To minimize the number of subscripts, hereafter, we will drop subscript v of Hv. Furthermore,

we concentrate on the the ”open-loop” transfer functions, Hvo, and denote this simply by

H(jω). The impact of loading is sperately discussed.

PassBand Band

Stop

| H(j ) | ω

ωωc

ω| H(j ) |

ωωc

Κ

0.7Κ

2.1 Low-Pass Filters

An ideal low-pass filter’s transfer function is shown. The

frequency between the pass- and-stop bands is called the

cut-off frequency (ωc). All of the signals with frequen-

cies below ωc are transmitted and all other signals are

stopped.

In practical filters, pass and stop bands are not clearly

defined, |H(jω)| varies continuously from its maximumtoward zero. The cut-off frequency is, therefore, defined

as the frequency at which |H(jω)| is reduced to 1/√

2 =

0.7 of its maximum value. This corresponds to signal

power being reduced by 1/2 as P ∝ V 2.

o

-

+

i

-

+

VV

L

R

Low-pass RL filters

A series RL circuit as shown acts as a low-pass filter. For

no load resistance (“open-loop” transfer function), Vo can

be found from the voltage divider formula:

Vo =R

R + jωLVi → H(jω) =

VoVi

=R

R + jωL=

1

1 + j(ωL/R)

We note

|H(jω)| = 1√1 + (ωL/R)2

ECE65 Lecture Notes (F. Najmabadi), Spring 2006 21

It is clear that |H(jω)| is maximum when denominator is smallest, i.e., ω → 0 and |H(jω)|decreases as ω is increased. Therefore, this circuit allows “low-frequency” signals to pass

through while “blocking” high-frequency signals (i.e., reduces the amplitude of the voltage

of the high-frequency signals). The reference to define the “low” and “high”-frequencies is

the cut-off frequency: “low”-frequencies mean frequencies much lower than ωc.

To find the cut-off frequency, we note that the |H(jω)|Max

= 1 occurs at ω = 0 (alterna-

tively find d |H(jω)| /dω and set it equal to zero to find ω = 0 which maximizes |H(jω)|).Therefore,

|H(jω)|max

= 1

|H(jω)|ω=ωc

=1√2|H(jω)|

max=

1√2

1√

1 + (ωcL/R)2=

1√2

−→ 1 +(

ωcL

R

)2

= 2 → ωcLR

= 1

Therefore,

ωc =R

Land H(jω) =

1

1 + jω/ωc

Input Impedance: Using the definition of the input impedance, we have:

Zi =ViIi

= jωL + R

The value of the input impedance depends on the frequency ω. For good voltage coupling,

we need to ensure that the input impedance of this filter is much larger than the output

impedance of the previous stage. Since we do not know the frequency of the input signal,

we need to ensure that good voltage coupling criteria is satisfied for all frequencies (or all

possible values of Zi). As such, the minimum value of Zi is an important number. Zi is

minimum when the impedance of the inductor is zero (ω → 0).

Zi|min = R

oZ

L

R

Output Impedance: The output impdenace can be

found by “killing” the source and finding the equivalent

impdenace between output terminals:

Zo = jωL ‖ R

ECE65 Lecture Notes (F. Najmabadi), Spring 2006 22

where the source resistance is ignored. Again, the value of the output impedance depends on

the frequency ω. For good voltage coupling, we need to ensure that the output impedance

of this filter is much smaller than the input impedance of the next stage for all frequencies,

the maximum value of Zo is an important number. Zo is maximum when the impedance of

the inductor is infinity (ω → ∞).

Zo|max = R

Bode Plots and Decibel

The voltage transfer function of a two-port network (and/or the ratio of output to input

powers) is usually expressed in Bel:

Number of Bels = log10

(

PoPi

)

or Number of Bels = 2 log10

∣

∣

∣

∣

VoVi

∣

∣

∣

∣

because P ∝ V 2. Bel is a large unit and decibel (dB) is usually used:

Number of decibels = 20 log10

∣

∣

∣

∣

VoVi

∣

∣

∣

∣

or∣

∣

∣

∣

VoVi

∣

∣

∣

∣

dB

= 20 log10

∣

∣

∣

∣

VoVi

∣

∣

∣

∣

There are several reasons why decibel notation is used:

1) Historically, the analog systems were developed first for audio equipment. Human ear

“hears” the sound in a logarithmic fashion. A sound which appears to be twice as loud

actually has 10 times power, etc. Decibel translates the output signal to what ear hears.

2) If several two-port network are placed in a cascade (output of one is attached to the input

of the next), the overall transfer function, H, is equal to the product of all transfer functions:

|H(jω)| = |H1(jω)| × |H2(jω)| × ...20 log10 |H(jω)| = 20 log10 |H1(jω)| + 20 log10 |H2(jω)| + ...|H(jω)|dB = |H1(jω)|dB + |H2(jω)|dB + ...

making it easier to find the overall response of the system.

3) Plot of |H(jω)|dB versus frequency has special properties that make analysis simpler. Forexample, the plot asymptotes to straight lines at low and high frequencies as is shown below.

ECE65 Lecture Notes (F. Najmabadi), Spring 2006 23

Also, using dB definition, we see that, there is a 3 dB difference between maximum gain and

gain at the cut-off frequency:

20 log |H(jωc)| − 20 log |H(jω)|max = 20 log[

|H(jωc)||H(jω)|

max

]

= 20 log

(

1√2

)

≈ −3 dB

Bode plots are plots of |H(jω)|dB (magnitude) and 6 H(jω) (phase) versus frequency in asemi-log format (i.e., ω axis is a log axis). Bode plots of first-order low-pass RL filters are

shown below (W denotes ωc).

|H(jω)|dB

6 H(jω)

At high frequencies, ω/ωc � 1,

|H(jω)| ≈ 1ω/ωc

→ |H(jω)|dB

= 20 log

[

1

ω/ωc

]

= 20 log(ωc) − 20 log(ω)

which is a straight line with a slope of -20 dB/decade in the Bode plot. It means that if ω

is increased by a factor of 10 (a decade), |H(jω)|dB

changes by -20 dB.

At low frequencies, ω/ωc � 1, |H(jω)| ≈ 1 which is also a straight line in the Bode plot.The intersection of these two “asymptotic” values is at 1 = 1/(ω/ωc) or ω = ωc. Because of

this, the cut-off frequency is also called the “corner” frequency.

The behavior of the phase of H(jω) can be found by examining 6 H(jω) = − tan−1(ω/ωc). Atlow frequencies, ω/ωc � 1, 6 H(jω) ≈ 0 and at high frequencies, ω/ωc � 1, 6 H(jω) ≈ −90◦.At cut-off frequency, 6 H(jω) ≈ −45◦.

ECE65 Lecture Notes (F. Najmabadi), Spring 2006 24

General first-order low-pass filters

As we discussed before, transfer functions characterize a two-port network. As such, it is

useful to group two-port networks into families based on their voltage transfer functions.

To facilitate this grouping, the convention is to simplify the voltage transfer function to a

form such that the “Real” part of the denominator of H(jω) is unity (i.e., the denominator

should be 1 + j · · · or 1 − j · · · ). As we will see later in this section, this grouping will alsohelp reduce the math that we do in analyzing various circuits.

The low-pass RL filter discussed before is part of the family of first-order low-pass filters

(first order means that ω appears in the denominator with an exponent of 1 or −1. Ingeneral, the voltage transfer function of a first-order low-pass filter is in the form:

H(jω) =K

1 + jω/ωc

The maximum value of |H(jω)| = |K| is called the filter gain. Note that the exponent of ωin the denominator is +1 so that |H(jω)| decreases with frequency (thus,a low-pass filter):

|H(jω)| = |K|√1 + (ω/ωc)2

6 H(jω) = − |K|K

tan−1(

ω

ωc

)

For RL filter, K = 1, and ωc = R/L. Note that K can be negative, and in that case, the

“minus” sign adds 180◦ phase shift to the transfer function as is denoted by |K|/K factorabove.

-

oi

+ +

-

V

R

CV

Low-pass RC filters

A series RC circuit as shown also acts as a low-pass filter.

For no load resistance (“open-loop” transfer function), Vocan be found from the voltage divider formula:

Vo =1/(jωC)

R + 1/(jωC)Vi =

1

1 + j(ωRC)Vi

H(jω) =1

1 + jωRC

We see that the voltage transfer function of this circuit is similar to transfer function of a

general first-order low-pass filter. So, this is a low-pass filter with K = 1 and ωc = 1/RC.

(Note: we identified the circuit and found the cut-off frequency without doing any math!).

ECE65 Lecture Notes (F. Najmabadi), Spring 2006 25

We could, of course, do the math following the procedure in analyzing the low-pass RL filter

to get the same answer. (Exercise: Show this.).

Following the same procedure as for RL filters, we find input and output Impedances

Zi = R +1

jωCand Zi|min = R

Zo = R ‖1

jωCand Zo|max = R

-

o

-

+

i L

+

VC

R

V R

Terminated RL and RC low-pass filters

Now let us examin the effect of a load on the perfor-

mance of our RL and RC filters. For this example,

a resistive load is considered but the analysis can be

easily extended to an impedance load. For example,

consider the terminated RC filter shown:

From the circuit,

H(jω) =VoVi

=1/(jωC) ‖ RL

R + [1/(jωC) ‖ RL]=

R′/R

1 + j(ωR′C)with R′ = R ‖ RL

This is similar to the transfer function for unterminated RC filter but with resistance R

being replaced by R′. Therefore,

ωc =1

R′C=

1

(R ‖ RL)Cand H(jω) =

R′/R

1 + jω/ωc

We see that the impact of the load is to reduce the filter gain (K = R′/R < 1) and to shift

the cut-off frequency to a higher frequency as R′ = R ‖ RL < R.

Input Impedance: Zi = R +1

jωC‖ RL Zi|min = R

Output Impedance: Zo = R ‖1

jωCZo|max = R

We could have arrived at the same results using the the relationship between open-loop,

Ho(jω), and terminated, H(jω), transfer functions of a two-port network:

H(jω) =ZL

ZL + ZoHo(jω) =

RL

RL + R ‖1

jωC

× 11 + jωRC

ECE65 Lecture Notes (F. Najmabadi), Spring 2006 26

(Exercise: show this.) Also, note that the output impdenace of the terminated circuit is

exactly the same as the open-loop version.

Furthermore, it can be seen that as long as RL � Zo or RL � Zo|max = R (our conditionfor good voltage coupling), R′ ≈ R and the terminated RC filter will look exactly like anunterminated filter – The filter gain is one, the shift in cut-off frequency disappears, and

input and output resistances become the same as before.

Terminated RL low-pass filters

The parameters of the terminated RL filters can be found similarly:

Voltage Transfer Function: H(jω) =VoVi

=1

1 + jω/ωc, ωc = (R ‖ RL)/L.

Input Impedance: Zi = jωL + R ‖ RL, Zi|min = R ‖ RLOutput Impedance: Zo = (jωL) ‖ R, Zo|max = R

Here, the impact of load is to shift the cut-off frequency to a lower value. Filter gain is not

affected. Again for RL � Zo or RL � Zo|max = R (our condition for good voltage coupling),the shift in cut-off frequency disappears and the filter will look exactly like an unterminated

filter.

Exercise: Derive above equations for the transfer function and input and output impde-

nacess.

ECE65 Lecture Notes (F. Najmabadi), Spring 2006 27

2.2 First-order high pass filters

In general, the voltage transfer function of a first-order high-pass filter is in the form:

H(jω) =K

1 − jωc/ω

It is a first-order filter because ω appears in the denominator with an exponent of −1. Itis a high-pass filter because |H| = 0 for ω = 0 and |H| is constant for high-freqnecies.Paramter ωc is the cut-off freqnecy of the filter (Exercise: prove that |H(jωc)| is 1/

√2 = 0.7

of |H(jω)|Max.)

The maximum value of |H(jω)| = |K| is called the filter gain.

|H(jω)| = |K|√1 + (ωc/ω)2

6 H(jω) = +|K|K

tan−1(

ωcω

)

Bode Plots of first-order high-pass filters (K = 1) are shown below. The asymptotic behavior

of this class of filters is:

At low frequencies, ω/ωc � 1, |H(jω)| ∝ ω (a +20dB/decade line) and 6 H(jω) = 90◦At high frequencies, ω/ωc � 1, |H(jω)| ∝ 1 (a line with a slope of 0) and 6 H(jω) = 0◦

|H(jω)|

6 H(jω)

ECE65 Lecture Notes (F. Najmabadi), Spring 2006 28

o

+

i

-

+

-

V V

C

R

High-pass RC filters

A series RC circuit as shown acts as a high-pass filter.

The open-loop voltage transfer function of this filter is:

H(jω) =VoVi

=R

R + 1/(jωC)=

1

1 − j(1/ωRC)

Therefore, this is a first-order high-pass filter with K = 1 and ωC = 1/RC. Input and output

impdenaces of this filter can be found similar to the procedure used for low-pass filters:

Input Impedance: Zi = R +1

jωCand Zi|min = R

Output Impedance: Zo = R ‖1

jωCand Zo|max = R

--

i

+

o

+

VL

R

V

High-pass RL filters

A series RL circuit as shown also acts as a high-pass filter.

Again, we find the open-loop tranfunction to be:

ωc =R

LH(jω) =

1

1 − jωc/ω

Input Impedance: Zi = R + jωL and Zi|min = R

Output Impedance: Zo = R ‖ jωL and Zo|max = R

Exercise: Compute the voltage transfer function and input and output impdenaces of

terminated RC and RL filters.

ECE65 Lecture Notes (F. Najmabadi), Spring 2006 29

2.3 Band-pass filters

A band pass filter allows signals with a range of frequencies (pass band) to pass through and

attenuates signals with frequencies outside this range.

Band

| H(j ) | ω

Pass

u ωωωl

ωl : Lower cut-off frequency;

ωu : Upper cut-off frequency;

ω0 ≡√

ωlωu : Center frequency;

B ≡ ωu − ωl : Band width;Q ≡ ω0

B: Quality factor.

As with practical low- and high-pass filters, upper and lower cut-off frequencies of practical

band pass filter are defined as the frequencies at which the magnitude of the voltage transfer

function is reduced by 1/√

2 (or -3 dB) from its maximum value.

Second-order band-pass filters:

Second-order band pass filters include two storage elements (two capacitors, two inductors,

or one of each). The transfer function for a second-order band-pass filter can be written as

H(jω) =K

1 + jQ(

ω

ω0− ω0

ω

)

|H(jω)| = |K|√1 + Q2

(

ω

ω0− ω0

ω

)2

6 H(jω) = − |K|K

tan−1[

Q(

ω

ω0− ω0

ω

)]

The maximum value of |H(jω)| = |K| is called the filter gain. The lower and upper cut-offfrequencies can be calculated by noting that |H(jω)|max = K, setting |H(jωc)| = K/

√2 and

solving for ωc. This procedure will give two roots: ωl and ωu.

|H(jωc)| =1√2|H(jω)|max =

K√2

=K

√

1 + Q2(

ωcω0

− ω0ωc

)2

Q2(

ωcω0

− ω0ωc

)2

= 1 → Q(

ωcω0

− ω0ωc

)

= ±1

ω2c− ω2

0± ωcω0

Q= 0

ECE65 Lecture Notes (F. Najmabadi), Spring 2006 30

The above equation is really two quadratic equations (one with + sign in front of fraction

and one with a − sign). Solving these equation we will get 4 roots (two roots per equation).Two of these four roots will be negative which are not physical as ωc > 0. The other two

roots are the lower and upper cut-off frequencies (ωl and ωu, respectively):

ωl = ω0

√

1 +1

4Q2− ω0

2Qωu = ω0

√

1 +1

4Q2+

ω02Q

Bode plots of a second-order filter is shown below. Note that as Q increases, the bandwidth

of the filter become smaller and the |H(jω)| becomes more picked around ω0.

|H(jω)|db

6 H(jω)

Asymptotic behavior:

At low frequencies, ω/ω0 � 1, |H(jω)| ∝ ω (a +20dB/decade line), and 6 H(jω) → 90◦

At high frequencies, ω/ω0 � 1, |H(jω)| ∝ 1/ω (a -20dB/decade line), and 6 H(jω) → −90◦

At ω = ω0, H(jω) = K (purely real) |H(jω)| = K (maximum filter gain), and 6 H(jω) = 0◦.

There are two ways to solve second-order filter circuits. 1) One can try to write H(jω) in

the general form of a second-order filters and find Q and ω0. Then, use the formulas above

to find the lower and upper cut-off frequencies. 2) Alternatively, one can directly find the

upper and lower cut-off frequencies and use ω0 ≡√

ωlωu to find the center frequency and

B ≡ ωu − ωl to find the bandwidth, and Q ≡= ω0/B to find the quality factor. The twoexamples below show the two methods. Note that one can always find ω0 and k rapidaly as

H(jω0) is purely real and |H(jω0)| = k

ECE65 Lecture Notes (F. Najmabadi), Spring 2006 31

o

-

+

-

+

i

C

VRV

L

Series RLC Band-pass filters

Using voltage divider formula, we have

H(jω) =VoVi

=R

R + jωL + 1/(jωC)

H(jω) =R

R + j(

ωL − 1ωC

)

There are two approaches to find filter parameters, K, ω0, ωu, and ωl.

Method 1: We transform the transfer function in a form similar to general form of the

transfer function for second order bandpass filters:

H(jω) =K

1 + jQ(

ω

ω0− ω0

ω

)

Note that the denominator of the general form is in the form 1 + j . . . Therefore, we divide

top and bottom of transfer function of series RLC bandpass filters by R:

H(jω) =1

1 + j(

ωL

R− 1

ωRC

)

Comparing the above with the general form of the transfer function, we find K = 1. To find

Q and ω0, we note that the imaginary part of the denominator has two terms, one positive

and one negative (or one that scales as ω and the other that scales as 1/ω) similar to the

general form of transfer function of 2nd-order band-pass filters (which includes Qω/ω0 and

−Qω0/ω). Equating these similar terms we get:

Qω

ω0=

ωL

R→ Q

ω0=

L

R

Qω0ω

=1

ωRC→ Qω0 =

1

RC

We can solve these two equations to find:

ω0 =1√LC

Q =ω0

R/L=

√

L

R2C

ECE65 Lecture Notes (F. Najmabadi), Spring 2006 32

The lower and upper cut-off frequencies can now be found from the formulas on page 31.

Method 2: In this method, we directly calculate the filter parameters similar to the proce-

dure followed for general form of transfer function in page 30. Some simplifications can be

made by noting: 1) At ω = ω0, H(jω) is purely real and 2) K = H(jω = jω0).

Starting with the transfer function for the series RLC filter:

H(jω) =R

R + j(

ωL − 1ωC

)

We note that the transfer function is real if coefficient of j in the denominator is exactly

zero (note that this happens for ω = ω0), i.e.,

ω0L −1

ω0C= 0 −→ ω0 =

1√LC

Also

K = H(jω = jω0) =R

R= 1

The cut-off frequencies can then be found by setting:

|H(jωc)| =K√2

=1√2

1 +(

ωcL

R− 1

ωcRC

)2

= 2

which can be solved to find ωu and ωl.

Input and Output Impedance of band-pass RLC filters

Zi = jωL +1

jωC+ R = j

(

ωL − 1ωC

)

+ R

Zi|min = R occurs at ω = ω0

Zo =

(

jωL +1

jωC

)

‖ R → Zo|max = R

ECE65 Lecture Notes (F. Najmabadi), Spring 2006 33

Wide-Band Band-Pass Filters

Band-pass filters can be constructed by putting a high-pass and a low-pass filter back to

back as shown below. The high-pass filter sets the lower cut-off frequency and the low-pass

filter sets the upper cut-off frequency of such a band-pass filter.

ω| H (j ) | 2

ω1

| H (j ) | X

ω =l ω =ω

uω

ω| H (j ) |

ω

ω| H (j ) |

ω

1

ω ω ω

2

c2 c1 c2 c1

o

−

+

i

−

+

−

+

1

High−PassLow−Pass

2 1

12 VV C V

R C

R

An example of such a band-pass filter is

two RC low-pass and high-pass filters put

back to back. These filters are widely

used (when appropriate, see below) instead

of an RLC filter as inductors are usually

bulky and take too much space on a cir-

cuit board.

In order to have good voltage coupling in the above circuit, the input impedance of the

high-pass filter (actually Zi|min = R1) should be much larger than the output impedance ofthe low-pass filter (actually Zo|max = R2), or we should have R1 � R2. In that case we canuse un-terminated transfer functions:

H(jω) = H1(jω) × H2(jω) =1

1 + jω/ωc2× 1

1 − jωc1/ωωc1 = 1/(R1C1) ωc2 = 1/(R2C2)

H(jω) =1

(1 + jω/ωc2)(1 − jωc1/ω)=

1

(1 + ωc1/ωc2) + j(ω/ωc2 − ωc1/ω)

Again, we can find the filter parameters by either of two methods above. Transforming the

transfer function to a form similar to the general form (left for students) gives:

K =1

1 + ωc1/ωc2Q =

√

ωc1/ωc2

1 + ωc1/ωc2ω0 =

√ωc1ωc2

ECE65 Lecture Notes (F. Najmabadi), Spring 2006 34

One should note that the Bode plots of previous page are “asymptotic” plots. The real

H(jω) differs from these asymptotic plots, for example, |H(jω)| is 3 dB lower at the cut-off frequency. A comparison of “asymptotic” Bode plots for first-order high-pass filters are

given in page 28. It can be seen that |H1(jω)| achieves its maximum value (1 in this case)only when ω/ωc1 < 1/3. Similarly for the low pass filter, |H2(jω)| achieves its maximumvalue (1 in this case) only when ω/ωc2 > 3. In the band-pass filter above, if ωc2 � ωc1 (i.e.,ωc2 ≥ 10ωc1), the center frequency of the filter will be at least a factor of three away fromboth cut-off frequencies and |H(jω)| = |H1| × |H2| achieves its maximum value of 1. If ωc2is not � ωc1 (i.e., ωc2 < 10ωc1), H1 and H2 will not reach their maximum of 1 and the filter|H(jω)|max = |H1| × |H2| will be less than one. This can be seen by examining the equationof K above which is always less than 1 and approaches 1 when ωc2 � ωc1.

More importantly, we can never make a “narrow” band filter by putting two first-order high-

pass and low-pass filters back to back. When ωc2 is not � ωc1, |H(jω)|max becomes smallerthan 1. Since the cut-off frequencies are located 3 dB below the maximum values, the cut-off

frequencies will not be ωc1 and ωc2 (those frequencies are 3 dB lower than |H(jω)|max = 1).The lower cut-off frequency moves to a value lower than ωc1 and the upper cut-off frequency

moves to a value higher than ωc2. This can be seen by examining the quality factor of this

filter at the limit of ωc2 = ωc1

Q =

√

ωc1/ωc2

1 + ωc1/ωc2=

1

1 + 1= 0.5

while our asymptotic description of previous page indicated that when ωc2 = ωc1, band-width

becomes vanishingly small and Q should become very large.

Because these filters work only when ωc2 � ωc1, they are called “wide-band” filters. Forthese wide-band filters (ωc1 � ωc2), we find from above:

K = 1 Q =√

ωc1/ωc2 ω0 =√

ωc2ωc1

H(jω) =1

1 + j(ω/ωc2 − ωc1/ω)

We then substitute for Q and ω0 in the expressions for cut-off frequencies (page 31) to get:

ωu = ω0

√

1 +1

4Q2+

ω02Q

=ω02Q

(

√

1 + 4Q2 + 1)

ωl = ω0

√

1 +1

4Q2+

ω02Q

=ω02Q

(

√

1 + 4Q2 − 1)

ECE65 Lecture Notes (F. Najmabadi), Spring 2006 35

Ignoring 4Q2 term compared to 1 (because Q is small),we get:

ωu =ω0Q

=

√ωc2ωc1

√

ωc1/ωc2= ωc2

For ωl, if we ignore 4Q2 term compared to 1, we will find ωl = 0. We should, therefore,

expand the square root by Taylor series expansion to get the first order term:

ωu ≈ω02Q

(

1 +1

24Q2 − 1

)

=ω02Q

× 2Q2 = ω0Q = ωc2

What are Wide-Band and Narrow-Band Filters? Typically, a wide-band filter is

defined as a filter with ωc2 � ωc1 (or ωc2 ≥ 10ωc1). In this case, Q ≤ 0.35 (prove this!). Anarrow-band filter is usually defined as a filter with B � ω0 (or B ≤ 0.1ω0). In this case,Q ≥ 10.

Example: Design a band-pass filter with cut-off frequencies of 160 Hz and 8 kHz. The load

for this circuit is 1 MΩ.

As this is wide-band, band-pass filter (ωu/ωl = fu/fl = 50 � 1), we use two low- andhigh-pass RC filter stages similar to circuit above. The prototype of the circuit is shown

below:

o

−

+

i

−

+

−

+

1

High−PassLow−Pass

2 1

12 VV C V

R C

RThe high-pass filter sets the lower cut-off

frequency, and the 1 MΩ load sets the out-

put impedance of this stage. Thus:

Zo|max = R1 � 1 MΩ → R1 ≤ 100 kΩ

ωc(High-pass) = ωl =1

R1C1= 2π × 160 → R1C1 = 1 × 10−3kΩ

One should choose R1 as close as possible to 100 kΩ (to make the C1 small) and R1C1 =

1×10−3 using commercial values of resistors and capacitors. A good set here are R1 = 100 kΩand C1 = 10 nF.

The low-pass filter sets the upper cut-off frequency. The load for this component is the input

resistance of the high-pass filter, Zi|min = R1 = 100 kΩ. Thus:

Zo|max = R2 � 100kΩ → R2 ≤ 10 kΩ

ωc(Low-pass) = ωu =1

R2C2= 2π × 8 × 103 → R2C2 = 2 × 10−5

ECE65 Lecture Notes (F. Najmabadi), Spring 2006 36

As before, one should choose R2 as close as possible to 10 kΩ and R2C2 = 2 × 10−5 usingcommercial values of resistors and capacitors. A good set here are R2 = 10 kΩ and C2 = 2 nF.

In principle, we can switch the position of low-pass and high-pass filter stages in a wide-

band, band-pass filter. However, the low-pass filter is usually placed before the high-pass

filter because the value of capacitors in such an arrangement will be smaller. (Try redesigning

the above circuit with low-pass and high-pass filter stages switched to see that one capacitor

become much smaller and one much larger.)

Exercise: Design an RLC filter with the specifications in the previous example. (Hint: Do

not set R = 100 kΩ as this would make the value of the inductor very large.)

2.4 Exercise Problems

Problem 1. Design a RLC bandpass filter with a lower cut-off frequency of 1 kHz and a

bandwidth of 3 kHz. What is the center frequency and Q of this filter?

Problem 2. We have an amplifier that amplifies a 1 kHz signal from a detector. The load

for this amplifier can be modeled as a 50 kΩ resistor. The amplifier output has a large

amount of 60 Hz noise. We need to reduce the amplitude of noise by a factor of 10. Design

a first-order passive filter which can be placed between the amplifier and the load and does

the job. Would this filter affect the 1 kHz signal that we are interested in? If so, by how

much?

Problem 3. The tuner for an FM radio requires a band-pass filter with a central frequency

of 100 MHz (frequency of a FM station) and a bandwidth of 2 MHz. a) Design such a filter.

b) What are its cut-off frequencies?

Problem 4. A telephone line carries both voice band (0-4 kHz) and data band (25 kHz to

1 MHz). Design a filter that lets the voice band through and rejects the data band. The

filter must meet the following specifications: a) For the voice band, the change in transfer

function should be at most 1 dB; and b) The transfer function should be as small as possible

at 25 kHz, the low end of the data band.

ECE65 Lecture Notes (F. Najmabadi), Spring 2006 37

2.5 Solution to Exercise Problems

Problem 1. Design a RLC bandpass filter with a lower cut-off frequency of 1 kHz and a

bandwidth of 3 kHz. What is the center frequency and Q of this filter?

o

-

+

-

+

i

C

VRV

L

The circuit prototype is:

For a 2nd order band-pass filter:

B(Hz) = fu − fl fu = 1 + 3 = 4 kHzB(rad/s) = 2πB(Hz) = 1.88 × 104

ωu = 2πfu = 2.51 × 104 ωl = 2πfl = 6.28 × 103

ω0 =√

ωuωl = 1.26 × 104

B(rad/s) =ω0Q

→ Q = 1.26 × 104

1.88 × 104 = 0.67

For the series RLC circuit:

ω0 =1√LC

→ LC = 1ω20

C =1

Lω20=

1

10 × 10−3 × (1.26 × 104)2 = 0.63 µF

Q =ω0

R/L→ R

L=

ω0Q

= B(rad/s)

R = LB = 10 × 10−3 × 1.88 × 104 = 188 Ω

Therefore, using commercial values, the design values are L = 10 mH, R = 180 Ω, and

C = 0.68 µF.

ECE65 Lecture Notes (F. Najmabadi), Spring 2006 38

Problem 2. We have an amplifier that amplifies a 1 kHz signal from a detector. The load

for this amplifier can be modeled as a 50 kΩ resistor. The amplifier output has a large

amount of 60 Hz noise. We need to reduce the amplitude of noise by a factor of 10. Design

a first-order passive filter which can be placed between the amplifier and the load and does

the job. Would this filter affect the 1 kHz signal that we are interested in? If so, by how

much?

We want to have 1 kHz signals to go through but reduce 60 Hz signals, so we need a high-pass

filter. The prototype of the circuit is shown below. For this circuit:

InvertingAmp. i Lo

-

+

-

+

RV

C

RV

H(jω) =VoVi

=1

1 − jωc/ω

ωc =1

RC

Zi|min = RZo|max = R

As the output impedance of the inverting amplifier circuit is “zero”, we do not need to worry

about the input impedance of our filter. The output impedance of the filter is restricted by

Zo|max = R � 50 kΩ → R ≤ 5 kΩ

This filter should reduce the amplitude of 60 Hz (ω60 = 2π × 60 = 120π rad/s) signal by afactor of 10, i.e.,

|H(jω = jω60)| =∣

∣

∣

∣

VoVi

∣

∣

∣

∣

60 Hz=

1√

1 + (ωc/ω60)2= 0.1

1 + (ωc/ω60)2 = 100 → 1

RC= ωc ≈ 10ω60 = 3751 rad/s → RC = 2.67 × 10−4

Reasonable choices are R = 3.9 kΩ (to keep it below 5 kΩ) and C = 68 nF (fc ≈ 600 Hz).

The impact on 1 kHz signal (ω1000 = 2000π rad/s) can be found from:

|H(jω = jω1000)| =1

√

1 + (ωc/ω1000)2=

1√

1 + (3751/6283)2= 0.86

So the amplitude of 1 kHz signal is reduced by 14% (or by -1.3 dB).

ECE65 Lecture Notes (F. Najmabadi), Spring 2006 39

Problem 3. The tuner for an FM radio requires a band-pass filter with a central frequency

of 100 MHz (frequency of a FM station) and a bandwidth of 2 MHz. a) Design such a filter.

b) What are its cut-off frequencies?

o

−

+

i

−

+

VV R

L CBecause this is not a wide-band filter, the

simplest filter will be an RLC filter as is

shown. For this filter:

ω0 =1√LC

= 2π100 × 106

Q =ω0B

=

√

L

R2C=

2π100 × 1062π × 2 × 106 = 50

Using a L = 1 µH inductor:

1

LC= 4π21016 → 1

C= 4π21016 × 10−6 → C = 2.5 × 10−12 F

Choose: C = 2.2 pF

L

R2C= 2, 500 → R2 = L

2, 500C=

10−6

2, 500 × 2.2 × 10−12 = 182 → R = 13.5 Ω

Choose: R = 13 Ω (L = 1 µH and C = 2.2 pF).

To find the cut-off freqnecies, we not:

B = fu − fl = 2 MHzf0 =

√

fufL = 100 MHz

Solution of the above two equations in two unknowns will give fl ≈ 99 MHz and fu ≈101 MHz.

ECE65 Lecture Notes (F. Najmabadi), Spring 2006 40

Problem 4. A telephone line carries both voice band (0-4 kHz) and data band (25 kHz to

1 MHz). Design a filter that lets the voice band through and rejects the data band. The

filter must meet the following specifications: a) For the voice band, the change in transfer

function should be at most 1 dB; and b) The transfer function should be as small as possible

at 25 kHz, the low end of the data band.

We need a low-pass filter as it should allow low-frequency signals (voice band) to go through

while eliminating high-frequency signals (data band). The prototype of an RC low-pass filter

is shown and its transfer function is:

H(jω) =1

1 + jω/ωc=

1

1 + jf/fci o

+

−

+

−

ω| H(j ) |

f (kHz)4 25

1

20

−1dB

V V

R

C

The cut-off frequency of the filter is not given and

it should be found from the specifications. First,

we need the change in transfer function to be at

most 1 dB for the frequency range of 0-4 kHz.

The transfer function of filters that satisfy this

constraint is the curve labeled “1” in the figure

and any transfer function located to the right of

this curve (such as transfer function labeled “2”).

Second, the transfer function should be as small as possible at 25 kHz. This requires that

we choose the cut-off frequency as small as possible. Therefore, the transfer function of our

filter should be curve labeled “1” as it has the smallest possible value at 25 kHz:

20 log (|H(jf = 4 kHz)|) = −1 dB → |H(jf = 4 kHz)| = 0.891

Using the expression for H(jω), we have:

|H(jf = 4 kHz)| = 1√1 + (f/fc)2

= 0.891

f/fc = 0.509 → fc =f

0.509= 7.85 kHz

fc =1

2πRC= 7.85 × 103 → RC = 2.0 × 10−5

Choosing C = 1 nF, we have R = 2.0 × 104. The commercial values then are C = 1 nF andR = 20 kΩ.

ECE65 Lecture Notes (F. Najmabadi), Spring 2006 41