EE 230 1st-order filters – 1

First-order filtersThe general form for the transfer function of a first order filter is:

T (s) = Go ⋅s + Zo

s + Po

There will always be a single pole at s = –Po. The pole must be real (there is only one, so no complex conjugates are not possible) and it must be negative (for stability). There will always be a zero, which can be at s = 0, as s → ±∞ (zero at infinity), or somewhere else, s = –Zo. (Note the zero can have a positive value.) There may be a gain factor, Go, which might be 1 or smaller (for a passive circuit with a voltage divider) or have a magnitude greater than 1 for an active circuit.

The two most important cases are the zero at infinity, which is a low-pass filter and the zero at zero, which is the high-pass filter.

T (s) =a1s + a0

b1s + bo

However, we will typically recast this into a standard form:

EE 230 1st-order filters – 2

Low-passIn the case were a1 = 0, we have a low-pass function.

T (s) = Go ⋅Po

s + Po

T (s) =ao

b1s + bo

In standard form, we write it as:σ

jω

xpole at –P0

s-plane

zero as s → ±∞

The reason for this form will become clear as we proceed. We will ignore the gain initially and focus on sinusoidal behavior by letting s = jω.

Po

s + Po s=jω

=Po

Po + jω

Re-expressing the complex value in magnitude and phase form:

Po

Po + jω= [ Po

P2o + ω2 ]exp (jθLP) θLP = − arctan ( ω

Po )

EE 230 1st-order filters – 3

By looking at the magnitude expression, we can see the low-pass behavior.

For low frequencies (ω << Po): .

At high frequencies (ω >> Po): .

At low frequencies, the magnitude is 1 (the output is equal to the input) and at high frequencies, the magnitude goes down inversely with frequency, consistent with the notion of a low-pass response.

We can also examine the phase at the extremes.

For low frequencies (ω ≈ 0): .

At high frequencies (ω → +∞): .

Po

P2o + ω2

≈Po

P2o

= 1

Po

P2o + ω2

≈Po

ω2=

Po

ω

θLP = − arctan ( ωPo ) ≈ 0

θLP = − arctan ( ωPo ) ≈ − 90∘

EE 230 1st-order filters – 4

M =Po

P2o + ω2

We can use the functions to make the magnitude and phase as frequency response plots.

θLP = − arctan ( ωPo )

mag

nitu

de

0.0

0.2

0.4

0.6

0.8

1.0

1.2

10 100 1000 10000 100000

mag

nitu

de (d

B)

-50

-40

-30

-20

-10

0

10

10 100 1000 10000 100000

phas

e (°

)

-90

-75

-60

-45

-30

-15

0

10 100 1000 10000 100000

angular frequency (rad/s) angular frequency (rad/s)

angular frequency (rad/s)

magnitude - linear scale magnitude - Bode plot

phase

EE 230 1st-order filters – 5

Use the standard definition for cut-off frequency, which is the frequency at which the magnitude is down by from the value in the pass-band. For our low-pass function, the pass band is at low frequencies, and the magnitude there is 1. (Again, we are “hiding” Go by assuming that it is unity. If Go ≠ 1, then everything is scaled by Go.)

2

Cut-off frequency

M =1

2=

Po

P2o + ω2

c

With a bit of algebra, we find that . The cut-off frequency is defined by the pole. Tricky! Thus, in all of our equations, we could substitute ωc for Po.

ωc = Po

We can also calculate the phase at the cut-off frequency.

θLP = − arctan ( ωc

Po ) = − 45∘

The cut-off frequency points are indicated in the plots on the previous slide.

EE 230 1st-order filters – 6

ս/3 = � arctan

�F

�

7/3 (M) = *R · FM + F

7/3 (V) = *R · FV+ F

7/3 (V) =*R

�+ VF

7/3 (M) =*R

�+ M�

F

�

ս/3 = � arctan

�F

�

Again, we could just as easily use real frequency rather than angular frequency. As an exercise: re-express all of the above formulas using f instead of ω.

To emphasize the importance of the corner frequency in the low-pass function, we can express all the previous results using ωc in place of Po. On the left are the functions in "standard" form. On the right, the functions are expressed in a slightly different form that is sometimes easier to use.

��7/3 (M)�� =

��*R�� F�

� + �F<latexit sha1_base64="ZcBktQQdyEg5AHKJu2kZSXSy48k=">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</latexit><latexit sha1_base64="ZcBktQQdyEg5AHKJu2kZSXSy48k=">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</latexit><latexit sha1_base64="ZcBktQQdyEg5AHKJu2kZSXSy48k=">AAAC53icbZFBaxNBFMdn16o12prq0ctgEFIsIRsEPSgUFOqhhwhNU8imy+zk7WbM7Mw681YSlv0M3sSrH8uLn8XZJC1J6oOFP7///83svBfnUljsdv94/r29+w8e7j9qPH5ycPi0efTs0urCcBhwLbW5ipkFKRQMUKCEq9wAy2IJw3j2sfaH38FYodUFLnIYZyxVIhGcoUNRcx7GIqWhiyC9iMrzfhVKSLD9NdQZpCw0Ip3icR1aZT7QjYazSNNbK0wM4+WqjUa8KkP7zeANuO69prfeda+qomar2+kui94VwVq0yLr60ZE3DCeaFxko5JJZOwq6OY5LZlBwCVUjLCzkjM9YCiMnFcvAjsvliCr6ypEJTbRxn0K6pJsdJcusXWSxS2YMp3bXq+H/vFGBybtxKVReICi+uigpJEVN63nTiTDAUS6cYNwI96+UT5kbFbqtbN2yPDsHvvWScl4owfUEdqjEORrmoAXMmFD1q8o+k26tSt9gd17N259EKtCenNfmyZkBmB1vZBtuFcHu4O+Ky14n6HaCL29ap+/XS9knL8hL0iYBeUtOyWfSJwPCyV9vzzvwDn3h//B/+r9WUd9b9zwnW+X//gdpMOvt</latexit><latexit sha1_base64="ZcBktQQdyEg5AHKJu2kZSXSy48k=">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</latexit>

��7/3 (M)�� =

��*R��

��+

�F

��

<latexit sha1_base64="tsruQKFNAqMSwf46/a2GHQ0xhLA=">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</latexit><latexit sha1_base64="tsruQKFNAqMSwf46/a2GHQ0xhLA=">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</latexit><latexit sha1_base64="tsruQKFNAqMSwf46/a2GHQ0xhLA=">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</latexit><latexit sha1_base64="tsruQKFNAqMSwf46/a2GHQ0xhLA=">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</latexit>

EE 230 1st-order filters – 7

Low-pass filter circuits: simple RC

Use a voltage divider to find the transfer function.

9R (V) ==&

=& + =59L (V)

Clearly, this is low-pass with Go = 1 and F =�5&

–

++–Vi(s)

ZR =R

Vo(s)=�V&

ZC

Resistor and capacitor in series — output taken across the capacitor.

7 (V) =9R (V)9L (V)

==&

=& + =5=

�V&

�V& + 5

=�5&

V+ �5&

=F

V+ F

The only real design consideration is choosing the RC product, which then sets the corner frequency.

EE 230 1st-order filters – 8

Low-pass filter circuits: simple RL

Use a voltage divider to find the transfer function.

Again, low-pass behavior with Go = 1, but now with

Inductor and resistor in series — output taken across the resistor.

7 (V) =9R (V)9L (V)

==5

=5 + =/=

55 + V/ =

5/

V + 5/

=F

V + F

9R (V) ==5

=5 + =/9L (V)

F =5/

+– –

+Vi(s) Vo(s)

ZL = sL

ZR

= R

As with the previous example choosing the “RL time constant” , we can define the pass-band of this low-pass filter.

EE 230 1st-order filters – 9

Z1=� = 5�

�����V& =

5��+ V5�&

–+

C

R1

R2

Vi(s) Vo(s)

7 (V) =9R (V)9L (V)

= �=�=�

= �5�

�+V5�&5�

=�5�

5��+ V5�&

At low frequencies: |T| → R2/R1, and θT → 180° (= – 180°)

At high frequencies: |T| → (ωR1C)–1, and θT → +90° (= – 270°)

=

��5�5�

� �5�&

V+ �5�&

Clearly, this is also low-pass with and *R = �5�5�

F =�

5�&

Be careful with the extra negative sign in the gain: –1 = exp( j180° )

Low-pass filter circuits: inverting op amp

EE 230 1st-order filters – 10

Magnitude and phase plots for an active low-pass filter with R1 = 1 kΩ, R2 = 25 kΩ, and C = 6.4 nF, giving fo = 1000 Hz and Go = –25 (|Go|= 28 dB).

EE 230 1st-order filters – 11

–+

R

CVo(s)

Vi(s)

R1R2

Low-pass filter circuits: non-inverting op amp

Note: It might slightly disingenuous to treat this as if it were some new type of filter — we can readily see that it is a simple RC filter cascaded with a simple non-inverting amp. However, it is still a useful circuit.

9+ (V) ==&

=& + =5=

�5&

V+ �5&

9R (V) =

��+

5�5�

�9+ (V) non-inverting amp

simple RC

Low-pass with and *R =

��+

5�5�

�F =

�5&

7 (V) =9R (V)9L (V)

=

��+

5�5�

���5&

V + �5&

�

EE 230 1st-order filters – 12

Low-pass filter circuits: another RC

–

++–Vi(s)

ZR1 = R1

Vo(s)=�V&

ZC ZR2

= R2

=3 = =5���=&

=5�

� �V&

�

5� + �V&

=5�

�+ V5�&

7 (V) =9R (V)9L (V)

==3

=3 + =5�

=5�

�+V5�&5�

�+V5�& + 5�

=5�

5� + 5� + V5�5�&

=

�5�

5� + 5�

� �53&

V+ �53&

= *R · FV + F

53 = 5���5�

*R =5�

5� + 5�Note the voltage divider. “Gain” < 1.

F =�

53&

Low-pass.

The corner depends on the parallel combination.

EE 230 1st-order filters – 13

High-passIn the case were a0 = 0, we have a high-pass function.

T (s) = Go ⋅s

s + Po

T (s) =a1s

b1s + bo

In standard form, we write it as:

We will ignore the gain initially (set Go = 1) and focus on sinusoidal behavior by letting s = jω.

ss + Po s=jω

=jω

Po + jω

Re-expressing the complex value in magnitude and phase form:

jωPo + jω

= [ ω

P2o + ω2 ]exp (jθHP) θHP = 90∘ − arctan ( ω

Po )

σ

jω

xpole at –P0

s-plane

zero at 0

EE 230 1st-order filters – 14

By looking at the magnitude expression, we can see the high-pass behavior.

For low frequencies (ω << Po): .

At high frequencies (ω >> Po): .

At low frequencies, the magnitude is increasing with frequency, and at high frequencies, the magnitude is 1 (the output is equal to the input). This behavior is consistent with a high-pass response.

We can also examine the phase at the extremes.

For low frequencies (ω << Po): .

At high frequencies (ω >> Po): .

ω

P2o + ω2

≈ω

P2o

=ωPo

ω

P2o + ω2

≈ω

ω2= 1

θHP = 90∘ − arctan ( ωPo ) ≈ 90∘

θHP = 90∘ − arctan ( ωPo ) ≈ 0∘

EE 230 1st-order filters – 15

M =ω

P2o + ω2

θHP = 90∘ − arctan ( ωPo )

angular frequency (rad/s) angular frequency (rad/s)

angular frequency (rad/s)

mag

nitu

de

0.0

0.2

0.4

0.6

0.8

1.0

1.2

10 100 1000 10000 100000

mag

nitu

de (d

B)

-50

-40

-30

-20

-10

0

10

10 100 1000 10000 100000ph

ase

(°)

0

15

30

45

60

75

90

10 100 1000 10000 100000

We can use the functions to make the magnitude and phase as frequency response plots.

magnitude - linear scale magnitude - Bode plot

phase

EE 230 1st-order filters – 16

Cut-off frequency

M =1

2=

ω

P2o + ω2

c

With a bit of algebra, we find that . The same result as for low-pass response, except that pass-band is above the cut-off frequency in this case. Once again, we see the importance of the poles in determining the behavior of the transfer functions.

ωc = Po

We can calculate the phase at the cut-off frequency.

θHP = 90∘ − arctan ( ωc

Po ) = 45∘

Use the standard definition for cut-off frequency, which is the frequency at which the magnitude is down by from the value in the pass-band. For our high-pass function, the pass band is at high frequencies, and the magnitude there is 1. (Again, we are “hiding” Go by assuming that it is unity. If Go ≠ 1, then everything is scaled by Go.)

2

The cut-off frequency points are indicated in the plots on the previous slide.

EE 230 1st-order filters – 17Exercise: Re-express all of the above formulas using f instead of ω.

7+3 (V) = *R · VV+ F

7+3 (M) = *R · MM + F

��7/3 (M)�� = *R · �

� + �F

ս+3 = arctan��

�� arctan

�F

�

= ��� � arctan

�F

�

7+3 (V) =*R

�+ FV

=*R

�� M�F

�

7+3 (M) =*R

�+�FM

�

ս+3 = + arctan�F

�

��7+3 (M)�� =

*R��+

�F

��

→*R��

=*R · 3���F + 3��

The corner frequency is the value of the pole.P0 = ωc

To emphasize the importance of the corner frequency in the high-pass function, we can express all the previous results using ωc in place of Po.

EE 230 1st-order filters – 18

9R (V) ==5

=& + =59L (V)

High-pass filter circuits: simple RC

–

++–Vi(s) ZR

=RVo(s)

ZC=�V&

Use a voltage divider to find the transfer function.

Capacitor and resistor in series — output taken across the resistor.

7 (V) =9R (V)9L (V)

==5

=5 + =&=

55+ �

V&=

VV+ �

5&=

VV+ F

Clearly, this is high-pass with Go = 1 and F =�5&

EE 230 1st-order filters – 19

9R (V) ==/

=/ + =59L (V)

F =5/Again, high-pass with Go = 1 but with

High-pass filter circuits: simple RL

+–

–

+Vi(s) Vo(s)ZL

= sL

ZR = R

7 (V) =9R (V)9L (V)

==/

=/ + =5=

V/V/ + 5 =

VV + 5

/=

VV + F

EE 230 1st-order filters – 20

Z2

Z1

–+

C R1

R2

Vi(s) Vo(s)

=� = 5� +�V&

We see that this is also high-pass with and *R = �5�5�

F =�

5�&

The same comments about the phase apply here: the –1 in the gain factor introduces an extra 180° (or –180°) of phase.

High-pass filter circuits: inverting op amp

7 (V) =9R (V)9L (V)

= �=�=�

= � 5�5� + �

V&=

��5�5�

�V

V + �5�&

EE 230 1st-order filters – 21

High-pass filter circuits: non-inverting op amp

Again, this is simple RC high-pass cascaded with a non-inverting amp.

9R (V) =

��+

5�5�

�9+ (V) non-inverting amp

simple RC high pass

High-pass with and *R =

��+

5�5�

�F =

�5&

–+

R

CVo(s)

Vi(s)

R1R2

9+ (V) ==5

=& + =5=

VV+ �

5&

7 (V) =9R (V)9L (V)

=

��+

5�5�

��V

V + �5&

�

EE 230 1st-order filters – 22

High-pass filter circuits: another RC

*R =5�

5� + 5�Note the voltage divider. “Gain” < 1.

High-pass.

The corner depends on the series combination.

7 (V) =9R (V)9L (V)

==5�

=5� + =5� + =&

=5�

5� + 5� + �V&

=

�5�

5� + 5�

�V

V+ �(5�+5�)&

= *R · VV + F

F =�

(5� + 5�)&

–

++–Vi(s) ZR2

= R2

Vo(s)

ZC=�V&ZR1 = R1

EE 230 1st-order filters – 23

Case study: inductors can be troubleCheap inductors can have a relatively large series resistance. This parasitic resistance can cause trouble in certain circumstances.

+– L

–

+Vo(s)Vi(s)

R

7 (V) =V

V+ 5/

=V

V+ R

+–

L

–

+

Vo(s)Vi(s)

R

Rs

7 (V) =V/+ 5V

V/+ 5V + 5�=

V+ 5V/

V+ 5V+5�/

=V+ =V+ 3

ideal

real

Pole and zero are both shifted!

As usual, zero at s = 0, pole at s = –ωc.

EE 230 1st-order filters – 24

7 (M) == + M3 + M

��7�� =

��= + �

��3 + �

ս7 = arctan

�=

�� arctan

�3

�

Effect of inductor parasitic resistance on high-pass filter: L = 0.027 H, R1 = 1 kΩ, and Rs = 60 Ω. The frequency responses for both magnitude and phase are quite different.