EE 205 Dr. A. Zidouri
Electric Circuits II
Frequency Selective Circuits (Filters)
Low and High Pass Filters
Lecture #37
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EE 205 Dr. A. Zidouri
The material to be covered in this lecture is as follows:
o
o
o
o
o
Series RC Circuit as a Low Pass filter
Relating The Frequency Domain to the Time Domain
Series RC Circuit as High-Pass filter
Series RC Circuit- Qualitative Analysis
The Series RC Circuit- Quantitative Analysis
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EE 205 Dr. A. Zidouri
After finishing this lecture you should be able to:
¾
¾
¾
¾
¾
Understand the Behavior of High-Pass Filters
Distinguish between Low-Pass and High-Pass Filters
Relate The Cutoff Frequency to the Time Constant
Analyze Qualitatively And Quantitatively A High-Pass Filter
Design A Simple High-Pass Filter
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EE 205 Dr. A. Zidouri
The Series RC Circuit Low-Pass Filter
The series RC Circuit shown in Fig. 37-1 also behaves as a low pass filter.
Note that the circuit’s output is defined as the output across the capacitor.
As we did in the previous qualitative analysis in Lec36, we use three frequency regions to develop
the behavior of the series RC circuit in Fig. 37-1:
1. Zero Frequency ω = 0 → impedance of the capacitor ZC = ∞ , the capacitor acts as an open
circuit. The input and output voltages are the same as shown in Fig. 37-2a
ω =0
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EE 205 Dr. A. Zidouri
2. Frequencies increasing from zero. The impedance of the capacitor Z C R , The output
voltage is smaller than the source voltage which divides between R and C.
3. Infinite frequency ω = ∞ → impedance of the capacitor ZC
circuit. The output voltage is zero as shown in Fig. 37-2b
= 0 , the capacitor acts as a short
R
Vi
C
Vo=0
Fig. 37-2b A Series RC
Circuit at ω = ∞
Based on this analysis we see that the series RC circuit functions as a low-pass filter.
The following example explores the quantitative analysis of the circuit.
Example 37-1
For the series RC circuit in Fig. 37-1 above:
Vo
™ Find the transfer function H ( jω ) =
Vi
™ Determine the cutoff frequency in the circuit
™ Choose values of R and C that will yield a low pass filter with
ω = 3kHz . Take C = 1µ F .
C
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EE 205 Dr. A. Zidouri
Solution:
To derive the transfer function, consider the s-domain circuit in Fig. 37-3:
1
H ( s ) = RC
s+ 1
RC
R
substituting
Vi(s)
1
RC
H ( jω ) =
s = jω we get
⎛
ω2 + ⎜⎜ 1
The cutoff frequency can be found by
2
⎞
1
sC
Vo(s)
Fig. 37-3 s-domain
equivalent for Fig. 37-1
⎟⎟
⎝ RC ⎠
solving
Given
( )
1
RC
H ωC =
⎛
ω 2 + ⎜⎜
⎝
C = 1µ F then R =
1 ⎞
⎟
RC ⎟⎠
2
=
1
2
for
ωC
we get:
ωC =
1
RC
1
1
=
= 53Ω
ωC C ( 2π ) 3×103 1×10−6
(
)(
)
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EE 205 Dr. A. Zidouri
Summary:
Fig. 37-4 summarizes the two low-pass filter circuits we have seen in Lec36 and this lecture. Look how
similar in form their transfer functions are:
H (s) =
L
R
Vi
Vo
R
Vi
C
Vo
ωC
s + ωC
R
H (s) = L
s+ R
L
ωC =
R
L
1
H ( s ) = RC
s+ 1
RC
1
ωC =
RC
(37-1)
Relating the Frequency Domain to the Time Domain
An other important relationship is relating the time
constant to the cutoff frequency.
Remember that for RL circuit the time constant is
τ = L and for an RC circuit the time constant is
R
τ = RC
Compare the time constants to the cutoff
frequencies for these circuits, notice that
τ
= 1
ω
(37-2)
C
Fig. 37-4 Two Low-Pass filters together with their
Transfer Functions and Cutoff Frequencies
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EE 205 Dr. A. Zidouri
High Pass Filter
The Series RC Circuit – Qualitative Analysis
Now we examine a High-Pass Filter, series RC Circuit shown in Fig. 37-5a
The circuit’s output is the voltage across R.
Behavior of R will not change with changing frequency
Behavior of capacitor C will change with changing frequency
Recall that the impedance of a capacitor is 1
at high frequencies the capacitor’s impedance
jωC
is very small compared with the resistor’s impedance,
The capacitor effectively functions as a short circuit. The term high frequencies thus refers to any
frequencies for which ωC R
The equivalent circuit for ω = ∞ is shown in Fig. 37-5b
At this frequency Vi = Vo both in magnitude and phase angle
Decreasing the frequency causes the capacitor’s impedance to increase.
Increasing the capacitor’s impedance causes a corresponding increase in the magnitude of the
voltage drop across C
Increasing the capacitor’s impedance causes a corresponding decrease in the magnitude of Vo
the output voltage drop across R
Increasing the capacitor’s impedance also introduces a shift in phase angle between VC the
voltage across the capacitor and IC the current in the capacitor
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EE 205 Dr. A. Zidouri
This results in a phase angle difference between the input and output voltage, Vo Leads Vi as
the frequency decreases this phase lead approaches 90D
The capacitor effectively functions as an open circuit. The term Low frequencies thus refers to any
frequencies for which ωC R
The equivalent circuit for ω = 0 is shown in Fig. 37-5c
C
Vi
ω =∞
R
Vo
Fig. 37-5c RC Circuit
Equivalent at ω = 0
Based on the behavior of the output voltage magnitude at low frequencies Vo = 0 at
series RC circuit selectively passes high-frequency inputs to the output.
This circuit response is shown in Fig. 37-6
ω = 0 this
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EE 205 Dr. A. Zidouri
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EE 205 Dr. A. Zidouri
The Series RC Circuit – Quantitative Analysis
Consider the circuit of Fig. 37-7
1
sC
R
Vi(s)
Vo(s)
Fig. 37-7 s-domain equivalent
for the Circuit of Fig. 37-5a
The voltage transfer function for this circuit is:
To study the frequency response we replace
H ( s) =
s = jω
s
1
s + RC
in equ. (37-3):
(37-3)
H ( jω ) =
jω
jω + 1
RC
(37-4)
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EE 205 Dr. A. Zidouri
The Series RL Circuit – Quantitative Analysis (cont)
Equation (37-4) expressed in polar form gives the magnitude and phase:
H ( jω ) =
θ ( jω )
ω
2 (37-5)
⎛ 1 ⎞
ω2 + ⎜⎜ ⎟⎟
⎝ RC ⎠
= 90D − tan−1ω RC
(37-6)
Close examination of these equations provide quantitative support for the plots seen in Fig. 37-6.
We can now compute C in terms of the circuit elements:
ω
(
Fig. 37-6 thus:
Solving we get:
)
H jωC = 1 Hmax , and for high-pass filter, Hmax = H ( j∞)
2
ω
H jωC = 1 =
(37-7),
2
2
⎛ 1 ⎞
ω2 + ⎜⎜ ⎟⎟
⎝ RC ⎠
Recall from (36-1) that
(
as seen in
)
ωC = 1
RC
(37-8).
Not surprising it is the same as for Low-pass filter as the RC circuit has the same time constant whether
configured as a low-pass or high-pass filter, as seen in equ. (37-2)
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EE 205 Dr. A. Zidouri
Example 37.2
Consider the circuit of Fig. 37-8, determine what type of filter is this and calculate its cutoff frequency.
Take R = 2k Ω, L = 2H , and C = 2µ F .
Solution:
The transfer function
1
R
&
sC
H ( s ) = Vo =
Vi sL + R & 1
sC
R & 1 = R sC = R substituting gives:
sC R +1 sC 1+ sRC
R
R
H ( s ) = Vo = 1+ sRC = 2
Vi sL + R
s RLC + sL + R
1+ sRC
(37-9)
But
(37-10)
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EE 205 Dr. A. Zidouri
Example 37.2 cont.
R
or H (ω ) =
(37-11)
−ω 2 RLC + jω L + R
since H 0 = 1 and H ( ∞ ) = 0 , we conclude that it is a Low-Pass Filter.
()
Cutoff frequency:
( )
H ωC =
R
⎛
⎞
2
⎜ R − ωC RLC ⎟
⎝
⎠
2
=
+ ωC2 L2
⎛ω L ⎞
⎛
⎞
2
2 = ⎜1− ωC LC ⎟ + ⎜⎜ C ⎟⎟
R
⎝
⎠
⎝
⎠
2
Or
1
2
2
Substituting the values given, we obtain:
2 = ⎛⎜1− ωC2 4×10−6 ⎞⎟
⎝
⎠
Assuming
2
+ ⎛⎜ ωC ×10−3 ⎞⎟
⎝
⎠
ωC = 0.742 krad s
2
or
16ωC4 − 7ωC2 −1 = 0 solving we get : ωC2 = 0.5509 or
ωC = 0.742
, therefore the cutoff frequency is
ωC = 742 rad s
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EE 205 Dr. A. Zidouri
Summary:
Similarly to RC we can also form a High-Pass filter using RL Circuit and take the output across the
inductor. Fig. 37-9 summarizes two High-pass filter circuits and their transfer functions. Look how
similar in form their transfer functions are:
H (s) =
R
(37-12)
H (s) = s
s+ R
L
L
Vi
Vo
C
Vi
s
s + ωC
ωC =
H (s) =
R
Vo
R
L
s
s+ 1
RC
ωC =
1
RC
Fig. 37-9 Two High-Pass filters together with their
Transfer Functions and Cutoff Frequencies
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EE 205 Dr. A. Zidouri
Self Test 37:
For the circuit in Fig. 37-10, obtain the transfer function
H (s) =
circuit represents and determine the cutoff frequency. Take
Vo (ω )
Vi (ω )
Identify the type of filter the
R1 = 100Ω = R2 L = 2mH ,
Answer:
The transfer function
sR2 L
H ( s ) = Vo =
Vi R1 R2 + sL( R1 + R2 )
(37-13)
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EE 205 Dr. A. Zidouri
⎛ R
⎞
R2 sL
2 ⎟s
⎜
⎜R +R ⎟
R2 + sL
=
= ⎝ 1 2⎠
= Ks
s + ωC
⎛ R
⎞R
R sL
2
1
R1 + 2
⎜
⎟
R2 + sL s + ⎜ R + R ⎟ L
1
2
⎝
or
since
R2
H (ω ) =
R1 + R2
(
)
⎛
⎜
⎜
⎜⎜
⎝
(37-14)
⎠
⎞
R1R2 ⎟
jω jω +
⎟
R1 + R2 L ⎟⎟
(
)
(37-15)
⎠
H ( 0) = 0 and H ( ∞ ) = 1 , we conclude that it is a High-Pass Filter.
Cutoff frequency:
ωC = K
R1
R1R2
=
= 25 krad s
L R +R L
1
2
(
)
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