DIT, Kevin St.
Electric circuits
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Chapter 5
Frequency Transformation
We need to apply a suitable frequency transformation, if we wish to design
bandpass, bandstop and high-pass filters, using the low-pass approximating
function analysis previously covered. The block diagram, shown in figure 5.25,
illustrates the procedure, which produces, from the specification supplied, the
required high-pass approximating function.
Figure 5.25: System transform approach.
Before we consider frequency transform techniques, lets consider the second order
series-tuned LCR circuit shown in figure 5.26. Here we obtain expressions for the
standard second-order transfer functions for low-pass, high-pass, and bandpass
filter configurations.
The standard low-pass transfer function
Consider the tuned circuit configured with the output voltage measured across the
capacitor C. The transfer function is:
Vc(s) =
1/sC
Vin(s)
1/sC + R + sL
(5.1)
Figure 5.26: Second - order bandpass tuned circuit.
Reworking the equation, multiply above and below by sC and divide both sides by
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DIT, Kevin St.
Electric circuits
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the input voltage.
T(s) =
Vc (s)
1
= 2
Vin ( s ) s LC + sCR + 1
(5.2)
As usual, making the coefficient of s 2 unity by dividing above and below by LC
yields:
1/LC
(5.3)
T(s) = 2
s + R/L s + 1/CL
This is a second-order system. Consider the generalized second-order transfer
function as:
2
Kω p
T ( s) = 2
(5.4)
2
s + (ω p /Q ) s + ω p
We can, by comparing both 5.3 and 5.4, define the pole frequency and the Q –
factor (=1/damping) as:
ωp2 =
ω L
1
and Q = p
R
LC
(5.5)
Here K is a constant for the generalised case.
The second-order bandpass transfer function.
The transfer function for the voltage measured across R is:
T(s) =
s R/L
s + (R/L)s + 1/CL
In standard form
T (s) =
(5.6)
2
K (ω p /Q) s
s + (ω p /Q ) s + ω p
2
2
(5.6a)
The standard high-pass transfer function
Consider the output voltage across the inductance L. The transfer function is:
2
s
T(s) = 2
(5.7)
s + (R/L)s + 1/CL
Comparing this second-order high-pass function to the standard high-pass transfer
function form:
T (s) =
Ks 2
s 2 + (ω p /Q ) s + ω p
2
(5.8)
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DIT, Kevin St.
Electric circuits
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1
to the low–pass seconds
order circuit, we see the transfer function has the form of a high-pass function.
If we apply the simple frequency transformation s =
Low-pass to high-pass frequency transformation
The specification for a high-pass filter includes the passband edge frequency, ωhp,
and the stopband edge frequency, ωhs The maximum passband attenuation is Amax
and the minimum stopband attenuation is Amin. The transformation of the
high-pass specification to an equivalent normalised low-pass specification is
achieved by applying the frequency transform SL = 1/s, where SL is the low-pass
normalised complex frequency variable. However to account for the process of
normalization we must replace SL by 1/(s/ω p) =ωp /s.
Figure 5.27: Normalised HPF response.
SL =
ωp
s
(5.9)
Or in terms of real frequencies:
SL =
ωp
ω
(5.10)
The equivalent low-pass specification:
Copyright: Paul Tobin School of Electronics and Comms. Eng.
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DIT, Kevin St.
Electric circuits
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Figure 5.28: Equivalent LPF response.
From figure 5.28, we define values for the transformed frequency Ω as
ω
Ω= p
(5.11)
ω
Let ω = ωp
ωp
Ω LP = 1 =
(5.12)
ωp
ω
Ω LS = p
(5.13)
ωs
We calculate the value for ε and n for the filter type chosen using the equivalent
normalised low-pass obtained previously. From these two quantities, we can then
obtain the normalised attenuation function. This must be frequency de-normalised
and the, low-pass to high-pass frequency transformation, performed. To deε 1/ n
nomalise a Butterworth approximation loss function we used F ( s ) =
s
ωp
(Chapter 2), but with frequency transformation, we substitute for ω/ωp with
$=
ε 1 / nω p
s
In the Chebychev case, we apply the substitution
$=
ωp
(5.14)
(5.15).
s
Example 5.1
Obtain the transfer function H(s) for a Chebychev high-pass filter, which satisfies
the following specification:
ωp = 3000 rs -1
Passband edge frequency
Stopband edge frequency
ωs = 1000 rs -1
Maximum passband attenuation Amax = 0.5 dB
Minimum stopband attenuation Amin = 22 dB
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DIT, Kevin St.
Electric circuits
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Draw an active circuit diagram, which would meet the above specification. Which
components determine the passband gain, passband frequency and damping
factor?
Solution
The order of the filter is calculated:
 100.1 Amin − 1 
cosh  0.1 Amax
− 1 
 10
n=
-1
cosh (Ω LS )
1/ 2
-1
Substituting values
 100.1.22 − 1 
cosh  0.1.0.5 
− 1
 10
n=
-1
cosh (3 / 1)
1/ 2
-1
= 2.42
A third-order Chebychev filter is required. From the table of Chebychev values
with ripple = 0.5 dB (i.e. Amax), we see the approximation loss function is:
A($) = ($ + 0.626)($2 + 0.626$ +1.142)
The low-pass to high-pass transformation function is:
$=
ωp
s
Where $ is the normalised LP complex frequency variable and s, the HP complex
frequency variable. Thus
A(s) = A($)

3000
 3000
  3000 2
A(s) = 
+ 0.626 (
) + 0.626(
) + 1.142
s
 s
 s


 3000 + 0.626 s   9.106 1879.2
A(s) = 
+
+ 1.142

2

s
s

 s

6
 3000 + 0.626 s   9.10
A(s) = 
 
s


+ 1879.2 s + 1.142 s 2 

2
s

The transfer function H(s) = 1/A(s) i.e.
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DIT, Kevin St.
Electric circuits
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2

s

 
s
H1(s) = 

6
2 

 3000 + 0.626 s   9.10 + 1.879.2 s + 1.142 s 
 1.602 s 
H1(s) =  2
 s + 4807 
2


s
 s 2 + 1644 s + 7.878.10 6 


Circuit implementation:
Figure 5.29: Third-order S & K active filter.
The first stage passband frequency is determined by C and R.
Figure 5.30: Third-order Chebychev high-pass filter
Copyright: Paul Tobin School of Electronics and Comms. Eng.
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DIT, Kevin St.
Electric circuits
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Example 5.2
The specification for a high-pass filter is:
Passband edge frequency
ωp = 3000 rs -1
Stopband edge frequency
ωs = 1500 rs -1
Maximum passband attenuation
Amax = 2.0 dB
Minimum stopband attenuation
Amin = 20 dB
Obtain the high-pass transfer function, which will meet this specification. (Use the
Butterworth tables).
ε = 10 0.1 Amax − 1 = ε = 10 0.1x2 − 1 = 0.764
n=
100.1 Amin − 1
100.1.20 − 1
log
10
100.1 Amax − 1 =
100.1.2 − 1 = 4
2 log(Ω LS )
2 log(3000 / 1500)
log10
From the table of Butterworth polynomials, we obtain a fourth-order polynomial.
A($) = ($2 + 0.765 $ +1)($2 +1.848$+ 1)
De-normalising and performing a LP to HP transforming, by making the
substituting for $ in the normalised function
$=
ε 1 / nω p
s
$=
0.7641 / 4 3000 2800
=
s
s
Substituting values:
Substituting this into the required high-pass approximating function which, when
inverted, gives the required H-P gain Inverting.
H(s) = 1/A(s)

2800
2800
 2800 2
  2800 2
A(s) = (
) + 0.765(
) + 1 (
) + 1.848(
) + 1
s
s
 s
 s

2
2



s
s
H(s) =  2
6  2
6 
 s + 2142 s + 7.84.10   s + 5174.4 s + 7.84.10 
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DIT, Kevin St.
Electric circuits
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Bandpass frequency transformation
The frequency transform technique, used in the previous analysis to design highpass filters, can also be used to design bandpass and bandstop filters using a
suitable transform. The bandpass specification is shown in figure 5.31.
Figure 5.31: Bandpass loss functions.
Where
ωs1 = the lower stopband edge frequency
ωs2 = the upper stopband edge frequency
ωp1 = the lower passband edge frequency
ωp2 = the upper passband edge frequency
If we consider the normalised low-pass specification in figure 5.32 we see that
A(ω) has a mirror image A(-ω) i.e. its an even function of A(ω). This is considered
a bandpass function whose centre frequency, ω is at ω = 0 and the frequency
range from - ∞ to + ∞. The frequency transform required to carry out this
frequency shift is:
s 2 + ω o2
$=
(5.16)
B .s
Here B is the –3 dB bandwidth, expressed as
B = ω p 2 − ω p1
(5.17)
s is the bandpass complex frequency variable and ωo is the centre frequency
defined as:
ω o = ω p1ω p 2 = ω s1ω s 2
(5.18)
It will be assumed that the filter is symmetrical around the centre frequency,
which is valid for bandpass filters with Q > 1. This makes the analysis much
easier.
Copyright: Paul Tobin School of Electronics and Comms. Eng.
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DIT, Kevin St.
Electric circuits
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Figure 5.32: LPF with mirror image.
The transformation, defined by (5.16), is now applied to the specification from
which we obtain an equivalent normalised low-pass specification. Using the
transformation but replacing the complex frequency with real frequencies and
then substituting the frequencies as defined in figure 5.32. This gives values for
the normalised frequencies in figure 5.33. To obtain the normalised values, we
consider values for ω at the four frequencies ωs1, ωs2, ωp1 and ωp2.
Note:
The Matlab code to produce figure 5.33 uses the matlab file h = freqs(b, a, w),
where the coefficients of the numerator polynomial of the transfer function are:
b=[160000 0 0]; and the denominator polynomial coefficients are a = [1 570
2.95e6 7.6e8 1.74e12]. The frequency range ω = 500:1:2000; plot(ω,
10*log10(h)). Note use conv function if the transfer function is in factored form
i.e. if ( s 2 + 2142s + 7.84.10 6 )( s 2 + 1865s + 7.84.10 6 ) , convolve the coefficients
of each factor to get a composite polynomial.
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DIT, Kevin St.
Electric circuits
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Figure 5.33: Chebychev bandpass filter.
We de-normalise and frequency transform, a Butterworth approximation function,
by replacing $ with the function defined below
s 2 + ω o2 1 / n
ε
(5.21)
Bs
To obtain the normalised values, we consider evaluating the transform at the
following frequencies: ωs1, ωs2, ωp1 and ωp2.
$=
( jω ) 2 + ω 0
(ω p 2 − ω p 1 ) jω
2
$=
But = jΩ so that
 − ω 2 + ω02 
jΩ = − j 

 (ω p 2 − ω p1ω 
Or
 ω02 − ω 2 
Ω = −

 (ω p 2 − ω p1 )ω 
Substituting the frequency values ωs1, ωs2, ωp1 and ωp2 into this equation.
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DIT, Kevin St.
Electric circuits
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Figure 5.34
1) At ω = ωo
Ω Lo
 ω02 − ω02 
= −
=0
 (ω p 2 − ω p1 )ω 0 
2) At ω =ωs1 and ω 2 o = ω s 1ω s 2
Ω Ls 1
 ω s1ω s 2 − ω s12 
 ω s 2 − ω s1 
= − = Ω Ls 2 − 
 =− 

 (ω p 2 − ω p1 )ω s1 
 (ω p 2 − ω p1 ) 
3)
At ω = ωs2 and ω 2 o = ω s 1ω s 2
 ω ω − ωs22 
 ω s1 − ω s 2   ω s 2 − ω s1 
Ω Ls 2 = −  s 1 s 2
 =−
=

 (ω p 2 − ω p1 )ω s 2 
 (ω p 2 − ω p 1 )   (ω p 2 − ω p1 ) 
4) At ω =ωp1 and ω 2 o = ω p1ω p 2
 ω p 1ω p 2 − ω p 12 
Ω Lp1 = −Ω Lp 2 = − 
 = −1
 (ω p 2 − ω p1 )ω p 1 
5) At ω =ωp2 and ω 2 o = ω p1ω p 2
Ω Lp 2
 ω p 1ω p 2 − ω p 2 2 
= −
 =1
 (ω p 2 − ω p 1 )ω p 2 
Example 5.3
The specification for a bandpass filter is
ω s1 = the lower stopband edge frequency = 1000 rs-1
ω s2 = the upper stopband edge frequency = 9000 rs-1
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DIT, Kevin St.
Electric circuits
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ω p1 = the lower passband edge frequency = 2000 rs-1
ω p2 = the upper passband edge frequency = 4500 rs-1
The maximum passband attenuation Amax = 0.5 dB. The minimum stopband
attenuation
Amin = 6 dB. Obtain the bandpass transfer function, which will
meet this specification. (Use Chebychev approximation loss function tables).
 ω ω − ω s12 
 ω s 2 − ω s1   9000 − 1000 
Ω Ls 1 = − = Ω Ls 2 −  s1 s 2
 =− 
=
 = 3.2
 (ω p 2 − ω p1 )ω s1 
 (ω p 2 − ω p1 )   4500 − 2000 
The maximum allowed ripple present in the passband is 0.5 dB so a Chebychev
design is considered. The order n is calculated as
1/ 2
 100.1x Amin − 1 
cosh  0.1.xAmax 
− 1
 10
n=
-1
cosh (Ω LS 1 )
-1
Substituting values
1/ 2
 100.1.6 − 1 
cosh  0.1.5 
 10 − 1
n=
-1
cosh (3.2)
-1
= 1.24
Example 5.4
A bandpass filter is required for a particular application, which will tolerate ripple
in the passband equal to 0.5 dB. The specification for a bandpass filter is:
ω s1 = the lower stopband edge frequency = 688 rs-1
ω s2 = the upper stopband edge frequency = 1930 rs-1
ω p1 = the lower passband edge frequency = 970 rs-1
ω p2 = the upper passband edge frequency = 1370 rs-1
The maximum passband attenuation Amax = 0.5 dB.
The minimum stopband attenuation Amin = 15 dB.
Obtain the bandpass transfer function, which will meet this specification. ((Use
Chebychev approximation loss function tables).
Solution
The maximum allowed ripple present in the passband is 0.5 dB so a Chebychev
design is considered. The order n is calculated as
1/ 2
 100.1 Amin − 1 
cosh  0.1 Amax 
− 1
 10
n=
-1
cosh (Ω S )
-1
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DIT, Kevin St.
Electric circuits
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 ω ω − ω s12 
 ω s 2 − ω s 1  1930 − 688 
Ω Ls 1 = − = Ω Ls 2 −  s1 s 2
 =−
 =
 = 3.1
 (ω p 2 − ω p1 )ω s1 
ω p 2 − ω p1  1370 − 970 
Substituting values
1/ 2
100.1.15 − 1
cosh  0.1.5 
 10 − 1 
n=
-1
cosh (3.1)
-1
= 1.93
A second order approximation function is required. From the Chebychev tables,
A($) = s2 + 1.426s + 1.516. The centre or resonant frequency is calculated using
the relationship ωo is the centre frequency defined as:
ω o = ω p1ω p 2 = ω s1ω s 2 = 688.1930 =1153 rs-1
s 2 + ω o2
in the loss function. Here B is the
Using the frequency transformation $ =
B.s
–3 dB bandwidth, which can be expressed:
B = ω p 2 − ω p1 = 400 rs-1.
.
$=
2
2
s 2 + ω o2 s + 1153
=
B.s
400s
Here s is the bandpass complex frequency variable.
2
 s 2 + 1153 2 
 s 2 + 1153 2 
A( s ) = 
+


1.426 + 1.516
 400 s 
 400 s 
H ( s) =
1
160000 s 2
= 4
A( s ) s + 570 s 3 + 2.95.10 6 s 2 + 7.5.10 8 s + 1.74.1012
 s 2 + ω o2  1 / n
As an exercise use the Butterworth $ = 
ε substitution on this
 Bs 
problem.
Exercise
Obtain the transfer function for a low-pass CR filter and a high-pass CR filter for C =
1F and R = 1 Ω. Substitute s = 1/s in the low-pass filter and compare with the hpf
transfer function.
Copyright: Paul Tobin School of Electronics and Comms. Eng.
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