INF5481 RF Circuit, Theory and Design Assignment #4INF5481: Assignment #4 Figure 1: Butterworth...

INF5481 RF Circuit, Theory and DesignAssignment #4

Use your favorite mathematical “spreadsheet” program for presenting thegraphs (Python, Matlab, ...).

I have written some rough python code to keep track of filter coefficientsand do varies scaling operations, it is not very pretty so I will not includethe code. All the filter responses are created by multipling the appropriateABCD representation and plotting abs(2/ABCD[0,0]).

Problem 1

Design a maximally flat high-pass filter with a cutoff frequency of 1 GHz, theattenuation at 2 GHz should be at least 20 dB at 2 GHz.

(a)

Start with a low pass Butterworth prototype filter, what filter order N isrequired for 20 dB at Ω = 2? Find the filter coefficients and plot the attenu-ation in dB.

(b)

Transform your normalized low-pass filter to the desired high pass filter. Plotthe attenuation in dB.

We start by creating a low-pass Butterworth prototype filter with 20 dBattenuation at f = 2f3 dB.

The attenuation or insertion loss of a Butterworth filter of order N canbe expressed as ([1] equation (5.41))

IL(Ω) = 10 log(1 + Ω2N

),

1

INF5481: Assignment #4

where we have set a = 1. Our required response is

IL(Ω = 2) ≤ 20 dB10 log

(1 + 22N

)≤ 20 dB

1 + 22N ≤ 10022N ≤ 100− 1

N ≤ log2(100− 1)2

N ≤ 3.314so we use N = 4, which is the closest integer fulfilling this requirement. Fromtable 5.2 in [1]

filt = [1, 0.7654, 1.8478, 1.8478, 0.7654, 1]

RG = 1 Ω

C1 = 0.765 F

L2 = 1.85 H

C3 = 1.85 F

L4 = 0.765 H

RL = 1 Ω

To confirm our results we plot the insertion loss for N = 4 and N = 3,see figure 1. We note that N = 4 is indeed the lowest order that satisfiesIL(Ω = 2) ≤ 20 dB.

We now convert our low pass filter into a high pass filter with cutoff ωcusing the transformations

R← R

L← 1Cωc

C ← 1Lωc

.

This results in

RG = 1 Ω

L1 = 2.08× 10−10 HC2 = 8.61× 10−11 FL3 = 8.61× 10−11 HC4 = 2.08× 10−10 FRL = 1 Ω

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0.0 0.5 1.0 1.5 2.0 2.5 3.0

Ω [rad/s]

0

5

10

15

20

25

30

35

40

3

IL [dB]

N=4

N=3

Figure 1: Butterworth insertion loss for order 3 and 4. At Ω = 2, N = 4 isabove the required insertion loss of 20 dB.

And the response is shown in figure 2 and we note the 3 dB cutoff at 1 GHz.

0.0 0.5 1.0 1.5 2.0

f [GHz]

0

10

20

30

40

50

60

70

80

90

3

IL [dB]

N=4

Figure 2: Insertion loss for high-pass Butterworth filter of order 4.

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Problem 2

In the design of an amplifier for cellular phone applications, it is discoveredthat the circuit exhibits excessive noise at 3 GHz.

Develop a bandstop filter with the minimum number of components withthe following properties:

• center frequency of fc = 3 GHz

• 30 dB stopband width of 10 % (from 2.85 GHz to 3.15 GHz)

• 3 dB stopband width of 36.7 %

3 dB ripple in the passband is acceptable. Plot the attenuation profile.We want to design a filter with a geometric center frequency

fc = 2.9491× 109 Hz

and upper and lower frequencies

fL = 2.4495× 109 Hz fU = 3.5505× 109 HzΩL = −1.0000 ΩU = 1.0000

at the frequencies

f30 dB = 2.8016× 109 Hz f30 dB = 3.0965× 109 HzΩ30 dB = −3.6377 Ω30 dB = 3.8245

we require an attenuation greater than 30 dB. The solution is visualized infigure 4.

Theory

We use the Chebyshev Polynomials Tn(Ω) to create a filter with an insertionloss

IL(Ω) = 10 log(1 + a2T 2N(Ω)

)where TN(Ω) =

{cos (N arccos (Ω)) for |Ω| ≤ 1cosh (N arccosh (Ω)) for |Ω| ≥ 1

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In the pass-band Ω ≤ 1 and T 2N is bound to the interval 0 to 1, so in thepass-band we can write

IL(Ω ≤ 1) = 10 log(1 + a2

)10IL(Ω≤1)/10 =

(1 + a2

)⇒ a2 = 10IL(Ω≤1)/10 − 1, (1)

which allows us to control the amount of ripple.

Solution

Low-pass

We start by designing a low-pass prototype Chebyshev filter normalized toa cutoff frequency of

ωc = 2.4495× 109 Hz

we require IL(Ω30 dB) ≥ 30 dB at

Ω30 dB = 3.8245

We start with the ripple, from (1)

a2 = 103/10 − 1 ≈ 1 .

We then find the order of the filter

IL(Ω30 dB) ≥ 30 dB10 log

(1 + a2T 2N(3.8245)

)≥ 30 dB

1 + T 2N(3.8245) ≥ 103

TN(3.8245) ≥√

103 − 1cosh (N arccosh (3.8245)) ≥

√103 − 1

N arccosh (3.8245) ≥ arccosh(√

103 − 1)

N · 2.0170 ≥ arccosh(√

103 − 1)

N ≥arccosh

(√103 − 1

)2.0170

N ≥ 2.0557

Which is satisfied with N = 3. Since a = 1 the filter coefficients can be foundin table 5.4(a) in [1]

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filt = [1, 3.3487, 0.7117, 3.3487, 1]

RG = 1 Ω

C1 = 3.35 F

L2 = 0.712 H

C3 = 3.35 F

RL = 1 Ω

We verify our progress by creating figure 3.

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.03.9

Ω [rad/s]

0

10

20

30

40

50

3

IL [dB]

N=3

Figure 3: Low pass prototype Chebyshev filter

Bandstop

We convert our filter with the following transformations

LP ←BWL

ω20

CP ←1

BWL

LS ←1

BWC

CS ←BWC

ω20,

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where

BW ≡ ωU − ωL= 6.9178× 109 rad/s

ω0 ≡√ωUωL

= 1.8529× 1010 rad/s.

RG = 1 Ω

CS1 = 6.75× 10−11 FLS1 = 4.32× 10−11 HCP2 = 2.03× 10−10 FLP2 = 1.43× 10−11 HLS3 = 4.32× 10−11 HCS3 = 6.75× 10−11 FRL = 1 Ω

The insertion loss is shown in figure 4, where figure 5 uses a logaritmicx-axis and figure 6 displays the magnitude.

1.0 2.0 2.4 2.8 3.1 3.6 4.0 5.0

f [GHz]

0

10

20

30

40

50

60

70

80

3

IL [dB]

N=3

Figure 4: Bandstop Chebyshev filter.

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100 101

f [GHz]

0

10

20

30

40

50

60

70

80

3

IL [dB]

N=3

Figure 5: Bandstop Chebyshev filter, symmetric with logaritmic x-axis.

100 101

f [GHz]

0.000

0.200

0.400

0.600

0.800

1.000

0.708

IL

N=3

Figure 6: Bandstop Chebyshev filter, magnitude (not db-y scale).

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Problem 3

(a)

Based on the maximally flat low-pass filter from problem 1a), convert it toa bandstop filter with a 3 dB bandstop at [2.22 GHz, 2.58 GHz]. Match yourfilter to 50 Ω on both ends. Plot the attenuation of your bandstop filter.

(b)

Transform your filter to realizable transmission line elements.

Theory

In this exercise we are going to replace the lumped elements L and C withmicrostriplines, writing out the impedance for a short-circuit transmissionline

Zshort = jZ0 tan(βl)

with l = λ0/4 and f0 ≡ vp/λ0 we can write

Zshort = jZ0 tan(2πf

vpλ0/4)

= jZ0 tan(2πf

f0).

This is inductive when

ZL = Zshort

jωL = jZ0 tan(πΩ/2)

jωL = jZ0 tan(π/2 ·ω/ω0)

this can not be fulfilled for all ω, but at ω = ω0, it reduces to

jωL = jZ0 tan(π/2)

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which goes to infinity, so we do little trick. We multiply our transmissionline with the so called “bandwidth factor”

1

tan(π/2 ·ωL/ω0)jωL = jZ0 tan(π/2 ·ω/ω0)

1

tan(π/2 ·ωL/ω0)1

tan(π/2 ·ωL/ω0)jωL = jZ0

tan(π/2 ·ωL/ω0)tan(π/2 ·ωL/ω0)

1

tan(π/2 ·ωL/ω0)jωLL = jZ0

ωLL

tan(π/2 ·ωL/ω0)= Z0

So we replace an inductor L with a transmission line with impedance Z0 =(bf)L. In a similar manner we can derive that a capacitor C should bereplaced with an open line of impedance Z0 = 1/((bf)C).

Solution

With the filter coefficents

filt = [1, 0.7654, 1.8478, 1.8478, 0.7654, 1]

and the bandwidth factor

bf =1

tan(π/2 ·ωL/ω0)

=1

tan(π/2 · 2.220/2.4)= 0.12

we obtain the transmission line elements

filt[1] = bf*filt[1] # series

filt[2] = 1/(bf*filt[2])# shunt

filt[3] = bf*filt[3] # series

filt[4] = 1/(bf*filt[4])# shunt

1, 0.0906, 4.57, 0.219, 11, 1

We now insert unit elements to each side of the filter

filt = [filt[0], 1, filt[1], filt[2], filt[3], filt[4], 1, filt[5]]

we now move these unit elements “inwards”, first the leftmost:

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Z_ue = filt[1]

Z_comp = filt[2]

N = 1 + Z_ue/Z_comp

filt[1] = N*Z_ue # shunt

filt[2] = N*Z_comp # ue

1, 12, 1.09, 4.57, 0.219, 11, 1, 1

then the rightmost:

Z_ue = filt[-2]

Z_comp = filt[-3]

N = 1 + Z_ue/Z_comp

filt[-3] = Z_ue/N # series

filt[-2] = Z_comp/N # ue

1, 12, 1.09, 4.57, 0.219, 0.917, 10.1, 1

and another move

Z_ue = filt[5]

Z_comp = filt[4]

N = 1 + Z_ue/Z_comp

filt[4] = Z_comp*N # ue

filt[5] = Z_ue*N # shunt

1, 12, 1.09, 4.57, 1.14, 4.76, 10.1, 1

We now insert another set of unit elements to each side of the filter

filt = [filt[0], 1, filt[1], filt[2], filt[3],

filt[4], filt[5], filt[6], 1, filt[7]]

and move the rightmost

Z_ue = filt[-2]

Z_comp = filt[-3]

N = 1 + Z_ue/Z_comp

filt[-3] = Z_ue*N # shunt

filt[-2] = Z_comp*N # ue

1, 1, 12, 1.09, 4.57, 1.14, 4.76, 1.1, 11.1, 1

1, 1, 12, 1.09, 4.57, 1.14, 4.76, 1.1, 11.1, 1

Which, de-normalized

50, 50, 602, 54.5, 229, 56.8, 238, 54.9, 556, 50

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0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Frequency (Hz) 1e10−80

−70

−60

−50

−40

−30

−20

−10

0

Magnitude (dB)

S11 S21

Figure 7: Simulated filter response

2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8

Frequency (Hz) 1e9−4

−3

−2

−1

0

1

Magnitude (dB)

S11 S21

Figure 8: Zoomed in view of figure 7

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Problem 4

Following example 2.5 in the book, you designed a 50 Ω transmission line.Your goal is to short out 2 GHz and pass 4 GHz. Your reasoning leads youto the open λ/4 stub in figure 9.

Z0 = 50 Ωl = 20.3 mm Cfringe

h = 1 mm

Figure 9: Open 50 Ω transmission line on a 1 mm FR4 substrate with anunkown fringe component at the end.

(a)

Express the input impedance of the ideal open transmission line stub assum-ing vp = 1.61× 108 m/s. Vertify that |Zin(f = 2 GHz)| ≈ 0 Ω.

(b)

Due to fringing fields some “leakage” will occur, model this as a fringe ca-pacitor Cfringe. What is the input impedance expression now?

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(c)

Since you find conformal mapping1 too difficult and do not know how to useEM simulation software, you decide to guess that Cfringe is between 0.1 pFand 1 pF.

Plot the magnitude of the input impedance |Zin| from DC to 8 GHz forthe following cases: ideal, Cfringe = 0.1 pF and Cfringe = 1 pF.Hint: Please limit the y axis to [0 Ω, 500 Ω].

We use the normal formula for a terminated transmission line, and alsostudy what happens with an open line

Zin = Z0ZL + jZ0 tan(βd)

Z0 + jZL tan(βd)

limZL→∞

Zin = limZL→∞

Z01 + jZ0/ZL tan(βd)

Z0/ZL + j tan(βd)= Z0

1

j tan(βd)

Writing these two equations in terms of frequency f we obtain

Zin = −jZ01− (2πfC)Z0 tan

(2πvpfd)

Z0(2πfC) + tan(

2πvpfd)

limZL→∞

Zin = −jZ01

tan(

2πvpfd)

For small values of fC the first equation will simplify to the last, for smallpositive values of f this goes rapidly from −∞ up to∞. Taking the absolutevalue this is turned around and becomes down to 0. For larger values we seethe periodicity of tan. The code is presented bellow and the plot in figure10.

import numpy as np

import pylab as plt

def plot_tl_C(f, C=np.inf, **kwargs):

omega = 2*np.pi*f

Z_0 = 50.

Z_C = 1/(1j*omega*C)

d = 80.67e-3/4

v_p = 1.61e8

beta = omega/v_p

1see for instance http://physics.usask.ca/~bradley/pll_plate_edge.pdf

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http://physics.usask.ca/~bradley/pll_plate_edge.pdf


t = np.tan(beta*d)

if C == np.inf:

Z_in = Z_0/(1j*t)

else:

Z_in = Z_0*(Z_C + 1j*Z_0*t)/(Z_0 + 1j*Z_C*t)

plt.plot(f/1e9, np.abs(Z_in), **kwargs)

plt.ylim(0, 500)

plt.xlabel(’freq GHz’)

plt.ylabel(’$|Z_{in}|$ [$\Omega$]’)

plt.figure()

f = np.linspace(0.1e9, 8e9, 1024)

plot_tl_C(f, label=’ideal’)#, ls=’--’)

plot_tl_C(f, C=0.1e-12, label=’$C=0.1$ pF’)#, ls=’--’)

plot_tl_C(f, C=1e-12, label=’$C=1$ pF’)#, ls=’:’)

plt.legend(loc=’best’)

0 1 2 3 4 5 6 7 8

freq GHz

0

100

200

300

400

500

|Zin

| [Ω]

idealC=0.1 pF

C=1 pF

Figure 10: Input impedance of an open transmission line, with load capaci-tance (shown in blue) and without (shown in green). Lower plot is a zoomedin view of the upper with linear x axis.

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References

[1] R. Ludwig and G. Bogdanov, RF circuit design: theory and applications,2nd ed. N.J.:Prentice-Hall, 2009.

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Problem 1(a)(b)

Problem 2TheorySolutionLow-passBandstop

Problem 3(a)(b)TheorySolution

Problem 4(a)(b)(c)

INF5481 RF Circuit, Theory and Design Assignment #4INF5481: Assignment #4 Figure 1: Butterworth...

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