INF5481 RF Circuit, Theory and Design Assignment #2

INF5481 RF Circuit, Theory and DesignAssignment #2

Drawing smith charts with sage

I found some sage code here: http://www.sagenb.org/home/pub/502/ todraw a smith chart, with some minor modifications it also plots the admit-tance chart. The code below produces figure 1.

from numpy import pi, angle, sqrt # dealing with exact pi is extremely slow!

def Gamma(z):

return (z-1)/(z+1)# return the (complex) reflection coefficient

def Gammaxy(z):

’’’return the real and imaginary part

of the reflection coefficient as a tuple’’’

r = Gamma(z)

return (real(r), imag(r))

def Z_real(x):

’’’generate smith chart coordinates of constant real part;

these are complete circles’’’

r = Gammaxy(x)

radius = 0.5*(1 - r[0])

return circle((1-radius, 0), radius, color=’red’)

def Z_imag(y):

’’’generate smith chart coordinates of constant imaginary part;

these are circular arcs’’’

rx = lambda t: Gammaxy(t + 1j*y)[0]

ry = lambda t: Gammaxy(t + 1j*y)[1]

return parametric_plot((rx,ry),(0,50), color=’red’)

1

http://www.sagenb.org/home/pub/502/

INF5481: Assignment #2

# draw an unlabeled Z smith chart

smithchart_Z = Z_real(0) + Z_real(1/3) + Z_real(1) + Z_real(3) + \

Z_imag(0) + Z_imag(1/3) + Z_imag(-1/3) + Z_imag(1) + \

Z_imag(-1) + Z_imag(3) + Z_imag(-3)

## Modified functions for admittance

def Y_real(x):

’’’generate smith chart coordinates of constant real part;

these are complete circles’’’

r = Gammaxy(x + 0j)

radius = 0.5*(1 - r[0])

return circle((radius-1, 0), radius, color=’blue’) # radius-1

def Y_imag(y):

’’’generate smith chart coordinates of constant imaginary part;

these are circular arcs’’’

rx = lambda t: -Gammaxy(t + 1j*y)[0] # -1 added

ry = lambda t: -Gammaxy(t + 1j*y)[1] # -1 added

return parametric_plot((rx,ry),(0,50), color=’blue’)

# draw an unlabeled Y smith chart

smithchart_Y = Y_real(0) + Y_real(1/3) + Y_real(1) + Y_real(3) + \

Y_imag(0) + Y_imag(1/3) + Y_imag(-1/3) + Y_imag(1) + \

Y_imag(-1) + Y_imag(3) + Y_imag(-3)

def Z(Gamma_d):

return (1 + Gamma_d)/(1 - Gamma_d)

def annote(Z, t, color=’black’):

xy = Gammaxy(Z)

p = point(xy, color=color)

p += text(’$’+t+’$’, xy, vertical_alignment=’bottom’,

horizontal_alignment=’left’, color=’black’,

fontsize=20)

return p

def Gamma(z):

try:

return (z - 1)/(z + 1)# return the (complex) reflection coefficient

except:

return (1 - 1/z)/(1 + 1/z)

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Figure 1: An empty Z and Y smith chart.

# p = text(’$%.2g$’ % (x), (r[0], 0), vertical_alignment=’top’,

# horizontal_alignment=’right’, color=’red’)

# p += circle((1-radius, 0), radius, color=’red’)

# if 0 < y: v_alignment = ’bottom’

# else: v_alignment = ’top’

# if 1 < abs(y): h_alignment = ’right’

# else: h_alignment = ’left’

# p = text(’$%.2g$’ % (y), (rx(0), ry(0)), vertical_alignment=v_alignment,

# horizontal_alignment=h_alignment, color=’red’)

# p = text(’$%.2g$’ % (x), (-r[0], -0.05), vertical_alignment=’top’,

# horizontal_alignment=’right’, color=’blue’)

# p += circle((radius-1, 0), radius, color=’blue’) # radius-1

# if 0 < -y: v_alignment = ’bottom’

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# else: v_alignment = ’top’

# if 1 < abs(y): h_alignment = ’right’

# else: h_alignment = ’left’

# p = text(’$%.2g$’ % (-y), (rx(0), ry(0)), vertical_alignment=v_alignment,

# horizontal_alignment=h_alignment, color=’blue’)

# parametric_plot((rx,ry),(0,50), color=’blue’)

In this exercise we will use Sage extensively, a few words of warnings: Sageuses i when outputting complex numbers (but accepts j in the input) andwhen outputting floating point numbers you specify the number of bits touse, instead of the number of significant digits, this can in some cases leadto odd rounding. In this exercise I have used 10 digits in the output. Tocreate the document I have used the LATEXpackage sagetex, which allowssage code to be embedded in the LATEXcode. Although it might not appearso, all exercises have been solved on paper using an ordinary smith chart.Sage is simply used to confirm the results and draw smith charts.

Problem 1

Z0 = 50 Ω

l0 = 0.4λZL = 60 + j75 Ω

Zin

Figure 2: Diagram for problem 1

We have a 50 Ω line connected to a complex load as shown in figure 2.Using a smith chart, we want to find ΓL, SWR, YL and Zin.

(a)

Normalize the load impedance and mark zL = ZL/Z0 on a smith chart. Readof the reflection coefficient at the load.

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(b)

Draw a circle that intersects with zL centered at z = 1 and read of thestanding wave ratio.

(c)

Use a ZY smith chart to read of the load admittance.

(d)

Rotate the load reflection coefficient from l = 0 to l = 0.4λ, read of the inputimpedance.

The solution is presented in section to .

Theory

We start with equation (3.2) and (3.3) from [?], which expresses the reflectioncoefficient and input impedance at a distance d from the load. Where we fornotation convenience refrain from dividing complex numbers into real andimaginary parts and using small letters for normalized impedance.

Γ(d) = ΓLe−j2βd (1)

Zin(d)

Z0

=1 + Γ(d)

1− Γ(d)(2)

from which is follows that

Γ(d) =

Zin(d)Z0− 1

Zin(d)Z0

+ 1(3)

(3) gives us a transformation from impedance to reflection coefficient and (2)goes back again. The transformation for admittance is straightforward from(2)

Yin(d)Z0 =1− Γ(d)

1 + Γ(d)(4)

(5)

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We also define transformation for SWR, from equation (3.12) and (3.13) in[?]

SWR =1 + |Γ(d)|1− |Γ(d)|

(6)

|Γ(d)| = SWR− 1

SWR + 1(7)

From the above formulas it is straightforward to go between impedance,admittance and SWR. The smith chart has the property that for every pointwe can read off the impedance, admittance and reflection coefficient in thefollowing manner

Impedance: Every point intersects with a constant resistance R and reactance Xline, these are shown as blue lines in 1. The normalized impedance isthen the sum of these components

Z

Z0

= R + jX

Admittance: The same is true for finding the admittance, but this time we follow theconstant conductance G and susceptance B. Normalized admittance isthen

Y Z0 = G+ jB

Reflection: We first measure the distance from the origin to the edge of the chart,on a A4 paper this is about 7.34 cm. Then measuring the distance dfrom the origin to the desired point. The angle θ can be found with aprotractor.

Γ =d

7.34 cmejθ

The above process is time consuming and prone to errors when carried outby hand, we will therefore let sage use the equations and show us the resultsusing the smith chart.

Solution

Reflection coefficient at load

We start by finding the load on the smith chart, to do this by hand wenormalize the impedance and follow the imaginary and real lines. In sage wefirst have to convert to impedance to a reflection coefficient.

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Z_0 = 50

Z_L = (60+75j)/Z_0

Gamma_L = Gamma(Z_L)

xy = Gammaxy(Z_L) # get real and imaginary parts of reflection coeff.

p1 = smithchart_Z # use Z chart

p1 += point(xy, color=’black’) # add point at load

p1 += text("$\Gamma_L$", xy, vertical_alignment=’bottom’,

color=’black’, fontsize=20)

p1 += Z_real(real(Z_L))

ΓL = 0.57ej0.84

= 0.38 + 0.42i

SWR

We now draw a circle through ΓL and measure the intersection where theimaginary part is 0, or according to (6)

def SWR(Gamma_d):

return (1 + abs(Gamma_d))/(1 - abs(Gamma_d))

r = abs(Gamma_L) # radius of circle

p1 += circle((0, 0), r)

Which results in figure 3 and SWR ≈ 3.6

ΓL

Figure 3: Smith chart for and

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Load admittance

We could just do YL = Z−1L , but lets do it the hard way. We start by plotting

the same point as we had before, but this time on the admittance chart.

p2 = smithchart_Y # use impedance chart

p2 += point(xy, color=’black’) # add point at load

p2 += text("$\Gamma_L$", xy, vertical_alignment=’bottom’,

color=’black’, fontsize=20)

We now have to find where this intersects with the constant conductance andsusceptance lines. Again, we could do this by hand, but using (4) we addthose lines to the chart

def Y(Gamma):

return (1 - Gamma)/(1 + Gamma)

Y_L = Y(Gamma_L)

p2 += Y_real(real(Y_L)) + Y_imag(imag(Y_L)) # add lines

which is YL ≈ (0.33−0.41i)/Z0 ≈ 0.0065−0.0081i ≈ 0.0065−0.0081i, whichis shown in figure 4

ΓL

Figure 4: Smith chart for

Input impedance

From (3) it is easy to see that this is just a rotation by the “electrical length”2βd, we can then go from reflection coefficient to impedance by using (2).

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def Z(Gamma_d):

return (1 + Gamma_d)/(1 - Gamma_d)

var(’lam’)

d = 0.4*lam

beta = 2*pi/lam

Gamma_in = Gamma_L*exp(-2j*beta*d) # rotate

Z_in = Z(Gamma_in)

xy = Gammaxy(Z_in)

p1 += point(xy)

p1 += text("$\Gamma_in$", xy, vertical_alignment=’top’,

horizontal_alignment=’left’, color=’black’,

fontsize=20)

p1 += Z_real(real(Z_in)) + Z_imag(imag(Z_in))

The alternative approach is to rotate by

2β · d = 22π

λ0.4λ

= 288,

which is a point at 360− (288−55) = 127 and read of the impedance fromthe constant resistance and reactance lines.

Zin = Z0(0.36 + 0.52i) = 18.+ 26.i

which is shown in figure 5

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ΓLΓin

Figure 5: Smith chart for

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Problem 2

l5 = λ/4

R1 = 85 ΩL2 = 2 nH

l3 = 0.15λl7 = λ/4

R4 = 33 ΩC8 = 2.2 pF

Zin

l6 = 0.4λ

Z0 = 50 Ω

Figure 6: Diagram for problem 2, all transmission lines are lossless and 50 Ω.

In a patent for a GSM900 receiver circuit working at 940 MHz, you findthe circuit diagram in figure 6. You decide to investigate the circuit usinga smith chart. Start at the load and work you way to the input, showintermediate steps and give the un-normalized input impedance Zin.

If you arrived around 30 Ω, can you think of a simpler circuit to transforma 85 Ω load to 30 Ω?

Theory

We first define the impedance of the different passive components

ZRZ0

=R

Z0

ZLZ0

=jωL

Z0

ZCZ0

=1

jωCZ0

,

transmission line 6 can be regarded as a transmission line without load

Zin(d) = Z0ZL + jZ0 tan(βd)

Z0 + jZL tan(βd)

Z6

Z0

=0 + jZ0 tan(βd)

Z0 + j0 · tan(βd)= j tan(βd) (8)

and transmission lines with length of λ/4 is equal to a rotation of 180.

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Solution

This is now an exercise in addition and taking the inverse of complex num-bers, sage is a lot better at this than me so

f = 940e6 # Hz, 1GHz

L_2 = 2e-9 # henry, L_2 = 4e-9

R_L = 85 # ohm

def annote(Z, t, color=’black’, vertical_alignment=’bottom’, horizontal_alignment=’left’):

xy = Gammaxy(Z)

p = point(xy, color=color)

p += text(’$’+t+’$’, xy, vertical_alignment=vertical_alignment,

horizontal_alignment=horizontal_alignment, color=color,

fontsize=20)

return p

p = Z_imag(0) + Z_real(0) + Y_real(0) + Y_imag(0)

omega = 2*pi*f # rad/s

Z_0 = 50 # ohm

#Z_A = 75/Z_0; Y_A = 1/Z_A

Z_A = R_L/Z_0; Y_A = 1/Z_A

p += annote(Z_A, ’A’)

p += Z_real(real(Z_A))

# B:

Z_2 = 1j*omega*L_2/Z_0

Y_2 = 1/Z_2

Y_B = Y_A + Y_2; Z_B = 1/Y_B

Gamma_B = Gamma(Z_B)

p += annote(Z_B, ’B’)

p += Y_real(real(Y_B)) + Y_imag(imag(Y_B))

# C:

#l_3 = 0.1

l_3 = 0.15

beta = 2*pi

Gamma_C = Gamma_B*exp(-2j*beta*l_3)

Z_C = Z(Gamma_C); Y_C = 1/Z_C

p += annote(Z_C, ’C’)

p += circle((0, 0), abs(Gamma_C))

# D:

R_4 = 33

Z_4 = R_4/Z_0; Y_4 = 1/Z_4 # = G_4 [Siemens]

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Y_D = Y_C + Y_4; Z_D = 1/Y_D

Gamma_D = Gamma(Z_D)

p += annote(Z_D, ’D’)

p += Y_real(real(Y_D)) + Y_imag(imag(Y_D))

We are now halfway, let’s see what we have done. The smith chart so far canbe found in figure 7.

We start by normalizing the load and finding it on the smith chart,

Z1

Z0

= 1.7ZAZ0

= 1.7

Y2Z0 =Z0

Z2

= −4.2i YBZ0 = Z0(YA + Y2) = 0.59− 4.2i

we can now read of the magnitude and phase of the reflection coefficient androtate it 2β · 0.15λ = 110. towards the generator

ΓB = 0.94ej2.7

ΓC = ΓBe−2jβl3 = 0.94ej0.80

this corresponds to

YCZ0 = 0.036− 0.42i

We view the resistor as parallel to YC , moving along the imaginary admit-tance line we obtain

Y4Y0

= 1.5 YDZ0 = (YC + Y4)Z0 = 1.6− 0.42i

ΓD = 0.27ej2.7

p = Z_imag(0) + Z_real(0) + Y_real(0) + Y_imag(0)

p += annote(Z_A, ’A’)

p += annote(Z_B, ’B’)

p += annote(Z_C, ’C’)

p += annote(Z_D, ’D’)

# # E:

l_5 = 1/4

Gamma_E = Gamma_D*exp(-2j*beta*l_5)

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A

B

C

D

Figure 7: Smith chart for the first part of exercise

Z_E = Z(Gamma_E);Y_E = 1/Z_E

p += annote(Z_E, ’E’)

p += circle((0, 0), abs(Gamma_E))

# F:

#l_6 = 0.1

l_6 = 0.4

Z_6 = 1j*tan(beta*l_6);Y_6 = 1/Z_6

Y_F = Y_E + Y_6; Z_F = 1/Y_F

Gamma_F = Gamma(Z_F)

p += annote(Z_F, ’F’)

p += Y_real(real(Y_F)) + Y_imag(imag(Y_F))

# G:

l_7 = 1/4

Gamma_G = Gamma_F*exp(-2j*beta*l_7)

Z_G = Z(Gamma_G)

p += annote(Z_G, ’G’)

p += circle((0, 0), abs(Gamma_G))

# In

#for C_8 in [2.2e-12, 2.21e-12]:

C_8 = 2.2e-12 # farad

Z_8 = 1/(1j*omega*C_8*Z_0)

Z_in = Z_8 + Z_G

p += annote(Z_in, ’Z_in’)

p += Z_real(real(Z_in)) + Z_imag(imag(Z_in))

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Continuing from point D to E, we see that we have a quarter wave trans-former, so we rotate ΓD by 180

ΓE = ΓDe−j180 = 0.27e−j0.49

Using (8) we find the impedance of the shorted transmission line, the admit-tance can then be read of the smith chart

YEZ0 = 0.60 + 0.16i

Z6

Z0

= j tan(βd) = −0.73i

Y6Z0 =Z0

Z6

= 1.4i YFZ0 = Z0(YE + Y6) = 0.60 + 1.5i

Finding the reflection coefficient of this point we rotate it by another 180

ΓF = 0.72ej−2.1

ΓG = ΓF e−j180 = 0.72ej1.1

ZGZ0

= 0.60 + 1.5i

We are finally ready to find the input impedance, as this is just C8 in serieswith ZG

Z8

Z0

=1

jωC8Z0

= −1.5i

Zin

Z0

=ZG + Z8

Z0

= 0.60 + 0.00071i

Zin = 30.+ 0.035iΩ

As an alterntive route to Zin = 30 Ω we could use a single λ/4 transformeror a FIXME: 45 Ω in shunt.

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A

B

C

D

E

F

G

Zin

Figure 8: Smith chart for the last part of exercise 2

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Problem 3

We now study the shorted and open transmission lines in figure 9.

(a)

Find the required length l0 to achieve an input impedance Zin = −25j Ω. Isyour answer unique?

(b)

Find the required length l1 to achieve an input admitance Yin = 0.04j Ω. Isyour answer unique?

Z0 = 50 Ω

l0 =?ZL = 0

Zin = −25j Ω

(a) Short circuit

Z0 = 50 Ω

l1 =?ZL =∞

Yin = 0.04j S

(b) Open

Figure 9: Diagrams for problem 3

Short circuit

Z_0 = 50

Z_L = 0/Z_0

Z_in = -25j/Z_0

Gamma_in = Gamma(Z_in)

angle = Gamma_in.argument()*180/pi

p = smithchart_Z

p += annote(Z_L, ’Z_L’)


r = -angle.n(10)+180

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Γin = 1.0e−j130.

When rotating from ΓL to Γin we are moving towards the generator, so wehave to rotate by 180 + 130. = 310.. This is shown in figure 10 This is

ZL

Zin

Figure 10: Smith chart for exercise 3a

achieved with the length

2βl = 310.

l ≈ 0.43λ

Short circuit

We have the short circuit transmission line shown in figure 11, the objectiveis to find an l so that Yin = 0.04j S.

Moving along the transmission line is a rotation of the reflection coefficienttoward the load.

Z_0 = 50

Z_L = infinity/Z_0; Y_L = 1/Z_L

Y_in = 0.04j*Z_0 #2j

Z_in = 1/Y_in

Gamma_in = Gamma(Z_in)

angle = Gamma_in.argument()*180/pi

p = smithchart_Y

p += annote(Z_L, ’Y_L’)

p += annote(Z_in, ’Y_in’)

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Z0 = 50 Ω

lZL =∞

Yin = 0.04j S

Figure 11: Diagram for exercise 3b

We have first found the input reflection coefficient

Γin = 1.0e−j130.

When rotating from ΓL to Γin we are moving towards the generator, so wehave to rotate by 130.. This is shown in figure 12. This is archived with the

YL

Yin

Figure 12: Smith chart for exercise 3b

length

2βl = 130.

l ≈ 0.181λ

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Problem 4

We have the complex load in figure 13, which we want to match to Zin = 50 Ω.Use a smith chart to find the required values for L and C at 100 MHz.

L

Zload = 10 + 40j Ω

Zin = 50 Ω

C

Figure 13: Diagram for problem 4

Theory

Impedance is generally a complex quantity, so we have two equations andtwo unknowns

< (Zin) = 50 Ω

= (Zin) = 0

<(Zin

Z0

)= 1 (9)

=(Zin

Z0

)= 0 (10)

Solution

f = 100e6

omega = 2*pi*f

Z_0 = 50

Z_load = (10 + 40j)/Z_0;Y_load = 1/Z_load

Y_C = var(’Y_C’)

Y_A = Y_load + 1j*Y_C; Z_A = 1/Y_A

eq = [real(Z_A) == 50/Z_0]

s = solve(eq, Y_C, solution_dict=True)

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Y_C = s[1][Y_C]*1j

C = Y_C/(1j*omega*Z_0)

Z_A = Z_A(Y_C=Y_C/1j).n(64); Y_A = 1/Z_A

The above sage code solves (9) using the solve function, we will bellow visu-alize this process using the smith chart. We start by normalizing the load

Zload

Z0

= 0.20 + 0.80i YloadZ0 = 0.29− 1.2i.

This is in parallel with a capacitor, so the impedance looking from A (seefigure 13) into the load is

ZCZ0

=1

jωCZ0

YCZ0 = jωCZ0

YAZ0 = (Yload + YC)Z0

= 0.29 + j (ωCZ0 − 1.2) .

From this we see that the imaginary part can be changed by choosing C,from the smith chart we can see that the condition

<(ZAZ0

)= 1

<(

1

0.29− 1.2i+ YC

)= 1

is fulfilled when =(YA) ≈ 0.45. Sage gets (this actually has two solutions,but the first gives a negative inductance value, so we went for the secondone.)

YAZ0 = 0.29 + 0.46i

ZAZ0

= 1.0− 1.5i

This can now be used to find the capacitance

C =YCjωZ0

=1.632j

jωZ0

≈ 51.95 pF

Again, we first show some sage code before we visualize the solution onthe smith chart.

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Z_L = var(’Z_L’)

Z_in = Z_A + 1j*Z_L

eq = [imag(Z_in) == 0]

s = solve(eq, Z_L, solution_dict=True)

Z_L = s[0][Z_L]*1j; Y_L = 1/Z_L

L = Z_L/(1j*omega)*Z_0

Z_in = Z_in(Z_L=Z_L/1j); Y_in = 1/Z_in

The input impedance can be expressed as ZA in series with the inductor ZL

Zin = ZA + ZL

Zin = 1.0 + j

(−1.5 +

ZLZ0

)We try to find an ZL where the imaginary part of this equation is 0, thismoves the input impedance to the center and gives a match.

0 = −1.5 +ZLZ0

ZLZ0

= 1.5

Which gives an inductance of

L = 123.3 nH

Figure 14 shows the smith chart for this exercise.

p = Z_real(0) + Z_imag(0)

p += Z_real(real(Z_in)) + Z_imag(imag(Z_in))


p += Z_real(real(Z_load)) + Z_imag(imag(Z_load))

p += Y_real(real(Y_load)) + Y_imag(imag(Y_load))

p += annote(real(Z_load), ’%.3g’ % (real(Z_load).n(10)), color=’red’)

p += annote(1j*imag(Z_load), ’%.3g’ % (imag(Z_load).n(10)), color=’red’)

p += annote(Z_load, ’Z_load’)

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p += Z_real(real(Z_A)) + Z_imag(imag(Z_A))

p += Y_real(real(Y_A)) + Y_imag(imag(Y_A))

p += annote(real(Z_A), ’%.3g’ % (real(Z_A).n(10)), color=’red’, vertical_alignment=’top’, horizontal_alignment=’right’)

p += annote(1j*imag(Z_A), ’%.3g’ % (imag(Z_A).n(10)), color=’red’)

p += annote(1/real(1/Z_A), ’%.3g’ % (real(1/Z_A).n(10)), color=’blue’, vertical_alignment=’top’, horizontal_alignment=’right’)

p += annote(-1j*1/imag(1/Z_A), ’%.3g’ % (imag(1/Z_A).n(10)), color=’blue’)

p += annote(Z_A, ’Z_A’)

Zin0.2

0.8

Zload

1

−1.55

0.294

0.456

ZA

Figure 14: Smith chart for problem 4. Where Z0 is arbitrarily chosen as 50 Ω

As a sanity check

#L = 123.3e-9

#C = 51.95e-12

omega = 2*pi*100e6

def p(z1, z2):

return z1*z2/(z1 + z2)

Z_in = 1j*omega*L + p(1/(1j*omega*C), 10+40j)

returns 50. + 0.12i.

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INF5481 RF Circuit, Theory and Design Assignment #2