ELEC 412 Assignment #1 Solutions 1

Assignment #1: Solutions P1.1 The phase velocity is

]/[104.16.4

103111 88

smcvrroo

p ×=×

====εεεµµε

and the wavelength is

mmf

vp 85.721092.1104.1

9

8

=××

==λ

P1.4 (a) From equation 1.2

⎟⎟⎠

⎞⎜⎜⎝

⎛==

ro

oy

ox ZHE

ε1377 and ]/[3.5

377mAk

EH rox

oy ==ε

(b) The phase velocity is 1.5 x 108 [m/s] and the wavelength λ = vp /f = 30 mm.

(c) Let cos( ωt1 – kz1 ) = 1 and solve ωt1 = kz1 ⇒ z1 = ( )( ) 4501031052 691 =

××=

−

pvkt

ωπω m

For cos( ωt2 – kz2 ) = 1 and solve ωt2 = kz2 ⇒ z2 = ( )( ) 10501031052 692 =

××=

−

pvkt

ωπω m

So z2 – z1 = 600 m .

P1.5 The impedance of the parallel LC circuit is

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−−=

+=

LC

jCj

Lj

Z

ωωω

ω1

11

11

The impedance of the series LC circuit is

⎟⎠⎞

⎜⎝⎛ −=

CLjZ

ωω 1

2

For the modified inductance we obtain for the parallel circuit

ELEC 412 Assignment #1 Solutions 2

( )LjRECj

Z

ωω

++

= 11

3

For the modified inductance in series we get

⎟⎠⎞

⎜⎝⎛ −+=

CLjRZ

ωω 1

4

The four plots are provided as MATLAB subplots as follows:

P1.8

( ) LjRLCRLj

LjCj

R

LjCj

VV

i

o

ωωω

ωω

ωω

+−=

++

+= 2

11

11

ELEC 412 Assignment #1 Solutions 3

% P1.8 % f=1e6:1e6:5e9; L=10e-9; C=1e-12; R=50; V=(j*2*pi*f*L)./(j*2*pi*f*L+(R-4*(pi^2)*(f.^2)*R*L*C)); plot(f/1e9,abs(V));grid title('Transfer Response |Vo/Vi|'); xlabel('Frequendcfy in GHz');ylabel('|Vo/Vi|'); plot -deps 'p1_8.eps'

P1.20 From the short circuit measurement, we can determine the series impedance Zs = Rs + jωL. We can record Rs from the DC measurement and find Ls at a fixed frequency from

ω

22ss

sRZ

L−

=

Next from the open circuit measurement we can find the impedance of the parallel circuit provided that the effect from the series impedance is negligible. We can record RP from the DC measurement and find CP at a fixed frequency from

ω

2211

PPP

RZC

−=

Put the total measured impedance Z in relation to two known cable impedances and the unknown ZDUT .

( )DUTPS ZZZZ ||+= so ( )PS

sPDUT ZZZ

ZZZZ−−−

= .

ELEC 412 Assignment #1 Solutions 4

P1.21 A capacitor C is connected in shunt with a resistor R followed by a series load inductance L. The

total impedance is of the form LjCj

R

Z ωω

++

= 11

P1.26 The input impedance of the circuit is

( ) RCjLCLCRLj

CjRLj

CjRLj

LjCj

RZωω

ωω

ωω

ωω

ωω +−

−=

++

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=⎟⎟⎠

⎞⎜⎜⎝

⎛+= 2

2

11

11

( ) ( )( ) 22222

2232222

111

CRLCRLCjLCLRCLRCLCLjZ

ωωωωωωωω

+−

+−−+−=

At resonance, X (ωr ) = 0 so ( ) 02222 =−+ CLRLCL rω

Solving for the resonance frequency: 222

1RCLC

fr−

=π

.

Solving for the capacitor

222 1

r

RCCLω

=+−

Rearranging so that we might use the quadratic formula to solve for C :

01222

2 =+⎟⎠⎞

⎜⎝⎛−

RRLCC

rω

Solving for C using the quadratic formula:

224

2

21

42 RRL

RLC

rω−+= =550.44 pF

Determine the bias voltage from:

CC

VV o

diff

bias =⎟⎟⎠

⎞⎜⎜⎝

⎛−

21

1

Resulting in

diffo

bias VCCV ⎟⎟

⎠

⎞⎜⎜⎝

⎛−= 2

2

1 =0.7497

ELEC 412 Assignment #1 Solutions 5