ECE 5324/6324 Spring 2018 Assignment 3 Solution

3.4-10 I really don’t like the book’s formulas that are unit system specific, like equations (3-52) through (3-55), though you are welcome to use them. Here I will start from equations (3-49), (3-50) and (2-30)

!

Then integrating to find the power, using equation (2-35) or (2-128)

!

Then the radiation resistance is

!

To add the effects of multiple windings we note that, as long as the current is constant along the entire wire (so that each loop adds with the same phase) each of the N loops will add field strength proportional to the area S. Thus S becomes NS. For adding the effects of a magnetic core we note that the magnetic permeability is present in the definition of both η and β. However, the factor in η arises from the ratio of electric to magnetic field in the far-field region, far from where the core is present, so only the β factors determine the exponent for an effective permeability. Thus we we make the following substitutions

! !

to obtain

!

Note that µeff is a relative (dimensionless) permeability.

E =ηβ 2S Ie− jβr

4πrsinθ φ̂ H = −β 2S Ie

− jβr

4πrsinθ θ̂

S = 12

Re E×H*( ) = 12

Re ηβ 2S Ie− jβr

4πrsinθ φ̂ × −β 2S Ie

− jβr

4πrsinθ θ̂

⎛⎝⎜

⎞⎠⎟

*⎛

⎝⎜⎞

⎠⎟ =1

32π 2 ηβ4S 2 I

2

r 2 sin2θr̂

ds = r 2 sinθdθdφr̂

P = S ⋅dssphere∫∫ = 1

32π 2 ηβ4S 2 I

2

r 2sin2θr̂

⎛⎝⎜

⎞⎠⎟ ⋅ r

2 sinθdθdφr̂( )0

π

∫−π

π

∫ = 132π 2 ηβ

4S 2 I2

sin3θ dθ dφ0

π

∫−π

π

∫

P = 132π 2 ηβ

4S 2 I22π 43= 112π

ηβ 4S 2 I2

Rr =2P

I2 =

16π

η β 2S( )2

S→ NS

β → µeff β

Rr =16π

η µeffβ2NS( )2

OK, now on to the problem. We must first confirm that the wavelength is long compared to the total length of the wire

!

To use the provided equation for effective permeability we must also confirm that the core length is much longer than the winding length. For the 30 gauge wire

!

Thus, we can use equation (3-56b), where we will use the diagonal of the rectangular rod as the effective diameter

!

Plugging in the numbers we find the effective permeability

!

Now plugging in the numbers !

Since we have thin wire, ! , we can use the simple result, equation (3-58), for the “ohmic” resistance

!

Note we indicated the vacuum permeability for determining the surface resistance. For all but the small contact area between the wire and the core, the copper has an interface with air, which is the usual case for surface currents on a metal. Plugging in the numbers

!

Not a great efficiency, but typical for this type of antenna.

f ! 1MHz c ! 2.9979×108 m/s ⇒ λ = cf! 300m

l1 = 10mm l2 = 3mm N = 500 ⇒ lwire = 2 l1 + l2( )N ! 13m ⇒ λ ≫ lwire

2a ! 0.255mm N = 500 ⇒ lwinding ! 2aN ! 128mm

lcore ! 250mm ⇒ lcore ≫ lwinding

µeff =µr

1+ D(µr −1) where D = 0.37R−1.44 and R =

lcore

dcore

and dcore = l12 + l2

2

dcore ! 10.44mm R ! 23.95 D ! 0.00382 µr ! 38 ⇒ µeff ! 33.29

Rr ! 0.962µΩ

a≪ l1,l2

Ro !lwire

2πaRs where Rs =

π f µ0

σ

Rs ! 261µΩ Ro ! 4.23Ω er =Rr

Rr + Ro! 2.3×10−7

3.5-5(b) For this problem we just have to do an integral which cannot be done symbolically, and must be done numerically. It is important to note that the antenna pattern given (3-74b) is the correct pattern everywhere in space for a pair of out-of-phase, half-wave dipoles spaced λ/2 apart. By image theory, this is also the correct pattern for a single half-wave dipole λ/4 above a PEC ground plane, but only for z > 0. For z < 0, the pattern is zero. Thus we need to find the beam solid angle for the pattern

!

Thus the beam solid angle is

!

Note the integration range for the angle θ. Mathematica (and I assume other systems) can easily do this integral.

!

If you integrate the full range of θ, you will get twice this beam solid angle and half this directivity.

F θ ,φ( ) =cos

π2sinθ cosφ⎛

⎝⎜⎞⎠⎟

1− sin2θ cos2φsin

π2cosθ

⎛⎝⎜

⎞⎠⎟ 0 <θ < π

2

0π2<θ < π

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

ΩA = F θ ,φ( ) 2 dΩsphere∫∫ = F θ ,φ( ) 2 sinθ dθ dφ

0

π /2

∫−π

π

∫

ΩA ! 2.24 and D = 4πΩA

! 5.60

4.2-1 (a) As above we use

!

to transform the single-loop, air-core electric field into the multi-loop, magnetic-core result

!

Then use equation (4-4), which we assume to be generally true (though no convincing arguments are given)

!

Note I have corrected this equation by inserting the current, I, which was missing in the book. Also, I swapped the ωµ for ηβ, as per usual. Equating these fields

!

and solving for the effective length vector, we find

!

(b) It is interesting that it says short dipole. I almost missed that. Anyway, if you look at (for example) equation (2-103), the radiation integral is essentially an integral of the current function (i.e. the area under the current curve). Comparing the current functions of the ideal and short dipole in Figure 2-16, we note that the area under the curve for the short dipole is one half that of the ideal dipole. This factor of one half will also be present in the electric field and thus the effective length h. The effective length magnitude for the short dipole is then

!

and the ratio of effective length magnitudes is

!

(c) Here we have a single loop, an effective permeability of one (since no other value is given), and a loop diameter equal to the dipole length.

!

S→ NS

β → µeff β

E =ηβ 2S Ie− jβr

4πrsinθ φ̂ ⇒ E =ηµeffβ

2NS Ie− jβr

4πrsinθ φ̂

E = jωµI4π

e− jβr

rh = jηβ I

4πe− jβr

rh

E =ηµeffβ2NS Ie

− jβr

4πrsinθ φ̂ = jηβ I

4πe− jβr

rh

h = − jµeffβNS sinθ φ̂

hshort dipole =12Δz sinθ

hloop

hshort dipole

=µeffβNS sinθ

12Δz sinθ

= µeffNβSΔz / 2

hloop

hshort dipole

=βπ d

2⎛⎝⎜

⎞⎠⎟

2

d / 2= π

2βd( )2

βd= π 2 d

λ

Plugging in the numbers

!

We require d/λ to be very small, for the small loop and short dipole results to be accurate. So, even with the pi-squared factor, we expect the dipole will be the winner. Basically, as we see above, the loop has two factors of βd and the dipole just one.

f ! 1.5MHz c ! 2.9979×108 m ⇒ λ ! 200md ! 0.3m

hloop

hshort dipole

= π 2 dλ! 0.015

4.3-X2 We use the following circuit to model the thermal noise

�

(a) Using the voltage divider formula we have for the transfer function

!

The squared magnitude is then

!

(b) The PSD of the two noise voltages in our circuit is

!

Then the PSDs of the two independent noise voltages across the load resistor

!

H ω( )= VLL ω( )VL ω( ) =

VLA ω( )VA ω( ) =

RLRL + RA⎡⎣ ⎤⎦ + j X L ω( )+ XA ω( )⎡⎣ ⎤⎦

H ω( ) 2 = RL2

RL + RA⎡⎣ ⎤⎦2+ XL ω( )+ XA ω( )⎡⎣ ⎤⎦

2

SvAvA ω( )= 4kTARA and SvLvL ω( )= 4kTLRL

SvLAvLA ω( )= H ω( ) 2 SvAvA ω( )= RL2

RL + RA⎡⎣ ⎤⎦2+ XL ω( )+ XA ω( )⎡⎣ ⎤⎦

2 4kTARA

SvLLvLL ω( )= H ω( ) 2 SvLvL ω( )= RL2

RL + RA⎡⎣ ⎤⎦2+ XL ω( )+ XA ω( )⎡⎣ ⎤⎦

2 4kTLRL

(c) The combined power dissipated in the load

!

(d) Assuming matched conditions

!

The above expression simplifies to the often quoted result

!

P = PLA + PLL

= 1RLSvLAvLA ω0( )Δf + 1

RLSvLLvLL ω0( )Δf

= 1RL

RL2

RL + RA⎡⎣ ⎤⎦2+ XL ω0( )+ XA ω0( )⎡⎣ ⎤⎦

2 4kTARAΔf +1RL

RL2

RL + RA⎡⎣ ⎤⎦2+ XL ω0( )+ XA ω0( )⎡⎣ ⎤⎦

2 4kTLRLΔf

= k4RL TARA +TLRL( )

RL + RA⎡⎣ ⎤⎦2+ XL ω0( )+ XA ω0( )⎡⎣ ⎤⎦

2 Δf

ZL = ZA* ⇒ RL = RA and XL + XA = 0

P = k TA +TL( )Δf

4.4-16 For reasons I don’t yet fully understand, the impedance mismatch factor for this circuit problem

! given by

! ! !

is the same as the impedance mismatch factor for this transmission line problem

! given by

!

where the reflection coefficient is

!

which is the standard formula, except for the complex conjugation in the numerator, which I have never seen presented anywhere. Normally, it would not matter, since transmission lines are frequently treated as lossless, with real valued impedances. Anyway, moving along, we write the complex impedances with real and imaginary parts

!

Then q factor is

!

q =PL

PL, matched

=4RARL

RA + RL( )2+ XA + XL( )2

q =PLPinc

=Pinc − PreflPinc

= 1−PreflPinc

= 1− Γ2

Γ =ZL − ZA

*

ZL + ZA

Γ =RL + jX L( )− RA + jX A( )*RL + jX L( )+ RA + jX A( ) =

RL − RA( )+ j X L + XA( )RL + RA( )+ j X L + XA( )

q = 1−RL − RA( )2 + XL + XA( )2RL + RA( )2 + XL + XA( )2

=RL + RA( )2 + XL + XA( )2 − RL − RA( )2 − XL + XA( )2

RL + RA( )2 + XL + XA( )2

=4RARL

RL + RA( )2 + XL + XA( )2

4.5-3 We will use the Friis equation

!

where I have written it, at first, in an intuitive form using the areas and directivities, and secondly in the usual form. In this problem we will not consider impedance mismatch or polarization loss, so the power attenuation is

!

Results are tabulated below. For part (c) we note that attenuations much greater than -100dB would require repeaters. For attenuations of -1000dB the signal is irretrievably lost in the noise.

PrPt

=Aem,iso

Asphere,R

DrDterrertqrqt p =λ

4πR⎛⎝⎜

⎞⎠⎟

2

GrGtqrqt p where Aem,iso =λ 2

4π and Asphere,R = 4πR2

PrPt

= λ4πR

⎛⎝⎜

⎞⎠⎟2

GrGt

medium attenuation [dB/m] f [Mhz] attenuation at 500m [dB] attenuation at 50km [dB]

coaxial cable -0.1 -50 -5000

wireless 150 -57 -97

coaxial cable -0.14 -70 -7000

wireless 300 -63 -103

fiber optic -0.001 -0.5 -50