CI: Elham Baladi
Winter 2020
Solution to Assignment 1
EECS 3604
January 21, 2020
Problem 1
Find the phasors of the following time domain functions:
(a) v(t) = 9 cos(ωt − π/3) (V)
Ṽ = 9 e−jπ/3
(b) v(t) = 4 sin(ωt + π/3) + 3 cos(ωt − π/6) (V)
= 4 cos(ωt + π/3 − π/2) + 3 cos(ωt − π/6) (V)
⇒ Ṽ = 4 e−jπ/6 + 3 e−jπ/6 = 7 e−jπ/6 (A)
(c) i(x, t) = 5e3x sin(ωt + π/6) (A) = 5e3x cos(ωt + π/6 − π/2) (A) ⇒
˜ = 5e3x e−jπ/3
I(x)
Problem 2
The instantaneous time domain functions:
(a) V(t) = -5 cos(ωt − π/3) (V)
(b) Ṽ = e−jπ/2 6 e−jπ/4 (V) = 6 ejπ/4 (V) ⇒ V(t)= 6 cos(ωt + π/4) (V)
◦
(c) I˜ = -3 + j2 (A) = 3.61 ej146.31 ⇒ I(t)= 3.61 cos(ωt + 146.31◦ ) (A)
Problem 3
Problem 2.1 in Pozar, 4th edition:
Z0 = 75 Ω
i(t,z) = 1.8 cos(3.77 × 109 t − 18.13z) mA ⇒ ω = 3.77 × 109 rad/s, k = 18.13 m−1
(a) f = ω/(2π) = 600 MHz
(b) Vph = ω/k = 2.08 × 108 m/s
(c) λ = Vph /f = 0.347 m
(d) Vph =
√1
µ
=
√
1
µr µ0 r 0
=
√C
µr r
(C= speed pf light= 3 × 108 )
⇒ r = (C/Vph )2 = 2.08 (Teflon)
˜ = 1.8 e−j18.13 (mA)
(e) I(z)
f) v(t,z) = 135 cos(3.77 × 109 t − 18.13z) mV
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CI: Elham Baladi
Winter 2020
Solution to Assignment 1
EECS 3604
January 21, 2020
Problem 4
Problem 2.3 in Pozar, 4th edition
(note: permittivity of Teflon= 2.08, loss tangent = 0.0004, conductivity of copper= 58.14×106
siemens per meter.)
L=
µ0
2π
C=
2π0 r
ln(b/a)
ln(b/a) = 2.4 × 10−7 H/m
= 9.64 × 10−11 F/m
Rs = 0.00825Ω ⇒
R=
Rs
2π
( a1 + 1b ) = 3.76 Ω/m
2πω0 r
ln(b/a)
tanδ = 2.42 × 10−4 S/m
p
For small loss: Z0 = L/C = 49.9Ω
G=
α = 12 (R/Z0 + GZ0 ) = 0.044 np/m = 0.38 dB/m (note: neper’s number= 2.71828 = 8.686
dB)
Manufacturer’s specifications: Z0 = 50 Ω, α = 0.426 dB/m
Problem 5
(a) γ =
p
(R + jLω)(G + jCω) = 0.707 + j88.838 m−1
(b) α = 0.707 np/m = (0.707 ×
8.686dB
)
np
dB/m = 6.14 dB/m
⇒ total attenuation= α × l = (6.14 dB/m)(0.15 m)= 0.92 dB
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