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Unformatted text preview: ECE 442 Summer 2010
HW#1 Solutions
1.14. 1.15. 1 Now, when a resistance of 1.5kΩ is connected between 4 and ground,
I= 0.77
= 0.1mA
6.15 + 1.5 1.67. Using the voltage divider rule 1
Zi = Ri  sCi
1
Yi = Ri + sCi Vi
Zi
1
1
=
=
=
=
1
Vs
Z i + Rs
1 + Rs Yi
1 + Rs ( Ri + sCi )
1+ = Rs
Ri 1/(1 + RS )
1
Ri
=
Rs
+ sCi Rs
1 + sCi 1+ Rs Ri
1
1
1
=
Rs
Rs Ri
Ri + Rs 1 + sCi (Ri Rs )
1 + Ri 1 + sCi ( Rs +Ri ) This transfer function is of the STC lowpass type with a dc gain K =
and a 3dB frequency ω = 1/C (R R ).
For
and C = 5pF, Ri /(Ri + Rs )
Rs = 20k Ω, Ri = 80k Ω,
ω0 = 5× 10−12 f0 = 0 i i s i 1
= 1.25 × 107 rad/s
× 20×80 × 103
20+80 ω0
1.25 × 107
=
≈ 2M Hz
2π
2π 2 Ri 1.68. Using the voltage divider rule, T (s) = R2
Vo
=
Vi
R2 + R1 + = 1
sC R2
R1 + R2 s + which from Table 1.2 is of the highpass type with
K= and
ω0 = s
1
C (R1 +R2 ) R2
R1 + R2 1
C (R1 + R2 ) As a further verication that this is a highpass network and T (s) is a highpass transfer function we observe as at s = 0,T (s) = 0; and that as s → ∞,
T (s) = R /(R + R ). Also, from the circuit observe as at s → ∞, (1/sC ) → 0
and V /V = R /(R + R ). Now, for R = 10kΩ, R = 40kΩ, and C = 0.1µF ,
2 o i 1 2 2 f0 = 1 2 1 2 ω0
1
=
= 31.8Hz
−6 (10 + 40) × 103
2π
2π × 0.1 × 10
1
K
40
√ = 0.57V /V
T (jω0 ) = √ =
10 + 40 2
2 1.69. Using the voltage divider rule, VL
RL
=
VS
RL + RS + 1
sC = 3 RL
RL + RS s + s
1
C (RL +RS ) which is of the highpass STC type (see Table 1.2) with
K=
,ω =
For f ≤ 10Hz,
RL
RL +RS 0 1
C (RL +RS ) 0 1
2πC (RL + RS ) ⇒C≥ ≤ 10 1
2π × 10(20 + 5) × 103 Thus, the smallest value of C that will do the job is C = 0.64µF. 4 ...
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 Spring '11
 NareshShanbhag
 Volt

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