This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ...
View
Full
Document
 Spring '11
 NareshShanbhag
 Volt

Click to edit the document details