Mail in this and all examinations promptly, as they are bornpletad. Theo start on the next lesson. 1. The relatived B amplitude responseo f an amplifier decreasesa t the rateo f 6 dB per octavea t...

I
I
l . What is the
( t ) z(s) :
(2) z(s) :
(3) z(s) :
t ransform
sC
I
'C
s
C
impedance o f an uncharged capac i to r '?
(4) Z(r) : 9
s
(5) Z(s) = RC
2. What is the t ransform impedance
( l ) Z (s ) : sZ
(2) Ztr) : #
(3) Z(r) : i
What is the t ransform impedance
Fig. E- I ?
of an unfluxed
(4) Z(s) =
(s) z(s) :
indu.ctor 'J
L
s
L
R
3. c i rcu i t o fa t te rmina lsaa' of the
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2l l r
Fig . E- l
( t) z(s)
(2) z (s)
(3\ z(s)
() z(s)
(s) z(s)
Fis. E-2
s2tC+sRC+l
sC
t / C
s2 + s / l?C + I ILC
IRs
l+RCs+s 'LC
s l ,+R
s'LC + sRC + I
_ s + I IRLC
s2 + s/RL + LIC
4. The switch in the circuit of Fig. E-2 is closed at r = 0. Find the
Thevenin equivalent transform voltage Z (s) at terminals a - a' .
(4 ) V(s) = r "
t *
4s '+ J + l0
rn
(5) V(s) = , ' t t
s '+25
(3) Z(s) = -.
tOt (Hint: Assume initial conditions to be
+s2 + 100 zero, use voltage divider rule, see 64)
5. For the circuit of Fig. E-3 determine an expression for ttre transform current
flowing through the resistor R. (Hint; Initial conditions : 0, write loop
equations for I, and 12 and solve using determinents)
Fi9. E-3
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Option 1 is correct
2I lu E (s)
soLzc + 2s2L - s + I IC
E (s)
L2CS3+RLCs2+2Ls+R
E(s)
L/Cs(r ' + 2 / LC)
E(s)
LC(sz + s lC + I ILC)
E(s)
(1) 1a(s) :
(2) /x(s) :
(3) /p(s) :
(4) /a(s) =
(5) ft(s) =
LC(s3 + Rs2 + s lC + RILC)
6. Deterrnine the Norton equivalent current source at terminals a -...