Franklin W. Olin College of Engineering
Needham, Massachusetts
Electrical and Computer Engineering
ENGR2410 – Signals and Systems
Assignment 3
Problem 1:
A. Find the equivalent impedance of a series RLC combination.
Solution:
Z = R + Ls +
s2 +
Z=
1
s
L/R
(
1
Cs
+
1
LC
s/L
1
1
s + LC
s2 + L/R
1
Z=
1
Cs
LC
)
(
1
1
s + LC
L s2 + L/R
Z=
s
)
B. Find an expression for the
in terms of the charac√ quality factor Q of the series RLC
ω0
teristic impedance Z0 = L/C. Recall that Q is defined as 2α .
Solution:
1
Q= √
L/R =
LC
√
L/C
R
C. Find a condition for R such that Q ≫ 1. In the limit, is R acting more like a short or
an open circuit?
Solution:
R≪
√
L/C
R is approaching a short.
D. Repeat the previous three parts for the parallel RLC combination.
Solution:
Z = R||Ls||
Z=
Ls
s/C
1
= R||
=
1
Cs
LCs2 + 1
s2 + RC
s+
1
·
C s2 +
1
LC
s
1
s
RC
+
1
LC
1
R
RC = √
LC
L/C
√
R ≫ L/C
Q= √
R is approaching an open.
Problem 2:
A. Find the transfer function for the circuit from last week’s problem 2 using impedances.
Solution:
1
Ls|| Cs
Vout
H(s) =
=
1
Vin
R + Ls|| Cs
H(s) =
H(s) =
H(s) =
R
Ls
LCs2 +1
+ LCsLs2 +1
Ls
RLCs2 + Ls + R
s2 +
1
s
RC
1
1
s + LC
RC
2
B. Show that at low frequencies a capacitor may be replaced with an open circuit and an
inductor may be replaced with a short circuit. Show that the inverse is true at high
frequencies.
Solution:
1
. The impedance of the capacitor is related
The complex impedance of a capacitor is Cjω
inversely to frequency. At low frequencies ω → 0, the impedance of the capacitor is
really large and the capacitor acts like an open circuit. At high frequencies ω → ∞, the
impedance of the capacitor is really small and the capacitor acts like a short circuit.
1
= ∞
ω→0 Cjω
1
lim
= 0
ω→∞ Cjω
lim
The complex impedance of an inductor is Ljω. The impedance of the inductor is
related directly to frequency. At low frequencies ω → 0, the impedance of the inductor
is really small and the inductor acts like a short circuit. At high frequencies ω → ∞,
the impedance of the inductor is really large and the inductor acts like an open circuit.
lim Ljω = 0
ω→0
lim Ljω = ∞
ω→∞
C. Draw equivalent circuits for low frequencies and high frequencies. Use them to verify
the extremes of the Bode plot from last week.
Solution:
The left circuit depicts how the original circuit behaves at low frequencies when the
capacitor acts like an open circuit and the inductor acts like a short. The right circuit
depicts the original circuit at high frequencies when the capacitor acts like a short
and the inductor acts like an open circuit. The result in either of these cases is that
vout = 0.
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D. What is the equivalent impedance of the parallel LC combination at resonance? What
can you replace it with?
Solution:
Z = Ls||
1
1
s
= · 2
1
Cs
C s + LC
)
(
1
1
j √LC
1
1
1 j √LC
1
= Ls||
Z(jω0 ) = Z j √
= ·
·
1
1 =
Cs
C − LC
C
0
+ LC
LC
A division by zero indicates that the current will be zero for any voltage across the
combination, therefore, the combination acts like an open circuit at resonance.
E. Draw an equivalent circuit at resonance. Does it correspond to your Bode plot?
Solution:
At resonance, vout = vin , or H(s) = 1 which corresponds to the Bode plot. Note
that it would be wrong to draw the L and the C open; neither is open, the parallel
combination acts like an open.
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