University of California, San Diego
Spring 2016
J. Connelly
ECE 45 Discussion 2 Notes
Frequency Response
The inputs and outputs of RLC circuits are generally either voltages or currents. We can model the
circuit as a system whose output depends on the frequency of the input.
• An input/output RLC circuit is defined by its transfer function (also called frequency response):
, where Out(ω) and In(ω) are the phasor transforms of out(t) and in(t), for an
H(ω) = Out(ω)
In(ω)
arbitrary frequency ω. The inputs and outputs of RLC circuits are generally either voltages or
currents.
• In general, out(t) 6= in(t) h(t) (a common mistake in this course).
• H(ω) is a complex number, which is a function of ω, since input/output relationships change
with different frequencies.
q
Im{H(ω)}
2
−1
2
|H(ω)| = Re {H(ω)} + Im{H(ω)} and ∠H(ω) = tan
.
Re {H(ω)}
• When x(t) is sinusoidal, x(t) = A cos(ω0 t + φ), and its phasor representation is A ej φ . If we
send it through a system, H(ω), then the output phasor Y = A |H(ω0 )| ej(φ+∠H(ω0 )) ,
so y(t) = A |H(ω0 )| cos(ω0 t + φ + ∠H(ω0 )).
P
• When x(t) = Ai cos(ωi + φi ),
P
by linearity, the output is y(t) = Ai |H(ωi )| cos(ωi + φi + ∠H(ωi )).
•
+
Vi
+
RLC Circuit
−
Please report any typos/errors to j2connelly@uscd.edu
Vo
−
Example 1:
1. Find the magnitude and phase of the frequency response, H(ω), of the circuit below.
2. How does the output behave as ω → 0 and ω → ∞
3. When vin (t) = cos2 (t), find vo (t).
4. For what vin (t) do we have vo (t) = 1/2?
R1
R2
Z1
+
+
v(t)
i
−
Vi +
vo(t)
L
C
Vc
Z2
ZC
−
+
Z L Vo
−
−
Where R1 = 1Ω, R2 = 3Ω, C = 1F , L = 1H
1. By using a voltage divider, we can write the output voltage as:
VO = VC
ZL
ZR2 + ZL
→ VC = VO
ZR2 + ZL
ZL
(1)
By KCL at the node VC :
VC
VC
Vin − VC
+
=
ZR1
ZC ZL + ZR2
Vin
1
1
1
→
+
= VC
+
ZR1
ZC ZL + ZR2 ZR1
Vin
.
→ VC =
1
1
1
ZR1 ZC + ZL +ZR + ZR
2
1
By setting equal 1 and 2:
VO
ZR2 + ZL
=
ZL
Vin
ZR1
ZC
+
ZR1
ZL +ZR2
+1
Then by definition of H(ω), we have:
ZL
jωL
VO
=
=
ZR1
Vin
(R2 + jωL)(1 + jωR1 C) + R1
(ZR2 + ZL )(1 + ZC ) + ZR1
jωL
jω
jω
=
=
=
2
2
R1 + R2 − ω R1 LC + jω(L + R1 R2 C)
4 − ω + 4 jω
(jω + 2)2
H(ω) =
and so, solving for magnitude and phase of H(ω) yields:
|H(ω)| =
|ω|
,
4 + ω2
∠H(ω) =
ω π
− 2 tan−1
2
2
(2)
2. H(0) =
lim
H(ω) = 0.
ω→∞
Since the transfer function is 0 at very low and very high frequencies, the output of the circuit
will also be 0. This can be verified by looking at how the components in the circuit behave at
high and low frequencies.
3. Note that
2
cos (t) =
ejt + e−jt
2
2
=
1 2 + ej2t + e−j2t
1 + cos(2t)
=
2
2
2
Then vin (t) = vi,1 (t) + vi,2 (t), where vi,1 (t) = 1/2 and vi,2 (t) = cos(2t)/2. Since vi,1 (t) and
vi,2 (t) are different frequencies we need to use super position to find vo (t).
We have:
H(0) = 0
π
1
|H(2)| = and ∠H(2) = − 2 tan−1 1 = 0
4
2
and so
1
|H(0)| = 0
2
|H(2)|
1
vo,2 (t) =
cos(2t + ∠H(2)) = cos(2t)
2
8
vo,1 (t) =
Thus by superposition vo (t) = vo,1 (t) + vo,2 (t) = 81 cos(2).
4. If vo (t) = 1/2, then by the linearity of the system, there must be a DC term in the input. However,
H(0) = 0, so any DC term in the input will be zeroed out. Thus we cannot have vo (t) = 1/2 for
this circuit.
Example 2:
Find the transfer function of an LTI system with input x(t) and output y(t) given by
d2 y(t)
dx(t)
+
x(t)
=
+ y(t)
dt2
dt
Assume x(t) is sinusoidal, then we can represent the differential equation using phasors:
(jω)2 Y + X = jωX + Y
and so
H(ω) =
jω − 1
1
Y
=
=
.
2
X
(jω) − 1
jω + 1
Example 3:
1. Find the frequency response, H(ω) = I0 /Ii , of the circuit below.
2. Find the value ω0 for which |H(ω)| is maximized.
√
3. Find the values ω1 such that |H(ω1 )| = |H(ω0 )|/ 2
√
4. With the value of R fixed, what value should L0 take so that we have |H(105 )| = |H(ω0 )|/ 2
+
ii(t)
R
L
vo(t)
−
R = 100 kΩ, L = 1 mH
1. We can use a current divider to find the current through the inductor:
Io = Ii
1
ZL
1
ZR
+
1
ZL
=
1
ZL
+1
ZR
and so
H(ω) =
2.
Io
1
=
Ii
1 + jω(L/R)
|ω|L
|H(ω)| = p
1 + (ωL/R)2
→ |H|max = |H(0)| = 1
3.
1
1
= √ → 1 + (ω1 L/R)2 = 2
|H(ω1 )| = p
2
1 + (ω1 L/R)2
ω1 = R/L = 105 /10−3 = 108
4.
105 = 105 /L → L = 1H