1
SIMON FRASER UNIVERSITY
School of Engineering Science
ENSC 320 Electric Circuits II
Solutions to Assignment 5
March 2005
1. Ideal Transformer
(a) The left-hand sketch refers the secondary to the primary. The right hand sketch does the opposite.
(b) We can use either of the equivalent circuits (or the original) to obtain the transfer function.
Arbitrarily, I'll choose the left hand sketch - and I'll remember to scale its transfer function by 3 to get
the true output voltage.
impedance of parallel combination
1
R'2⋅
s⋅ C
1
R'2 +
s⋅ C
=
R'2
R'2⋅ s⋅ C + 1
R'2
R'2⋅ s⋅ C+1
H ( s) = 3⋅
R1 +
R'2
R'2⋅ s⋅ C + 1
=
3
⋅
R1⋅ C ⎛
1
R1 + R'2 ⎞
⎜s +
⎝ R1⋅ R'2⋅ C ⎠
=
3
⋅
R1⋅ C ⎛
1
9⋅ R1 + R2 ⎞
⎜s +
R1⋅ R2⋅ C
⎝
⎠
(c) We have component values
R1 := 2⋅ kΩ
R2 := 18⋅ kΩ
C := 10⋅ nF
The transfer function has no zeros and it has a single pole at
p := −
9⋅ R1 + R2
R1⋅ R2⋅ C
5 rad
p = −1 × 10
s
(d) The bandwidth of this first order circuit (i.e., the 3 dB point) is numerically equal to the (negative
of) the pole location:
ω 3dB := −p
5 rad
ω 3dB = 1 × 10
s
2
The time constant is, as usual, the negative reciprocal of the real part of the pole. Here the pole is real,
so
τ :=
1
−p
−5
τ = 1 × 10
s
(e) The voltage gain at DC (note this is an ideal transformer) is H(0), or
3
1
3
3
⋅
=
⋅ R2 =
R1⋅ C ⎛ 9⋅ R1 + R2 ⎞
9⋅ R1 + R2
2
⎜
⎝ R1⋅ R2⋅ C ⎠
(d) The frequency response, sketched from the pole-zero diagram, is
2. Transformer With Finite Permeability
The coupled coils model of the transformer has parameters
L1 := 0.1⋅ mH
L2 := 0.9⋅ mH
M := 0.3⋅ mH
(a) The first step is to check if we have unity coupling.
L1⋅ L2 = 0.3 mH
which is the same value as M, so unity coupling (k = 1).
All we have to model is finite permeability. Using an embedded ideal transformer with turns ratio
sqrt(L1):sqrt(L2) or 1:3, we draw the circuit as
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(b) To obtain the transfer function, we'll refer the secondary to the primary again (fewer components to
transform, that way). The result is shown below
We could obtain the transfer function as a voltage divider, first combining the parallel branches with C,
L1 and R'2. Alternatively, we could combine the source and the resistors into a Thevenin equivalent,
producing the classic "tank circuit," for which we have obtained the transfer function many times (e.g.
page 1.2.2.f in the notes). Hmm. Which approach to use? I'll do the voltage divider, just to remind
you of the admittance form of the divider equation. And I'll remember to scale the result by 3, to get
the correct output voltage. Also, I'll substitute the transformed value R'2 = R2/9 right from the start.
H ( s) = 3⋅
1
R1
1
1
9
+ s⋅ C +
+
R1
s⋅ L1 R2
3⋅ s
C⋅ R1
=
⎛ R2 + 9⋅ R1 ⎞
2
s +⎜
⎝ C ⋅ R 1⋅ R 2 ⎠
⋅s +
1
L1⋅ C
(c) We now have a zero, located at the origin. As for the poles, we have to solve a quadratic with
coefficients
R2 + 9⋅ R1
C⋅ R1⋅ R2
Solve
2
1
12 rad ⎞
= 1 × 10 ⎛⎜
L1⋅ C
⎝ s ⎠
5 rad
= 1 × 10
5
12
s + 10 ⋅ s + 10
s
2
to obtain the poles
= 0
p1 = 5000⋅ ( −1 + j ⋅ 399)
p2 = 5000⋅ ( −1 − j ⋅ 399)
(d) Since there is a zero at the origin, the DC gain is zero. Physically, this is because non-zero voltages
at the terminals of coils require changing currents (remember, the voltages are proportional to the
derivatives of the currents). The steady state response to DC input is a constant current, so no voltage
is induced at either the primary or the secondary terminals of the transformer.
(e) For the frequency response, we first write the transfer function with numerical values
5
H ( s) =
5
1.5⋅ 10 ⋅ s
2
5
s + 10 ⋅ s + 10
12
=
(
1.5⋅ 10 ⋅ s
s − p1 ⋅ s − p2
)(
)
p1, p2 defined above
4
It's a bandpass filter, differing from the standard bandpass form only by the factor 1.5 in the
numerator. We have natural frequency
5
6
ω o := 10
and damping factor
ζ :=
10
2⋅ ω o
ζ = 0.05
5
So it's resonant (as we could infer from the pole values, too). Then β = 10
Q = 10
The maximum voltage gain is 1.5, by comparison with the standard bandpass form, which has unit
gain at resonance. The frequency response sketch is given below.
(f) Will it go back to the Question 1 circuit with ideal transformer if the permeability goes back to
infinity? The transfer function was obtained as
3⋅ s
C⋅ R1
H ( s) =
2
⎛ R2 + 9⋅ R1 ⎞
s +⎜
⎝ C⋅ R1⋅ R2 ⎠
⋅s +
1
L1⋅ C
As the permeability increases, and the transformer inductances scale with it, the transformer turns ratio
remains constant. Only the natural frequency decreases - and if we take the limit, the natural frequency
becomes zero, and the transfer function function becomes
3⋅ s
H ( s) =
C⋅ R1
R + 9⋅ R1 ⎞
2 ⎛ 2
s +⎜
⋅s
C⋅ R1⋅ R2
⎝
⎠
3
=
C⋅ R1
⎛ R2 + 9⋅ R1 ⎞
s+⎜
⎝ C⋅ R1⋅ R2 ⎠
which is exactly the transfer function we obtained in Question 1.
pole-zero cancellation
5
But what if the permeability is very large, but not infinite? This is more realistic, and it involves some
subtleties. I didn't expect you to get into this, but it's interesting. To see exactly how the finite
permeability transfer function morphs into the first order version in Question 1, rewrite the transfer
function again as
H ( s) =
3⋅ s
C⋅ R1
β=
where
2
s + β⋅s + ε
−1
1
2
⋅ β − ⋅ β − 4⋅ ε
2
2
= −β
approximately,
as ε becomes
small wrt
β2/4.
C⋅ R1⋅ R2
(the negative of the pole
location in Question 1)
1
2
ε = ωo =
L1⋅ C
and
It has poles
p1 =
R2 + 9⋅ R1
p2 =
=
−1
1
2
⋅ β + ⋅ β − 4⋅ ε
2
2
−β ⎛
⋅ 1−
2 ⎜
1 − 4⋅
−β ⎛ ε ⎞
−ε
⋅ ⎜ 2⋅
=
2
2
β
β
⎝
2
β ⎠
⎝
=
ε ⎞
⎠
approximately,
as ε becomes
small wrt
β2/4.
So the pole-zero diagram has one pole almost on the pole of Question 1, and the other creeps closer to
the origin as ε/β => 0. So we don't really get rid of the zero, even for huge, but finite, permeability.
Therefore the frequency response differs from that of the ideal transformer circuit of Question 1 only in
a range of very low frequencies, a range that decreases as ε/β => 0.