# April 4

```ENSC 320 Problem of the Week, 4 April 2005
1. DC&amp;L, Chapter 19, Problem 16
(a) Calculate the y-parameters of the circuit above. Use circuit reduction or matrix
partioning, whichever is easier.
(b) The z-parameters of a circuit do the reverse of y-parameters - they give the voltages in
terms of currents.
⎛ z11
⎜
⎜ z21
⎝
z12 ⎞ ⎛ I1 ⎞ ⎛ V1 ⎞
⋅⎜
= ⎜
⎜
z22 ⎜ I2
⎠ ⎝ ⎠ ⎝ V2 ⎠
Having solved (a), find the z-parameters of this circuit.
2. DC&amp;L, Chapter 19, Problem 14
Consider the circuit above.
(a) Compute the short-circuit admittance parameters.
(b) If port 1 is short-circuited and V2 = 2 K / (s2+4), find I1(s) and I2(s).
Question 1
(a) Calculate the y-parameters of the circuit above. Use circuit reduction or matrix
partioning, whichever is easier.
Analyzing this one by circuit reduction methods, like series and parallel combination,
voltage and current dividers and the like, looks like a lot of work. So we will use good old
loop or node equations. But which one? There are five loops and only three nodes, so we'll
use node equations - and node equations are the natural choice for y-parameters,
anyway, since they give currents in terms of voltages.
At each node we calculate the sum of the currents leaving and equate it to any external
currents entering. Also, for each node, we write the sum of currents leaving as the node
voltage times the sum of admittances impinging minus each of the adjacent node voltages times
the admittance connecting it to the node in question. The result is
( 2 + 2⋅ s) ⋅ V1 − V2 − 2⋅ s⋅ V3 = I1
−V1 + ( 2 + s) ⋅ V2 − s⋅ V3 = I2
−2⋅ s⋅ V1 − s⋅ V2 + 5⋅ s⋅ V3 = 0
or
V
⎛⎜ 2 + 2⋅ s −1 −2⋅ s ⎞ ⎛⎜ 1 ⎞ ⎛⎜ I1 ⎞
⎜ −1 2 + s −s ⎟ ⋅ ⎜ V2 ⎟ = ⎜ I2 ⎟
⎜ −2⋅ s −s 5⋅ s ⎜
⎜
⎝
⎠ ⎝ V3 ⎠ ⎝ 0 ⎠
We could go through the matrix partioning formalism, but we have only one interior (and
therefore unneeded) node. So we'll just solve the third equation for V3 and substitute it into
the other two equations - and that is precisely what matrix partitioning would do, anyway.
Equation 3 gives
V3 =
2
1
⋅ V1 + ⋅ V2
5
5
Substituting it into the other two equations gives
2
1
⎞
( 2 + 2⋅ s) ⋅ V1 − V2 − 2⋅ s⋅ ⎛⎜ ⋅ V1 + ⋅ V2 = I1
5
⎝5
⎠
2
1
⎞
−V1 + ( 2 + s) ⋅ V2 − s⋅ ⎛⎜ ⋅ V1 + ⋅ V2 = I2
5
⎝5
⎠
After tidying, that's
⎛ 2 + 6 ⋅ s⎞ ⋅ V + ⎛ −1 − 2 ⋅ s⎞ ⋅ V = I
⎜
1 ⎜
1
5 ⎠ 2
⎝ 5 ⎠
⎝
⎛ −1 − 2 ⋅ s⎞ ⋅ V + ⎛ 2 + 4 ⋅ s⎞ ⋅ V = I
⎜
⎜
2
5 ⎠ 1 ⎝
5 ⎠ 2
⎝
which is in the desired y-parameter form
⎛⎜ 2 + 6 ⋅ s −1 − 2 ⋅ s ⎞
5
5 ⎟ ⎛⎜ V1 ⎞ ⎛⎜ I1 ⎞
⎜
⋅
=
⎜
⎜ I2
2
4 ⎟ ⎜ V2
⎝
⎠
⎝ ⎠
−
1
−
⋅
s
2
+
⋅
s
⎜
5
5
⎝
⎠
⎛ y11
⎜
⎜ y21
⎝
y12 ⎞ ⎛ V1 ⎞ ⎛ I1 ⎞
⋅⎜
= ⎜
⎜
y22 ⎜ V2
⎠ ⎝ ⎠ ⎝ I2 ⎠
As a check, note that Y is symmetric, which it should be, since there are no internal sources.
We could also have solved by circuit reduction. Just to get a sense of what it would be
like, let's get the first column of Y, that is, the admittances with the output short circuited. The
circuit then looks like this:
To get y11, the short circuit input admittance, calculate I1. First, do a parallel, then series
combination of the capacitors, giving 6/5 F, with an admittance 6s/5. Combine with the two
resistors (1 S each), to get .
y11 = 2 +
6
⋅s
5
To get y21, a short circuit transfer admittance, calculate I2. It's the negative of the sum of
currents flowing down through the 1 F capacitor and the 1 Ω resistor that was originally the
bridge over the T circuit. For V1 = 1, the total current through the caps is 6s/5, and the 1 F
cap's share is 1/3 of it (current divider), so I2 = -(2s/5 +1), so
y21 =
−2
⋅s − 1
5
Both y11 and y21 agree with our first derivation.
(b) Having solved part (a), find the z-parameters of this circuit.
If we were to do a ground-up solution for this one, we would probably use loop equations,
because loop equations are quite natural for z-parameters (they give voltages in terms of
currents). But there are five loops in this circuit! A lot of work, and definitely a job for
partitioned matrix solution. But wait a minute - from part (a) we have the y-parameters
⎛⎜ 2 + 6 ⋅ s −1 − 2 ⋅ s ⎞
5
5 ⎟ ⎛⎜ V1 ⎞ ⎛⎜ I1 ⎞
⎜
⋅
=
⎜
⎜ I2
2
4 ⎟ ⎜ V2
⎝
⎠
⎝ ⎠
−
1
−
⋅
s
2
+
⋅
s
⎜
5
5
⎝
⎠
⎛ y11
⎜
⎜ y21
⎝
y12 ⎞ ⎛ V1 ⎞ ⎛ I1 ⎞
⋅⎜
= ⎜
⎜
⎜
y22
⎠ ⎝ V2 ⎠ ⎝ I2 ⎠
To get the voltages from the currents, we just multiply the equation by the inverse of the matrix.
How hard can that be? A little work, and we get
1
⎛ 2
⎞
⎜ 2⋅ s + 3
2⋅ s + 3
⎛I ⎞ ⎛V ⎞
⎜
⎟⋅⎜ 1 = ⎜ 1
10 + 6⋅ s
⎜ 1
⎜
⎟ ⎜ I2
⎝ ⎠ ⎝ V2 ⎠
⎜ 2⋅ s + 3
2
15 + 16⋅ s + 4⋅ s ⎠
⎝
or
⎛ z11
⎜
⎜ z21
⎝
z12 ⎞ ⎛ I1 ⎞ ⎛ V1 ⎞
⋅⎜
= ⎜
⎜
z22 ⎜ I2
⎠ ⎝ ⎠ ⎝ V2 ⎠
That is, the matrix Z = Y-1. Caution - not all circuits have Z parameters, and not all
circuits have Y parameters, as we will see in class.
(cont'd next page)
Question 2
(a) Compute the short-circuit admittance parameters.
It looks as though the easiest way to get the admittance parameters is circuit reduction. First,
short circuit port 2 so we can get y11 and y21. The short, referred to the primary, is another
short, so
I1 = G1⋅ V1
==&gt;
y11 = G1
Now for the current flowing in port 2. Note that all of I1 flows through the primary, since it still
looks like a short. The dot on the secondary shows that the induced voltage V2 in response to
increases in primary current is negative, which in turn causes I2 to be positive. Then
y11
1
I2 = ⋅ I1 =
⋅ V1
a
a
==&gt;
y21 =
y11
a
=
G1
a
Now short circuit port 1, in order to get y12 and y22. Refer the primary to the secondary:
scale the primary impedances by a2 (admittances by 1/a2) and currents by 1/a. Then the
y22 =
G1 + G2
2
a
For the transfer admittance, we must calculate I1 in response to V2. We already have I2 (from
y22). A positive V2 causes I2 to be positive, but it's going into the undotted terminal. The dot
on the primary shows that its voltage induced by increasing I2 is nominally negative, from the
dotted to undotted terminal. This makes I1 positive. I1 also takes a current divider share of
the current. So
G1
G1
G1
G1 + G2
G1
I1 =
⋅ a⋅ I2 =
⋅ a⋅ y22⋅ V2 =
⋅ a⋅
⋅ V2 =
⋅ V2
2
G1 + G2
G1 + G2
G1 + G2
a
a
so that
y22 = G1
This part would have been easier if I had just referred the secondary to the primary! Oh well,
at least we have
G1 ⎞
⎛⎜
G1
a
⎜
⎟
or
Y= ⎜
G1 G1 + G2 ⎟
⎜
⎟
2
⎜ a
a
⎝
⎠
G1 ⎞
⎛⎜
G1
a
⎜
⎟ ⎛⎜ V1 ⎞ ⎛⎜ I1 ⎞
=
⎜ G G + G ⎟⋅⎜
⎜
V
1
1
2
2
⎜
⎟ ⎝ ⎠ ⎝ I2 ⎠
2
⎜ a
a
⎝
⎠
(b) If port 1 is short-circuited and V2 = 2 K / ( s2+4), find I1(s) and I2(s).
If port 1 is short-circuited, then V1=0. Using the matrix Y, we see
G1 2⋅ K
I1 ( s) =
⋅
a s2 + 4
I2 ( s) =
G1 + G2 2⋅ K
⋅
2
2
s +4
a
```