Solution:

← Back to Fundamentals of Electric Circuits Overview Making I2 the subject of Eq. (19.37b),

$$\mathbf{I}_2 = -\frac{\mathbf{z}_{21}}{\mathbf{z}_{22}}\mathbf{I}_1 + \frac{1}{\mathbf{z}_{22}}\mathbf{V}_2$$

(19.38)

Substituting this into Eq. (19.37a),

Eq. (19.3/8),
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$$\mathbf{V}_1 = \frac{\mathbf{z}_{11}\mathbf{z}_{22} - \mathbf{z}_{12}\mathbf{z}_{21}}{\mathbf{z}_{22}}\mathbf{I}_1 + \frac{\mathbf{z}_{12}}{\mathbf{z}_{22}}\mathbf{V}_2$$

\n(19.39)

Putting Eqs. (19.38) and (19.39) in matrix form,

$$\begin{bmatrix}\nV_1 \\ I_2\n\end{bmatrix} = \begin{bmatrix}\n\frac{\Delta_z}{\mathbf{z}_{22}} & \frac{\mathbf{z}_{12}}{\mathbf{z}_{22}} \\ -\frac{\mathbf{z}_{21}}{\mathbf{z}_{22}} & \frac{1}{\mathbf{z}_{22}}\n\end{bmatrix} \begin{bmatrix}\nI_1 \\ \overline{V}_2\n\end{bmatrix}$$

\n(19.40)\n
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From Eq. (19.15),

$$\begin{bmatrix} \mathbf{V}_1 \\ \mathbf{I}_2 \end{bmatrix} = \begin{bmatrix} \mathbf{h}_{11} & \mathbf{h}_{12} \\ \mathbf{h}_{21} & \mathbf{h}_{22} \end{bmatrix} \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{V}_2 \end{bmatrix}$$

Comparing this with Eq. (19.40), we obtain

$$\mathbf{h}_{11} = \frac{\Delta_z}{\mathbf{z}_{22}}, \qquad \mathbf{h}_{12} = \frac{\mathbf{z}_{12}}{\mathbf{z}_{22}}, \qquad \mathbf{h}_{21} = -\frac{\mathbf{z}_{21}}{\mathbf{z}_{22}}, \qquad \mathbf{h}_{22} = \frac{1}{\mathbf{z}_{22}} \quad (19.41)$$

Table 19.1 provides the conversion formulas for the six sets of twoport parameters. Given one set of parameters, Table 19.1 can be used to find other parameters. For example, given the T parameters, we find the corresponding h parameters in the fifth column of the third ro w. Also,

TABLE 19.1

Conversion of two-port parameters.

		z		y	h		g		T		t
z	z11	z12	y22 ___ ∆y	y12 − ___ ∆y	∆h ___ h22	h12 ___ h22	___1 g11	g12 − ___ g11	A __ C	∆T ___ C	d __ c	__1 c
	z21	z22	y21 − ___ ∆y	y11 ___ ∆y	h21 − ___ h22	___1 h22	g21 ___ g11	∆g ___ g11	__1 C	D __ C	∆t __ c	__a c
y	z22 ___ ∆z	z12 − ___ ∆z	y11	y12	___1 h11	h12 − ___ h11	∆g ___ g22	g12 ___ g22	D __ B	∆T − ___ B	__a b	− __1 b
	z21 − ___ ∆z	z11 ___ ∆z	y21	y22	h21 ___ h11	∆h ___ h11	g21 − ___ g22	___1 g22	− __1 B	A __ B	∆t − __ b	d __ b
h	∆z ___ z22	z12 ___ z22	___1 y11	y12 − ___ y11	h11	h12	g22 ___ ∆g	g12 − ___ ∆g	B __ D	∆T ___ D	b __ a	__1 a
	z21 − ___ z22	___1 z22	y21 ___ y11	∆y ___ y11	h21	h22	g21 − ___ ∆g	g11 ___ ∆g	− __1 D	C __ D	∆t __ a	__c a
g	___1 z11	z12 − ___ z11	∆y ___ y22	y12 ___ y22	h22 ___ ∆h	h12 − ___ ∆h	g11	g12	C __ A	∆T − ___ A	__c d	− __1 d
	z21 ___ z11	∆z ___ z11	y21 − ___ y22	___1 y22	h21 − ___ ∆h	h11 ___ ∆h	g21	g22	__1 A	B __ A	∆t __ d	b − __ d
T	z11 ___ z21	∆z ___ z21	y22 − ___ y21	− ___1 y21	∆h − ___ h21	h11 − ___ h21	___1 g21	g22 ___ g21	A	B	__d ∆t	__b ∆t
	___1 z21	z22 ___ z21	− ∆y ___ y21	y11 − ___ y21	h22 − ___ h21	− ___1 h21	g11 ___ g21	∆g ___ g21	C	D	__c ∆t	__a ∆t
t	z22 ___ z12	∆z ___ z12	y11 − ___ y12	− ___1 y12	___1 h12	h11 ___ h12	− ∆g ___ g12	g22 − ___ g12	___ D ∆T	___B ∆T	a	b
	___1 z12	z11 ___ z12	− ∆y ___ y12	y22 − ___ y12	h22 ___ h12	∆h ___ h12	g11 − ___ g12	− ___1 g12	___ C ∆T	___ A ∆T	c	d

∆z = z11z²² − z12z21, ∆h = h11h²² − h12h21, **∆**T = AD − BC

∆y = y11y²² − y12y21, ∆g = g11g²² − g12g21, **∆**t = ad − bc

given that z21 = z12 for a reciprocal netw ork, we can use the table to express this condition in terms of other parameters. It can also be shown that

$$[g] = [h]^{-1}$$

(19.42)

but

$$[t] \neq [T]^{-1}$$

(19.43)

Example 19.10 Find [z] and [g] of a two-port network if

$$[\mathbf{T}] = \begin{bmatrix} 10 & 1.5 \ \Omega \\ 2 \ \mathrm{S} & 4 \end{bmatrix}$$

Solution:

If A = 10, B = 1.5, C = 2, D = 4, the determinant of the matrix is

$$\Delta_T = \mathbf{AD} - \mathbf{BC} = 40 - 3 = 37$$

From Table 19.1,

$$\mathbf{z}_{11} = \frac{\mathbf{A}}{\mathbf{C}} = \frac{10}{2} = 5, \qquad \mathbf{z}_{12} = \frac{\Delta_T}{\mathbf{C}} = \frac{37}{2} = 18.5$$

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$$\mathbf{z}_{21} = \frac{1}{\mathbf{C}} = \frac{1}{2} = 0.5, \qquad \mathbf{z}_{22} = \frac{\mathbf{D}}{\mathbf{C}} = \frac{4}{2} = 2$$

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$$\mathbf{g}_{11} = \frac{\mathbf{C}}{\mathbf{A}} = \frac{2}{10} = 0.2, \qquad \mathbf{g}_{12} = -\frac{\Delta_T}{\mathbf{A}} = -\frac{37}{10} = -3.7$$

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$$\mathbf{g}_{21} = \frac{1}{\mathbf{A}} = \frac{1}{10} = 0.1, \qquad \mathbf{g}_{22} = \frac{\mathbf{B}}{\mathbf{A}} = \frac{1.5}{10} = 0.15$$

Thus,

[z] = [ 5 0.5 18.5 ² ] ^Ω, [g] = [ 0.2 S 0.1 −3.7 0.15 Ω]

Practice Problem 19.10 Determine [y] and [T] of a two-port network whose z parameters are

$$\begin{aligned} \n\left[\mathbf{z}\right] &= \begin{bmatrix} 6 & 4 \\ 4 & 6 \end{bmatrix} \Omega\\ \n\text{Answer: } \n\left[\mathbf{y}\right] &= \begin{bmatrix} 0.3 & -0.2 \\ -0.2 & 0.3 \end{bmatrix} \text{S}, \quad \n\left[\mathbf{T}\right] = \begin{bmatrix} 1.5 & 5 \Omega \\ 0.25 \text{ S} & 1.5 \end{bmatrix}. \n\end{aligned}$$

Example 19.11 Obtain the y parameters of the op amp circuit in Fig. 19.37. Show that the circuit has no z parameters.

Solution:

Because no current can enter the input terminals of the op amp, I1 = 0, which can be expressed in terms of V1 and V2 as

$$I_1 = 0V_1 + 0V_2 \tag{19.11.1}$$

Comparing this with Eq. (19.8) gives

$$\mathbf{y}_{11} = 0 = \mathbf{y}_{12}$$

Also,

$$\mathbf{V}_2 = R_3 \mathbf{I}_2 + \mathbf{I}_o (R_1 + R_2)$$

where Io is the current through R1 and R2. But Io = V1∕R1. Hence,

$$\mathbf{V}_2 = R_3 \mathbf{I}_2 + \frac{\mathbf{V}_1 (R_1 + R_2)}{R_1}$$

which can be written as

$$\mathbf{I}_2 = -\frac{(R_1 + R_2)}{R_1 R_3} \mathbf{V}_1 + \frac{\mathbf{V}_2}{R_3}$$

Comparing this with Eq. (19.8) shows that

$$\mathbf{y}_{21} = -\frac{(R_1 + R_2)}{R_1 R_3}, \quad \mathbf{y}_{22} = \frac{1}{R_3}$$

The determinant of the [y] matrix is

$$\Delta_{y} = \mathbf{y}_{11}\mathbf{y}_{22} - \mathbf{y}_{12}\mathbf{y}_{21} = 0$$

Since ∆y = 0, the [y] matrix has no inverse; therefore, the [z] matrix does not exist according to Eq. (19.34). Note that the circuit is not reciprocal because of the active element.

Find the z parameters of the op amp circuit in Fig. 19.38. Sho w that the Practice Problem 19.11 circuit has no y parameters.

Answer: [ ^z] = [ R1 −R² 0 0] . Because [ z] −1 does not e xist, [ y] does not exist.

Figure 19.38 For Practice Prob. 19.11.