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1164 ⏐⏐⏐ SYSTEM ANALYSIS: AN INTRODUCTION<br />

I i1<br />

+<br />

E i1<br />

–<br />

A v1 = 0.986<br />

A i 1 = –27.397<br />

E o1 = E i2<br />

Solutions:<br />

RL Zi2 1.8 k�<br />

a. Av1 � AvNL1� � AvNL1� � (1) ��<br />

RL � Ro Zi2 � Ro1 1.8 k��25 �<br />

� 0.986<br />

1.2 k�<br />

� � (�600) �� � �17.476<br />

Zi3 � Ro2 1.2 k��40 k�<br />

RL 3.3 k�<br />

� � (�1200) �� � �74.296<br />

RL � Ro3 3.3 k��50 k�<br />

Ri Zi1 50 k�<br />

� ��AvNL1� ��(1) ��<br />

RL � Ro Zi2 � Ro1 1.8 k��25 �<br />

� �27.397<br />

A v2 � A vNL2<br />

A v3 � A vNL3<br />

A i1 ��A vNL<br />

A i2 ��A vNL2<br />

A i3 ��A vNL3<br />

Z i3<br />

Note Fig. 26.21.<br />

A v2 = –17.476<br />

A i 2 = 26.214<br />

Z i2<br />

1.8 k�<br />

� ��(�600) �� � 26.214<br />

Zi3 � Ro2 1.2 k��40 k�<br />

Zi3 1.2 k�<br />

� ��(�1200) �� � 27.017<br />

RL � Ro3 3.3 k��50 k�<br />

E o2 = E i3<br />

FIG. 26.21<br />

Solution to Example 26.6.<br />

A v3 = –74.296<br />

A i 3 = 27.017<br />

Eo3 b. AvT ��� Av1 ⋅ Av2 ⋅ Av3 � (0.986)(�17.476)(�74.296)<br />

Ei1<br />

� 1280.22<br />

Io3 �<br />

Ii1<br />

A iT � �A i1 ⋅ A i2 ⋅ A i3 � (�27.397)(26.214)(27.017)<br />

I o<br />

+<br />

E o3<br />

–<br />

R L = 3.3 k�<br />

� �19,403.20<br />

c. AGT ��AvT ⋅ AiT ��(1280.22)(�19,403.20) � 24.84 � 10 6<br />

d. A′ v2 � A v1 ⋅ A v2 � (0.986)(�17.476) � �17.231<br />

A′ i2 � Ai1 ⋅ Ai2 � (�27.397)(26.214) � �718.185<br />

Zi Zi1 50 k�<br />

e. A′ i2 � Av� � A′ v2� � (�17.231) �<br />

RL Zi 1.2 k�<br />

3<br />

� �717.958 versus �718.185<br />

with the difference due to the level of accuracy carried through the<br />

calculations.<br />

f. A′ G2 � A′ v2 ⋅ A′ i2 � (�17.231)(�718.185) � 12,375.05<br />

g. A′ G2 � A 2 v � (A′ v2 ) 2 � (�17.231) 2<br />

Ri Ri1 � �<br />

RL Zi3<br />

50 k�<br />

� � 12,371.14<br />

1.2 k�

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