Chapter 17_Two-Port Networks

# Chapter 17_Two-Port Networks - The Analysis and Design of...

• Homework Help
• 23

This preview shows pages 1–3. Sign up to view the full content.

The Analysis and Design of Linear Circuits Seventh Edition 17 Two-Port Networks 17.1 Exercise Solutions Exercise 17–1. Find the impedance parameters of the circuit in Figure 17–3. Let port 2 have an open circuit so that I 2 = 0. Solve for z 11 and z 21 . z 11 = V 1 I 1 vextendsingle vextendsingle vextendsingle I 2 =0 = 75 + 50 = 125 Ω z 21 = V 2 I 1 vextendsingle vextendsingle vextendsingle I 2 =0 = 75 I 1 I 1 = 75 Ω Let port 1 have an open circuit so that I 1 = 0. Solve for z 12 and z 22 . z 12 = V 1 I 2 vextendsingle vextendsingle vextendsingle I 1 =0 = 75 I 2 I 2 = 75 Ω z 22 = V 2 I 2 vextendsingle vextendsingle vextendsingle I 1 =0 = 100 + 75 = 175 Ω Exercise 17–2. The impedance parameters of a two-port network are z 11 = 25 Ω, z 12 = 50 Ω, z 21 = 75 Ω, and z 22 = 75 Ω. Find the port currents I 1 and I 2 when a 15-V voltage source is connected at port 1 and port 2 is short circuited. The i - v relationships are: V 1 = z 11 I 1 + z 12 I 2 = 25 I 1 + 50 I 2 V 2 = z 21 I 1 + z 22 I 2 = 75 I 1 + 75 I 2 With port 2 short circuited, we have V 2 = 0: 0 = 75 I 1 + 75 I 2 I 1 = I 2 Apply the voltage at port 1 and solve for the currents. 15 = 25 I 1 + 50 I 2 = 25 I 1 50 I 1 = 25 I 1 I 1 = 0 . 6 A I 2 = I 1 = 0 . 6 A Exercise 17–3. Find the admittance parameters of the circuit in Figure 17–5. Let port 2 be short circuited so that V 2 = 0. Solve for y 11 and y 21 . y 11 = I 1 V 1 vextendsingle vextendsingle vextendsingle V 2 =0 = V 1 /j 50 V 1 = j 20 mS y 21 = I 2 V 1 vextendsingle vextendsingle vextendsingle V 2 =0 = V 1 /j 50 V 1 = j 20 mS Let port 1 be short circuited so that V 1 = 0. Solve for y 12 and y 22 . y 12 = I 1 V 2 vextendsingle vextendsingle vextendsingle V 1 =0 = V 2 /j 50 V 2 = j 20 mS y 22 = I 2 V 2 vextendsingle vextendsingle vextendsingle V 1 =0 = 1 j 50 + 1 200 = 5 j 20 mS Solution Manual Chapter 17 Page 17-1

This preview has intentionally blurred sections. Sign up to view the full version.

The Analysis and Design of Linear Circuits Seventh Edition Exercise 17–4. The admittance parameters of a two-port network are y 11 = 20 mS, y 12 = 0, y 21 = 100 mS, and y 22 = 40 mS. Find the output voltage V 2 when a 5-V voltage source is connected at port 1 and port 2 is connected to a 100-Ω load resistor. The i - v relationships are: I 1 = y 11 V 1 + y 12 V 2 = 0 . 02 V 1 + (0) V 2 = 0 . 02 V 1 I 2 = y 21 V 1 + y 22 V 2 = 0 . 1 V 1 + 0 . 04 V 2 With a 100-Ω load at port two, we have V 2 = 100 I 2 . Substitute and solve. I 2 = (0 . 1)(5) + (0 . 04)( 100 I 2 ) = 0 . 5 4 I 2 I 2 = 100 mA V 2 = 100 I 2 = 10 V Exercise 17–5. Find the h -parameters of the circuit in Figure 17–8. Let port 2 be short circuited so that V 2 = 0. Write an expression for V 1 in terms of current I 1 . V 1 = 100000 I 1 + (1000)( I 1 + 50 I 1 ) = 151000 I 1 h 11 = V 1 I 1 vextendsingle vextendsingle vextendsingle V 2 =0 = 151 kΩ Solve for h 21 . h 21 = I 2 I 1 vextendsingle vextendsingle vextendsingle V 2 =0 = 50 I 1 I 1 = 50 Let port 1 have an open circuit so that I 1 = 0. We have I 2 = 50 I 1 = 0, so no current flows in the circuit, regardless of the value of V 2 . Solve for h 12 and h 22 .
This is the end of the preview. Sign up to access the rest of the document.
• Fall '15
• Scattering parameters, Two-port network, Admittance parameters, Impedance parameters, Design of Linear Circuits

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern