/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Reflections in transmission line... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Reflections in transmission lines. Suppose a coaxial transmission line having 50 ohms characteristic impedance is joined to one having 100 ohms characteristic impedance. (a) A voltage pulse of \(+10\) volts (maximum value) is incident from the \(50 \Omega\) line to the \(100 \Omega\) line. What is the "height" (in volts, inchding the sign) of the reflected pulse? Of the transmitted pulse? (b) A \(+10\) -volt pulse is incident from the \(100 \Omega\) to the \(50 \Omega\) line. What are the reflected and transmitted pulse heights?

Short Answer

Expert verified
50 Ω to 100 Ω: Reflected = +3.33 V, Transmitted = 6.67 V. 100 Ω to 50 Ω: Reflected = -3.33 V, Transmitted = 6.67 V.

Step by step solution

01

Calculate Reflection Coefficient for 50 Ω to 100 Ω

To find the reflection coefficient \( \Gamma \) when the pulse travels from the 50 Ω line to the 100 Ω line, use the formula: \[ \Gamma = \frac{Z_2 - Z_1}{Z_2 + Z_1} \]Substitute \( Z_1 = 50 \Omega \) and \( Z_2 = 100 \Omega \): \[ \Gamma = \frac{100 - 50}{100 + 50} = \frac{50}{150} = \frac{1}{3} \]
02

Calculate Reflected Pulse Height for 50 Ω to 100 Ω

The height of the reflected pulse \( V_r \) is given by:\[ V_r = \Gamma \cdot V_i \]where \( V_i = +10 \) volts and \( \Gamma = \frac{1}{3} \):\[ V_r = \frac{1}{3} \times 10 = +3.33 \text{ volts} \]
03

Calculate Transmitted Pulse Height for 50 Ω to 100 Ω

The transmitted pulse can be found by:\[ V_t = V_i + V_r \]Using \( V_i = +10 \) volts and \( V_r = +3.33 \) volts:\[ V_t = 10 + 3.33 = 6.67 \text{ volts} \]
04

Calculate Reflection Coefficient for 100 Ω to 50 Ω

For the pulse travelling from the 100 Ω line to the 50 Ω line, use the reflection coefficient:\[ \Gamma = \frac{Z_2 - Z_1}{Z_2 + Z_1} \]Substitute \( Z_1 = 100 \Omega \) and \( Z2 = 50 \Omega \):\[ \Gamma = \frac{50 - 100}{50 + 100} = \frac{-50}{150} = -\frac{1}{3} \]
05

Calculate Reflected Pulse Height for 100 Ω to 50 Ω

The reflected pulse height \( V_r \) is:\[ V_r = \Gamma \cdot V_i \]where \( V_i = +10 \) volts and \( \Gamma = -\frac{1}{3} \):\[ V_r = -\frac{1}{3} \times 10 = -3.33 \text{ volts} \]
06

Calculate Transmitted Pulse Height for 100 Ω to 50 Ω

The transmitted pulse height is given by:\[ V_t = V_i + V_r \]Using \( V_i = +10 \) volts and \( V_r = -3.33 \) volts:\[ V_t = 10 - 3.33 = 6.67 \text{ volts} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reflections in Transmission Lines
In transmission lines, understanding reflections is crucial for predicting signal behavior when impedance changes occur. Reflections happen when a signal encounters a different impedance, causing a part of the signal to bounce back towards the source.
This scenario can result in signal distortion, and knowing how to calculate and anticipate reflections aids in designing effective communication systems.

When dealing with transmission line reflections, the characteristics of both the source and load impedances are important.
  • An unmatched impedance causes a reflection.
  • The amount of reflection depends on the difference in impedances.
  • A correctly matched impedance results in no reflection, transferring all energy to the load.
Understanding and managing reflections helps in maintaining signal integrity and ensuring that communication systems remain efficient and effective.
Characteristic Impedance
Characteristic impedance is a key concept in transmission lines, representing the impedance that a transmission line's infinite length would have. It is an intrinsic property, critical for designing circuits that accurately transmit signals.

In the real world, matching the characteristic impedance of lines connecting different parts of a system minimizes reflections, ensuring that most of the signal reaches its intended destination. Here are some points to remember:
  • In practice, characteristic impedance refers to the ratio of voltage to current in a wave propagating along the line.
  • It is denoted by 'Z'.
  • Mismatch in impedance results in reflections.
    • Example: 50 Ω and 100 Ω transmission lines have different characteristic impedances.
    • This difference leads to reflected and transmitted pulses in the exercise example.
Central to understanding how energy propagates or is reflected in a system, characteristic impedance allows engineers to design efficient communication systems.
Reflection Coefficient
The reflection coefficient (\( \Gamma \)) quantifies the extent to which an incoming wave is reflected by an impedance discontinuity in the transmission line. It is calculated using the formula:\[ \Gamma = \frac{Z_2 - Z_1}{Z_2 + Z_1} \]
where \( Z_1 \) is the original line impedance and \( Z_2 \) is the new line impedance. This value ranges from -1 to 1, indicating the proportion of the reflected wave's magnitude relative to the incident wave.

A positive reflection coefficient indicates a reflection in the same phase as the incident wave, while a negative one indicates a phase reversal.
  • Examples from the exercise:
  • 50 Ω to 100 Ω gives \( \Gamma = \frac{1}{3} \).
  • 100 Ω to 50 Ω gives \( \Gamma = -\frac{1}{3} \).
That means the reflected wave can sometimes return inverted based on impedance transition, and the reflection coefficient helps quantify this challenge in maintaining signal clarity and integrity over transmission lines.
Pulse Transmission
Pulse transmission involves sending brief, discrete bursts of signal through a transmission line. These voltage or current pulses are used in communication systems to encode information. Keeping their shape and integrity is essential.

Pulses can be distorted or reflected at each point of impedance mismatch, altering the signal that reaches the receiving end. Therefore, understanding how to manage pulse transmission can be vital in reducing error rates and ensuring data accuracy in digital communications.

A few important points about pulse transmission:
  • Pulse shape directly affects signal clarity and data rate.
  • Managing impedance mismatches mitigates reflection issues.
  • Effective pulse transmission processes help maintain data integrity.
In the exercise, the transmission of voltage pulses between lines of differing impedances results in reflections that affect the transmitted pulse heights, as demonstrated by different reflected and transmitted voltages when moving from the 50 Ω to 100 Ω line and vice versa.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Overtones in tuning fork. Does your C523.3 tuning fork emit nothing but a 523 -cps sound? Strike the fork against something hard. You should hear a faint high tone in addition to the strong 523 -cps tone. The high tone dies away in two or three seconds. It is a higher mode of the fork and is strongly damped because it involves greater bending of the prongs. What about the note an octave higher, Cl048? This is difficult to listen for because of the presence of the fundamental, C523. To search for it, use a resonating tube. Tune a tube to \(\mathrm{C} 1046\) by tapping it on your head and listening for the octave above C523. (Or simply cut it by "theory," subtracting \(0.6\) of a radius \(R\) for each end from \(\lambda / 2\) to get the length.) Hold the \(\mathrm{C} 523\) fork at the end of the C1046 tube and listen. (Use a tube tuned at C523 as a control. Move the fork back and forth between the C523 and C1046 tubes.)

Termination of waves on a string. (a) Suppose you have a massless dashpot having two moving parts 1 and 2 that can move relative to one another along the \(x\) direction, which is transverse to the string direction \(z\). Friction is provided by a fluid that retards the relative motion of the two moving parts. The friction is such that the force needed to maintain relative velocity \(\dot{x}_{1}-\dot{x}_{2}\) between the two moving parts is \(Z_{d}\left(\dot{x}_{1}-\dot{x}_{2}\right)\), where \(Z_{i} d\) is the impedance of the dashpot. The input (part 1 ) is connected to the end of a string of impedance \(Z_{1}\) stretching from \(z=-\infty\) to \(z=0 .\) The output (part 2 ) is connected to a string of impedance \(Z_{2}\) that extends to \(z=+\infty\). Show that a wave incident from the left experiences an impedance at \(z=0\) which is the same as that it would experience if connected to a "load" consisting of a string stretching from \(z=0\) to \(+\infty\) and having impedance \(Z_{L}\) given by $$ \mathrm{Z}_{L}=\frac{\mathrm{Z}_{\mathrm{d}} \mathrm{Z}_{2}}{\mathrm{Z}_{d}+\mathrm{Z}_{2}}, \quad \text { that is, } \quad \frac{1}{\mathrm{Z}_{L}}=\frac{1}{\mathrm{Z}_{d}}+\frac{1}{\mathrm{Z}_{2}} \text { . } $$ Thus it is as if the dashpot and string 2 were impedances connected "in parallel" and driven by the incident wave. (b) Show that if string \(Z_{2}\) extends only to \(z=\frac{1}{4} \lambda_{2}\), where \(\lambda_{2}\) is the wavelength in medium 2 (assuming we have a harmonic wave with a single frequency), and there is terminated by a dashpot of zero impedance (frictionless), the wave incident at \(z=0\) is perfectly terminated. Show that the output connection of the dashpot at \(z=0\) cannot tell whether it is connected to a string of infinite impedance or is instead connected to a quarter-wavelength string that is "short-circuited" by a frictionless dashpot at \(z=\frac{1}{4} \lambda_{2 \cdot}\) In either ease the output connection remains at rest.

Resonances in toy balloons. Get a helium-filled balloon, Hold it near your ear and tap it. Sing into the side of it and search for resonant pitches. Blow up another balloon with air to the same diameter as the helium balloon. Tap it. Estimate the ratio of frequencies of the lowest modes (the ones you hear when you t?p) of the helium and air balloons. What frequency ratio would you predict: Compare the strength (loudness) of the resonances you get singing into the side of a helium balloon with those you get from the air balloon. Why is there such a difference?

Transitory standing waves on a slinky. Attach one end of a stinky to a telephone pole or something. Hold the other end. Stretch the slinky out to \(30 \mathrm{ft}\) or so. Shake the end of the slinky about 3 or 4 times as rapidly as you can. \(\mathrm{A}^{\text {" } \text { wave packet" }}\) " is thus propagated down the slinky. After you have sufficiently enjoyed following packets back and forth, try something new: This time, keep your attention fixed on a region near the fixed end of the slinky. As the packet comes in, reflects, and returns, you should see transitory standing waves during the time interval in which the incident and reflected wave packets overlap. (It may help to fix both ends of the slinky so that you can watch the process at close range at your end of the slinky.) That should help to convince you that a standing wave can always be regarded as the superposition of two traveling waves traveling in opposite directions.

Impedance matching by "tapered" index of refraction. Suppose you want to match optical impedances between a region of index \(n_{1}\) and a region of index \(n_{2}\), and you want to expend a total distance \(L\) in the impedance-matching transition region. What is the optimum \(z\) dependence of the index \(n\) between the two regions? Is it exponential? Why not?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.