Signal strength of a wave packet

In summary, the conversation discusses a wave packet with contributions from various frequencies, represented by g(ω) = C for |ω|<ω0 and g(ω) = 0 for elsewhere. The signal strength as a function of time, V(t), is determined using the equation V(t) = 2C/√(2π) * sin(ω0t)/ω0t. Sketches of g(ω) and V(t) are provided and the values of Δω and Δt are calculated using the standard deviation formula. It is found that the product of Δω and Δt satisfies ΔωΔt > 1/2.
  • #1
Elvis 123456789
158
6

Homework Statement


Assume a wave packet is has contributions from various frequencies, give by g(ω)=C for |ω|<ω0, and g(ω) =0 for elsewhere.

a)What is the signal strength as a function of time, i.e., V(t)=?

b) Sketch g(ω) and V(t); You can use fooplots.com, for example, or python.

c) Indicate Δω and Δt in the above plots; Does the products of these two satisfy ΔωΔt>1/2?

Homework Equations


V(t) = 1/√(2π) * integral from -∞ to +∞ of [ g(ω)*exp(iωt)] dω

exp(iωt) = cos(ωt) + isin(ωt)

The Attempt at a Solution



a.) V(t) = 1/√(2π) * integral from -ω0 to +ω0 of [ C*exp(iωt)] dω

using eulers formula and the properties of even and odd functions

V(t) = 2/√(2π) * integral from 0 to +ω0 of [ C*cos(ωt)] dω

V(t) = 2C/√(2π) * sin(ω0t)/t = √(2/π)*ω0C * sin(ω0t)/ω0t

b.) the sketches for g(ω) and V(t) are in the attachments

c.) I am not really sure what Δω and Δt in these graphs
 

Attachments

  • phy graph.jpg
    phy graph.jpg
    29.4 KB · Views: 463
  • phy graph 2.jpg
    phy graph 2.jpg
    27.4 KB · Views: 454
  • signal strength vs time.png
    signal strength vs time.png
    11.1 KB · Views: 549
Physics news on Phys.org
  • #2
Elvis 123456789 said:
V(t) = 2/√(2π) * integral from 0 to +ω0 of [ C*cos(ωt)] dω
Shouldn't the lower integral limit be ##-\omega_0##?
Elvis 123456789 said:
c.) I am not really sure what Δω and Δt in these graphs
Use the definition of standard deviation. For example for V(t), you will use
$$
\Delta t = \sqrt{E[t^2]-(E[t])^2}
$$
where ##E[\,\,]## means taking average over the intensity ##|V(t)|^2##. Similar arguments for ##g(\omega)##.
 
  • #3
blue_leaf77 said:
Shouldn't the lower integral limit be −ω0−ω0-\omega_0?
I have a factor of 2 in the front to account for that. Is that not right?
 
  • #4
Elvis 123456789 said:
I have a factor of 2 in the front to account for that. Is that not right?
Ah sorry I missed that, you are right.
 

FAQ: Signal strength of a wave packet

What is a wave packet?

A wave packet is a short burst of energy that travels through a medium as a wave. It is a combination of many different frequencies, or wavelengths, that work together to create a single pulse of energy.

How is the signal strength of a wave packet measured?

The signal strength of a wave packet is measured by the amplitude of the wave. Amplitude refers to the maximum displacement of a wave from its resting position. The greater the amplitude, the stronger the signal.

What factors affect the signal strength of a wave packet?

The signal strength of a wave packet can be affected by several factors, such as the distance the wave has traveled, the medium it is traveling through, and any obstacles or interference it may encounter along its path.

How does the signal strength of a wave packet relate to its frequency?

The frequency of a wave packet does not directly affect its signal strength. However, the frequency does determine the energy of the wave, which can indirectly impact the signal strength. Higher frequency waves tend to have more energy and therefore stronger signals.

What is the role of the receiver in determining the signal strength of a wave packet?

The receiver plays a crucial role in determining the signal strength of a wave packet. The receiver must be able to detect and interpret the wave packet in order to measure its signal strength accurately. Factors such as the sensitivity and capabilities of the receiver can also affect the measurement of signal strength.

Similar threads

Replies
2
Views
2K
Replies
6
Views
1K
Replies
5
Views
2K
Replies
21
Views
3K
Replies
10
Views
5K
Replies
3
Views
1K
Back
Top