Envelope function for two sine curves

In summary, the author is trying to find an envelope function for two sinusoidal curves (A, B) added up. Case-1 (using an envelope function) results in a nice envelope. Case-2 (using a envelope function) results in an envelope that can be represented by a set of sinusoids. The author is looking for a way to combine the two conditions so that they can have one generalized envelope function that does not miss the peaks.
  • #1
kmsiva
2
0

Homework Statement



Hi,[/B]
I am trying to get an envelope function for two sinusoidal curves (A, B) added up.
a1,a2 are the amplitudes(metre), T1,T2 are the periods (hrs), B lags by dt from A at t=0.
case-1:
a1= 1, a2=0.5, p1=11,p2=10,dt=0
w1=2*pi/T1,w2=2*pi/T2
A = a1*cos(w1*t), B=a2*cos(w2*t+dt)
t varies from 0 to 650 hours

Homework Equations


Case-1[/B]

using a envelope function sqrt(a1^2 + a2^2+2*a1*a2*(cos(w2-w1)*t+dt))
i am getting a nice envelope as in the figure 1

env12.jpg

Case-2:

a1= 1, a2=0.5, T1=60, T2=10, dt=0
using a envelope function

a2 – a1 +2*a1*a2*(cos(w1)*t+dt))
I can manage an envelope as in the figure 2

The Attempt at a Solution



Is there a way to combine these two conditions ( or any combination of T1 and T2)
so that I can have one generalized envelope function that does not miss the peaks.


Thanks in anticipation for any help
 
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  • #2
Just a reminder of a trigonometric identity: [itex]\cos(x)+\cos(y)= 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2}) [/itex].
 
  • #3
Thanks Svein,
Could you please elaborate little more.
From what ever I have tried out, it works only for a range and not for all combinations of T1 and T2.
I want to identify the location of extreme amplitude analytically from added up sine curves.
I believe the envelope could be represented by a set of sinusoids

sorry if it is something too simple and i am getting lost.
 
  • #4
OK. We start out with [itex]A_{1}\cos(\omega_{1} t)+A_{2}\cos(\omega_{2}t+\phi) [/itex]. Assume A1>A2. Then we can transform it into [itex](A_{1}-A_{2})\cos(\omega_{1} t)+A_{2}(\cos(\omega_{1} t)+\cos(\omega_{2}t+\phi))=(A_{1}-A_{2})\cos(\omega_{1} t)+2A_{2}\cos(\frac{\omega_{1}+\omega_{2}+\phi}{2})\cos(\frac{\omega_{1}-\omega_{2}-\phi}{2}) [/itex].
In radio technology this shows us that adding an audio signal (ω2) to a carrier wave (ω1) gives us a carrier wave plus two side bands [itex] \frac{\omega_{1}+\omega_{2}+\phi}{2}[/itex] and [itex] \frac{\omega_{1}-\omega_{2}-\phi}{2}[/itex]. You can see this best when ω1>>ω2. In your case (when ω1≈ω2), the result is harder to see.
 
  • #5
I don't know how you arrived at the envelope function you posted, so I may be just telling you what you already know here.
First, lose the phase constant. If the two frequencies are different then the phase constant is equivalent to a shift along the time axis, so is not interesting.
I would then apply the identity Svein mentioned, but in a more symmetric manner.

Get F(t) into the form ##C\sin(\bar \omega t)\cos(\hat \omega t)+D\cos(\bar \omega t)\sin(\hat \omega t)## where ##\bar\omega = \frac{\omega_1+\omega_2}2## and ##\hat\omega = \frac{\omega_1-\omega_2}2##. Clearly the ##\bar \omega## factors are the higher frequency of the two, so we want to find the maximum of F over a cycle of that.

If this frequency is much higher than ##\hat \omega##, we can treat the ##\hat \omega## as constant over such a cycle and find the max of F on that basis. This leads to the equation you posted.

If the original frequencies are very different then ##\hat \omega## and ##\bar \omega## will be of similar magnitude, so this method does not work. Instead, you can treat the lower frequency function as roughly constant over a cycle of the higher frequency and arrive at your other equation.

So each solution is just an approximation that works well in its own domain. The worst case, I guess, would be where the ratio of the original frequencies is the same as the ratio of the sum and difference frequencies: ##\omega_1:\omega_2=\bar\omega:\hat\omega##. Try plotting one of those to see if it gives any insight.
 

Related to Envelope function for two sine curves

1. What is an envelope function?

An envelope function is a mathematical tool used to describe the amplitude variations of a signal or waveform. It is often represented as a curve that outlines the maximum and minimum values of the signal over time.

2. How is an envelope function used in the context of two sine curves?

In the case of two sine curves, the envelope function represents the upper and lower boundaries of the amplitude of the combined signal. It can be used to visualize the amplitude modulation of the two sine curves and to determine the maximum and minimum values of the combined signal.

3. What is the mathematical equation for an envelope function for two sine curves?

The equation for an envelope function for two sine curves is given by: A(t) = (A1 + A2) ± (A1 - A2) * cos(ω1 - ω2)t, where A1 and A2 are the amplitudes of the two sine curves, ω1 and ω2 are their frequencies, and t is the time variable.

4. What is the significance of the envelope function in signal processing?

The envelope function is an important tool in signal processing as it allows us to analyze the amplitude variations of a signal and extract useful information. It is commonly used in tasks such as demodulation, filtering, and feature extraction.

5. Can the envelope function be used for more than two sine curves?

Yes, the envelope function can be extended to more than two sine curves. In this case, the equation becomes more complex, but the concept remains the same - to describe the amplitude variations of the combined signal. It can be particularly useful in the analysis of complex signals with multiple components.

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