FM Signal Complex Envelope: Understanding & Resolving Integral

  • Thread starter Thread starter FrankJ777
  • Start date Start date
  • Tags Tags
    Complex Fm
Click For Summary
SUMMARY

The discussion centers on the complex envelope of a Frequency Modulated (FM) signal, specifically the integral representation of the phase θ(t) derived from the modulating signal m(t) = Am cos(ωt). Participants clarify that the integral can be evaluated from either -∞ or 0, depending on whether the signal is causal. The standard formula for the phase is given as φ(t) = 2πk_f ∫0t m(τ) dτ, which simplifies the evaluation of the integral. The conversation emphasizes the importance of understanding the nature of the signal when determining the limits of integration.

PREREQUISITES
  • Understanding of Frequency Modulation (FM) principles
  • Familiarity with complex envelopes in signal processing
  • Knowledge of improper integrals and their evaluation
  • Basic concepts of causal and non-causal signals
NEXT STEPS
  • Study the derivation of the complex envelope for various modulation techniques
  • Learn about the properties of causal signals and their implications in signal processing
  • Explore the differences between phase modulation and frequency modulation
  • Investigate the role of the unit step function in signal definitions
USEFUL FOR

Electrical engineers, signal processing students, and researchers interested in understanding the mathematical foundations of Frequency Modulation and its applications in communications.

FrankJ777
Messages
140
Reaction score
6
Hey guys, I'm trying to understand the complex envelope of a Frequency Modulated signal. According to the Textbook by Leon Couch, the complex envelope of an FM signal, g(t) modulated by the signal m(t) is given by:

g(t) = Ae^{jθ(t)}

where

A = |m(t)|

and

θ(t) = D_{f} \int^{t}_{-∞} m(σ) dσ

Now if i want to determine the angle θ(t) of the complex envelope of the FM signal modulated by m(t) = Am cos (ωt)

i get:
θ(t) = D_{f} \int^{t}_{-∞} Am cos (ωσ) dσ

= D_{f} Am \frac{sin(ωσ)}{ω}|^{t}_{-∞}

= D_{f} Am ( \frac{sin(ωt)}{ω} - \frac{sin(ω(-∞)}{ω} ) This is where i get stuck. I'm not sure what to do with the -∞.

I though I knew how to deal with improper integrals, but I'm not sure where to go from here. Also I'm not sure why the lower limit of integration should be -∞, it seems to me like it should be zero? Any one know how I can resolve that integral? I would like to really understand this concept, but this part has me stuck.

Thanks a lot.
 
Engineering news on Phys.org
It should be 0 to t
 
Here is the equation for the FM complex envelope given by a tutorial from the Illinois Institute of Technology, also my textbook by Leon Couch uses an integral with the lower limit at negitive infinty. I did notice however that there are some tutorials that use 0 as the lower limit. Anyways, I'm not sure what to do with this.

99o0eo.jpg


Here's a link to the whole tutorial.

http://www.ece.iit.edu/~biitcomm/research/references/Other/Tutorials%20in%20Communications%20Engineering/Tutorial%2017%20-%20Frequency%20Modulation%20(FM)%20,%20FSK,%20MSK%20and%20more.pdf
 
Read the wikipedia page for fm modulation. Also, try to understand phase modulation vs frequency modulation.

I'll take a stab at it, but I'm not an expert. Since, in the above equation, the intregal represents the phase, you must integrate the frequency since the beginning of time to determine the phase. Since you are dealing with a sinewave and it repeats you can assume an arbitrary starting time. I'm not sure about the difference between starting at 0 or pi/4 or any other initial time. But I think you just exploit the periodic nature of the signal.
 
I read the whole tutorial: it tries to keep things simple, which is a good (since it is just an introductory tutorial), but I think this may lead to some inconsistencies if you aren't careful. However, I noted a slightly abused notation; thus, let me restate things a bit.

Given the following signal: s(t) = A_c\cos(2\pi f_c t + \varphi(t)), you can define the so-called instantaneous phase \theta(t) \triangleq 2\pi f_c t + \varphi(t), and the so-called instantaneous frequency f_i(t) = \frac{1}{2\pi} \frac{d}{dt} \theta(t). We know that in FM: k_fm(t)=\frac{1}{2\pi} \frac{d}{dt} \varphi(t). Now, if you try to solve this equation, giving an explicit expression of \varphi(t), you end up with the integral you wrote. Usually, that integral goes from t_0 (which can be set equal to zero) to a generic t. However, it isn't technically incorrect to say that it goes from -\infty to t if you let m(t) be a causal signal, i.e. there must exist t_0 (which, again, can be zero) such as that m(t)=0 \ \ for \ \ t<t_0. More precisely, the generic expression is: \varphi(t) = 2\pi k_f \int^t_{t_0} m(\tau) d\tau + \varphi_0 where \varphi_0 \triangleq \varphi(t_0) is the initial phase (which is irrelevant and can be neglected). By choosing t_0=0 and \varphi_0=0, you obtain:
\varphi(t) = 2\pi k_f \int^t_0 m(\tau) d\tau and this is the standard formula. If you (re)do all the computations with this formula, you can "forget" the fact that m(t) must be causal, since the integration domain is now a finite time interval. However, if you want to keep using your notation, then you must write: m(t)=A_m cos(\omega t) u(t), where u(t) is the unit step function (zero for t<0, 1 otherwise), where I assumed (for simplicity) t_0=0. Hope this helps.
 
Nice explanation.
 
Thanks for your guy's explanations. If i can sum up what you said...
When we consider signals, m(t) that begin at time t=0 or t> 0, which are realizable signals, we evaluate the integral from the lower limit of 0?
That seems to make sense to me. Hope I understand it correctly.
Thanks.
 
So the integral with the lower limit of neg infinity, is a more general expression, that could include non causal signals?
 
Last edited:
FrankJ777 said:
So the integral with the lower limit of neg infinity, is a more general expression, that could include non casual signals?
You mean "causal", right? Yes, it may be thought as such (theoretically).

meBigGuy said:
Nice explanation.
Thanks!
 
  • #10
Oops! Causul. Just edited it!
Thanks.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
5
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K