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In about ten signal processing texts I have see this thrown around as if it is self explanatory but I am not getting it. Can someone spell this out for me? Thanks!

-xerxes73

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In summary, a cosine wave in the frequency domain is represented by a 1/2 magnitude vector at the positive frequency and a 1/2 magnitude vector at the negative frequency.f

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In about ten signal processing texts I have see this thrown around as if it is self explanatory but I am not getting it. Can someone spell this out for me? Thanks!

-xerxes73

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Actually I meant, euler's inverse formula: cos x = e^ix / 2 + e^-ix / 2

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okay xerxes, it's a curious combination of what is evident that you know and then ostensibly what you don't know, judging from the question asked. I'm not sure what is the missing connection. hmmmmm.

maybe it's two concepts: 1. eigenfunctions and 2. basis functions. it turns out that for Linear, Time-Invariant (LTI) systems exponential functions make for nice eigenfunctions. and a combination of the discoveries of Fourier and Euler show that exponentials make good basis functions to decompose a general signal function into.

do you know about eigenfunctions and basis functions?

the fact is that [itex]e^{i \omega t}[/itex] and [itex]e^{-i \omega t}[/itex] make for better**basis functions** than just cos() or sin(). (we can add up a bunch of exponential functions with imaginary exponents to get any general function.)

and exponential functions are**eigenfunctions** for LTI systems. if [itex] e^{s t} [/itex] is input to an LTI system, then something times [itex] e^{s t} [/itex] is what comes out (that's what it means to be an eigenfunction). and that something is [itex]H(s)[/itex], where *H(s)* is called the *transfer function* of the LTI system and is a function of that complex exponent rate factor *s* in *e*^{st}.

if you put those two concepts together, when you have a single cosine function, you*might* think that is a single component, represented by a single frequency, but mathematically, they really are **two** basis components, with two different frequencies (that are negatives of each other).

maybe it's two concepts: 1. eigenfunctions and 2. basis functions. it turns out that for Linear, Time-Invariant (LTI) systems exponential functions make for nice eigenfunctions. and a combination of the discoveries of Fourier and Euler show that exponentials make good basis functions to decompose a general signal function into.

do you know about eigenfunctions and basis functions?

the fact is that [itex]e^{i \omega t}[/itex] and [itex]e^{-i \omega t}[/itex] make for better

and exponential functions are

if you put those two concepts together, when you have a single cosine function, you

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As a simple counterexample, if you assume e^(ix) = cos(x), for real x you would still produce complex numbers from the left side but the right side would be real, for real x inputted. Are you able to visualize the complex plane? cos(x) with respect to e^(ix) (or e^(-ix)) can be seen as a projection of those complex functions onto the real axis, and summing e^(ix) and e^(-ix) - which are conjugate functions - together can be seen to cancel out the imaginary parts of both functions and double the real part. If you were to plot both functions as vectors in the complex plane it might make the most sense.

Euler's identity is most easily seen from the expansion of e^(ix) and I believe the "inverse formulas" are working backwards from this. Since e^(ix)=cos(x)+i*sin(x) and e^(-ix)=cos(x)-i*sin(x) (once again you'd have to inspect the power series expansion to see this), e^(ix)+e^(-ix)=cos(x)+i*sin(x)+cos(x)-i*sin(x)=2*cos(x).

I hope I answered your question, but I'm with rbj on this one in that I don't know what you know and what you don't so I tried to keep the argument as simple as possible.

Euler's identity is most easily seen from the expansion of e^(ix) and I believe the "inverse formulas" are working backwards from this. Since e^(ix)=cos(x)+i*sin(x) and e^(-ix)=cos(x)-i*sin(x) (once again you'd have to inspect the power series expansion to see this), e^(ix)+e^(-ix)=cos(x)+i*sin(x)+cos(x)-i*sin(x)=2*cos(x).

I hope I answered your question, but I'm with rbj on this one in that I don't know what you know and what you don't so I tried to keep the argument as simple as possible.

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So...a cosine wave when shown in the frequency domain is represented by a 1/2 magnitude vector at the positive frequency and a 1/2 magnitude vector at the negative frequency. I am surprised by this because I cannot understand why a vector of magnitude 1 at the positive frequency of the cosine wave is wrong?

Keep in mind that the time-domain signal must be real-valued (i.e., no imaginary part).

A frequency spectrum of 1 at only the positive frequency w gives a time domain signal of

f(t) = exp(i w t)

which has an imaginary component [i sin(w t)], so that is wrong.

The frequency spectrum of 1/2 at both +w and -w gives the required real-valued time signal:

f(t) = (1/2)exp(i w t) + (1/2)exp(-i w t)

= cos(w t)

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-xerxes73

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Thanks for the patient explanations. I can't say I have intuitively grasped this yet but I am not finished playing with these new ideas provided by you all on basis, eigenfunctions, and vectors in the s plane.

i wouldn't call them "vectors" on the

there's even division with complex numbers, but i don't know of what division with

anyway, the wikipedia article on basis functions doesn't look so good. do you want the 5 minute spiel on that? then, after that, we could talk about eigenfunctions of LTI systems. it really is an important mathematical gift to signal processing engineers that exponential functions have the properties that they make for an elegant family of basis functions for either periodic functions or finite-energy aperiodic functions.

and exponentials are eigenfunctions of LTI operators. that is because:

1. when you add two exponentials with the same complex rate scaler

2. when you multiply an exponential by a constant, it's another exponential (with the same

3. when you delay and exponential, it's another exponential function (with the same

4. when you differentiate an exponential function, it's another exponential (with the same

and LTI systems are sort of the combination of the above 4 operations. so if

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So...a cosine wave when shown in the frequency domain is represented by a 1/2 magnitude vector at the positive frequency and a 1/2 magnitude vector at the negative frequency. I am surprised by this because I cannot understand why a vector of magnitude 1 at the positive frequency of the cosine wave is wrong?

Lots of good responses in this thread already, but perhaps you can still benefit from yet another point of view. The easiest way to see that a cosine can't be just a single (postive) frequency component is to note that then the spectrum would not be symmetric, and so the time-domain waveform would be complex-valued, which a cosine is not. The spectrum of a real-valued signal is always (conjugate) symmetric.

I understand euler's inverse formula that cos ix = e^ix / 2 + e^-ix / 2. When I look at this equation, I see a one positive and negative associated with 1/2 magnitudes but I cannot make the jump from e^ix and e^-ix locating these 1/2 magnitude vectors at the positive x frequency and -x frequency.

Well, there's a couple of problems here. First, Euler's formula is [itex] \cos (x) = \frac{1}{2}(e^{ix} + e^{-ix}) [/itex]. There is no imaginary component in the argument to the cosine. Second, simply applying Euler's formula doesn't get you to the frequency domain: this is still an expression for a time-domain waveform, so some further interpretation is required to see how this related to the spectrum. In particular, there are no components at a frequency of "x;" x is a time index. The frequency of [itex]\cos (x)[/itex] is 1 (on a normalized scale), and so that is where the components are. To make this more clear, consider writing Euler's formula with a more general argument to the cosine:

[tex]

\cos (2\pi f t) = \frac{1}{2}\left(e^{i2\pi f t} + e^{-i2\pi f t} \right)

[/tex]

Where [itex] f[/itex] is the frequency in Hertz and [itex]t[/itex] is the time variable. Perhaps this expression makes it more explicit that the two frequency components are at plus and minus [itex]f[/itex]?

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just a short comment. Please correct me if wrong.

I think that the cosine is represented by the two complex exponentials because we have chosen to represent it using the complex Fourier series or transform which are based on these complex exponentials as basis functions.

The complex Fourier series can synthesize both real and complex functions. In the case of real functions, the resulting spectrum is always symmetric (you need a basis and its conjugate to make real: Re(z)=(z+z*)/2 ).

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