What are negative frequencies?

[SherlockBenedict]

I know that the concept of negative frequencies seems to be absurd. I have been searching for the answer for quite a long time.

One answer that I mostly saw was that it was because of the Euler's theorem.

But now I have another reason-

Here's one real negative frequency as given in Proakis-

There are two waves one is 1/8 Hz and other one is -7/8 Hz.



I can even prove this mathematically.

Consider a sine wave y=sin wt.

Now consider an initial phase shift of Π

This means that

y= sin (wt + Π)= -sin wt = sin (-wt).

So clearly the frequency has become negative!

Now I believe that this is the real meaning of negative frequency.


What's your answer?

Awaiting for your reply

Thanks a lot.......

 

[iVenky]

I think both are one and the same. According to Proakis, he has considered that -ve sign. As in the example, if there was no phase shift the frequency would have been 7/8. We usually don't care about the initial phase shift (as this is not going to affect the frequency) and leave out negative sign and write 7/8 instead of -7/8. Instead what we do is that we have a mirror image in the negative frequency side.(which is in accordance to the Euler's theorem) I mean both 7/8 and -7/8 are one and the same (in terms of frequencies) though they have different initial phase shift.

I understand that just knowing the frequency is not enough. You have to know the initial phase shift too if you have to recover the data. This is made sure if you have sampling frequency more than twice the maximum frequency of the wave. Then you will recover the data. You can even find out the initial phase shift from that recovered one if you wish.:grin:

 

[permute]

I know that the concept of negative frequencies seems to be absurd. I have been searching for the answer for quite a long time.

One answer that I mostly saw was that it was because of the Euler's theorem.

But now I have another reason-

Here's one real negative frequency as given in Proakis-

There are two waves one is 1/8 Hz and other one is -7/8 Hz.



I can even prove this mathematically.

Consider a sine wave y=sin wt.

Now consider an initial phase shift of Π

This means that

y= sin (wt + Π)= -sin wt = sin (-wt).

So clearly the frequency has become negative!

Now I believe that this is the real meaning of negative frequency.


What's your answer?

Awaiting for your reply

Thanks a lot.......

the use of negative frequencies becomes more important when the input signal is complex. This comes up in communications systems where real data samples are multiplied by a complex exponential (sine and cosine). this shifts the spectrum and results in complex values. but the positive and negative frequencies can have different magnitudes and phases (and are no longer conjugate symmetric).

If you know the input signal is real, then there isn't a strong need to show the negative portion of the spectrum -- it will be conjugate symmetric. As a result, many applications that only use real valued signals will leave out the negative spectrum. However, if the input is complex-valued, then you must show the full spectrum, as there is information in both negative and positive frequencies.

 

[milind.a.kulkarni]

In a reality let say frequency = 1/ time and time is non-negative quantity then how come the frequency is negative? this is a traditional question comes in the mind of many signal processing persons at initial stage....so here is some explanation for that....First of all we we say negative frequency it is only in case of analysis...i.e. either the spectrum plotting or in the output of mathematical solution.....the point which I will like to make is it is not a physical quantity.... it is fictitious quantity.... now speaking to the negative frequency concept....as premute said in the explanation that the signal have very complex in nature....so in order to analysis them one apply the transform like Frequency transform like Fourier transform....now these transform changes the time domain signal to the frequency domain based on certain rules of mapping or transform...these rule of the transform or mapping introduces the concept of negative frequency .... like in case of one want to do the analysis of simple cosine sign signal using Fourier transform say then let consider the x(t) = a*cos(wc*t) where wc is say frequency of signal then what we know if we solve this for Fourier transform we get two answers on the w(omega ) axis one on positive side and second at negative side

refer -

https://mathworld.wolfram.com/FourierTransform.html

Good Luck

 

[D.A.(Tony)Stewart]

Consider the Spectral density meter, A.k.a. Spectrum Analyzer. The entire inoput bandwidth is up converted to an IF above the maximum instrument BW then down converted to 2nd IF 75MHz and then 3rd IF1@ 10.7MHz then log detected to baseband.

Here negative frequencies are shown on display, which as you say seems illogical but actually occur at the 1st IF . e,g, a 0~4GHz SA might have 1st IF at 4.1GHz or >8GHz.

Image rejection is a key aspect of the conjugate frequencies from intermodulation in receiver design. this includes + and - products.

In your case the -7/8 Hz is shown as an inverted phase to the +7/8 Hz and means the same thing.

 

[iVenky]

Consider the Spectral density meter, A.k.a. Spectrum Analyzer. The entire inoput bandwidth is up converted to an IF above the maximum instrument BW then down converted to 2nd IF 75MHz and then 3rd IF1@ 10.7MHz then log detected to baseband.

Here negative frequencies are shown on display, which as you say seems illogical but actually occur at the 1st IF . e,g, a 0~4GHz SA might have 1st IF at 4.1GHz or >8GHz.

Image rejection is a key aspect of the conjugate frequencies from intermodulation in receiver design. this includes + and - products.

In your case the -7/8 Hz is shown as an inverted phase to the +7/8 Hz and means the same thing.

I have a doubt here. If we assume that IF= fo - input frequency and also input frequency is lesser than fo then IF will be positive only. In case we have an image which would be at some frequency higher than fo we would be having negative IF. Is that what you are trying to say? But we usually don't pass image frequency into the mixer. This means that we should not have negative frequency in the spectrum analyzer output. Also the positive and negative frequency value should not be same.I have seen negative frequency in spectrum analyzer and both positive and negative frequency have the same value. I just couldn't understand how we get negative frequency in spectrum analyzer. Could you please help me?

Thanks in advance.

 

[Kichuguu]

Simple is this:

Frequency is defined as the number of cycles per second. The cycle can be a clockwise cycle or a counterclockwise cycle. By convention, a clockwise cycle results in a positive frequency and a counterclockwise cycle results in a negative frequency.

Most analytical tools, such as spectral analysis, do treat both counterclockwise and clockwise cycles equally, therefore they always take both +ve and -ve signs.

 

[D.A.(Tony)Stewart]

I have a doubt here. If we assume that IF= fo - input frequency and also input frequency is lesser than fo then IF will be positive only.
Thanks in advance.

SA's can easily sweep to zero Hz but not all log amps are stable to DC. WHen the When mixing signals whenever the LO -RF difference sweep is centered at the IF, one is symmetrically sweeeping negative & positive frequencies. But since this is redundant. For viewing baseband start to stop freq. [KHz/MHz/GHz] one normally choose zero Hz on the far left so is offset in its sweep such that it only looks at a single sideband. and does not sweep past the IF center. There is rarely any practical use for this, but thats how it can work. Normally SA's which cannot detect down to DC will indicate a large spike at 0Hz with no signal applied, because of the DC bias in the log amp is extremely temperature sensitive and requires accurate compensation well over 6 decades. But they do exist.

In either case , one can sweep past zero Hz on the left by offsetting the center sweep of start sweep to the left.


Note the dynamic range of this SA is 7 decades at widest BW spurious free.. Naturally narrower BW can detect sidebands 15 decades down from the carrier (in this example --150dBc @1GHz)

- wiki-def'n... "dBc (decibels relative to the carrier) is the power ratio of a signal to a carrier signal"

So again negative F amplitude = positive F amplitude + 180deg.

 

[FvM]

y= sin (wt + Π)= -sin wt = sin (-wt).

So clearly the frequency has become negative!

Now I believe that this is the real meaning of negative frequency.

The identity of negative frequency and negative magnitude only works for scalar signals. For "analytical" or complex signals, negative frequency has the meaning of a reverse rotation direction of the phase vector, which can be clearly distinguished from a negtative magnitude or a 180° phase shift.