Harmonics vs Max Frequency

Hej!

Isnt it the maximum frequency of a signal actually means(or equal to) the highest/largest harmonics of that signal?

Hello

Did you mean to ask fundamental frequency and harmonics? if so

fundamental frequency is the minimum resonant frequency or the first harmonic, then multiples of of the first resonances are harmonics.

Regards
ChachitoEL

Hej
When we need to know Sampling freq. value, which is, we know, equal to Fs= 2 * Max. Freq.
In such a case if we had known the value of highest harmonics, would the highest harmonics value equal to the Max freq. value for this problem?

max frequency in this case is the upperlimit of the band you want to sample.like for instance human voice is 20 hz to 20KHZ a sampling frequency of 40KHZ or more will result in perfect reconstruction of the original signal.

hope that helps

Regards
ChachitoEL

Yah, I got you now. You and few like people are good only at memorizing some book staffs and always deliver those without understanding well the question. Both of your answers are known to all people, you dont need to teach us those. if you dont really know the answer , then its good not to answer. I am wasting my time here too.

Well unfortunately,i am sorry that i was of not much help.But if there is a specific answer you are looking for,please let us know and enlighten me as well.so i can definitely learn from you,who knows things which are not in the books.

Regards
ChachitoEL.

sajib333 said:

Hej
When we need to know Sampling freq. value, which is, we know, equal to Fs= 2 * Max. Freq.
In such a case if we had known the value of highest harmonics, would the highest harmonics value equal to the Max freq. value for this problem?

Click to expand...

In simple terms, yes. In reality, it's not that easy. Most real signals can have harmonics reaching in to the MHz, and GHz frequency range (just ask anyone that's had to troubleshoot a SMPS that's failing EMI conducted emissions testing). The question for a real-world scenario is, how high of frequency do you have to sample/capture in order to replicate the desired signal with sufficient accuracy. In the case of a square wave, you'd need an infinite number of harmonics to make a perfectly square corner (see Fourier series expansion of a squarewave). However, you can obtain a pretty good looking squarewave with somewhere on the order of 15-20 harmonics (try this applet, adjust the number of terms using the slider at the bottom-right).

In practice, everything's a trade-off... sufficient performance vs. cost/time/complexity/size/weight/power/color/smell/etc. :wink:

enjunear said:

In simple terms, yes. In reality, it's not that easy. Most real signals can have harmonics reaching in to the MHz, and GHz frequency range (just ask anyone that's had to troubleshoot a SMPS that's failing EMI conducted emissions testing). The question for a real-world scenario is, how high of frequency do you have to sample/capture in order to replicate the desired signal with sufficient accuracy. In the case of a square wave, you'd need an infinite number of harmonics to make a perfectly square corner (see Fourier series expansion of a squarewave). However, you can obtain a pretty good looking squarewave with somewhere on the order of 15-20 harmonics (try this applet, adjust the number of terms using the slider at the bottom-right).

In practice, everything's a trade-off... sufficient performance vs. cost/time/complexity/size/weight/power/color/smell/etc. :wink:

Click to expand...

Thanks.....

Just one more thing: if you take less harmonics than the original signal it has to be filtered before, otherwise all the harmonics higher than sampling frequency/2 will fold back to your signal causing aliasing. The filter can be avoided if the energy of the harmonics not taken into account is negligible.

enjunear said:

In simple terms, yes. In reality, it's not that easy. Most real signals can have harmonics reaching in to the MHz, and GHz frequency range (just ask anyone that's had to troubleshoot a SMPS that's failing EMI conducted emissions testing). The question for a real-world scenario is, how high of frequency do you have to sample/capture in order to replicate the desired signal with sufficient accuracy. In the case of a square wave, you'd need an infinite number of harmonics to make a perfectly square corner (see Fourier series expansion of a squarewave). However, you can obtain a pretty good looking squarewave with somewhere on the order of 15-20 harmonics (try this applet, adjust the number of terms using the slider at the bottom-right).

In practice, everything's a trade-off... sufficient performance vs. cost/time/complexity/size/weight/power/color/smell/etc. :wink:

Click to expand...

Dear sir,
The applet you provided is extremely useful, I have just checked that after a long time. Thanks a lot.

I have one more query, I would appreciate your time.

The following text is not only defined in an weblink (i.e **broken link removed** ) , but also written in a number of other sites (having the same meaning).

''" AM radio stations transmit audio signals, with a range from 20 Hz to 20 kHz, using carrier waves with a range from 500 kHz to 1.7 MHz. If they were to transmit audio signals directly they would need an antenna that is around 10,000 km! ""
Where the antenna height is calcualted using the formula, f=c/wavelength and c= velocity of light.

However, my question isnt it "' The f=c/wavelength is only for electromagnetic waves ?? "' , If so, then why all of them are using the velocity of light as the value of C in the equation??
And, if the velocity should not be the light speed in such a case (I mean If the web sources who written this are wrong), what would be actually the value of C here to calculate the wavelength?
If you are not sure, would you please refer some link/blogs where I may get my answer.


Regards

sajib333 said:

The f=c/wavelength is only for electromagnetic waves ?? "' , If so, then why all of them are using the velocity of light as the value of C in the equation??

Click to expand...

Because they're talking about antennas to transmit electromagnetic waves.

sajib333 said:

Dear sir,
''" AM radio stations transmit audio signals, with a range from 20 Hz to 20 kHz, using carrier waves with a range from 500 kHz to 1.7 MHz. If they were to transmit audio signals directly they would need an antenna that is around 10,000 km! ""
Where the antenna height is calcualted using the formula, f=c/wavelength and c= velocity of light.

However, my question isnt it "' The f=c/wavelength is only for electromagnetic waves ?? "' , If so, then why all of them are using the velocity of light as the value of C in the equation??
And, if the velocity should not be the light speed in such a case (I mean If the web sources who written this are wrong), what would be actually the value of C here to calculate the wavelength?
If you are not sure, would you please refer some link/blogs where I may get my answer.

Click to expand...

They are trying to show that transmitting the audio directly with EM waves is not feasible (i.e. driving an antenna directly with 20 Hz -20 kHz signals). Rather, it's easier to build a transmitter & antenna that uses a frequency much higher than the desired signal (audio), and modulate the desired signal onto the high frequency carrier.

In the equation \[\lambda=c/f\], they are still talking about sending signals via antenna and EM propagation; this is to illustrate the point that sending very low frequency EM is a poor method due to physical implementation, and that modulating your desired signal onto a higher frequency carrier is a much more efficient/practical solution.

That being said, ultra-low frequency EM communication has been used. Case in point, communication with submarines via VLF and ELF. Granted, the data rate is typically in the sub-Hertz range, but for a short "come to the surface so we can talk on VHF/UHF/other radio band" message, it does the job.