The negative frequencies are only for mathematical purposes. In real life there is no such as negative frequencies it is only applicable for math ONLY.

Hello,
Can you please tell me how to find HILBERT transform from a given complex envelope and also the solution for the following example.

My question is as follows :

1)The complex envelope of a signal s( t ) with respect to f0 = 50 Hz is sin ( πt )/(πt).Find the complex envelope of Z(t)= s(t-0.01) with respect to 50HZ

The first signal is a perfect passband filter centered at 50 Hz. A .01 sec time shift in this signal is equivalent to as half Hz frequency shift in the envelope of the signal or a phase shift of pi.

The given first signal sin(πt)/(πt) is the complex envelope of the s(t).So i think the approach to that question is to first find the original signal s(t) and then find the complex envelope of the delayed signal.
I was able to find the original signal as s(t)=square root of 2 multiplied by
cos(100πt)(sin(πt)/(πt)) But i was not able to find the complex envelope of the delayed signal.
Can you please help me in finding the complex envelope of the delayed signal.

I am not clear as to how a complex envelop is useful in simulation for low pass as well as modulated signals. Specifically, the explanation given that we no longer need to do simulation at carrier frequencies and just required to do the same at highest frequency of signal, needs clarification

Dear Madam,
Thanks a lot for your excellent explanation about the Hilbert Transform (HT) in communicatoin systems!
Here, after reading your paper, I have a question and hope to discuss it with you.
Since
cos(wt) -> sin(wt) -> -cos(wt) -> -sin(wt) -> cos(wt) -> …
holds in HT, you obtain your result as
HT(cos(2-100)t)=sin(2-100)t.
However, in my opinion,
HT(cos(2-100)t)=-sin(2-100)t=sin(100-2)t.
Because only in this case can we have cos(2-100)t+jsin(100-2)t=exp(j(100-2)t), attaining that the positive domain is left. Otherwise,g+(t)=2cos(2+100)t+…+j3cos(3-100)t cannot be rewritten as g+(t)=(4cos2t-6sin3t)exp(j100t).

Dear Madam,

In Hilbert Transform -90 dgree phase shift is produced if we the signal frequency is f>0

and produces +90 degree phase shift for the signal if the signal frequency is f<0.If I have a bandwidth

of frequencies 0 Hz to B Hz how could the Hilbert transform will produce the +90 degree phase shift since

I don't have negative frequency

Thanks

regards

Arunpradhap Natarajan

The negative frequencies are only for mathematical purposes. In real life there is no such as negative frequencies it is only applicable for math ONLY.

Hello,

Can you please tell me how to find HILBERT transform from a given complex envelope and also the solution for the following example.

My question is as follows :

1)The complex envelope of a signal s( t ) with respect to f0 = 50 Hz is sin ( πt )/(πt).Find the complex envelope of Z(t)= s(t-0.01) with respect to 50HZ

The first signal is a perfect passband filter centered at 50 Hz. A .01 sec time shift in this signal is equivalent to as half Hz frequency shift in the envelope of the signal or a phase shift of pi.

The given first signal sin(πt)/(πt) is the complex envelope of the s(t).So i think the approach to that question is to first find the original signal s(t) and then find the complex envelope of the delayed signal.

I was able to find the original signal as s(t)=square root of 2 multiplied by

cos(100πt)(sin(πt)/(πt)) But i was not able to find the complex envelope of the delayed signal.

Can you please help me in finding the complex envelope of the delayed signal.

Dear Madame

Good Day

I am not clear as to how a complex envelop is useful in simulation for low pass as well as modulated signals. Specifically, the explanation given that we no longer need to do simulation at carrier frequencies and just required to do the same at highest frequency of signal, needs clarification

Dear Madam,

Thanks a lot for your excellent explanation about the Hilbert Transform (HT) in communicatoin systems!

Here, after reading your paper, I have a question and hope to discuss it with you.

Since

cos(wt) -> sin(wt) -> -cos(wt) -> -sin(wt) -> cos(wt) -> …

holds in HT, you obtain your result as

HT(cos(2-100)t)=sin(2-100)t.

However, in my opinion,

HT(cos(2-100)t)=-sin(2-100)t=sin(100-2)t.

Because only in this case can we have cos(2-100)t+jsin(100-2)t=exp(j(100-2)t), attaining that the positive domain is left. Otherwise,g+(t)=2cos(2+100)t+…+j3cos(3-100)t cannot be rewritten as g+(t)=(4cos2t-6sin3t)exp(j100t).