# Tutorial 7 – Hilbert Transform and the Complex Envelope

Description: Hilbert Transform and the Complex Envelope

###### 8 comments on “Tutorial 7 – Hilbert Transform and the Complex Envelope”
1. Arunpradhap Natarajan says:

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

• Ahmed Nabeel Musa says:

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.

2. Franklin says:

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

• Charan L. says:

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.

• Franklin says:

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.

3. Imran Shaf says:

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

4. Cai Hengbin says:

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).

5. Marco Pilas says:

Hello,
Thank you very much for sharing with us your knowledge, it is a very interesting and useful article.
In the article you wrote: “An analytic signal (composed of a real signal and its Hilbert transform) has a spectrum that exists only in the positive frequency domain” but according to the calculation the results of the analytic signal of a sine function is exp(-jwt), which is mathematically represented by a pure tone a the negative frequency plane. Can you please explain?

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