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From: | Md. Atiqur Rahman |
Subject: | Re: Excess bandwidth in constellation modulator |
Date: | Wed, 26 Feb 2020 12:52:43 +0100 |
Hello,That comes down to baseband signal and passband as shown below. Traditionally, we only quote positive frequencies. For example, for the SRRC pulse you mentioned, the baseband signal extends from -Rs(1+alpha)/2 to +Rs(1+alpha)/2. Thus, we say the baseband bandwidth is Rs(1+alpha)/2. When transmitting over the air, we have a passband signal which extends from Fc -Rs(1+alpha)/2 to Fc +Rs(1+alpha)/2 where Fc is the carrier frequency. In this case, the positive bandwidth is Rs(1+alpha), which is twice the bandwidth at baseband. As a general rule, passband frequency (the frequency you transmit over the air) is twice the baseband signal.Regards,Moses.PS- The attached pdf is quite a good introduction to gnuradio and signal processing basics.On Wed, Feb 26, 2020 at 5:09 AM Md. Atiqur Rahman <address@hidden> wrote:Thank you for your reply with the details.I do already read from the GNU Radio tutorial but wasn't fully sure.The equation now clear, just want to make sure one point from you.After the modulation(e.g qpsk) signal upconverted to RF, then the signal bandwidth still will be the same?Thank you again.On Wed, Feb 26, 2020 at 12:31 AM Moses Browne Mwakyanjala <address@hidden> wrote:I saw your post on the gnuradio mailing list.The constellation modulator block uses a Square-Root Raised-Cosine filter with occupied bandwidth (double-sided a.k.a passband) given by BW = (1 + alpha)Rs, where Rs is the symbol rate. The bit rate for M-ary linear modulation is given by Bit Rate = Rs*Log(M)base2. Thus, for BPSK you will achieve Bit Rate = Rs, and for QPSK it will be Bit Rate = 2 * Rs.The transmission over-the-air is called passband, which for the SRRC filter is given as described above.You could read more on the SRRC filter from this Wikipedia article:You could get more information here:Good luck.--Sincerely,Md Atiqur Rahman
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