Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronic Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

Bandwidth in frequency spectrum

Status
Not open for further replies.

luckyvictor

Member level 4
Joined
Dec 1, 2009
Messages
78
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,286
Activity points
1,903
Why in general, QAM uses twice the bandwidth as PAM?? is it because of its dimension happens to be twice as PAM?

Also, does cos(2*pi*f*t) look the same as sin(2*pi*f*t) in the frequency spectrum? assuming f is the same in both. is f the bandwidth?
 

enjunear

Advanced Member level 3
Joined
Dec 21, 2010
Messages
960
Helped
309
Reputation
618
Reaction score
303
Trophy points
1,343
Location
USA, midwest
Activity points
9,749
A modulated signal will have a bandwidth approximately equal to the bandwidth of the baseband signal being carried. Some modulation formats will be wider because they are designed for spectral spreading.

By inspection, your examples of cos(2*pi*f*t) and sin(2*pi*f*t) will simply create a pure sinusoid at frequency f, if you sweep the time variable, t. If you took an FFT of cos(...) and sin(...), you'd get a single tone at frequency = f Hz.
 

luckyvictor

Member level 4
Joined
Dec 1, 2009
Messages
78
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,286
Activity points
1,903
Thanks for the reply.

I asked the second question is because, I think the reason of QAM uses twice the bandwidth as PAM is due to the fact that is has two time waveforms, and one is 90 degree shift of the other, so cos and sin. Therefore I am trying to find a way to visualize this fact on the frequency spectrum and hence explain why it uses twice the bandwidth
 

enjunear

Advanced Member level 3
Joined
Dec 21, 2010
Messages
960
Helped
309
Reputation
618
Reaction score
303
Trophy points
1,343
Location
USA, midwest
Activity points
9,749
It's more complex than that. Sine and cosine are 90 degrees out of phase with one another, so they are not (fully) acting at the same time. If you sweep them together in the time domain, you simply draw a circle of constant radius (unless you are modulating either component). In effect, you can put two baseband signals on one carrier, which is the concept of baseband I & Q signals (Inphase and Quadrature, i.e. 90 degrees out of phase with the "in-phase" signal). You can put two completely independent signals onto one carrier, and easily demodulate them at the receiver.

I'm not sure how to put the bandwidth concept for QAM vs PAM into words... perhaps a signals guru can jump in and give some more insight?
 

Status
Not open for further replies.
Toggle Sidebar

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top