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Unformatted text preview: 2 Time (Amplitude) Modulation 2.1 Delaying a signal in time Now that we understand how to convert a time signal into the frequency domain, we can think about how we can combine multiple signals for simultaneous transmission in a communication channel. This was our motivation, given in section 1.2, where we saw that if two signals occupy separate frequency ranges, they can be sent at the same time but still recovered. The question now is how we translate a signal from one frequency range to another. Figure 36 shows an example of a frequency domain signal (left) and the same f X f ( ) f X f- ( ) f f Figure 36: signal shifted to a different frequency (right). We can think of this of a delay in frequency by f , just as you might picture a signal delayed in time by some amount. This delay is reflected in the change in the function X ( f ) to X ( f − f ), like you might expect it to change for a time domain function. We will use a triangular signal like the one shown for most frequency domain representations, just because it is simple to draw. It just depicts an arbitrary frequency domain representation of a function, like the ones we have seen previously. To get some insight into this shift, we will first look to what happens when you shift a signal in the time domain. Lets start with a rectangular function x 1 ( t ) = Π( t ) as defined earlier and delay it in time by half a second as in Fig. 37 to get a function x 2 ( t ) = x 1 ( t − . 5). We found earlier that the FT of the pulse function is the sinc function, that is x 1 ( t ) = Π( t ) ⇔ X 1 ( f ) = sin( πf ) πf . To find the FT of x 2 ( t ), we can evaluate the integral X 2 ( f ) = integraldisplay ∞ −∞ x 2 ( t ) e − j 2 πft dt = integraldisplay 1 (1) e − j 2 πft dt 35 1 x t 1 ( ) t ½-½ 1 x t x 2 1 ( )= ( ) t-0.5 t 1 Figure 37: = − 1 j 2 πf bracketleftBig e − j 2 πft bracketrightBig t =1 t =0 = − 1 j 2 πf bracketleftBig e − j 2 πf − 1 bracketrightBig This would be an answer be could stop at, but ... = − 1 j 2 πf bracketleftBig e − jπf − e jπf bracketrightBig e − jπf = 1 πf bracketleftBigg e jπf − e − jπf 2 j bracketrightBigg e − jπf = sin( πf ) πf e − jπf = sin( πf ) πf e − j 2 π 1 2 f . We can see that the sinc function shows up in the FT of the shifted function as well but is now multiplied by a complex exponential: X 2 ( f ) = X 1 ( f ) e − j 2 π 1 2 f . The ‘frequency’ of the exponential is equal to the time shift in the delay. It turns out that this is true for any FT pair and any delay. That is, if we know a transform pair x ( t ) ⇔ X ( f ) , then we can calculate the FT of a shifted version as x ( t − t ) ⇔ X ( f ) e − j 2 πt f . (28) This is known as the time shift property of the FT. A delay in time is equivalent to “modu- lation in frequency” where the modulating (multiplying) function is a complex exponential with a frequency t . This terminology may be confusing because t is a time, but in this context it defines the frequency of a modulating wave.context it defines the frequency of a modulating wave....
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This note was uploaded on 07/16/2009 for the course SYSC 3501 taught by Professor Osama during the Summer '09 term at Carleton CA.
- Summer '09