am - 2 2.1 Time(Amplitude Modulation Delaying a signal in...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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 0 f 0 Figure 36: signal shifted to a different frequency (right). We can think of this of a delay in frequency by f 0 , 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 0 ), 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 0 . 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 0 (1) e j 2 πft dt 35
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
1 x t 1 ( ) t ½ 1 x t x 2 1 ( )= ( ) t-0.5 t 1 0 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 0 ) X ( f ) e j 2 πt 0 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 0 . This terminology may be confusing because t 0 is a time, but in this context it defines the frequency of a modulating wave.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern