{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# lecture12n - Stanford University Winter 2009-2010 Signal...

This preview shows pages 1–5. Sign up to view the full content.

Stanford University Winter 2009-2010 Signal Processing and Linear Systems I Lecture 12: Moduation and Demodulation February 8, 2010 EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 1 Modulation The modulation theorem tells us how to place a message signal m ( t ) on a carrier cos( ω c t ) . How do we recover m ( t ) from the modulated signal? What are the problems? Are there better receivers? Can I use other modulation methods to make the receiver easier? EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 2

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

View Full Document
Double-Sideband, Suppressed Carrier (DSB-SC) Modulation This is a complicated way of saying ”multiply by a cosine.” We have a message (baseband) signal, and cosine cos( ω c t ) at a frequency ω c . The modulated message signal and its Fourier transform are m ( t ) cos( ω c t ) 1 2 [ M ( j ( ω + ω c )) + M ( j ( ω ω c ))] where m ( t ) M ( j ω ) . The block diagram is × m ( t ) cos ( ! c t ) m ( t ) cos ( ! c t ) This signal and its spectrum are illustrated on the next page: EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 3 m ( t ) m ( t ) m ( t ) cos ( ! c t ) m ( t ) t t ! c ! c ! ! M ( j ! ) 1 2 M ( j ( ! ! c )) 1 2 M ( j ( ! + ! c )) EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 4
We can think of modulation as frequency domain convolution. F [ m ( t ) cos( ω c t )] = 1 2 π [ M ( j ω ) ( πδ ( ω + ω c ) + πδ ( ω ω c ))] ! ! c ! c ! ! c ! c ! = 1 2 ! M ( j ! ) 1 2 M ( j ( ! + ! c )) 1 2 M ( j ( ! ! c )) !" ( # # c ) !" ( # + # c ) EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 5 To demodulate this signal, consider what happens if we multiply again by cos( ω c t ) . Again, we can think of this as a convolution in the frequency domain: ! c ! c ! ! c ! c ! = ! c ! c ! 2 ! c 2 ! c 0 0 0 Lowpass Filter 1 2 ! 1 2 M ( j ( ! + ! c )) 1 2 M ( j ( ! ! c )) 1 4 M ( j ( ! 2 ! c )) 1 4 M ( j ( ! + 2 ! c )) 1 2 M ( j ! ) !" ( # # c ) !" ( # + # c ) EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 6

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

View Full Document
After the convolution there is a replica of the spectrum centered at ω = 0 , which we can extract with a lowpass filter. The modulated signal spectrum is F [ m ( t ) cos( ω c t )] = 1 2 M ( j ( ω + ω c )) + 1 2 M ( j ( ω ω c )) Multiplying this by cos( ω c t ) corresponds to convolving in frequency, F m ( t ) cos 2 ( ω c t ) = 1 2 π 1 2 M ( j ( ω + ω c )) + 1 2 M ( j ( ω ω c )) [ πδ ( ω + ω c ) + πδ ( ω ω c )] = 1 4 [ M ( j ( ω + ω c )) + M ( j ( ω ω c ))] [ δ ( ω + ω c ) + δ ( ω ω c )] = 1 4 M ( j ( ω + 2 ω c )) + 1 2 M ( j ω ) + 1 4 M ( j ( ω 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}