Complex Baseband Representation Saravanan Vijayakumaran - PowerPoint PPT Presentation

Complex Baseband Representation Saravanan Vijayakumaran sarva@ee.iitb.ac.in Department of Electrical Engineering Indian Institute of Technology Bombay July 23, 2013 1 / 19

Baseband Signals S ( f ) = 0 , | f | > W | S ( f ) | f − W W 2 / 19

Passband Signals S ( f ) � = 0 , | f ± f c | ≤ W , f c > W > 0 . | S p ( f ) | f − f c − W − f c − f c + W f c − W f c f c + W 3 / 19

Sampling Theorem Theorem If a signal s ( t ) is bandlimited to B, S ( f ) = 0 , | f | > B then a sufficient condition for exact reconstructability is a uniform sampling rate f s where f s > 2 B . Baseband Signals B = W Passband Signals B = f c + W (Can we do better?) 4 / 19

Fourier Transform for Real Signals Im [ s ( t )] = 0 ⇒ S ( f ) = S ∗ ( − f ) (Conjugate Symmetry) ⇒ | S ( f ) | = | S ( − f ) | , arg ( S ( f )) = − arg ( S ( − f )) Re ( S ( f )) Im ( S ( f )) 1 1 − W W f − W W f -1 5 / 19

Fourier Transform of a Real Passband Signal Re ( S p ( f )) − f c f c f Im ( S p ( f )) − f c f c f 6 / 19

Positive Spectrum of a Real Passband Signal Re ( S + p ( f )) − f c f c f Im ( S + p ( f )) − f c f c f S + p ( f ) = S p ( f ) u ( f ) 7 / 19

Complex Envelope of a Real Passband Signal Re ( S ( f )) f Im ( S ( f )) f √ √ 2 S + S ( f ) = p ( f + f c ) = 2 S p ( f + f c ) u ( f + f c ) 8 / 19

Complex Envelope in Time Domain Frequency Domain Representation √ √ 2 S + S ( f ) = p ( f + f c ) = 2 S p ( f + f c ) u ( f + f c ) Time Domain Representation of Positive Spectrum S + p ( f ) = S p ( f ) u ( f ) s p ( t ) ⋆ F − 1 [ u ( f )] s + p ( t ) = Time Domain Representation of Frequency Domain Unit Step j 2 π f + 1 1 − ⇀ u ( t ) 2 δ ( f ) ↽ − − j 2 π t + 1 1 − ⇀ u ( f ) 2 δ ( − t ) ↽ − 2 π t + 1 j = 2 δ ( t ) 9 / 19

Complex Envelope in Time Domain Time Domain Representation of Positive Spectrum � 1 � j s + p ( t ) = s p ( t ) ⋆ 2 δ ( t ) + 2 π t 1 2 [ s p ( t ) + j ˆ = s p ( t )] Time Domain Representation of Complex Envelope √ 1 − [ s p ( t ) + j ˆ 2 S p ( f ) u ( f ) ⇀ √ s p ( t )] ↽ − 2 √ 1 s p ( t )] e − j 2 π f c t − ⇀ √ [ s p ( t ) + j ˆ 2 S p ( f + f c ) u ( f + f c ) ↽ − 2 1 s p ( t )] e − j 2 π f c t − ⇀ [ s p ( t ) + j ˆ S ( f ) √ − ↽ 2 1 s p ( t )] e − j 2 π f c t [ s p ( t ) + j ˆ s ( t ) = √ 2 10 / 19

Passband Signal in terms of Complex Envelope Complex Envelope s ( t ) = s c ( t ) + js s ( t ) s c ( t ) In-phase component s s ( t ) Quadrature component Time Domain Relationship � √ 2 s ( t ) e j 2 π f c t � s p ( t ) = Re � √ 2 { s c ( t ) + js s ( t ) } e j 2 π f c t � = Re √ √ = 2 s c ( t ) cos 2 π f c t − 2 s s ( t ) sin 2 π f c t Frequency Domain Relationship S p ( f ) = S ( f − f c ) + S ∗ ( − f − f c ) √ 2 11 / 19

Upconversion √ √ s p ( t ) = 2 s c ( t ) cos 2 π f c t − 2 s s ( t ) sin 2 π f c t s c ( t ) × √ 2 cos 2 π f c t + s p ( t ) √ − 2 sin 2 π f c t s s ( t ) × 12 / 19

Downconversion √ 2 s c ( t ) cos 2 2 π f c t − 2 s s ( t ) sin 2 π f c t cos 2 π f c t 2 s p ( t ) cos 2 π f c t = = s c ( t ) + s c ( t ) cos 4 π f c t − s s ( t ) sin 4 π f c t × s c ( t ) LPF √ 2 cos 2 π f c t s p ( t ) √ − 2 sin 2 π f c t × s s ( t ) LPF 13 / 19

Inner Product and Energy Let s ( t ) and r ( t ) be signals. Definition (Inner Product) � ∞ � s , r � = s ( t ) r ∗ ( t ) dt −∞ Definition (Energy) � ∞ E s = � s � 2 = � s , s � = | s ( t ) | 2 dt −∞ 14 / 19

I and Q Channels of a Passband Signal √ √ s p ( t ) = 2 s c ( t ) cos 2 π f c t − 2 s s ( t ) sin 2 π f c t � �� � � �� � I Component Q Component √ x c ( t ) = 2 s c ( t ) cos 2 π f c t √ x s ( t ) = 2 s s ( t ) sin 2 π f c t I and Q Channels of a Passband Signal are Orthogonal � x c , x s � = 0 15 / 19

Passband and Baseband Inner Products � u p , v p � = � u c , v c � + � u s , v s � = Re ( � u , v � ) Energy of Complex Envelope = Energy of Passband Signal � s � 2 = � s p � 2 16 / 19

Complex Baseband Equivalent of Passband Filtering s p ( t ) Passband signal h p ( t ) Impulse response of passband filter y p ( t ) Filter output y p ( t ) = s p ( t ) ⋆ h p ( t ) Y p ( f ) = S p ( f ) H p ( f ) S + ( f ) = S p ( f ) u ( f ) H + ( f ) = H p ( f ) u ( f ) Y + ( f ) = Y p ( f ) u ( f ) Y + ( f ) = S + ( f ) H + ( f ) √ √ 1 Y ( f ) = 2 Y + ( f + f c ) = 2 S + ( f + f c ) H + ( f + f c ) = √ S ( f ) H ( f ) 2 17 / 19

Complex Baseband Equivalent of Passband Filtering 1 y ( t ) = √ s ( t ) ⋆ h ( t ) 2 1 √ y c = ( s c ⋆ h c − s s ⋆ h s ) 2 1 y s = √ ( s s ⋆ h c + s c ⋆ h s ) 2 18 / 19

Thanks for your attention 19 / 19