Of michigan fall 2012 september 6 2012 55 130 lecture

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Unformatted text preview: ichigan) Fall 2012 September 6, 2012 53 / 130 Lecture Notes 5 Up and Down Conversion Signal Decomposition Yc (f ) = Ys (f ) = Y (f ) = Y (f ) = EECS 455 (Univ. of Michigan) √ 2 [Xc (f − fc ) + Xc (f + fc )] 2 √ j2 [Xs (f − fc ) − Xs (f + fc )] 2 [Yc (f ) + Ys (f )] √ 2 [Xc (f − fc ) + Xc (f + fc ) 2 +j (Xs (f − fc ) − Xs (f + fc ))] Fall 2012 September 6, 2012 54 / 130 Lecture Notes 5 Up and Down Conversion Signal Decomposition (Frequency Domain) Zc (f ) = = = LPF[Zc (f )] = √ 2 [Y (f − fc ) + Y (f + fc )] 2 1 [Xc (f − 2fc ) + Xc (f ) + j (Xs (f − 2fc ) − Xs (f )) 2 +Xc (f ) + Xc (f + 2fc ) + j (Xs (f ) − Xs (f + 2fc ))] 1 [Xc (f − 2fc ) + 2Xc (f ) + Xc (f + 2fc ) 2 +j (Xs (f − 2fc ) − Xs (f + 2fc ))] 1 LPF[Xc (f − 2fc ) + 2Xc (f ) + Xc (f + 2fc ) 2 +j (Xs (f − 2fc ) − Xs (f + 2fc ))] = Xc (f ) EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 55 / 130 Lecture Notes 5 Up and Down Conversion Signal Decomposition (Time Domain) Suppose s(t ) is a bandpass signal with representation √ √ s(t ) = xc (t ) 2 cos(2π fc t ) − xs (t ) 2 sin(2π fc t ) √ zc (t ) = s(t ) 2 cos(2π fc t ) = 2 [xc (t ) cos(2π fc t ) cos(2π fc t ) − xs (t ) sin(2π fc t ) cos(2π fc t )] = xc (t )[1 + cos(2π 2fc t )] − xs (t )[1 sin(2π 2fc t ) − sin(0)] = xc (t ) + [xc (t ) cos(2π 2fc t ) − xs (t ) sin(2π 2fc t )] The first term above is the desired signal xs (t ) while the second term is called the double frequency term. If zI (t ) is then filtered with a filter that removes the double frequency term what remains is the message. LPF[zc (t )] = sI (t ) EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 56 / 130 Lecture Notes 5 Up and Down Conversion Signal Decomposition Similarly the spectrum of zs (t ) is √ j2 [Y (f − fc ) − Y (f + fc )] Zs (f ) = 2 j = [Xc (f − 2fc ) + Xc (f ) + j (Xs (f − 2fc ) − Xs (f )) 2 −Xc (f ) − Xc (f + 2fc ) − j (Xs (f ) − Xs (f + 2fc ))] j = [Xc (f − 2fc ) + Xc (f + 2fc ) 2 +j (Xs (f − 2fc ) − 2Xs (f ) + Xs (f + 2fc ))] j LPF[Xc (f − 2fc ) + Xc (f + 2fc ) LPF[Zs (f )] = 2 +j (Xs (f − 2fc ) − 2Xs (f ) + Xs (f + 2fc ))] = Xs (f ) EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 57 / 130 Lecture Notes 5 Up and Down Conversion Signal Decomposition Similarly √ zs (t ) = s(t )[− 2 sin(2π fc t )] = 2[−xs (t ) cos(2π fc t ) sin(2π fc t ) + xs (t ) sin(2π fc t ) sin(2π fc t )] = xc (t )[sin(2π 2fc t ) − sin(0)] + xs (t )[1 − cos(2π 2fc t )] = xs (t ) + [xc (t ) sin(2π 2fc t ) − xs (t ) cos(2π 2fc t )] LPF[zs (t )] = xs (t ) EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 58 / 130 Lecture Notes 5 Up and Down Conversion Signal Decomposition Re[Zc (f )] −2fc EECS 455 (Univ. of Michigan) −W W Fall 2012 2fc September 6, 2012 f 59 / 130 Lecture Notes 5 Up and Down Conversion Signal Decomposition Im[Zc (f )] −2fc −W 2fc W EECS 455 (Univ. of Michigan) Fall 2012 f September 6, 2012 60 / 130 Lecture Notes 5 Up and Down Conversion Signal Decomposition Re[Zs (f )] −2fc 2fc f −W EECS 455 (Univ. of Michigan) W Fall 2012 September 6, 2012 61 / 130 Lecture Notes 5 Up and Down Conversion Signal Decomposition Im[Zs (f )] −2fc −W W EECS 455 (Univ. of Michigan) Fall 2012 2fc September 6, 2012 f 62 / 130 Lecture Notes 5 Up and Down Conversion Signal Decomposition: Example xc (t ) = ac ,1 cos(2π f1 t ) + ac ,2 sin(2π f1 t ) + ac ,3 cos(2π f2 t ) xs (t ) = as,1 cos(2π f1 t ) + as,2 sin(2π f1 t ) + as,3 sin(2π f2 t ) where ac ,1 = 0.25, ac ,2 = 0.5, ac ,3 = 1, as,1 = −1.0, ac ,2 = 0.25, ac ,3 = 1, f1 = 1, f2 = 2. The signals are upconverted with a quadrature modulator to produce √ √ = xc (t ) 2 cos(2π fc t ) − xs (t ) 2 sin(2π fc t ) = xe (t ) cos(2π fc t + θ (t )) where fc = 16. EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 63 / 130 Lecture Notes 5 Up and Down Conversion Example of Up and Down Conversion of Signals 3 2 xc(t) 1 0 -1 -2 -3 0 0.2 0.4 0.6 0.8 1 time 1.2 1.4 1.6 1.8 2 0.2 0.4 0.6 0.8 1 time 1.2 1.4 1.6 1.8 2 3 2 xs(t) 1 0 -1 -2 -3 0 EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 64 / 130 Lecture Notes 5 Up and Down Conversion Example of Up and Down Conversion of Signals 2.5 xe(t) 2 1.5 1 0.5 0 0 0.2 0.4 0.6 0.8 1 time 1.2 1.4 1.6 1.8 2 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 4 phi.(t) 2 0 -2 -4 0 EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 65 / 130 Lecture Notes 5 Up and Down Conversion Example of Up and Down Conversion of Signals 2 1.5 1 xs(t) 0.5 0 -0.5 -1 -1.5 -2 -2 EECS 455 (Univ. of Michigan) -1.5 -1 -0.5 0 xc(t) Fall 2012 0.5 1 1.5 2 September 6, 2012 66 / 130 Lecture Notes 5 Up and Down Conversion Example of Up and Down Conversion of Signals 2 1.5 1 y(t) 0.5 0 -0.5 -1 -1.5 -2 0 0.2 EECS 455 (Univ. of Michigan) 0.4 0.6 0.8 1 time Fall 2012 1.2 1.4 1.6 1.8 2 September 6, 2012 67 / 130 Lecture N...
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This note was uploaded on 02/12/2014 for the course EECS 455 taught by Professor Stark during the Fall '08 term at University of Michigan.

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