Eecs 455 univ of michigan fall 2012 september 6 2012

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Unformatted text preview: otes 5 Up and Down Conversion Example of Up and Down Conversion of Signals 3 2 zc(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 zs(t) 1 0 -1 -2 -3 0 EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 68 / 130 Lecture Notes 5 Up and Down Conversion Signal Decomposition: Imperfect Phase √ 2 cos(2π fc t + φ) × zc (t ) GLP (f ) wc (t ) y (t ) × zs (t ) GLP (f ) ws (t ) √ − 2 sin(2π fc t + φ) EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 69 / 130 Lecture Notes 5 Up and Down Conversion Quadrature-Phase (xc (t ), xs (t )) wc (t ) = xc (t ) cos(φ) + xs (t ) sin(φ) φ (wc (t ), ws (t )) ws (t ) = −xc (t ) sin(φ) + xs (t ) cos(φ) wc (t ) + jws (t ) = (xc (t ) + jxs (t ))e−j φ In-Phase Note that this is equivalent to a phase rotation by angle −φ. EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 70 / 130 Lecture Notes 5 Up and Down Conversion We can recover the orignial signal by rotating the signal. xc (t ) + jxs (t ) = (wc (t ) + jws (t ))e+j φ EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 71 / 130 Lecture Notes 5 Up and Down Conversion cos(φ) wc (t ) × + xc (t ) − × sin(φ) × ws (t ) × + xs (t ) cos(φ) EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 72 / 130 Lecture Notes 5 Up and Down Conversion Frequency Offset √ xc (t ) √ 2 cos(2π fd t ) 2 cos(2π fc t ) yc (t ) × + xs (t ) × √ − 2 sin(2π fc t ) EECS 455 (Univ. of Michigan) × zc (t ) GLP (f ) wc (t ) y (t ) ys (t ) × zs (t ) GLP (f ) ws (t ) √ − 2 sin(2π fd t ) Fall 2012 September 6, 2012 73 / 130 Lecture Notes 5 Up and Down Conversion Frequency Offset For frequency offset (that is sufficiently small relative to the bandwidth of the low pass filters) the output of the down converstion part is related to the input by (wc (t ) + jws (t )) = (xc (t ) + jxs (t ))ej 2π(fc −fd )t . Conceptually we can think about shifting up in frequency by fc and down in frequency by fd . Using a technique similar to the phase offset correction we can correct for frequency offset. EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 74 / 130 Lecture Notes 5 Up and Down Conversion Typical IQ Modulator: Texas Instruments TRF3703 TRF3703 I x2 x2 DAC Phase Gain/ Offset Adjust Q x2 x2 DAC DAC5687 EECS 455 (Univ. of Michigan) TRF3761 Fall 2012 September 6, 2012 75 / 130 Lecture Notes 5 Up and Down Conversion Typical IQ Demodulator: Analog Devices AD6620 FUNCTIONAL BLOCK DIAGRAM I REAL, DUAL REAL, OR COMPLEX INPUTS Q COS AD6620 EECS 455 (Univ. of Michigan) I CIC FILTERS Q I FIR FILTER Q OUTPUT FORMAT SERIAL OR PARALLEL OUTPUTS –SIN COMPLEX NCO EXTERNAL SYNC CIRCUITRY Fall 2012 JTAG PORT P OR SERIAL CONTROL September 6, 2012 76 / 130 Lecture Notes 5 Up and Down Conversion Bandpass/Baseband Relations Assume that s0 (t ), s1 (t ), ..., are bandpass signals. That is, the bandwidth occupied is [−fc − W , −fc + W ] and [fc − W , fc + W ]. Let ˆ ˆ s0 (t ), s1 (t ), ... be the lowpass complex representation of the bandpass waveforms. T ˆ si (t ) = 0 √ si (τ ) 2 cos(2π fc τ )gLP (t − τ )d τ √ −j 2 T 0 si (τ ) sin(2π fc τ )gLP (t − τ )d τ √ si (τ ) 2 exp{−j 2π fc τ }gLP (t − τ )d τ 0 √ ˆ si (t ) = Re[ 2si (τ ) exp{j 2π fc t }] T = EECS 455 (Univ. of Michigan) Fall 2012 September 6, 2012 77 / 130 Lecture Notes 5 Up and Down Conversion Signals Claim: s0 (t )s1 (t )dt = Re[ ˆ ˆ∗ s0 (t )s1 (t )dt ], ˆ ˆ s0 (t )s1 (t )dt ≈ 0 Proof: 1 First note that if a and b are any complex numbers then Re[a]Re[b] = 2 Re[ab + ab∗ ]. To show this just examine the following ab = = (ar + jai )(br + jbi ) ar br − ai bi + j ( ar bi + ai br ) ab ∗ = ab + ab ∗ = 2ar br + j 2( ai br ) ∗ = 2ar br ∗ = ar br = Re[a]Re[b] = Re[ab + ab ] 1 2 EECS 455 (Univ. of Michigan) Re[ab + ab ] (ar + jai )(br − jbi ) a r b r + a i b i + j ( −a r b i + a i b r ) Fall 2012 September 6, 2012 78 / 130 Lecture Notes 5 Up and Down Conversion Signals Proof (cont.): Now consider Z s0 (t )s1 (t )dt = = = Z Z √ √ s s Re[ 2ˆ0 (t ) exp{j 2π fc t }]Re[ 2ˆ1 (t ) exp{j 2π fc t }]dt Re[ˆ0 (t ) exp{j 2π fc t }ˆ1 (t ) exp{j 2π fc t } s s ∗ ˆ s +s0 (t ) exp{j 2π fc t }ˆ1 (t ) exp{−j 2π fc t }]dt Z ∗ Re[ˆ0 (t )ˆ1 (t ) exp{j 2π (2fc )t } + ˆ0 (t )ˆ1 (t )]dt s s s s s s Now the first integral is zero since ˆ0 (t ) and ˆ1 (t ) are low pass functions while exp{j 2π (2fc )t } is a double frequency term. Thus Z s0 (t )s1 (t )dt = = EECS 455 (Univ. of Michigan) Z ∗ Re[ ˆ0 (t )ˆ1 (t )dt ] s s Z ∗ Re[ ˆ0 (t )ˆ1 (t )dt ] s s Fall 2012 September 6, 2012 79 / 130 Lecture Notes 5 Up and Down Conversion Signals As a special case of the above result consider s0 (t ) = s1 (t ). Equal Energy 2 s0 (t )dt = EECS 4...
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