Lecture 15: Introduction to Mixers - Ali M. Niknejad - University of ...

Frequency Translationdown-conversionIFIFRFLOAs shown above, an ideal mixer translates themodulation around one carrier to another. In a receiver,this is usually from a higher RF frequency to a lower IFfrequency. In a transmitter, it’s the inverse.We know that an LTI circuit cannot perform frequencytranslation. Mixers can be realized with eithertime-varying circuits or non-linear circuitsA. M. Niknejad University of California, Berkeley EECS 142 Lecture 15 p. 3/22

Ideal MultiplierSuppose that the input of the mixer is the RF and LOsignalv RF = A(t) cos (ω 0 t + φ(t))v LO = A LO cos (ω L0 t)Recall the trigonometric identitycos(A + B) = cosAcos B − sin A sin BApplying the identity, we havev out= v RF × v LO= A(t)A LO{cos φ (cos(ω LO + ω 0 )t + cos(ω LO − ω 0 )t)2− sin φ (sin(ω LO + ω 0 )t + sin(ω LO − ω 0 )t)}A. M. Niknejad University of California, Berkeley EECS 142 Lecture 15 p. 4/22

Mixer + FilterLO-RFRFLOLO+RFNote that the LO can be below the RF (lower sideinjection) or above the RF (high side injection)Also note that for a given LO, energy at LO ± IF isconverted to the same IF frequency. This is a potentialproblem!A. M. Niknejad University of California, Berkeley EECS 142 Lecture 15 p. 6/22

Upper/Lower Injection and ImageExample: Downconversion MixerRF = 1GHz = 1000MHzIF = 100MHzLet’s say we choose a low-side injection:LO = 900MHzThat means that any signals or noise at 800MHz willalso be downconverted to the same IFA. M. Niknejad University of California, Berkeley EECS 142 Lecture 15 p. 7/22

Receiver ApplicationRF+IMAGEIFLNALOThe image frequency is the second frequency that alsodown-converts to the same IF. This is undesirablebecuase the noise and interferance at the imagefrequency can potentially overwhelm the receiver.One solution is to filter the image band. This places arestriction on the selection of the IF frequency due tothe required filter QA. M. Niknejad University of California, Berkeley EECS 142 Lecture 15 p. 8/22

Image RejectionIMAGE REJECTRF+IMAGEIFLNALOSuppose that RF = 1000MHz, and IF = 1MHz. Thenthe required filter bandwidth is much smaller than 2MHzto knock down the image.In general, the filter Q is given byQ = ω 0BW = RFBWA. M. Niknejad University of California, Berkeley EECS 142 Lecture 15 p. 9/22

Image Reject FilterIn our example, RF = 1000MHz, and IF = 1MHz. TheImagine is on 2IF = 2MHz away.Let’s design a filter with f 0 = 1000MHz andf 1 = 1001MHz.A fifth-order Chebyshev filter with 0.2 dB ripple is downabout 80 dB at the IF frequency.But the Q for such a filter isQ = 103 MHz1MHz = 103Such a filter requires components with Q > 10 3 !A. M. Niknejad University of California, Berkeley EECS 142 Lecture 15 p. 10/22

RF FilteringIF 1IF 2LNALO 1LO 2The fact that the required filter Q is so high is related tothe problem of filtering interferers. The very reason wechoose the superheterodyne architecture is to simplifythe filtering problem. It’s much easier to filter a fixed IFthan filter a variable RF.The image filtering problem can be relaxed by usingmulti-IF stages. Instead of moving to such a low IFwhere the image filtering is difficult (or expensive andbulky), we down-convert twice, using successively lowerIF frequencies.A. M. Niknejad University of California, Berkeley EECS 142 Lecture 15 p. 11/22

Direct Conversion ReceiverRFIF=DCLO leakageLNALO=RFA mixer will frequency translate two frequencies,LO ± IFWhy not simply down-convert directly to DC? Anotherwords, why not pick a zero IF?This is the basis of the direct conversion architecture.There are some potential problems...A. M. Niknejad University of California, Berkeley EECS 142 Lecture 15 p. 12/22

Phase of LOIn a direction conversion system, the LO frequency isequal to the RF frequency.Consider an input voltage v(t) = A(t) cos(ω 0 t). Since theLO is generated “locally”, it’s phase is random relativeto the RF input:v LO = A LO cos(ω 0 t + φ 0 )If we are so unlucky that φ 0 = 90 ◦ , then the output of themixer will be zero∫A(t)A LO sin(ω 0 t) cos(ω 0 t)dtT≈ A(t)A LO∫Tsin(ω 0 t) cos(ω 0 t)dt = 0A. M. Niknejad University of California, Berkeley EECS 142 Lecture 15 p. 15/22

IQ-Mixercos(ω LO t)IRFQsin(ω LO t)To avoid this situation, we can phase lock the LO to theRF by transmitting a pilot tone. Alternatively, we canuse two mixersAs we shall see, there are other benefits to such amixer.A. M. Niknejad University of California, Berkeley EECS 142 Lecture 15 p. 16/22

AM ModulationWe can see that an upconversion mixer is a naturalamplitude modulatorIf the input to the mixer is a baseband signal A(t), thenthe output is an AM carrierv o (t) = A(t) cos(ω LO t)How do we modulate the phase? A PLL is one way todo it. The IQ mixer is another way.Let’s expand a sinusoid that has AM and PMv o (t) = A(t) cos(ω 0 t + φ(t))= A(t) cosω 0 t cos φ(t) + A(t) sin ω 0 t sin φ(t)= I(t) cosω 0 t + Q(t) sin ω 0 tA. M. Niknejad University of California, Berkeley EECS 142 Lecture 15 p. 17/22

I-Q PlaneQφII(t) = A(t) cos φ(t)Q(t) = A(t) sin φ(t)We can draw a trajectory of points on the I-Q plane torepresent different modulation schemes.The amplitude modulation is given byI 2 (t) + Q 2 (t) = A 2 (t)(cos 2 φ(t) + sin 2 φ(t)) = A 2 (t)A. M. Niknejad University of California, Berkeley EECS 142 Lecture 15 p. 18/22

General ModulatorThe phase modulation is given byQ(t)I(t)=sin φ(t)cosφ(t) = tanφ(t)or−1 Q(t)φ(t) = tanI(t)The IQ modulator is a universal digital modulator. Wecan draw a set of points in the IQ plane that representsymbols to transmit. For instance, if we transmitI = 0/A and Q = 0, then we have a simple ASK system(amplitude shift keying).A. M. Niknejad University of California, Berkeley EECS 142 Lecture 15 p. 19/22

Digital Modulation: BPSK/QPSKQQ0110I1000I11For instance, if we transmit I(t) = ±1, this representsone bit transmission per cycle. But since the I and Qare orthogonal signals, we can improve the efficiency oftransmission by also transmitting symbols on the Q axis.If we select four points on a circle to represent 2 bits ofinformation, then we have a constant envelopemodulation scheme.A. M. Niknejad University of California, Berkeley EECS 142 Lecture 15 p. 20/22

10.750.50.25-0.25-0.5-0.75-1Modulation Waveforms1 2 3 4 5 6The data wave form(left) is modulatedonto a carrier (below).The first plot showsa simple on/off keying.The second plotshows binary phaseshift keying on onechannel (I).110.750.750.50.50.250.251 2 3 4 5 61 2 3 4 5 6-0.25-0.25-0.5-0.5-0.75-0.75-1-1A. M. Niknejad University of California, Berkeley EECS 142 Lecture 15 p. 21/22

More Bits Per CycleQQ011010001100000II101110111Eventually, the constellation points get very closetogether. Because of noise and distortion, the receivedspectrum will not lie exactly on the constellation points,but instead they will form a cluster around such points.If the clusters run into each other, errors will occur inthe transmission.We can increase the radius but that requires morepower.A. M. Niknejad University of California, Berkeley EECS 142 Lecture 15 p. 22/22