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142 EECS Lecture 18: Introduction to Mixers Prof. Ali M. Niknejad University of California, Berkeley Copyright c 2008 by Ali M. Niknejad A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 1/31 Mixers IF LO RF RF LO IF An ideal mixer is usually drawn with a multiplier symbol A real mixer cannot be driven by arbitrary inputs. Instead one port, the LO port, is driven by an local oscillator with a xed amplitude sinusoid. In a down-conversion mixer, the other input port is driven by the RF signal, and the output is at a lower IF intermediate frequency In an up-coversion mixer, the other input is the IF signal and the output is the RF signal A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 2/31 Frequency Translation down-co nve rsio n IF IF RF LO As shown above, an ideal mixer translates the modulation around one carrier to another. In a receiver, this is usually from a higher RF frequency to a lower IF frequency. In a transmitter, it s the inverse. We know that an LTI circuit cannot perform frequency translation. Mixers can be realized with either time-varying circuits or non-linear circuits A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 3/31 Ideal Multiplier Suppose that the input of the mixer is the RF and LO signal vRF = A(t) cos ( 0 t + (t)) vLO = ALO cos ( L0 t) Recall the trigonometric identity cos(A + B) = cos A cos B sin A sin B Applying the identity, we have vout = vRF vLO A(t)ALO {cos (cos( LO + 0 )t + cos( LO 0 )t) = 2 sin (sin( LO + 0 )t + sin( LO 0 )t)} A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 4/31 Ideal Multiplier (cont) Grouping terms we have vout = A(t)AL O 2 {cos (( LO + 0 )t + (t)) + cos (( LO 0 )t + (t))} We see that the modulation is indeed translated to two new frequencies, LO + RF and LO RF . We usually select either the upper or lower sideband by ltering the output of the mixer high-pass or low-pass RF LO IF A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 5/31 Mixer + Filter IF=LO-RF RF=LO-IF LO IM=LO+IF Note that the LO can be below the RF (lower side injection) or above the RF (high side injection) Also note that for a given LO, energy at LO IF is converted to the same IF frequency. This is a potential problem! A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 6/31 Upper/Lower Injection and Image Example: Downconversion Mixer RF = 1GHz = 1000MHz IF = 100MHz Let s say we choose a low-side injection: LO = 900MHz That means that any signals or noise at 800MHz will also be downconverted to the same IF A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 7/31 Receiver Application RF+IMAGE LNA LO IF The image frequency is the second frequency that also down-converts to the same IF. This is undesirable becuase the noise and interferance at the image frequency can potentially overwhelm the receiver. One solution is to lter the image band. This places a restriction on the selection of the IF frequency due to the required lter Q A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 8/31 Image Rejection IMAGE REJECT RF+IMAGE LNA LO IF Suppose that RF = 1000MHz, and IF = 1MHz. Then the required lter bandwidth is much smaller than 2MHz to knock down the image. In general, the lter Q is given by Q= 0 RF = BW BW A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 9/31 Image Reject Filter In our example, RF = 1000MHz, and IF = 1MHz. The Imagine is on 2IF = 2MHz away. Let s design a lter with f0 = 1000MHz and f1 = 1001MHz. A fth-order Chebyshev lter with 0.2 dB ripple is down about 80 dB at the IF frequency. But the Q for such a lter is 103 MHz = 103 Q= 1MHz Such a lter requires components with Q > 103 ! A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 10/31 RF Filtering IF 1 IF 2 LNA LO 1 LO 2 The fact that the required lter Q is so high is related to the problem of ltering interferers. The very reason we choose the superheterodyne architecture is to simplify the ltering problem. It s much easier to lter a xed IF than lter a variable RF. The image ltering problem can be relaxed by using multi-IF stages. Instead of moving to such a low IF where the image ltering is dif cult (or expensive and bulky), we down-convert twice, using successively lower IF frequencies. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 11/31 Direct Conversion Receiver RF LNA LO l IF=DC e LO=RF A mixer will frequency translate two frequencies, LO IF Why not simply down-convert directly to DC? In other words, why not pick a zero IF? This is the basis of the direct conversion architecture. There are some potential problems... k ag ea A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 12/31 Direction Conversion First, note that we must down-convert the desired signal and all the interfering signals. In other words, the LNA and mixer must be extremely linear. Since IF is at DC, all even order distortion now plagues the system, because the distortion at DC can easily swamp the desired signal. Furthermore, CMOS circuits produce a lot of icker noise. Before we ignored this source of noise becuase it occurs at low frequency. Now it also competes with our signal. Another issue is with LO leakage. If any of the LO leaks into the RF path, then it will self-mix and produce a DC offset. The DC offset can rail the IF ampli er stages. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 13/31 Direct Conversion (cont) Example: If the IF ampli er has 80 dB of gain, and the mixer has 10 dB of gain, estimate the allowed LO leakage. Assume the ADC uses a 1V reference. To rail the output, we require a DC offset less than 10 4 V. If the LO power is 0 dBm (316mV), we require an input leakage voltage < 10 5 V, or an isolation better than 90 dB! A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 14/31 Phase of LO In a direction conversion system, the LO frequency is equal to the RF frequency. Consider an input voltage v(t) = A(t) cos( 0 t). Since the LO is generated locally , it s phase is random relative to the RF input: vLO = ALO cos( 0 t + 0 ) If we are so unlucky that 0 = 90 , then the output of the mixer will be zero A(t)ALO sin( 0 t) cos( 0 t)dt T A(t)ALO T sin( 0 t) cos( 0 t)dt = 0 A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 15/31 IQ-Mixer cos( LO t) I RF Q sin( LO t) To avoid this situation, we can phase lock the LO to the RF by transmitting a pilot tone. Alternatively, we can use two mixers As we shall see, there are other bene ts to such a mixer. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 16/31 AM Modulation We can see that an upconversion mixer is a natural amplitude modulator If the input to the mixer is a baseband A(t), signal then the output is an AM carrier vo (t) = A(t) cos( LO t) How do we modulate the phase? A PLL is one way to do it. The IQ mixer is another way. Let s expand a sinusoid that has AM and PM vo (t) = A(t) cos( 0 t + (t)) = A(t) cos 0 t cos (t) + A(t) sin 0 t sin (t) = I(t) cos 0 t + Q(t) sin 0 t A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 17/31 I-Q Plane Q I(t) = A(t) cos (t) I Q(t) = A(t) sin (t) We can draw a trajectory of points on the I-Q plane to represent different modulation schemes. The amplitude modulation is given by I 2 (t) + Q2 (t) = A2 (t)(cos2 (t) + sin2 (t)) = A2 (t) A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 18/31 General Modulator The phase modulation is given by Q(t) sin (t) = = tan (t) I(t) cos (t) or (t) = tan 1 Q(t) I(t) The IQ modulator is a universal digital modulator. We can draw a set of points in the IQ plane that represent symbols to transmit. For instance, if we transmit I = 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 18 p. 19/31 Digital Modulation: BPSK/QPSK Q Q 01 1 0 I 10 00 I 11 For instance, if we transmit I(t) = 1, this represents one bit transmission per cycle. But since the I and Q are orthogonal signals, we can improve the ef ciency of transmission by also transmitting symbols on the Q axis. If we select four points on a circle to represent 2 bits of information, then we have a constant envelope modulation scheme. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 20/31 Modulation Waveforms 1 0.75 0.5 0.25 1 2 3 4 5 6 -0.25 -0.5 -0.75 -1 The data wave form (left) is modulated onto a carrier (below). The rst plot shows a simple on/off keying. The second plot shows binary phase shift keying on one channel (I). 1 0.75 1 0.75 0.5 0.5 0.25 0.25 1 2 3 4 5 6 1 2 3 4 5 6 -0.25 -0.25 -0.5 -0.5 -0.75 -0.75 -1 -1 A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 21/31 More Bits Per Cycle Q Q 010 011 100 001 000 I I 101 110 111 Eventually, the constellation points get very close together. Because of noise and distortion, the received spectrum 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 in the transmission. We can increase the radius but that requires more power. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 22/31 I/Q Hartley Mixer sin( LO t) 90 RF cos( LO t) 90 IF An I/Q mixer implemented as shown above is known as a Hartley Mixer. We shall show that such a mixer can be designed to select either the upper or lower sideband. For this reason, it is sometimes called a single-sideband mixer. We will also show that such a mixer can perform image rejection. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 23/31 Delay Operation Consider the action of a 90 delay on an arbitrary signal. Clearley sin(x + 90 ) = cos(x). Even though this is obvious, consider the effect on the complex exponentials ejx j /2 e jx+j /2 sin(x ) = 2 2j ejx ( j) e jx (j) ejx e j /2 e jx ej /2 = = 2j 2j ejx + e jx = = cos(x) 2 Positive frequencies get multiplied by j and negative frequencies by +j. This is true for a narrowband signal when it is delayed by 90 . A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 24/31 Complex Modulation Consider multiplying a waveform f (t) by ej t and taking the Fourier transform F ej 0 t f (t) = f (t)ej 0 t e j t dt Grouping terms we have = f (t)e j( 0 )t dt = F ( 0 ) It is clear that the action of multiplication by the complex exponential is a frequency shift. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 25/31 Real Modulation Now since cos(x) = (ejx + e jx )/2, we see that the action of time domain multiplication is to produce two frequency shifts F {cos( 0 t)f (t)} = 1 1 F ( 0 ) + F ( + 0 ) 2 2 These are the sum and difference (beat) frequency components. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 26/31 Image Problem (Again) Complex Modulation (Positive Frequency) RF ( 0 ) e RF ( ) j LO t IM ( ) IM + ( 0 ) RF + ( 0 ) e j LO t RF + ( ) IM ( 0 ) LO IM + ( ) LO Complex Modulation (Positive Frequency) RF ( ) RF ( + 0 ) ej LO t IM ( + 0 ) IM + ( ) LO RF + ( ) ej LO t RF + ( + 0 ) LO IM ( ) IM + ( + 0 ) Real Modulation LO IF IF LO We see that the image problem is due to to multiplication by the sinusoid and not a complex exponential. If we could synthesize a complex exponential, we would not have the image problem. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 27/31 Sine/Cosine Modulation Cosine Modulation LO LO IF IF Sine Modulation /j /j /j LO IF /j IF LO Delayed Sine Modulation LO IF IF LO Using the same approach, we can nd the result of multipling by sin and cos as shown above. If we delay the sin portion, we have a very desirable situation! The image is inverted with respect to the cos and can be cancelled. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 28/31 Image Rejection The image rejection scheme just described is very sensitive to phase and gain match in the I/Q paths. Any mismatch will produce only nite image rejection. The image rejection for a given gain/phase match is approximately given by 1 IRR( dB) = 10 log 4 A A 2 + ( ) 2 For typical gain mismatch of 0.2 0.5 dB and phase mismtach of 1 4 , the image rejection is about 30 dB - 40 dB. We usually need about 60 70 dB of total image rejection. A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 29/31 45 Delay Element cos( LO t) RF IF The passive R/C and C/R lowpass and highpass lters are a nice way to implement the delay. Note that their relative phase difference is always 90 . sin( LO t) Hlp = 1 = arctan RC 1 + j RC Hhp = j RC = arctan RC 1 + j RC 2 A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 30/31 Gain Match / Quadrature LO Gen But to have equal gain, the circuit must operate at the 1/RC frequency. This restricts the circuit to relatively narrowband systems. Multi-stage polyphase circuits remedy the situation but add insertion loss to the circuit. The I/Q LO signal is usually generated directly rather than through an high-pass and low-pass network. Two ways to generate the I/Q LO is through a divide-by-two circuit (requires 2 LO) or a quadrature oscillator (requires two tanks). A. M. Niknejad University of California, Berkeley EECS 142 Lecture 18 p. 31/31

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