Digital IF Receivers

Required A/D converter performance :

 Requires sampling at least at the Nyquist rate 1 / Ts  2  f IF :
 The signal level is typically as small as  100  V(-80 dBV) and the
quantization and thermal noise of the ADC must be smaller than
10  V(-100 dBV).

100 μV

10μV

Jayanta Mukherjee
 ADC is affected by the clock jitter:
the higher the IF frequency sampled the higher
the signal slew rate and the error voltage.
• ADC dynamic range should be large enough to accommodate
variations in the signal level due to path loss and multipath fading.
  2 N  1 LSB 
ADC DR = 20log10    6.02  N (dB )
 LSB 
 

• BW of ADC should be sufficient to the IF BW. e.g. Say an ADC has

Analog i/p bw of 500 MHz (this is the analog 3 dB BW than can enter
the ADC before sampling) with sample rate 100 Msps. This means
that it can sample frequencies till 50 MHz .
 IF

Jayanta Mukherjee
Sampling IF
• If the sampling is at a rate fs = 1/Ts slightly smaller than fIF the signal is
downconverted to fIF - fs.
• This relaxes the sampling rate by a factor 2 since the Nyquist rate if 2 x fIF .
• Requires high-speed and high linearity sampling and hold circuits.
• Works only if |fIF – fs |is greater than Δf.

fS Δf
fIF –fS > Δf

IF Amplifier

Jayanta Mukherjee
Subsampling Receivers
• Signal is sampled at a rate fs equal or larger than 2 x Δf.
• Time and frequency representation of sampling:

1 
 n
s(t )    t  mTs   S ( f )     f  
m   Ts n    Ts 

• If the IF signal is xIF(t) the sampled output is in the time and frequency domain:

x (t )  s(t )  X ( f ) * S ( f )

• This generates multiple replica of the spectrum spaced every 1/Ts. (see Figure 5.33).
• No aliasing arises since the sampling rate fs is equal or larger than 2 x Δf.

Jayanta Mukherjee
 x(t)

x
Multiple mixing
s(t)
Ts

X(f)
N0/2 (noise floor)
f0
* S(f)
fs=1/Ts
Theoretically we will have
Infinite no of sampled products. 2Δf (minimum)
However all products above f0 are This one is chosen
filtered out
2m (N0/2) (noise floor)

Each product will add N0/2 noise psd.

Total 2m no of products between dc and f0 2Δf

Jayanta Mukherjee
Problem with Subsampling Receivers
Drawbacks

• The noise is multiplied by a factor 2m with m = f0/Δf (the factor 2 is due to the
folding from the negative frequency spectra).
• The clock phase noise power is also amplified by m2 (see reference below).

D.H Shen et al., “A 900 MHz RF Front End with Integrated Discrete Time Filters,” IEEE Journal of Solid State Circuits, vol 31, pp 1945-1954, Dec 1996

Jayanta Mukherjee
Transmitter Architectures
• Less variety of approaches because noise, interference rejection and band selectivity are more relaxed.
• Types of architectures:
 Direct conversion: homodyne
 Two-step conversion: heterodyne
• Signal conditioning. Examples:
 Baseband pulse shaping, e.g. raised-cosine pulses (Figure 5.37)
 GMSK baseband generation (Fig 5.38)
Issues
• Mismatch in I and Q paths → gives rise to cross talk and beam forming problems.
• PA efficiency → duplexer can cause 2-3 dB loss (corresponding to 370 mW for 1W power consumed.
Since PA efficiency rarely exceeds 50% this can lead to around 700 – 800 mW additional power wastage.

Jayanta Mukherjee
Fig 5.37

Jayanta Mukherjee
 • First level modulation (recall MSK)
• Phase continuity needs to be maintained
• Digital implementation more accurate

LPF

Jayanta Mukherjee
Direct Conversion Transmitter
• Modulation and up conversion is performed in one step. ISI is more
• Problem the PA output corrupts the oscillator spectrum by frequency pulling(due
to injection locking when frequencies are close).
Solution:
• PA o/p frequency must be sufficiently different from the LO frequency.
• This can be achieved by using a mixer to create the carrier ωc by adding two LO
frequencies. ωc=ω1+ω2
NF of mixer not so critical since I and Q are sufficiently high

Jayanta Mukherjee
Effect of I/Q Mismatch of QPSK at o/p of transmitter
– spurious beams in beam forming

Consider the symbol b1  1, b2  1, so that

x(t )  b1 (1   / 2) cos(c t   / 2)  b2 (1-  / 2) sin(c t -  / 2).
Say, b1 =1 and b 2 =1,
 x(t )  (1   / 2) cos(c t   / 2)  (1-  / 2) cos(c t -  / 2   / 2)
Jayanta Mukherjee
 Original beam

Distorted beam

/2

        
Array Factor (AF) = 2cos  cos      2   sin  cos    
2 2 4  2  2 2 4
        
+2cos  cos      2   sin  cos    
2 2 4   2  2 2 4
 is the beam direction w.r.t the x  axis

When,  =0 and  =0,

    
AF=2cos  cos    +2cos  cos    , AF is max for  
2 4 2 4 2
Two Step Transmitter
See Figure 5.42

• First step: the baseband I and Q undergo quadrature modulation at ω1.

• The first BPF suppresses the harmonics of the IF.
• Second step: the IF is up converted to the carrier ω2 + ω1.
• The second BPF removes the undesired image frequency. (50 to 60 dB
rejection typical).
• A single-sideband mixer (same as image rejection mixer) can also be used.

Jayanta Mukherjee
Channel filter can also be used instead of BPF
Two Step Transmitter
Advantages
• Cross talk is less since I/Q modulation takes place at lower frequency and
hence lower chance of I/Q imbalance.
• Channel filter may be used at first IF to limit unwanted signals in adjacent
channel.
• Reduces LO pulling by PA since they are at different frequencies.

Disadvantages
• Second BPF needs to be high Q to reject unwanted signals.
• More power consumption, more mixer spurs.

Jayanta Mukherjee