EEO 401 Digital Prof. Mark Fowler

Note Set #15 • Bandpass Signal Representations • Reading Assignment: Sect. 6.4.3 of Proakis & Manolakis

1/37 Definition: Bandpass Signal A Bandpass Signal is a signal x(t) whose X( f ) is nonzero only in some small band around some “central”

frequency fo. For example:

X( f ) = 0 for | f - fo| > W where W< fo. |X( f )|

foW  fo  fo+W foW fo fo+W f B

The bandwidth B of the bandpass signal = the width of the positive-frequency interval on which the signal is nonzero. (Note: this is consistent with the bandwidth definition for lowpass signals).

2/37 Definition: Bandpass Signal (cont.)

Note that the choice of fo is arbitrary: |X( f )|

 f o fo f |X( f )|

 fo fo f Bandpass signals are encountered when receiving radio frequency (RF) signals such as communication and radar signals. In the analysis and actual processing of BP signals it is convenient to work with a related, equivalent signal called the Equivalent Lowpass Signal. This is a natural generalization of the idea of used in sophomore-level circuits. 3/37 Recall: Phasor Idea Used in Circuits j Idea: Replace Acos(2 fo t+) by complex DC value Ae  X( f ) x(t)  Acos(2 fot ) First, the sinusoid A A   e j(2 fot )  e j(2 fot ) 2 2  fo fo f gets represented by a complex-valued signal called the analytic signal: Xa( f )

xa(t) = A exp{ j(2 fo t+)} = A cos(2 fo t+) + jAsin(2 fo t+) fo f Then to get the phasor, we frequency-shift the analytic signal down by f to get: o Xl( f )

xl = exp{–j2 fot} xa(t) = Aej 4/37 f Recall: Phasor Idea Used in Circuits (cont.) phasor = equivalent lowpass signal representing the sinusoid (that’s why we used the subscript l – for lowpass). Note that this equivalent lowpass signal is complex valued, whereas the bandpass signal (the sinusoid) it represents is real valued. Alternate View – Frequency Domain: 1. Suppress the negative frequency part of the sinusoid:

Xa( f )

f f 2. Frequency-shift the positive ofrequency part down to DC:

Xl( f )

f 5/37 Frequency-Domain View of Equiv. LP Signal Now… use this FD view to do the same thing for a general bandpass signal that consists of more than one frequency. ….Then after that we interpret the results in the time domain.

Bandpass Signal’s Fourier Transform: X( f )

 fo fo f Now to get the FT of the so-called Analytic Signal we suppress the negative frequencies:

Xa( f ) Note: Since |Xa( f )| is NOT even-symmetric,

the TD signal xa(t) is fo f complex-valued. (see Porat p. 12, #9)6/37 F-D View of Equiv. LP Signal (cont.) System View of Generating Analytic Signal: define a system frequency response H( f ) such that

 j f  0  X a ( f )  X ( f )  jH ( f )X ( f ) H ( f )  0 f  0 then ˆ   X ( f )  jX ( f )  j f  0 where Xˆ ( f )  H ( f )X ( f )

X( f ) Xa( f )  H( f ) j

Called Hilbert Transformer 7/37 F-D View of Equiv. LP Signal (cont.)

Then to get the FT of the Equivalent Lowpass Signal, frequency- shift the analytic signal down by fo to get:

Xl( f ) Also called: Equivalent Signal f

Note that because |Xl( f )| does not necessarily have even symmetry, the equivalent lowpass signal is complex valued, ….. whereas the bandpass signal it represents is real valued.

Now… how do we describe the ELP signal in the Time-Domain?

8/37 T-D View of ELP Signal

Consider the IFT of H( f )X( f ): xˆ(t)  F 1{Xˆ ( f )}  F 1{H ( f )X ( f )}

xa (t)  x(t)  jxˆ(t) ()

Let xl(t) be the time-domain signal that corresponds to Xl( f ). Because it is the frequency-shifted version of xa(t) …. using the frequency-shift property of FT gives:

 j2fot xl (t)  e xa (t) () (Note: this is the same as an equation above for the phasor case!) This, can then be written as:

 j2fot xl (t)  e x(t)  jxˆ(t) 9/37 I&Q Form of ELP Signal An extremely useful viewpoint for the ELP signal is the I&Q form:

… since xl(t) is complex-valued (see comment above in frequency- domain discussion), we can write its real and imaginary parts, which we will denote as xl (t)  xi (t)  jxq (t) () where subscripts i and q are for In-phase (I) and Quadrature (Q).

We’d now like to find relationships between the bandpass signal x(t) and the I-Q components of the lowpass equivalent signal.

10/37 Relationship: I&Q Parts and BP Signal Solving () for the analytic signal gives

j2fot xa (t)  e xl (t) ()

(Makes sense… xa(t) is xl(t) shifted up.)  Using the I-Q form given in () gives:  j2fot  xa (t)  e xi (t)  jxq (t)  cos(2 fot)  jsin(2 fot) xi(t)  jxq (t)   xi (t)cos(2 fot)  xq (t)sin(2 fot)

 j xi (t)sin(2 fot)  xq (t)cos(2fot) By ()  x(t)  jxˆ(t) 11/37 Relationship: I&Q Parts and BP Signal (cont.) This shows how the I&Q components are related to the BP signal:

x(t)  xi (t)cos(2fot)  xq (t)sin(2fot) ()

Similarly – but less important – we have:

xˆ(t)  xi (t)sin(2fot)  xq (t)cos(2fot)

12/37 Envelope/Phase Form of ELP Signal This is an alternate form (but equally important to IQ form) of the ELP signal. Note in () that the I&Q form is a “rectangular form” for the complex ELP signal.

So… converting to a “polar form” gives: j (t) () xl (t)  A(t)e

where… Note Similarity to Phasor!! 2 2 But… A(t)  xi (t)  xq (t)  0 Time-varying Envelope &   Magnitude  xq (t)  θ(t)  arctan   xi (t)  13/37 Relationship: Env/Phase and I&Q Often we need to convert between the two forms (rect & polar). If in () we expand the complex exponential: j (t) xl (t)  A(t)e  A(t)cos[(t)]  jA(t)sin[ (t)] By () xi (t) xq (t)

Thus….

xi (t)  A(t)cos[ (t)] xq (t)  A(t)sin[ (t)]

14/37 Envelope/Phase Form of BP Signal We already saw Env/Phase form for the ELP signal… Do we get something similar for the original BP signal ??

Using () and () we can write  j2 fot j (t) xa (t)  e [A(t)e ]   A(t)e j[2 fot (t)]   A(t)cos[2 fot  (t)]  jA(t)sin[2 fot  (t)]  x(t)  jxˆ(t) By ()

x(t)  A(t)cos[2fot  (t)] (.) 15/37 Envelope/Phase Form of BP Signal (cont.)

So what we have just shown is: Any BP signal can be expressed as:

x(t) = A(t) cos[2 fo t + (t)] where A(t)  0.

Note: A(t) and (t) vary slowly compared to cos(2 fo t).

The LPE signal has the same envelope and phase as the BP signal … compare (.) and (). j (t) xl (t)  A(t)e

16/37 Analog Generation of I&Q Components

As stated earlier… processing for radar & communication is actually implemented using the ELP signal. • Thus we need some way to get the ELP signal from a received BP signal… • The I&Q form is the most commonly used

So… given the BP signal x(t)  A(t)cos[2fot  (t)]

we need to be able to extract through processing the I&Q signals:

xi (t)  A(t)cos[ (t)] xq (t)  A(t)sin[ (t)] 17/37 Analog Generation of I&Q Components (cont.) These give the clue as to how to extract the I-Q signals by using analog techniques. Using trigonometric identities: 

2x(t)cos(2fot)  2A(t)cos(2fot  (t))cos( 2fot)  A(t)cos[ (t)]  A(t)cos[2 (2 f )t  (t)]    o xi (t) x(t).... but centered at 2 fo

-2fo -fo 0 fo 2fo f

18/37 Analog Generation of I&Q Components (cont.) Similarly, we also get….

2xt ( )[ sin(2 ftooo )] 2 At ( )cos(2 ft  ( t )) [ sin(2 ft )] A()sin[()]ttAtftt ()sin[2(2 )  ()]  o  xtq () xtˆ( ).... but centered at 2 fo  Analog Circuitry to Generate I&Q

xi(t) LPF

2cos(2 f t) x(t) Osc.  o

fo 90º

-2sin(2fot) LPF xq(t) 19/37 Digital Generation of I&Q Components

Analog Devices One Technology Way PO Box 9106 Norwood, MA 02062-9106 20/37 www.analog.com Uses of These Ideas • Bandpass Signal Model – usually used to model RF signals in radar and communications – also often used to model acoustic signals in sonar – not generally used for audio/speech signals • Lowpass Equivalent Signal – used as a conceptual tool to aid analysis/design – used as the actual representation in real processing

Receiver From Our Case Study

Estimate Detect Classify Signal Signal Emitter Parameters

RF Sampling Digital Compress Front-End Sub-System “Front-End” & Data Link Analog   Digital 21/37 Uses of These Ideas (cont.) • Analytic Signal – generally used as a conceptual tool to prove results – usually applied directly to the continuous-time RF bandpass signal – There are occasions where we actually compute the analytic signal of a real-valued digital signal, • but usually applying it some real-valued lowpass signal. – See MATLAB Warning Below • of a Signal – generally used as a conceptual tool to prove results – There are occasions where we actually compute the Hilbert transform of a real-valued digital signal, • but usually applying it some real-valued lowpass signal. – See MATLAB Warning Below

22/37 MATLAB Warning MATLAB has a command that is called “hilbert”. The Help entry on MATLAB for this command is: “HILBERT(X) is the Hilbert transform of the real part of vector X. The real part of the result is the original real data; the imaginary part is the actual Hilbert transform.”

Thus, executing hilbert(x) does NOT return the Hilbert transform of x; • It gives the analytic signal – see ().

23/37 Summary of Relationships Equiv. Lowpass Model

BP Signal & Its Hilbert Transform  jft2 o xtla() xte () xt() At ()cos[2 ft ()] t Analytic Signal jt() o xtl () Ate () xtˆ() At ()sin[2 ft ()] t x (t)  x(t)  jxˆ(t) o a  xtli() xt () jxt q () j2 fot  xa (t)  xl (t)e *****************  xt() Re{ x ()} t j[2 fot (t)] xt() Re{()} xt a xa (t)  A(t)e il

xtˆ() Im{ x ()} t *********** xtql() Im{()} xt a   A(t)  xa (t)  xt() At ()cos[()] t xt() xioqo ()cos(2 t ft ) x ()sin(2 t ft ) (t)  [xa (t)]  2fot i

 xtq () At ()sin[()] t xtˆ() xioqo ()sin(2 t ft ) x ()cos(2 t ft )  ******************************* xioo()txt ()cos(2 ftxt )ˆ ()sin(2 ft) At() x22 () t xˆ () t  xqoo()txtˆ ()cos(2 ftxt ) ()sin(2 ft ) xtˆ() ***************** (tft ) arctan 2 o xt() 22 At() xtliq () xt ()  xt ()  xtq () (txt )l ( )  arctan  xti ()

X( f ) Xa( f ) Xl( f )

 f f f –B/2 B/2 o o  fo fo f f 24/37 B B Sampling Rate Needed for ELP Signal

Given a complex-valued equivalent lowpass signal, what is an appropriate sampling rate to use? To answer this question… look at the ELP signal’s Fourier transform: Xl( f )

–B/2 B/2 f

Sampling this signal is no different than sampling some real-

valued lowpass signal: choose Fs > 2fmax

…. in this case gives Fs > 2(B/2) = B. Now does this make sense?

25/37 Sampling Rate Needed for ELP Signal (cont.) Now does this make sense? Bandpass Sampling on the corresponding bandpass signal (BPS)

….would require Fs > 2B, BUT… need only half that rate for the ELP signal!!! Do we really need only half the amount of information to represent the ELPS as we need for the BPS? Would that even make sense? Since ELPS  BPS ? It doesn’t at first!!!! BUT … the ELPS is complex it requires a real sample value and an imaginary sample value for each signal sample

[(I+Q) @ Fs = B] = [BPS @ Fs = 2B]

# of Real Values for ELPS # of Real Values for BPS 26/37 Examples of IQ Signals

27/37 Binary Phase Shift Keying (BPSK) xRF (t)  Acos(ct  (t))

/2  (t) -/2 t

“1”“0” “1” “1”

Acos[ t   / 2] xRF (t) c

t

Acos[ct   / 2] Acos[ct   / 2] 28/37 Binary Phase Shift Keying (Cont.) What is the LPE signal for BPSK? Applying the General Result to BPSK gives: x (t)  Acos[ (t)] xl (t)  xi (t)  jxq (t) i xq (t)  Asin[ (t)] xi (t) A

xq (t) A

-A “1”“0” “1” “1”

29/37 Quad Phase Shift Keying (QPSK) xRF (t)  Acos(ct  (t))

 /2  (t) -/2 t

“10”“01” “11” “00”

30/37 Quad Phase Shift Keying (cont.)

  /2 xi (t)  Acos[ (t)]  (t) xq (t)  Asin[ (t)] -/2 t

“10”“01” “11” “00”

xi (t) “Signal Constellation” A Q “10” -A “11” “00” xq (t) A I “01” -A

31/37 Radar Pulse Train - Linear FM Most radars use signals that are trains of pulses. Those pulses have a carrier frequency that also has frequency imparted on it – and often the frequency is varied linearly with time (during the pulse).

Sinusoid with: x(t)  A(t)cos[ t  (t)]  c • Carrier Frequency c • Time-Varying Phase (t) d (t)  [ t (t)] i dt c  Gives Inst. Frequency of: d • Center c d  c   (t) • Frequency Deviation  (t) dt dt

Want this to vary linearly

32/37 Radar Pulse Train - Linear FM (cont.) To get a linear variation of the frequency we need a quadratic variation of the phase:

x(t)  A(t)cos[ct  (t)] A(t)

t

(t)

t d  (t) dt t

33/37 Radar Pulse Train - Linear FM (cont.) The LPE Signal for the radar signal is:   x (t)  A(t)e j[ lt (t)] l   A(t)cos[ t  (t)]  j A(t)sin[ t  (t)]  l  l xi (t) xq (t)

Small Value: Due to Uncertain

Knowledge of c

d  [ t  (t)] dt l l t

34/37 Radar Pulse Train - Linear FM (cont.) The IQ parts of Linear FM Radar Pulse:

xi(t)

0 20 40 60 80 100 120 140 time samples

xq(t)

0 20 40 60 80 100 120 140 time samples 35/37 Radar Pulse Train - Linear FM (cont.) Say you intercepted a radar signal and want to measure its modulation rate (i.e rate of frequency change). You could do it like this:

Antenna lt  (t) x (t) RF i Get Front atan[x (t)/x (t)] d2/dt2 I&Q q i End xq(t)

Rate of Change of Frequency

36/37 Radar Pulse Train - Linear FM (cont.) Say you wanted to simulate this freq-rate-measurement processing to test how well it works…. How would you write code??? Would you need to simulate the RF signal and the RF Front-End? NO!!!!! Would need a HUGE Sampling Antenna Rate to Simulate the RF signal x (t) RF i Get Front atan[x (t)/x (t)] d2/dt2 I&Q q i End xq(t) Simulate using this:

xi[n] Generate atan{x [n]/x [n]} d2/dt2 I&Q q i

xq[n] 37/37