EE456 Digital Communications

EE456 Digital Communications Professor Ha Nguyen September 5 EE456 Digital Communications

Block Diagram of Binary Communication Systems m ( t { b k } b k = s( t b = s ( t k m ˆ ( t { bˆ } k r( t Bits in two different time slots are statistically independent. a priori probabilities: P[b k = ] = P, P[b k = ] = P. w( t Signals s (t and s (t have a duration of seconds and finite energies: E = s (tdt, E = s (tdt. Noise w(t is stationary Gaussian, zero-mean white noise with two-sided power spectral density of N / (watts/hz: E{w(t} =, E{w(tw(t +τ} = N ( δ(τ, w(t N, N. EE456 Digital Communications

m ( t { k }!" b k &%$' ($!*+ '', b = s ( t b = s ( t k #$!!%.!*''! m ˆ ( t { bˆ } k -+%$' (. /, r( t w( t Received signal over [(k,k ]: r(t = s i (t (k +w(t, (k t k. Objective is to design a receiver (or demodulator such that the probability of making an error is minimized. Shall reduce the problem from the observation of a time waveform to that of observing a set of numbers (which are random variables. EE456 Digital Communications 3

Can you Identify the Signal Sets {s (t,s (t}? NRZ (Tx NRZ (Rx OOK (Tx OOK (Rx.5.5.5 3 3.5 4 4.5 5 t/ 5 5.5.5.5 3 3.5 4 4.5 5 t/.5.5.5 3 3.5 4 4.5 5 t/ 5 5.5.5.5 3 3.5 4 4.5 5 t/ EE456 Digital Communications 4

Geometric Representation of Signals s (t and s (t (I Wish to represent two arbitrary signals s (t and s (t as linear combinations of two orthonormal basis functions φ (t and φ (t. φ (t and φ (t form a set of orthonormal basis functions if and only if: φ (t and φ (t are orthonormal if: Tb φ (tφ (tdt = (orthogonality, Tb Tb φ (tdt = φ (tdt = (normalized to have unit energy. If {φ (t,φ (t} can be found, the representations are s (t = s φ (t +s φ (t, s (t = s φ (t +s φ (t. It can be checked that the coefficients s ij can be calculated as follows: s ij = Tb s i (tφ j (tdt, i,j {,}, An important question is: Given the signal set s (t and s (t, can one always find an orthonormal basis functions to represent {s (t,s (t} exactly? If the answer is YES, is the set of orthonormal basis functions UNIQUE? EE456 Digital Communications 5

Geometric Representation of Signals s (t and s (t (II Provided that φ (t and φ (t can be found, the signals (which are waveforms can be represented as vectors in a vector space (or signal space spanned (i.e., defined by the orthonormal basis set {φ (t,φ (t}. φ ( t s s s ( t s s (t = s φ (t+s φ (t, d = d s (t = s φ (t+s φ (t, Tb E s ( t s ij = s i (tφ j (tdt, i,j {,}, Tb E i = s i E (tdt = s i +s i, i {,}, s φ ( t Tb d = d = [s (t s (t] dt Tb s i (tφ j (tdt is the projection of signal s i (t onto basis function φ j (t. The length of a signal vector equals to the square root of its energy. It is always possible to find orthonormal basis functions φ (t and φ (t to represent s (t and s (t exactly. In fact, there are infinite number of choices! EE456 Digital Communications 6

Gram-Schmidt Procedure Gram-Schmidt (G-S procedure is one method to find a set of orthonormal basis functions for a given arbitrary set of waveforms. Let φ (t s (t E. Note that s = E and s =. Project s (t = s (t E onto φ (t to obtain the correlation coefficient: Tb s (t ρ = φ (tdt = E Tb s (ts (tdt. E E 3 Subtract ρφ (t from s (t to obtain φ (t = s (t E ρφ (t. 4 Finally, normalize φ (t to obtain: φ (t = = Tb ρ φ (t [ = φ (t] dt φ (t ρ [ s (t ρs ] (t. E E EE456 Digital Communications 7

Gram-Schmidt Procedure: Summary φ ( t φ ( t s ( t φ (t = s (t s ( t φ (t = s E α E s d ρφ ( t s ( t ρ = cos( α φ ( t s = s = d = E, [ s (t ρs ] (t, ρ E E Tb s (tφ (tdt = ρ E, ( ρ E, Tb [s (t s (t] dt = E ρ E E +E. EE456 Digital Communications 8

Example s ( t s ( t V Tb 3 V φ ( t Tb 567 4 s ( t E 89: s ( t E φ ( t (a Signal set. (b Orthonormal function. (c Signal space representation. EE456 Digital Communications 9

Example s ( t s ( t V V ; Tb ; φ ( t V <=> φ ( t φ ( t E s ( t? @AB Tb? s ( t E φ ( t EE456 Digital Communications

Example 3 s ( t s ( t V V Tb C α C V φ( t, α I E 3E F G D, H E α = ( E, s ( t α = 4 E T α = b ρ = increasingα, ρ = s ( t ( E, φ ( t Tb s (ts (tdt E [ ] V α V (T V b α = α EE456 Digital Communications

Example 4 s ( t V s ( t 3V b φ ( t J T KLM Tb φ ( t 3 J Tb N N 3 OPQ EE456 Digital Communications

φ ( t s ( t E E 6 o s ( t φ ( t 3E ρ = Tb s (ts (tdt = Tb / E E ( 3 3 V t V dt =, [ s(t ρ s(t ] = [s (t E E E φ (t = ( 3 4 3 s = E, s = E. [ Tb d = [s (t s (t] dt] ( = 3 E. 3 s(t ], EE456 Digital Communications 3

Example 5 θ = 3π ρ = φ ( t s (t = E cos(πf ct, s (t = E cos(πf ct + θ. where f c = k, k an integer. θ = π ρ = E θ s ( t φ ( t locus of s ( t as θ varies from to π. s ( t θ = π ρ = EE456 Digital Communications 4

Obtaining Different Basis Sets by Rotation φ ( t φ ( t s ( t φ ( t s s φ ( t φ ( t s s ( t s φ ( t = s s s θ s s s s ( t φ ( t s s θ s s s ( t φ ( t s s [ ˆφ (t ˆφ (t ] [ = cosθ sinθ sinθ cosθ ][ φ (t φ (t ]. s = s (a Show that, regardless of the angle θ, the set {ˆφ (t, ˆφ (t} is also an orthonormal basis set. (b What are the values of θ that make ˆφ (t perpendicular to the line joining s (t to s (t? For these values of θ, mathematically show that the components of s (t and s (t along ˆφ (t, namely ŝ and ŝ, are identical. Remark: Rotating counter-clockwise for positive θ and clock-wise for negative θ. EE456 Digital Communications 5

Example: Rotating {φ (t,φ (t} by θ = 6 to Obtain {ˆφ (t, ˆφ (t} [ ˆφ(t ˆφ (t ] [ = cosθ sinθ sinθ cosθ φ ( t ][ φ(t φ (t ] { ˆφ(t = cosθ φ (t + sinθ φ (t ˆφ (t = sinθ φ (t + cosθ φ (t φ ( t.5 ( + 3 φ ( t φ ( t ( 3. 5 ( 3. 5 ( + 3 EE456 Digital Communications 6

Gram-Schmidt Procedure for M Waveforms {s i (t} M i= φ (t = s (t s (tdt, φ i (t = φ i (t [, i =,3,...,N, φ i (t] dt φ i (t = s i(t i ρ ij φ j (t, Ei ρ ij = j= s i (t Ei φ j (tdt, j =,,...,i. If the waveforms {s i (t} M i= form a linearly independent set, then N = M. Otherwise N < M. EE456 Digital Communications 7

Example: Find a Basis Set for the Following M = 4 Waveforms s ( t s ( t 3 s ( t 3 s ( t 4 3 EE456 Digital Communications 8

Answer Found by Applying the G-S Procedure φ ( t φ ( t 3 3 φ ( t s (t s (t s 3 (t s 4 (t = φ (t φ (t φ 3 (t EE456 Digital Communications 9

Representation of Noise with Walsh Functions 5 x (t 5 5 x (t 5 φ (t φ (t φ 3 (t φ 4 (t...3.4.5.6.7.8.9 t Exact representation of noise using 4 Walsh functions is not possible. EE456 Digital Communications

The First 6 Walsh Functions...3.4.5.6.7.8.9 t Exact representation might be possible by using many more Walsh functions. EE456 Digital Communications

The First 6 Sine and Cosine Functions Can also use sine and cosine functions (Fourier representation..5.5.5.5.75 t EE456 Digital Communications

Representation of Noise To represent noise w(t, need to use a complete set of orthonormal functions: w(t = w i φ i (t, where w i = i= Tb w(tφ i (tdt. The coefficients w i s are random variables and understanding their statistical properties is imperative in developing the optimum receiver. Of course, the statistical properties of random variables w i s depend on the statistical properties of the noise w(t, which is a random process. In communications, a major source of noise is thermal noise, which is modelled as Additive White Gaussian Noise (AWGN: White: The power spectral density (PSD is a constant (i.e., flat over all frequencies. Gaussian: The probability density function (pdf of the noise amplitude at any given time follows a Gaussian distribution. When w(t is modelled as AWGN, the projection of w(t on each basis function, w i = w(tφ i (tdt, is a Gaussian random variable (this can be proved. For zero-mean and white noise w(t, w, w, w 3,... are zero-mean and uncorrelated random variables: Tb } E{w i} = E{ w(tφ i(tdt = E{w(t}φ i(tdt =. } = E{w iw j} = E{ Tb dλw(λφ i(λ dτw(τφ j(τ { N, i = j, i j Since w(t is not only zero-mean and white, but also Gaussian {w,w,...} are Gaussian and statistically independent!!! EE456 Digital Communications 3.

Need to Review Probability Theory & Random Processes Chapter 3 Slides EE456 Digital Communications 4

Observing a waveform Observing a set of numbers t = r( t = s ( t + w( t i AWGN φ ( t φ ( t φ ( t 3 φ n ( t ( dt ( dt ( dt ( dt t = t = r t = r = s i + w = s i + w r = + w 3 3 r = + w n n Choose φ (t and φ (t so that they can be used to represent the two signals s (t and s (t exactly. The remaining orthonormal basis functions are simply chosen to complete the set in order to represent noise exactly. The decision can be based on the observations r,r,r 3,r 4,... Note that r j, for j = 3,4,5,..., does not depend on which signal (s (t or s (t was transmitted. EE456 Digital Communications 5

Optimum Receiver The criterion is to minimize the bit error probability. Consider only the first n terms (n can be very very large, r = {r,r,...,r n} Need to partition the n-dimensional observation space into two decision regions, R and R. Decide a "" was transmitted if r R falls in this region. R R Observation space R Decide a "" was transmitted if r R falls in this region. R EE456 Digital Communications 6

P[error] = P[( decided and transmitted or ( decided and transmitted]. = P[ D, T ]+P[ D, T ] = P[ D T ]P[ T ]+P[ D T ]P[ T ] = P[ r R T ]P +P[ r R T ]P = P f( r T d r +P f( r T d r R R = P f( r T d r +P f( r T d r R R R = P f( r T d r + [P f( r T P f( r T ]d r R R = P + [P f( r T P f( r T ]d r R = P [P f( r T P f( r T ]d r. R The minimum error probability decision rule is { P f( r T P f( r T decide ( D P f( r T P f( r T < decide ( D. EE456 Digital Communications 7

r( t = s ( t + w( t i AWGN, PSD N φ ( t φ ( t φ 3 ( t ( dt ( dt ( dt t = t = t = t = r = + w r s i = s i + w r = + w 3 3 ( g r > < Pr[error] = P + [ P f ( r P f ( r ]dr T R R g ( r = P + [ P f ( r P f ( r ]dr T g ( r > Optimal decision rule: g( r < R D g( r > T T D D R n φ n ( t ( dt r = + w n n g( r = R D g( r < EE456 Digital Communications 8

Equivalently, f( r T f( r T The expression f( r T is called the likelihood ratio. f( r T D D P P. ( The decision rule in ( was derived without specifying any statistical properties of the noise process w(t. Simplified decision rule when the noise w(t is zero-mean, white and Gaussian: (r s +(r s D D (r s +(r s +N ln( P P. For the special case of P = P (signals are equally likely: (r s +(r s D D (r s +(r s. minimum-distance receiver! EE456 Digital Communications 9

Minimum-Distance Receiver (r s +(r s }{{} d d r φ ( t D (r s +(r s D }{{} D d D d s ( t r( t ( r, r Choose ( s, s s ( t ( s, s s ( Choose s ( t t d d φ ( t r EE456 Digital Communications 3

Correlation Receiver Implementation t = r( t = s ( t + w( t i φ ( t T ( dt S ( dt t = r r Compute ( r s + ( r s i i N ln( P for i =, and choose the smallest i Decision φ ( t t = r( t φ ( t V ( dt U ( dt t = r r Form the dot product a a r s i N E ln( P WXYYZ[ \X[ ]^_`[Z\ Decision φ ( t N E ln( P EE456 Digital Communications 3

Receiver Implementation using Matched Filters t = r( t ( dt r Decision φ ( t t = ( dt r r( t φ ( t t = Tb h t = φ ( T ( b t t = r Decision h ( t = φ ( T b t r EE456 Digital Communications 3

Example 5.6 s ( t s ( t.5.5. 5 b.5 b cde φ ( t φ ( t f.5 f ghi EE456 Digital Communications 33

s ( t φ ( t.5 s ( t.5.5 φ ( t s (t = φ (t+ φ (t, s (t = φ (t+φ (t. EE456 Digital Communications 34

For each value of the signal-to-noise ratio (SNR, Matlab simulation was conducted for transmitting/receiving 5 equally-likely bits. 3 (E +E / N =.99 (db; P[error]=. 3 (E +E / N =6.7 (db; P[error]=. φ(t φ(t 3 3 φ (t 3 3 φ (t EE456 Digital Communications 35

s ( t φ ( t r s ( t φ ( t r jkl.5 s ( t mno.5 s ( t Choose s ( t Choose s ( t φ ( t Choose s ( t Choose s ( t φ ( t.5.5 r.5.5 r φ ( t r pqr s ( t Choose s ( t.5 s ( t Choose s ( t φ ( t.5.5 r (a P = P =.5, (b P =.5, P =.75. (c P =.75, P =.5. EE456 Digital Communications 36

Example 5.7 s (t = φ (t +φ (t, s (t = φ (t φ (t. φ ( t φ ( t 3 s s EE456 Digital Communications 37

φ ( t r s ( t N y P v ln w t 4 x P u Choose s ( t φ ( t Choose s ( t r s ( t EE456 Digital Communications 38

r(t r φ ( t 3 ( dt t = z{ }~~ { N = P T ln 4 P ƒ r T choose s ( t r < T choose s ( t Tb ˆ ~Š r(t h ( t 3 Tb t = r Œ Ž r T choose s ( t r < T choose s ( t š œ N = P T ln 4 P EE456 Digital Communications 39

Implementation with One Correlator/Matched Filter Always possible by choosing ˆφ (t and ˆφ (t such that one of the two basis functions is perpendicular to the line joining the two signals. φ ( t s ( t φ ( t φ ( t s s = s s s θ s s s ( t φ ( t The optimum receiver is still the minimum-distance receiver. However the terms (ˆr ŝ and (ˆr ŝ are the same on both sides of the comparison and hence can be removed. This means that one does not need to compute ˆr! (ˆr ŝ + (ˆr ŝ }{{} ˆr d D ŝ + ŝ D }{{} midpoint of two signals s D D (ˆr ŝ + (ˆr ŝ (ˆr ŝ (ˆr ŝ D }{{} D ( N/ + ŝ ŝ ln d ( P T. P } {{ } equal to if P =P EE456 Digital Communications 4

T r(t b ˆr ( dt t = žÿ Ÿ rˆ T rˆ < T D D ( φˆ t Threshold T r(t h( t = ˆ φ( T t b t = ˆr ª«ª rˆ T ³ rˆ < T ³ D D ±² s (t s (t ŝ + ŝ ˆφ (t = (E ρ E E + E, T Threshold T ( N/ + ln ŝ ŝ ( P P. EE456 Digital Communications 4

Example 5.8 ( φˆ t φ ( t s ( t E s ˆ = s ˆ ( φˆ t θ = π / 4 E s ( t φ ( t ˆφ (t = ˆφ (t = [φ (t + φ (t], [ φ (t + φ (t]. EE456 Digital Communications 4

T r(t b (» dt t = rˆ T ˆr µ ¹ ºµ¹ rˆ < T D D ( φˆ t Threshold T Tb ¾ ¼ ½ h(t r(t Ê t = rˆ T Ë D ˆr ÀÁÂÃÄÅÄÆÁÅ rˆ < T Ë Threshold T ÇÈÉ D EE456 Digital Communications 43

Receiver Performance ktb To detect b k, compare ˆr = r(tˆφ (tdt to the threshold ((k T = ŝ+ŝ N + (ŝ ŝ ln P. P ÍÎÏÐÑÐÒÓ ÔÒÕÓÖ ØÙ ( rˆ f ( r f T ˆ T ŝ ŝ choose Ì choose T T T P[error] = P[( transmitted and decided or ( transmitted and decided] = P[( T, D or ( T, D]. ˆr EE456 Digital Communications 44

ßàááàâãä ßàåæ ìàâíë ìàâíë f r çàèéâêåë çàèéâêåë îãéïåéâä ðññéòàåéâä ˆ f ( r ( T ˆ T ðññéòàåéâä ìàâíë ìàâíë ßàåæ îãéïåéâä óôõæ çàèéâêåë çàèéâêåë ßàááàâãä öõàä ó óôõæ öõàä ó ôæøõñéíéáë îãéïåêéñ ùâéøæää ôæøõñéíéáë îãéïåêéñ ùâéøæää ŝ ŝ ˆr ÚÛÜÝ Þ choose ÚÛÜÝ û ú choose T T T P[error] = P[ T, D] + P[ T, D] = P[ D T]P[ T] + P[ D T]P[ T] T = P f(ˆr Tdˆr +P f(ˆr Tdˆr T }{{}}{{} Area B Area A ( [ ( ] T ŝ T ŝ = P Q + P Q. N/ N/ EE456 Digital Communications 45

Q-function λ e π ý üÿý ÿ üýþÿ þý λ x Area = Q( x Q(x ( exp λ dλ. π x 4 Q(x 6 8 3 4 5 6 x EE456 Digital Communications 46

Performance when P = P ( r T f ( r T f s s choose choose T T = Q s s ( N / s + s T = ( ( ŝ ŝ distance between the signals P[error] = Q = Q. N / noise RMS value Probability of error decreases as either the two signals become more dissimilar (increasing the distances between them or the noise power becomes less. To maximize the distance between the two signals one chooses them so that they are placed 8 from each other s (t = s (t, i.e., antipodal signaling. The error probability does not depend on the signal shapes but only on the distance between them. EE456 Digital Communications 47 r

Example 5.9 φ ( t t s ( t T φ ( t t T s ( t (a Determine and sketch the two signals s (t and s (t. EE456 Digital Communications 48

(b The two signals s (t and s (t are used for the transmission of equally likely bits and, respectively, over an additive white Gaussian noise (AWGN channel. Clearly draw the decision boundary and the decision regions of the optimum receiver. Write the expression for the optimum decision rule. (c Find and sketch the two orthonormal basis functions ˆφ (t and ˆφ (t such that the optimum receiver can be implemented using only the projection ˆr of the received signal r(t onto the basis function ˆφ (t. Draw the block diagram of such a receiver that uses a matched filter. (d Consider now the following argument put forth by your classmate. She reasons that since the component of the signals along ˆφ (t is not useful at the receiver in determining which bit was transmitted, one should not even transmit this component of the signal. Thus she modifies the transmitted signal as follows: s (M (component (t = s(t of s (t along ˆφ (t s (M (t = s(t (component of s (t along ˆφ (t Clearly identify the locations of s (M (t and s(m (t in the signal space diagram. What is the average energy of this signal set? Compare it to the average energy of the original set. Comment. EE456 Digital Communications 49

s ( t s ( t 3 t t 3 EE456 Digital Communications 5

!"# D φ ( t M s ( t t ˆ φ ( t D M s ( t π θ = 4 s ( t T φ ( t t T s ( t ˆ φ ( t EE456 Digital Communications 5

[ ˆφ (t ˆφ (t ] [ = cos( π/4 sin( π/4 sin( π/4 cos( π/4 ][ φ (t φ (t ] [ = ] [ φ (t φ (t ]. ˆφ (t = [φ (t φ (t], ˆφ (t = [φ (t+φ (t]. ˆ ( φ t ˆ φ ( t / t / t r(t h( t = ˆ φ ( t / t t = rˆ $ rˆ < $ D D EE456 Digital Communications 5

Antipodal Signalling w( t PSD N ( Tb T b ( ( t + h ( t = p( t x( t T h ( ( R t = κ p Tb t s ( t = s ( t = p( t E = E = E y y( t t =k = ± ( κ E + w ( k ( k The pulse shaping filter h T = p(t defines the power spectrum density of the transmitted signal, which can be shown to be proportional to P(f. (, E The error performance, P[error] only depends on the energy E of p(t and noise PSD level N. Specifically, the distance between s (t and s (t is E (you should show this for yourself, algebraically or geometrically. Therefore ( E P[error] = Q. N N For antipodal signalling, the optimum decisions are performed by comparing the samples of the matched filter s output (sampled at exactly integer multiples of the bit duration with a threshold. Of course such an optimum decision rule does not change if the impulse response of the matched filter is scaled by a positive constant. EE456 Digital Communications 53

Scaling the matched filter s impulse response h R (t does not change the receiver performance because it scales both signal and noise components by the same factor, leaving the signal-to-noise ratio (SNR of the decision variable unchanged! In the above block diagram, h R (t = κp( t. We have been using κ = / E in order to represent the signals on the signal space diagram (which would be at ± E and to conclude that the variance of the noise component is exactly N /. For an arbitrary scaling factor κ, the signal component becomes ±κe, while the variance of the noise component is N κ E. Thus, the SNR is Signal power SNR = Noise power = (±κe N κ E = E, (indepedent of κ! N In terms of the SNR, the error performance of antipodal signalling is ( E P[error] = Q N ( = Q SNR In fact, it can be proved that the receive filter that maximizes the SNR of the decision variable must be the matched filter. It is important to emphasize that the matching property here concerns the shapes of the impulse responses of the transmit and receive filters. EE456 Digital Communications 54

Outputs of the Matched/Mismatched Filters (No-Noise Scenario Clean received signals for rect and half-sine (HS shaping filters Output of a matched filter: rect/rect matching Outputs of HS/HS matched filter (red and HS/rect mismatched filter (pink When the matched filter is used, sampling at exact multiples of the bit duration maximizes the power of the signal component in the decision variable, hence maximizing the SNR. A timing error (imperfect sampling would reduce the power of the signal component, hence reducing the SNR, hence degrading the performance, i.e., increasing P[error]. When the receive filter is not matched to the transmit filter, the power of the signal component and the SNR are not maximized, even under perfect sampling! EE456 Digital Communications 55

Antipodal Baseband Signalling with Rectangular Pulse Shaping b k h( t = p( T t b EE456 Digital Communications 56

Antipodal Baseband Signalling with Half-Sine Pulse Shaping b k h( t = p( T t b EE456 Digital Communications 57

PSD Derivation of Arbitrary Binary Modulation Applicable to any binary modulation with arbitrary a priori probabilities, but restricted to statistically independent bits. s ( t T s ( t s ( t T b Tb T b 3T b 4T b t s ( 3 t s T (t = k= { s (t kt g k (t, g k (t = b, with probability P. s (t k, with probability P The derivation on the next slide shows that: ( ( S st (f = P P S (f S (f P S n +P T S n + b n= T δ (f ntb. b EE456 Digital Communications 58

s T(t = E{s T(t} +s T(t E{s T(t} = v(t + q(t }{{}}{{} DC AC v(t = E{s T(t} = [P s (t k + P s (t k ] S v(f = k= D n δ (f ntb, D n = [ ( ( ] n n P S + P S, n= ( ( n n P S T + P S v(f = b S n= T δ (f ntb. b To calculate S q(f, apply the basic definition of PSD: E{ G T(f } S q(f = lim = = PP S (f S (f. T T ( ( n n S st (f = PP S (f S (f P S T + P + b S n= T δ (f ntb. b For the special, but important case of antipodal signalling, s (t = s (t = p(t, and equally likely bits, P = P =.5, the PSD of the transmitted signal is solely determined by the Fourier transform of p(t: S st (f = P(f EE456 Digital Communications 59

Baseband Message Signals with Different Pulse Shaping Filters.5.5 Information bits or amplitude levels 4 6 8 4 Output of the transmit pulse shaping filter Rectangular 4 6 8 4 Half-sine pulse shaping filter 4 6 8 4 SRRC pulse shaping filter (β =.5 4 6 8 4 t/tb Magnitude (db Magnitude (db Magnitude (db PSD - Rectangular 4 6 f Tb PSD - Half-sine 4 6 f Tb PSD - SRRC (3 symbols long 4 6 Normalized frequency, f Tb EE456 Digital Communications 6

Building A Binary (Antipodal Comm. System in Labs #4 and #5 low-rate sequence, index k high-rate sequences, index n......... t data bits S/P bits-tolevels mapping N T sym N T pulse shaping filter h[ n] pnt ( T in [ ] i( nt ak [ ] ( nt kt k sn [ ] in [ ] hn [ ] ak [ ] pnt ( kt sym cos( ˆ c n NCO fˆc k DAC correction filter Figure : Block diagram of the transmitter. sym F ˆ c c s DAC st ( a( k pt ( kts cos( ct k T EE456 Digital Communications 6

s( t T x[ n ] x[ n d] ω d c = θ h [ n] = p( nt R x [ ] c n cos( ω n c θ sin( ω c n θ x [ ] s n θ fc T sym k = n T Figure : Block diagram of the receiver. EE456 Digital Communications 6