a0= ( Ep-tq)

Communication System II By Engr. Elemuwa E.P.

6.0 Threshold in Frequency modulation

When the noise power at the demodulator input is comparable to the carrier power, the threshold occurs.

However, in FM this effect is much more than in AM. As showed in fig 5.1 previously, the phasor En To, delays the decision-making instant tm an unnecessary length of time. The case tm = To gives the minimum delay for decision-making using a realizable filter.

Notice that both P(t) and h(t) have a width of To Seconds. Hence, po(t), which is convolution of P(t) and h(t) has a width of 2To Seconds, with its peak occurring at t = To. Also, P0( () = P(( )H(() = K1/P(()/2 ejwTo, but Po(t) is symmetrical about t = To.

Thus, K1 in eq. (7.7) multiplies both the signal and the noise by the same factor and does not affect the ratio (. The error probability, or the system performance is independent of the value of K1. For simplicity, K1 = 1. This equation becomes,

h(t) = P(To t)

(7.8a)

and H() = P (-()e-j (To

.. (7.8b)

The optimum filter is given in equ. (7.8) and is known as the match filter. Hence, at the output of this filter, the signal to rms noise amplitude ratio is maximum at the decisionmaking instant t = To. The matched filter is optimum in the sense that it maximizes the signal amplitude to rms noise ratio at the decision-making instant. Although, it is reasonable to assume that maximization of this particular signal to noise ratio will minimize the detection error probability. Thus, the maximum value of this signal to rms noise ratio attained by the matched filter is given in eg. (7.7a), and the peak amplitude Po(tm) = AP is found by substituting eq.(7.8b) into eg. (7.3), we have,

(7.9a)

Then, the noise power (2n is obtained by substituting eq.(7.8b) into eq. (7.4);

(2n = .(7.9b)

Hence,

..(7.9c)

and Pe = Q((max) = Q...(7.9d)Equation 7.9 shows that at the decisionmaking instant, the signal amplitude, and the rms noise amplitude depend on the waveform P(t) only through its energy Ep, As far as the system performance is concerned, when the matched filter receiver is used, all the waveforms used for P(t) are equivalent as long as they have the same energy. The alternative arrangement of the matched filter is shown in fig. 7.3, if the input to the matched filter is r(t), then the output y(t) is given by

where,

h(t) = P[(To t) and

h(t x) = P[(To (t x )] = P(x +To - t)

Then,

y(t) = (7.10a)

But, at the decision making instant t = To, we have,

y(TO) = ...(7.10b)

Input r(x) is assumed to start at x = 0 and P(x) = 0 for x > To,

(7.10c)

p(t)

t = T0r(t) =

y(t)

y(T0)

Fig. 7.3. Correlation detector

This type of arrangement is known as the correlation receiver and is equivalent to the matched filter receiver. The right hand side of eg. 7.10a is (rp (To = t), where (rp() is the cross correlation of the received pulse with p(t). Thus, the optimum detector measures the similarity of the received signal with the pulse p(t). Based on this similarity measure, it decides whether p(t) was transmitted or not.

7.2. Optimum Binary Receiver

In binary communication, we use two distinct pulses p(t) and q(t) to represent the two symbols. In this scheme, symbols are transmitted every Tb seconds. Let p(t) and q(t) be the two pulses used to transmit 1 and 0. The optimum receiver structure is shown in fig. 7.4.a. The incoming pulse is transmitted through a filter H(), and the output r(t) is sampled at Tb. The decision as to whether 1 or 0 was present at the input which depends on whether r(Tb) < or > ao.

Where,

ao = Optimum threshold

Let po(t) and qo(t) be the response of H() to inputs p(t) and q(t) respectively. Equation (7.3) becomes,

..(7.11a)

q0(Tb) = ..(7.11b)

Hence, (2n , the variance, or power, of the noise at the filter output is

(2n (7.11c)

t = Tb

r(t)

r(Tb)

Decision:m=0 if r(Tb)a0

(a)

(b)Fig. 7.4 Optimum binary threshold detection If n is the noise output at Tb, then the sampler output r(Tb) = qo(Tb) + n or Po(Tb) +n which depends on whether m = 0 or 1 is received, r is a gaussian RV of variance 6U2n and mean q0(Tb) or Po(Tb)depending on whether m = 0 or 1. Hence, the conditional PDFs of the sampled output r(Tb) are

This too PDFs of the sampled output r(Tb) are

This two PDFs are showed in fig 7.4b, if ao is the optimum threshold of detection, then the decision is m = O if r < a0 and m = 1 if r > a0. The conditional error probability p(E\m = O ) is the probability of making a wrong decision when m = O This is simply the area AO under Pr\m(r\0) from aO to . Similarly, p(E\m =1) is the area A1 under pr/m (r/1) from - to ao.

Pe =

Assuming Pm(0) = Pm(1) = 0.5

Notice that the sum Ao + A1 of the shaded areas is minimised by choosing ao at the intersection of the two PDFs. So,

.(7.12a)

and the corresponding Pe is

Pe = P(E10) = P(E/1)

=

.(7.12b)

=Q (7.12c)

Where,

(7.13)

Substituting eqs. (7.11) in to eg . (7.13) , we have,

This equation is similar with eg (7,5) with P(w) replaced by P(w) Q(w), Hence,

..(7.14a)

Thus, the optimum filter H(w) is given by

..(7.14b)

For white Noise,

and the optimum filter H(w) is given by H(w) = [P(-w) Q(-w)]e-jwTb..(7.15a)

and,

h(t) = P(Tb - t) q(Tb - ) (7.15b)

This is a filter matched to the pulse p(t) q(t) and the corresponding

in eg. 7.14a becomes,

.(7.16b)

...(7.16c)

where,

Ep and Eq are the energies of p(t) and q(t) respectively,

But,

Epq = .(7.17)

In binary, the error probability is the bit error probability or bit error rate (BER), and will be denoted by Pb (not Pe). Thus, from equs (7.12c) and (7.16c), we have,

(7.18b)

=Q (7.18b)

Hence, the optimum threshold ao is obtained by Substituting equs. (7.11a, b) and (7.15a) into equ. (7.12a),

( a0 = ( Ep - Eq)

.

(7.19)

Equivalent Optimum Binary Receivers

From equ. (7.15a), as shown in fig 7.4, which indicate that the filter can be achieved by parallel combination of two filters matched to p(t) and q(t) respectively as shown in fig 7.5a. Its equivalent is shown in fig7.5b, since the threshold is (Ep Eq)/2 respectively, which is from the two matched filter outputs, which implies shifting the threshold to 0. When Ep = Eq, the receiver is shown in fig 7.5c.

Fig. 7.5 Realization of the Optimum Binary Threshold Detector

Polar signaling: when q(t) = -p(t), and

Ep = Eq and Epq = -

Substituting into equ (7.18b), we have,

Pb = Q

. (7.20a)

and from equ. 7.15b,

h(t) = 2p(Tb - t)

Recall, that the multiplication of h(t) by any constants multiplies both the signal and the noise by the same factor, and does not affect the system performance.

Then, if we multiply h(t) by 0.5; we obtain,

h(t) = p(Tb - t)

.. (7.20b)

Also , from equ (7.19), the threshold is

a0 = 0

.(7 .20c)

In this case, the receiver in fig 7.5a reduces to that shown in fig 7.6 with threshold 0. Then, the error probability can be expressed in terms of a more basic parameter Eb, the energy per bit: Assuming 1 and 0 are equally likely. Then,

Eb =

and from equ (7.20a),

Pb = Q(

This shows that for optimum threshold detection, the system performance depends on pulse energy, and not on pulse shape. Using an asymptotic approximation equ (7.21b) becomes ,

Pb (e-Eb/N Eb/N>> 1.(7.21c)

On off signaling: Also Fig 7.5a reduces to Fig 7.6a except that the threshold, as shown in equ (7.19), is Ep. Hence,

q(t) = 0

From equs. (7.17) and (7.18)

Eq =0; Epq = 0, and Pb = Q

( Eb =

( Pb = Q

.

(7.22a)

= e-Eb/2N Eb/N >>/

...

(7.22b)

This shows that on off signaling requires twice as much energy per bit (3dB more power) to achieve the same performance as polar signaling.

Orthogonal signaling: As shown in fig 7.7, p(t) and q(t) are selected to be orthogonal over the interval (0, Tb).

Epq =

from eq (7.18),

Pb = Q

..

(7.23)

Assuming 1and 0 to be equiprobable, then,

Eb =

Hence, eq 7.23 becomes,

Pb =Q

...

(7.24a)

e-Eb/2NEb/N >> 1

(7.24b)

This implies that, it is the same with on-off signaling but inferior to that of polar signaling by 3dB.

Fig. 7.7. Examples of orthogonal signals.

Bipolar signaling

Although, this is a binary Scheme, but it uses three symbols: p(t), -p(t) and 0, Hence, the result needs some modifications. Using the receiver in fig 7.6a with threshold ( a0 = ( Ep/2.

Thus,

If \r(Tb)\ < Ep/2, the decision is 0, and if \r(Tb)\ > Ep/2 the decision is 1. When 0 is transmitted by no pulse, the receiver output is just the noise n with variance.

2n = NEp/2. Hence,

P(E/0) = probability ( \n\ > Ep/2),

= 2Q

= 2Q

when 1 is transmitted, the filter output at Tb is Ap + n, where

Ap = Ep when p(t) is transmitted

Similarly,

Ap = - Ep when -p(t) is transmitted.

Thus,

P(E\1) = probability (n < - ) when p(t) is used, or probability (n > ) when P(t) is used

= Q

= Q

( the average error probability is

Pb = [P(E/0) + P(E/1)] = 1.5 Q

This shows that Pb is 50% higher than for on-off signaling.

7.3 Carrier Systems In all digital carrier systems, baseband pulses modulate a high-frequency carrier. It involves amplitudeshift Keying (ASK), frequency - shift keying (FSK), pulse shift keying (PSK) and DPSK.

The first three schemes are shown in fig 7.8, using a rectangular base- band pulse. The baseband pulse may be specifically shaped to eliminate inter-symbol interference and to have a finite bandwidth.

Recall that, the error probability of the optimum detector depends only on the pulse energy, not on the pulse shape. Hence, the performance of a modulated scheme will be identical to that of the baseband scheme of the same energy. So, the incoming modulated pulses can be demodulated either coherently (synchronously) or non-coherently (by envelope detection). The former method is the optimum and give a better performance than the later.

Fig. 7.8. Digital Modulated Waveforms

Coherent Detection

Fig. 7.9. Coherent Detection of Digital Modulated Signals

The RF pulse can be detected by a filter matched to the RF pulse p(t) followed by a sampler as shown in fig 7.9a.

Let the RF pulse p(t) = (2 p1(t) Cos (ct ,

Applying equ 7.7a,

where,

p1(t) = Baseband pulse

Ep = Energy of P(t)

The RF pulse can also be detected by first demodulating it coherently by multiplying it by (2 Cos (ct. Hence, the product is the baseband pulse p(t) plus a baseband noise with PSD N/2. Since P(t) = (2 p(t)Cos (ct, Ep = Ep, also in this case,

(2 =

Therefore, the two schemes are equivalent. Now, let us consider the cases of PSK , ASK and FSK.

Phase shift keying: Like in the polar signaling, the optimum detector is shown in fig 7.9 with threshold 0. Applying equ. 7.21b, we have,

EMBED Equation.3 Amplitude - shift keying: This is case of on-off signaling. Thus, the optimum detector for ASK is the same as that for PSK with threshold Ep/2 as shown in fig 7.9.

Applying equ. (7.22); we have,

This shows that for the same performance, the pulse energy in ASK must be twice that in PSK. This implies that in coherent detection, PSK is always preferable to ASK. ASK is useful in non-coherent.

Frequency- Shift Keying: In FSK, binary 0 and 1 are transmitted by RF pulses (2 p1(t) Cos [(c-(((/2)]t and (2p(t)Cos [(c+(((/2)]t respectively. The receiver in fig 7.5c is the optimum receiver. The filter p(Tb - t) and q(Tb - t) are matched to the 2 RF pulses and can be replaced by respective synchronous demodulators followed by filters matched to the baseband pulse p1(t).

Let q(t) = (2 A Cos ((c - )t

P(t) = (2 A Cos ((c - )t

From equ (7.18b),

In practice (CTb>> 1, and the second term on the right-hand side can be ignored.

( Epq = A2 Tb Sinc((()TbThis implies that to minimize Pb in equ. (7.18b), Epq must be minimized. If the minimum value of Epq is -0.217A2 Tb and occurs at ((()Tb = 1.43( or

(f =

Assuming 1 and 0 equiprobable, then, Eb = Ep = Eq = A2 Tb , and

Epq = - 0.217 A2 Tb. Hence,

Pb = Q

= Q

If Epq = 0, it is orthogonal signaling.

Larger (f means wider separation between the signaling frequencies which result to larger transmission bandwidth. Hence, to minimize the bandwidth, (f should be as small as possible. The minimum value of (f that can be used for orthogonal signaling is Tb. The FSK using this value of (f is known as minimum-shift keying (msk) or fast frequencyshift keying. FSK schemes where phase continuity is maintained are known as continuousphase FSK (CP-FSK) e.g. MSK. These schemes have rapid spectral roll-off and improved efficiency.

MSK being an orthogonal scheme, its error probability is given by

Pb =

..

(7.27b)

Non-coherent Detection

This is used when the phase Q in the received RF pulse (2 p1(t) cos ((ct + () is un known. When the phase ( of the received pulse is random and uniformly distributed over (0,2(), then, the optimum detector is a filter matched to the RF pulse(2 p1(t) cos (ct followed by an envelope detector, a sampler and a comparator to make the decision as shown in fig. 7.10.

Fig.7.10. Noncoherent detection of digital modulated signals.

Amplitude shift keying: As shown in fig 7.10, the filter H(() is a filter matched to the RP Pulse ignoring the phase. This implies that the filter output amplitude Ap will not necessarily be maximum at the sampling instant, but the envelope will be close to maximum at the sampling instant. Thus, the matched filter output is now detected by an envelope detector and the envelope is sampled at t = Tb for making the decision.

Since the matched filter is used, Ap = Ep and 2n = NEp/2 as in equs. (7.9b). For ASK there are on the average, only Rb/2 pulses per second. Thus,

Eb = Ep/2.

The optimum threshold is not constant but depends on the value of Eb/N, which is a serious drawback in a fading channel. For a strong signal, Eb/N >> 1.

...(7. 28b)

Using the approximation to find pb, we have,

For a coherent detector,

This is similar at to eg. (7.29b). Hence, for a large Eb/N, the performance of the coherent detector and the envelope detector are the some.

Also the error probability of FSK in non coherent detection is similar to that of non coherent ASK. Again for Eb/N >> 1, the performance of coherent and non coherent are essentially similar:

Synchronization

In synchronous or coherent detection, we can achieve synchronization at three different levels.

(a) carrier synchronization

(b) Bit synchronization

(c) Word synchronization

While for non coherent detection, we use only the bit synchronization and word synchronization. Carrier synchronization is similar to bit synchronization but more difficult. In bit synchronization, the problem is to achieve synchronism from bit interval- which is of the order Tb. while in carrier synchronization, we must achieve synchronism within a fraction of a cycle, and because the aeration of one carrier cycle is 1/f0