EBPSK modulation [8] is defined as follows: (1) g 0 ( t ) = A   sin   w c t , 0 ≤ t < T g 1 ( t ) = { A   sin ( w c t + θ ) , 0 ≤ t < τ , 0 ≤ θ ≤ π , A   sin   w c t , 0 ≤ τ ≤ t < T ,where g₀(t) and g₁(t) is modulation waveform corresponding to bit “0” and bit “1”, respectively, T is the bit duration, τ is the phase modulation duration, and θ is the modulating angle. Obviously, if τ = T and θ = π, Eq.(1) degenerates to: (2) g 0 ( t ) = A   sin   w c t ,         g 1 ( t ) = − A   sin   w c t ,         0 ≤ t < T

It is just the classical Binary Phase Shift Keying (BPSK) modulation, so is named as the Extended BPSK. Figure 1 depicts the normalized the power spectral density (PSD) of the typical BPSK of (2) and the EBPSK of (1) at the same bit rate R, where bit duration T = 20T_c = 20*2π / w_c, the phase modulation duration τ for EBPSK modulation lasts one period of carrier, i.e., τ = T_c = 2π / w_c, the modulation angle θ = π, the amplitude A = 1, and the carrier frequency f_c = w_c / 2π = 50KHz,

As we know, if A = 2 E b T, the bit error ratio (BER) of the optimal BPSK receiver is calculated by: (3) P e − BPSK = Q ( 2 E b N 0 ) = 1 2 π ∫ 2 E b N 0 ∞ exp ( − u 2 2 ) duwhere E_b is the signal energy used for transmitting one bit, and N₀ is the PSD of Addition Gaussian White Noise (AWGN).

However, if the same optimal BPSK receiver using traditional matched filter was utilized as the EBPSK demodulator, the BER performance would be much poorer since the difference in EBPSK modulated waveforms corresponding to “0” and “1” is very tiny and hard to detect, although in Figure 1 more centralized PSD of the EBPSK appears. Therefore a special infinite impulse response (IIR) filter as given in Figure 2 is used in EBPSK receiver to produce high impulse at the phase jumping points τ of bit 1s, so as to transform phase modulation into amplitude changes [10]. And by this way it amplifies the signal characters as much as possible and removes utmost noise. Figure 3 depicts the response of this filter to EBPSK modulated signals. Obviously, a simple amplitude detector followed would perform the demodulation of EBPSK signals because of the existence of high impulse in coded 1 s. Therefore, we gave this kind of filter the name of “impacting filter” [10].

In this section, the BER performance of the impacting filter based EBPSK receiver is analyzed in an AWGN channel.

The EBPSK modulated waveform corresponding to “0” is a pure sine wave, after passing the AWGN channel and the special impacting filter at the receiver, then the envelope r₀ of the filter output is with Rice distribution [11], whose probability density function (pdf) [12] is as follows: (4) p ( r 0 ) = r 0 σ 2 exp ( − r 0 2 + A 0 2 2 σ 2 ) · I 0 ( r 0 A 0 σ 2 )where A₀ is the amplitude of the filter output, σ² the noise variance, and I₀(z) the zero-order modified Bessel function, defined as follows: (5) I 0 ( z ) = ∑ n = 0 ∞ z 2 n 2 2 n · n ! · n !

A similar analysis can also be done aiming at code “1”. If we only consider those periods with phase jumping during τ, the jumping information can be transformed into parasitic amplitude information after receiving filter, i.e., producing the higher amplitude impulse. At this time, its envelope r₁ still is Rice distribution, the corresponding pdf is: (6) p ( r 1 ) = r 1 σ 2 exp ( − r 1 2 + A 1 2 2 σ 2 ) · I 0 ( r 1 A 1 σ 2 )

This derivation of the BER is based on the special and linear impacting filter as given in Figure 2, which can produce the amplitude impulse at the point of the phase jumping corresponding to code “1” that is much higher than the background of code “0”, i.e., A₁ > A₀. Let A₁ = A₀ + ΔA = A₀ + kA₀ and k > 0

Therefore, assuming that code “0” and “1” be transmitted with equal probability and the decision threshold be U_T, then the BER of the EBPSK system should be: (7) P e-EBPSK = 1 2 [ P ( 1 | 0 ) + P ( 0 | 1 ) ] = 1 2 [ ∫ U T ∞   p ( r 0 ) dr 0 + ∫ − ∞ U T   p ( r 1 ) dr 1 ] = 1 2 [ ∫ U T ∞ r 0 σ 2 exp ( − r 0 2 + A 0 2 2 σ 2 ) · I 0 ( r 0 A 0 σ 2 ) dr 0 + ∫ − ∞ U T r 1 σ 2 exp ( − r 1 2 + A 1 2 2 σ 2 ) · I 0 ( r 1 A 1 σ 2 ) dr 1 ] = 1 2 [ 1 − F 0 ( u T ) + F 1 ( u T ) ] = 1 2 [ 1 + Q 1 ( A 0 σ , U T σ ) − Q 1 ( ( 1 + k ) A 0 σ , U T σ ) ]where F₀(x) and F₁(x) is cumulative distribution function of Ricean random variable r₀ and r₁, respectively, and Q₁(a,b) is Marcum’s fuction, defined as follows: (8) Q 1 ( a , b ) = e − ( a 2 + b 2 ) / 2 ∑ k = 0 ∞ ( a b ) k   I k ( ab )

We will now discuss in detail how to ascertain the parameters A₀, k, σ² and U_T in the BER formula (7) in this subsection.

(1) The amplitude A₀ is evaluated as follows:

In EBPSK modulation and via Fourier transform: (9) g 0 ( t ) = A   sin   w c t ↔ F G 0 ( w ) = A π j [ δ ( w − w c ) − δ ( w + w c ) ]

Let the frequency response of the filter be H(w), then the signal filtered in frequency domain can be written as: (10) Y 0 ( w ) = G 0 ( w ) · H ( w ) = A π j [ δ ( w − w c ) − δ ( w + w c ) ] · H ( w ) = A π j · H ( w c ) · δ ( w − w c ) − A π j · H ( − w c ) · δ ( w + w c )

So the waveform in time domain is: (11) y 0 ( t ) = F − 1 [ Y 0 ( w ) ] = A 2 j [ H ( w c ) · e jw c t − H ( − w c ) · e − jw c t ] = A 2 j [ | H ( w c ) | · e j ( w c t + φ 1 ) − | H ( − w c ) | · e − j ( w c t − φ 2 ) ]where φ₁ = ∠H(w_c), φ₂ = ∠H(−w_c). The impacting filter has conjugate zero points and poles, so φ₁ = −φ₂, and |H(w_c)| = |H(−w_c)|, then the received signal after filter is as follows: (12) y 0 ( t ) = | H ( w c ) | · A   sin   ( w c t + φ 1 )

So A₀ = A · |H(w_c)|, A₁ = (1+k) · A · |H(w_c)|, and in accordance with Eq.(3), let A = 2 E b T. Finally we get: (13) A 0 =   | H ( w c ) | · 2 E b T

(2) The variance σ² is evaluated as follows:

Let the PSD of the input AWGN be N₀ / 2, measured in Watt/Hz, then the variance σ² at the output of the filter should be: (14) σ 2 = N 0 2 π ∫ 0 ∞ | H ( w ) | 2 dw = N 0 ∫ 0 ∞ | H ( f ) | 2 df

(3) The threshold U_T is evaluated as follows.

Based on the binary detection model, the sketch map of the detection performance illustrated by pdf is shown in Figure 4, where shaded area indicates the BER. P(1|0) and P(0|1) represent shaded area at the right and left of the threshold, respectively. Obviously, when threshold increases, P(1|0) will decrease, while P(0|1) will increase, and vice versa. Therefore, in order to obtain a minimum BER, the optimal decision threshold U_T should be located at the intersection point of curve p(r₀) and p(r₁), i.e., the theoretical minimum BER should be equal to half of the area in the intersection part of curve p(r₀) and p(r₁).

(4) The parameter k is evaluated as follows:

According to Figure 4, the larger is the parameter k, the further is the distance of the curve p(r₀) and p(r₁), i.e., the smaller is the area of the shaded part, i.e., the smaller is the system BER. Therefore, we hope to choose the larger k value in order to improve BER performance.

Among all the parameters discussed above, how to select k, the increment in amplitude at the beginning of bit “1” after filtering, needs improving:

Conceptually, the maximum of kA₀ could reach the peak, say ±20dB, which is formed by a very narrow frequency selector and trap locating at the center of impacting filter as shown in Figure 2. In practice, the peak of kA₀ might be limited by filter design and waveform selection (such as τ, T and θ), or even the status of channel, which still needs revealing. In fact, as an example given in Figure 3, where the peak of kA₀ is about two times of its background, so at this situation, we may roughly get the estimation as k = 2.

In this subsection we will calculate the various signal to noise ratios (SNRs) so as to evaluate the EBPSK-MODEM.

(1) SNR at the input of the impacting filter

Similar as Eq.(14), let the PSD of the input AWGN be N₀ / 2, then from the autocorrelation function R ( τ ) = N 0 2 ⋅ δ ( τ ) we get its variance as following: (15) ɛ 2 = R ( 0 ) ↔ F 1 2 π ∫ − ∞ ∞ N 0 2 · δ ( 0 ) · e jwt dw = N 0 2where the N₀ / 2 is measured in Watt since it is the average power of the input noise, although in numerical value it is the same as its PSD.

Let A = 2 E b T again, the average power of the signal at the input of the impacting filter is: (16) A 2 2 = E b T = E b · R bwhich results in the SNR calculation at the input of the impacting filter as below: (17) SNR in = A 2 2 · ε 2 = E b · R b N 0 / 2where N₀ / 2 is the average power of input noise.

(2) SNR at the output of the impacting filter

The average SNR at the output of the impacting filter for bit “0” during T (and in theory for bit “1” during T − τ also) is: (18) SNR 0 = 1 2 A 0 2 / σ 2 = E b N 0 T · | H ( w c ) | 2 ∫ 0 ∞ | H ( f ) | 2 df = R b · E b N 0 · | H ( f c ) | 2 ∫ 0 ∞ | H ( f ) | 2 df = E b · R b N 0 · B neqwhere f_c is the carrier frequency, E_b is the average energy for transmission of one bit, R_b = 1/T is the bit rate because both the BPSK and the EBPSK is binary modulation, so R_b · E_b is the average signal power transmitted, N₀ / 2 is still the PSD, and: (19) B neq = ∫ 0 ∞ | H ( f ) | 2 df / | H ( f c ) | 2is the noise equivalent bandwidth of filters [13].

Therefore, the gain in SNR of impacting filter for bit “0” should be: (20) G 0 = SNR 0 SNR in = E b · R b N 0 · B neq / E b · R b N 0 / 2 = 1 2 B neq

From Eq. (20) it is clear that B_neq is the SNR “controlling factor” for bit “0” determined by the shape of frequency response of the impacting filter used.

Similarly as Eq.(18), the peak SNR at the output of the impacting filter for bit “1” during its beginning is given as follows: (21) SNR 1 = A 1 2 σ 2 = A 0 2 · ( 1 + k ) 2 σ 2 = 2 SNR 0 + A 0 2 · ( 2 k + k 2 ) σ 2 > SNR 0

And the gain in SNR of impacting filter for bit “1” should be: (22) G 1 = SNR 1 SNR in = 1 B neq + A 0 2 · ( 2 k + k 2 ) 2 E b R b · ∫ 0 ∞ | H ( f ) | 2 df