E4702 HW#4-5 solutions by Anmo Kim

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1 E70 HW#-5 solutions by Anmo Kim (P3.7) Midtread type uniform quantizer (figure 3.0(a) in Haykin) Gaussian-distributed random variable with zero mean and unit variance is applied to this quantizer input (a) the probability that the amplitude of the input lies outside the range - to This problem can be described more clearly with the complementary error function. (see eqn.(.9) in Haykin) erfc(u) = e z dz π For the quantizer input, m, with the normal distribution, P [ m > ] = u π e t / dt = erfc( ) = (b) output signal-to-noise ration of the quantizer ( input located between [,-]) The average power of the message signal is P = x π e x / dx x π e x / dx = We can evaluate the above integral by using Matlab with the following commands: >> syms x >> eval(int(x^*exp(-x^/)/sqrt(*pi),-,)) Then, we will get as a result. Here, we will use the approximated average power.(i.e. P = ) Putting this and m max = into equation(3.3) in Haykin, The signal-to-noise ratio is In db scale, σ Q = 3 R = 6 3 R (SNR) O = 3 R 6 = 3 R 6 (SNR) O(dB) = 0 log 0 (3/6) + 0 log 0 R = R We can compare this result to that of sinusoidal modulating signal, which is given in Example3. of Haykin. (SNR) O(dB) of the sinusoidal modulating signal is.8 + 6R, which is 9dB greater than Gaussian input. The reason can be directly observed from the average power: P sinusoidal = A m/ = 8 where A m = and P Gaussian =. This means that more number of signals in the Gaussian input signal come with lower amplitude compared to the sinusoidal signal. (This can be also seen on the bell-shaped curve of Gaussian distribution.) As a following result, in case of Gaussian input signal, non-uniform quantizer - dividing the lower amplitude part in a finer scale - would outperform the uniform quantizer.

2 . (P3.9) mean-square quantization error for a nonuniform quantizer - i : ith step size of the given nonuniform quantizer - p i : probability that the input signal amplitude lies within the ith interval - i is small compared with the excursion of the signal The expected mean-square quantization error can be represented in terms of the quantization error in each interval. where L is the number of quantization levels. σ Q = L i= σ Q,ip i In the ith interval, the quantization error for input signal, q i, and the mean-square quantization error, σ Q i, is defined as follows. q i = m m i + m i+ σ Q,i = E m [ (qi E[q i ]) ] From the condition that the quantization interval is small compared with the excursion of the signal, we can assume that the input signal is uniformly distributed in an interval. Then, σ Q,i = E[q i ] = 0 i / qi i / σ Q = i dq i = i L i p i i= 3. (P3.(d)) Power spectral density of bipolar return-to-zero signals From the lecture, power spectral density is described as follows. where s N (t) = N N a ng(t nt b ). t S s (f) = G(f) T b <k< For the bipolar RZ, g(t) = rect( T b / ) and G(f) = T b sinc(ft b /) The probabilistic description of a n is: 0, with P (a n = 0) = / a n = A, with P (a n = ) = / A, with P (a n = ) = / R(k)e jkπt b () The autocorrelation function of a n is: R(k) = E[a n a n+k ]

3 R(0) a n a n { 0, with p = / (when a n = 0) A, with p = / (when a n = A or A) R(0) = 0 + A = A R() Here, notice that P (a n+ = m a n = 0) = P (a n+ = m), m {0, A, A} /, m = 0 P (a n+ = m a n = A or A) = /, m = a n 0, m = a n { 0, with p = 3/ (when a n = 0 or a n+ = 0) a n a n+ A, with p = / (otherwise) R() = ( A ) = A R(k), k > a n a n+k (k > ) has non-zero value with the probability of only when both bits are non-zero. Given that a n a n+k is non-zero, it is either or -. Here, we define b n,k = n+k j=n+ a j which is zero when the number of non-zero bits between a n+ and a n+k is even. P (a n a n+k = ) = P (b n,k = 0), P (a n a n+k = ) = P (b n,k 0) Notice here that P (a n a n+k = ) = P (a n a n+k = ). Applying the equation () to be given in (P3.3), R(k) = E[a n a n+k ] P (a n a n+k = ) = P (a n a n+k = ) = = P (a n a n+k = ) + ( ) P (a n a n+k = ) + 0 P (a n a n+k = 0) = P (a n a n+k = ) P (a n a n+k = ) = 0 For the negative index of k, remind that R(k) is an even function. By putting G(f) and R(k) into (), S s (f) = T b sinc(ft b /) T b A = A T b sinc (ft b /) [ ( e jπft b + e jπft ) ] b [ ] cos(πft b) = A T b sinc (ft b /) cos(πft b) ( ) = A T b ftb sinc sin (πft b ) 3

4 . (P3.3) a chain of (n-) regenerative repeaters with n sequential decisions made on a binary PCM wave p p -p repeaters source -p 0 n- n- destination p : probability of error on each decision procedure (a) The error probability on the nth stage can be represented in a recursive form. This equation can be expanded as p n+ = p n ( p ) + ( p n )p = p + ( p )p n p n+ = p + ( p )p n = p + p ( p ) + ( p ) p n = p + p ( p ) + p ( p ) + ( p ) 3 p n n = p ( p ) k + ( p ) n p k=0 = p ( p ) n ( p ) + ( p ) n p = [ ( p ) n+ ] p n = [ ( p ) n ] () (b) ( p ) n = n k=0 ( ) n ( p ) k k If p is very small, we can approximate this equation by taking only the constant term and the first polynomial term. ( p ) n np 5. (P3.3) A one-step linear predictor - input : sin(π f 0 0f 0 n) = sin(π 0.n) p n [ ( np ] = np

5 x[ n] Asin( 0. n) Z w x [ n] w Asin( 0.( n )) (a) the optimum value of w for the minimizing the prediction error variance The prediction error is defined as: e[n] = x[n] ˆx[n] The index of performance (or the prediction error variance) is defined as: J = E[e [n]] Then, J = E[x [n]] + E[ˆx [n]] E[x[n]ˆx[n]] = A E[sin(π0.n) ] + wa E[sin(π0.(n )) ] A w E[sin(π0.n) sin(π0.(n ))] = A + w A + w A E[cos(π0.n π0.n)] w A E[cos(π0.)] [ = A w cos(π0.)w + ] Taking derivative of J over w, dj dw = A w A cos(π0.) J has its minimum when this derivative is zero. (b) the minimum value of the prediction error variance w = cos(π0.) = when w = 0.809, J min = 0.77A 5