12.10. The message signal y[n] in Figure P12.10 is to be encrypted and transmitted across a...

  

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12.10. The message signal y[n] in Figure P12.10 is to be encrypted and transmitted across a noisy channel, then decrypted and filtered at the receiver. We model y[n] as a zero-mean WSS random process with autocorrelation function Ryy[m] and corresponding PSD Syy(e). The signal p[n] is used for both the encryption at the transmitter and the decryption at the receiver, and is an i.i.d. process that takes the values +1 or -1 with equal probability at each time; it is independent of the process y[-]. Note that p²[n] = 1 for all n. The transmitted signal q[n] is the product p[n]y[n]. Message signal y[n] Encryption/decryption signal p[n] q[n] Channel noise Noncausal. Wiener filter H(e) y[n] Figure P12.10 (a) Determine the respective means up and μg of the processes p[n] and q[n], their respective autocorrelations Rpp[m] and Rag[m] (expressed in terms of Ryy[-]), and also the cross-correlation Ryg[m] between the message signal and the transmitted signal. Would an intruder who was able to intercept the transmitted process q[-] have any use for a (possibly noncausal) linear estimator of y[n] based on measurements of q[-]? Explain your answer. The channel adds a noise signal v[n] to the transmitted signal, so that the received signal is q[n]+v[n] = p[n]y[n]+v[n]. Assume v[n] is a zero-mean and white WSS process, with Ry[m] = o²8[m]; sup- pose it is uncorrelated with y[-], and both processes are independent of p[-]. We assume, as indicated in Figure P12.10, that the intended receiver knows the specific encryption signal p[n], i.e., the specific sample function from the ensem- ble that was used for encryption. If there was no channel noise (i.e., if we had v[n] = 0), the decryption would then simply involve multiplying the received signal by p[n] because p[n]q[n] = p[n] (p[n]y[n]) =p²[n]y[n] = y[n], where the last equality is a consequence of having p²[n] = 1. In the presence of noise, we can still attempt to decrypt in the same manner, but will follow it up by a further stage of filtering. The signal to be filtered is thus (b) Determine x, Rx[m], and Ryx[m]. (c) Suppose the filter at the receiver is to be a (stable) noncausal Wiener filter, constructed so as to produce the LMMSE estimate [n] of y[n]. Determine the frequency response H(e) of this filter, and explicitly check that it is what you would expect it to be in the two limiting cases of o2 = 0 and o →∞o. Also write an expression, in terms of Syy(e) and o?, for the mean square error obtained with this filter, and explicitly check that it is what you would expect it to be in the preceding two limiting cases. 12.10. The message signal y[n] in Figure P12.10 is to be encrypted and transmitted across a noisy channel, then decrypted and filtered at the receiver. We model y[n] as a zero-mean WSS random process with autocorrelation function Ryy[m] and corresponding PSD Syy(e). The signal p[n] is used for both the encryption at the transmitter and the decryption at the receiver, and is an i.i.d. process that takes the values +1 or -1 with equal probability at each time; it is independent of the process y[-]. Note that p²[n] = 1 for all n. The transmitted signal q[n] is the product p[n]y[n]. Message signal y[n] Encryption/decryption signal p[n] q[n] Channel noise Noncausal. Wiener filter H(e) y[n] Figure P12.10 (a) Determine the respective means up and μg of the processes p[n] and q[n], their respective autocorrelations Rpp[m] and Rag[m] (expressed in terms of Ryy[-]), and also the cross-correlation Ryg[m] between the message signal and the transmitted signal. Would an intruder who was able to intercept the transmitted process q[-] have any use for a (possibly noncausal) linear estimator of y[n] based on measurements of q[-]? Explain your answer. The channel adds a noise signal v[n] to the transmitted signal, so that the received signal is q[n]+v[n] = p[n]y[n]+v[n]. Assume v[n] is a zero-mean and white WSS process, with Ry[m] = o²8[m]; sup- pose it is uncorrelated with y[-], and both processes are independent of p[-]. We assume, as indicated in Figure P12.10, that the intended receiver knows the specific encryption signal p[n], i.e., the specific sample function from the ensem- ble that was used for encryption. If there was no channel noise (i.e., if we had v[n] = 0), the decryption would then simply involve multiplying the received signal by p[n] because p[n]q[n] = p[n] (p[n]y[n]) =p²[n]y[n] = y[n], where the last equality is a consequence of having p²[n] = 1. In the presence of noise, we can still attempt to decrypt in the same manner, but will follow it up by a further stage of filtering. The signal to be filtered is thus (b) Determine x, Rx[m], and Ryx[m]. (c) Suppose the filter at the receiver is to be a (stable) noncausal Wiener filter, constructed so as to produce the LMMSE estimate [n] of y[n]. Determine the frequency response H(e) of this filter, and explicitly check that it is what you would expect it to be in the two limiting cases of o2 = 0 and o →∞o. Also write an expression, in terms of Syy(e) and o?, for the mean square error obtained with this filter, and explicitly check that it is what you would expect it to be in the preceding two limiting cases.