If the variance $\sigma_d^2$ of $d(n)=x(n)-x(n-1)$ is one-fifth the variance of a stationary zero mean discrete time signal x(n). If the variance of $x(n)$ is $4$. Then the value of auto-correlation function $R_{XX}(k)$ at $k=1$ is ____________
a) $7.2$
b) $4$
c) $3.6$
d) $8$
Correct Answer: Option C
Solution:
Mean of discrete time signal $x(n)$ = $0$
Variance of $d(n)$ is ${\sigma_d}^2$ = ${\sigma_x}^2/5$
Mean of discrete time signal $d(n)$ is $$E[d(n)]=E[x(n)-x(n-1)]$$ $$=E[x(n)]-E[x(n-1)]$$ $$=0-0$$ $$E[d(n)]=0$$
Variance of $d(n)$ is $${\sigma_d}^2=E[d(n)^2]-(E[d(n)])^2$$ $$=E[d(n)^2-0$$ $$=E[d(n)^2]$$ $$=E[x(n)-x(n-1)]^2$$ $$=E[x(n)^2+x(n-1)^2-2x(n)x(n-1)]$$ $$=E[x(n)^2]+E[x(n-1)^2]\\-2E[x(n)x(n-1)]$$ $$={\sigma_x}^2+{\sigma_x}^2-2R_{xx}(1)$$ $${\sigma_d}^2=2{\sigma_x}^2-2R_{xx}(1)$$ $$2R_{xx}(1)=2{\sigma_x}^2-{\sigma_x}^2/5$$ $$2R_{xx}(1)=8-4/5$$ $$2R_{xx}(1)=7.2$$ $$R_{xx}(1)=3.6$$
