## Appendix 1A Solutions to Exercises

To obtain the results described below initialize the random number generators to `rand(’state’,0)` and `randn(’state’,0)` at the beginning of any code. These commands are for the uniform and Gaussian random number generators, respectively.

**1.1.** For part a, run the code with `randn(’state’,0)` at the start of the program to intialize the random number generator. You should observe 92 estimates that meet the specifications, i.e., lie within the interval [9.75, 10*.*25]. For part b, we now require so that from the CRLB var(*Â*_{N}) = *σ*^{2}/*N* = 1/400 and thus, we require that *N* = 400. Modifying the program and running it, produces 95 estimates that meet the specification, i.e., lie within the interval [9.9, 10.1].

**1.2.**

Now let *z* = exp[*−**j*2*π*(*f* − *f*_{0})] and note that *|z|* = 1. Dropping the second sum we have

and taking its complex magnitude

Next letting *α* = −2*π*(*f* − *f*_{0}), we have *z* = exp(*jα*) and

Since *α/*2 = − π(*f* − *f* _{0}) we have finally that

Note that this is approximate since we have neglected the second term in the sum, i.e., the term given by (1A.2). Since this term is the mirror image of the positive frequency term (the one we retained) and is centered about *f* = *f*_{0}, its contribution will be small as long as it does not “interfere” with the positive frequency term. This will be the case if *f*_{0} is not near 0 or 1/2. This is illustrated in Figure 1A.1 in which we have plotted the magnitude of the terms (1A.1) (positive frequency component) and (1A.2) (negative frequency

Figure 1A.1. Magnitude of discrete-time Fourier transform of frequency components with *f*_{0} = 0*.*2 and *N* = 20. The heavier line indicates the Fourier transform magnitude of the positive frequency component given by (1A.1) while the lighter line is that for the negative frequency component given by (1A.2). Significant interactions between the two components occur when 0 *< f*_{0} *<* 1/ *N* = 0.05 or 0.45 = 1/2 − 1/*N* < *f*_{0} < 1/2. component) for the entire frequency band − 0.5 ≤ *f* ≤ 0.5. The signal parameters are *N* = 20, *A* = 1, *φ* = 0, and *f*_{0} = 0.2. It is seen that there is little interference in this case.