# Chapter 10

# Hints and Solutions

# H1 Signal fundamentals

# H2 Discrete time signals and sampling

## H2.1 (An illustration of the sampling theorem) (see p. 78)

1.Because *F*_{s} = 500 Hz is greater than twice the signal’s frequency (that is, 2 × 200 Hz), the sampling makes it possible to perfectly reconstruct the signal. Hence we end up with the same sine at the 200 Hz frequency;

2. because

*F*_{s} = 250 Hz is smaller than twice the signal’s frequency, the sampling introduces aliasing. The ±

*F*_{s} shifts in the spectrum (corresponding to

*n* = ±1 in formula 2.6) contribute to the frequency with –250 + 200 = 50 Hz. Since the spectrum is symmetrical, everything happens as if the 200 Hz frequency were “aliased” by symmetry about the frequency

*F*_{s}/2 = 125 Hz. The result of the reconstruction is a sine with the frequency 50 Hz (

Figure H2.1);

## H2.2 (Time domain hermitian symmetry) (see p. 86)

1. If we take the conjugate complex of

*X*(

*f*) and use

*x*(

*n*) =

*x*^{*}(–

*n*):

If, furthermore, *x*(*n*)is real, then we know that *X*(*f* ) = *X*^{*} (–*f* ), hence *X*(*f* ) = *X*(–*f* ) = *X*^{*} (*f* ). The conclusion ...