Seonglae Cho2017143020 조성래
4.21
Except , is repeated per . We need which means
(b). To make available the signal can be recovered, the frequency should not be duplicated.
, which means is area which is not duplicated with original signal.
When the signal is not duplicated with perfect harmonics. In this case, we must choose like below because another area is filled by value.
4.25
. We can derive that after frequency, this value will be 0 because of convolution property.
Based on Nyquist frequency, the maximum sampling interval
4.27
(a). We can only consider because is Nyquist frequency and the ideal D/C has lowpass filter under .
Then we can represent using and
After D/C filter, gain to the signal
Therefore we can compute
(b). Then which are sinc function in CT but delta function in DT and within
4.28
(a).
(b). When then . Plot is like this because of sampling and is bigger than Nyquist frequency which means there is no aliasing so
(c). Then which is smaller than Nyquist frequency so
based on below plot
4.31
(a). Since is Nyquist frequency of this signal, the diagram is like below
(b). Since filter above we can ignore aliasing above which means
Also T should not be oversampled more than the largest frequency is smaller than which means
finally
(c). When , and when , and it increase linearly as increase.
4.37
(a). 5 elements require each division. So we need 5 multiples per output sample.
(b). and according to z transform.
Therefore,
(c). Unlike (a), new implementation requires 6 multiples ( with 3 for and 2 for each ).
4.39
(a). Nyquist frequency is and is half of that. Since the has form of triangle, even after the lowpass filter, the aliased signal would become a constant value because it is half.
and
(b). We need lowpass filter and DT one is cheaper. We don’t need to scale so and choose to match after C/D with cost .
(c). In this case, we also need up-sampling expander with along with the case (b). Therefore, the total cost is