Seonglae Cho2017143020 조성래
2.4
Take Fourier transform to both sides.
Because the system function convolution is multiplication for in frequency domain.
We know that so and we knows that
2.23
(a).
- It has only 1 or -1 value, with magnitude un-changing so stable.
- It is causal because it only depend on nth value
- It is linear because
- It is not time-invariant because
(b).
- It is stable because is stable
- It is not causal because if the response dependent on large
- It is linear because and x is linear
- It is not time-invariant because
(c).
- It is stable because unit step function value is all 1 and x is stable
- It is causal because it only dependent on past values.
- It is linear because
- It is not time-invariant because
(d).
- It is not stable because we can’t ensure the boundness
- It is not causal because the all future value is considered
- It ls linear because
- It is time-invariant because
2.36
(a). we knows that
(b). find the relation equation between X, Y and apply inverse FT
(c). so for given x
Since y value is constant for all means that cos part’s coefficient |H| is zero.
finally
2.46
(a).
Let then, when or or to exist DTFT of
(b).
2.48
(a). Is is not linear because but
(b). according to ,
(c). We can use shifted inputs only because it is not linear
2.49
(a). It is not time-invariance because
- ,
- It does not have same shape on summarization
(b). Since it is linear,
We can represent by sum of shifted delta functions and take sum of it then
2.51
(a). It is not time-invariant system because if change then value should be changed
(b). Since , the system is not linear
(c). (a)’s answer is same because it is related to first property but (b)’s answer becomes linear when
2.56
- Since FT of delta function is 1, after LFW the frequency domain’s value is
- so shifted then
- The sum is which is 1
- After inverse FT,