ENGN 1570 Fall 2014 Homework 3 (Solution) Problem 1

ENGN 1570 Fall 2014 Homework 3
(Solution)
Example 1.2
Problem 1
We are given the signal x(t):
x(t)
1
−1
0
1
2
3
4
t
5
Let us transform the given x(t) to x(−2t + 6). We need to use all three types of transformations (a shift, a
scale and a flip), but what in what order shall we do them? How do we do it? The following guide explains:
For the signal x(t) above. Sketch x ∗ h when
(a)
What to do when you need to do multiple transformations:



0
t<0


h(t) =of 2 0 < t < 1
• We can think of cascaded transformations as repeated substitutions

the independent variable t.


 0
t>1
• We can do the transformations in any order. However, the amount and
direction of the shift depends on whether it is performed before or after
the reversal and the scale.
To demonstrate these principles, let’s do the transformation in three different orders.
• x(t)
Advance by 6
−→
Reverse
−→
x(t + 6)
x(−t + 6)
Compress by 2
−→
x(−2t + 6).
3
2.5
• x(t)
Compress by 2
−→
x(2t)
2
1.5
Advance by 3
−→
Reverse
−→
x(2(t + 3))
x(2(−t + 3)).
1
0.5
0
−0.5
• x(t)
Reverse
−→
−1
−2
x(−t)
−1
Delay by 6
−→
0
1
2
3
4
5
x(−(t − 6))
Compress by 2
−→
x(−(2t − 6)).
6
9
1
(b) h(t) = δ(t) + δ(t − 1)
3
2.5
2
1.5
1
0.5
0
−0.5
−1
−1
0
1
2
3
4
5
6
7
Problem 2
Let Ch be the system that has output x ∗ h on input x. For each of the choices for h below,
determine if Ch is invertible. If so give the signal h0 such that Ch0 is the inverse of Ch . If not
give two input signals that map to the same output signal.
(a) h(t) = e−j2t
Ch can not be invertible because h(t) is periodic. Let T = π be a period of h(t).
Take any non-periodic signal x1 (t) and let x2 (t) be x1 (t) delayed by T . Since x1
is not periodic we have x2 6= x1 . But since h is periodic with period T we have
Ch (x1 ) = Ch (x2 ).
(b) h(t) = jδ(t)
Ch is invertible, take h0 = −jδ(t).
(c) h(t) = u(t) + u(t − 5) (u is the unit step function).
We will leave a discussion of this for later in the semester.
2
Problem 3
Let y = x ∗ h. Let y 0 , x0 and h0 denote y, x and h delayed by 1 respectively. Determine
which of the statements below are true and which are false. Justify your answers.
(a) If x[n] = 0 for n < N1 and h[n] = 0 for n < N2 then y[n] = 0 for n < N1 + N2 .
True.
Suppose y[n] 6= 0.
y[n] =
X
k
x[k]h[n − k]
For y[n] to be non-zero there must exist a k for which x[k] 6= 0 and h[n − k] 6= 0.
This means k ≥ N1 and n − k ≥ N2 . Since k + (n − k) = n we have n ≥ N1 + N2 .
(b) If x[n] = 0 for n > N1 and h[n] = 0 for n > N2 then y[n] = 0 for n > N1 + N2 .
True.
Suppose y[n] 6= 0.
y[n] =
X
k
x[k]h[n − k]
For y[n] to be non-zero there must exist a k for which x[k] 6= 0 and h[n − k] 6= 0.
This means k ≤ N1 and n − k ≤ N2 . Since k + (n − k) = n we have n ≤ N1 + N2 .
(c) y 0 = x0 ∗ h0
False
(d) y 0 = x0 ∗ h
True (by time invariance of Ch )
(e) y 0 = x ∗ h0
True (by time invariance of Cx )
Problem 4
NOTE: Part (a) of this problem is the same as Problem 6 from Homework 2 but with more
non-zero values in the definition of h.
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One important use of inverse systems is in situations in which one wishes to remove
distortions of some type. A good example is the problem of removing echoes from acoustic
signals. For example if a room has echo then an initial acoustic impuse will be followed by
attenuated versions of the sound at regularly spaced intervals. A model for this is an LTI
system with impuse response given by a train of impuses
h(t) =
∞
X
k=0
hk δ(t − kT )
Here the echos occur T seconds apart, and hk represents the gain factor on the k-th echo.
We can think of T and the hk ’s as parameters describing the acoustics of a room.
Suppose that x(t) represents an original acoustic signal and that y(t) = x(t) ∗ h(t) is the
signal recorded by a microphone. Note that y(t) is the superposition of delayed versions of
x(t), scaled by the coefficients hk .
The system defined by convolution with h is invertible, therefore we can remove the echos
from y(t) by convolving it with another signal g(t). The signal g(t) is also a train of impulses
g(t) =
∞
X
k=0
gk δ(t − kT )
(a) Suppose that h0 = 1, h1 = 1/2, h2 = 1/4, and hi = 0 for all i ≥ 3. Determine the
values for the gk ’s in this case.
If convolution with g removes echos to restore the original signal we must have
that h ∗ g = δ. For each value of t = kT the fact that (h ∗ g)(t) = δ(t) leads to
one linear equation constraining the gk .
If we look at h ∗ g at t = 0 we get that a certain linear combination of the gk
must equal to 1. If we look at h ∗ g at t = kT for k 6= 0 we get that a certain
linear combination of the gk must equal to 0.
When k = 0 we get h0 g0 = 1 so g0 = 1.
When k = 1 we get h0 g1 + h1 g0 = 0 so g1 = −1/2
When k = 2 we get h0 g2 + h1 g1 + h2 g0 = 0 so g2 = −1/4 + 1/4 = 0.
In general we get h0 gk + h1 gk−1 + h2 gk−2 so gk = −1/2gk−1 − 1/4gk−2 .
(b) With the hk ’s defined in part (a), is Ch stable? What about Cg ? Justify your answer.
4
Problem 5
It is often important to transform between continuous and discrete signals. We can define
a discrete signal from a continuous signal by “sampling” the continuous signal at integer
times. The problem of defining a continuous signal y(t) from a discrete signal x[n] is more
complicated because we have to fill in the missing values. A common approach involve using
spline basis functions.
One of the simplest spline functions is the “triangular shaped” signal defined by



0
t < −1




 t + 1 −1 < t < 0
h(t) =


1−t
0<t<1




 0
t>1
(a) Sketch h(t)
Let x[n] be a discrete signal. We can define a continuous signal y(t) by taking linear
combinations of shifted and scaled versions of h(t).
y(t) =
∞
X
k=−∞
x[k]h(t − k)
Sketch the resulting y(t) for the following cases:
(b) x[n] = 2
In this case y(t) = 2.
(c) x[n] = n
In this case y(t) = t.
5
(d)

 3 n=0
x[n] =
 0 n=
6 0
3
2.5
2
1.5
1
0.5
0
−5
−4
−3
−2
−1
0
1
2
3
4
6
5
(e)



0






5




 2
x[n] =


2






−1




 0
n≤0
n=1
n=2
n=3
n=4
n≥5
5
4
3
2
1
0
−1
−2
0
2
4
6
8
7
10
(f ) In general, what can you say about the resulting y(t) at integer values for t? What
about non-integer values for t?
At integer t we have y(t) = x[t]. Between two consecutive integer values t and
t + 1 we have that y(t) is a linear function. So y(t) is continuous and piecewise
linear. It goes exactly “through” the values defined by x at integer t and fills in
the missing parts using linear interpolation.
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