ELEN 4810 Homework 5
Analytical Questions
10.24 (a) x[n] = x(nT ) = ej(3π/8)n . Xw [k] will be nonzero at exactly one value whenever (3π/8) =
2πk/N for some N and integer j. This occurs if and only if 3/16 = k/N . Because 3 does not divide
16, the smallest N for which this is possible is N = 16.1
(b) Top (Figure 10.24-2): w1 , Bottom (Figure 10.24-3): w2 . The N point DFT samples the
frequency-domain convolution of X(ejω ) and W (ejω ). W1 (ejω ) exhibits a narrower mainlobe and
more small sidelobes; W2 (ejω ) has a wider mainlobe, and more prominent sidelobes away from ω = 0.
(c) The largest entry in the DFT occurs at k = 6. This corresponds to discrete-time frequency
ω = 2π × 6/32 = 3π/8. The discrete-time frequency ω is related to the continuous-time frequency
Ω via ω = ΩT , and so we can estimate Ω = ω/T = (3π/8) × 104 rad/s. This estimate is not exact.
The actual maximum of |Xw (ejω )| could occur at some ω which is not an integer multiple of 2π/N .
A reasonable estimate assumes that the maximum occurs in some interval of width 2π/N centered
1
π
4
about ω. When this assumption is valid, the maximum error in Ω is 21 2π
N × T = 32 × 10 rad/s.
(d) The most direct approach is to take the inverse DFT, to obtain xw1 [n] = x[n]w1 [n]. We can
then divide by w1 [n] to obtain x[n] for n = 0, . . . , 31. This signal is of the form
x[n] = ejω0 n ,
(1)
with ω0 = Ω0 T . By inspecting x[1], we obtain
Re (x[1])
=
cos(ω0 ),
(2)
Im (x[1])
=
sin(ω0 ).
(3)
This determines ω0 up to integer multiples of 2π. Since the sampling rate is assumed to be such
that no aliasing occurs, we may simply take whichever integer multiple of 2π yields ω0 ∈ (−π, π].
We then have Ω0 = ω0 T .
10.32
(a) Spectrograms (a) and (c) were computed with rectangular windows.
(b) The pairs (a & b) and (c & d) have approximately the same frequency resolution.
(c) Spectrogram (c) has the shortest time window. Spectrogram (a) clearly uses a longer window
than (c); spectrogram (b) clearly uses a longer window than (d). So we need only compare (c) and
(d). For (c), the mainlobe width is roughly 0.08 π. With a rectangular window, this corresponds
1 Strictly speaking, N = 1 also works, because the 1-point DFT can only be nonzero at at most one location. This
answer is correct, but a little bit contrary to the spirit of the problem. No value between 1 and 16 works.
to a length of M = 49. For (d), the mainlobe width is again roughly 0.08 π; however, this is a
Hamming window, and so this corresponds to a length of roughly M = 100 samples.
(d) ≈ 350 samples. The window length is roughly the width of the transition regions that occur
around 1,000 samples and around 2,000 samples. Answers that are close to 350 samples and/or
exhibit correct reasoning are acceptable.
(e) As follows:


cos(7, 000πt + φ2 ) 0 ≤ t ≤ 0.1s
cos(4, 000πt + φ1 ) + 0
0.1s < t < 0.2s


cos(5, 000πt + φ3 ) 0.2s ≤ t ≤ 0.3s
(4)
The phases φ1 . . . φ3 cannot be determined from the given information. Answers noting some uncertainty in the amplitude will also be accepted as correct.
3.21 (a) The system has poles at z = 1/4 and z = −1/2. Because the system is causal, the ROC
extends outward from the largest magnitude pole, including |z| = ∞:
ROC = {z | |z| > 1/2} .
(5)
(b) The ROC contains the unit circle, and so the system is stable.
(c) Y (z) = H(z)X(z), and so (1 − 0.25z −1 )(1 + 0.5z −1 )Y (z) = (4 + 0.25z −1 − 0.5z −2 )X(z). Multiplying out the left hand side, we obtain
(1 + 0.25z −1 − 0.125z −2 )Y (z) = (4 + 0.25z −1 − 0.5z −2 )X(z).
(6)
This corresponds to the difference equation
y[n] + 0.25y[n − 1] − 0.125y[n − 2] = 4x[n] + 0.25x[n − 1] − 0.5x[n − 2].
(7)
(d) Using polynomial division, we obtain
H(z) = 4 −
(3/4)z −1
(1 − 0.25z −1 )(1 + 0.5z −1 )
(8)
H(z) = 4 +
A
B
+
1 − 0.25z −1
1 + 0.5z −1
(9)
We can express this as
where
−(3/4)z −1 1 + 0.5z −1 z=0.25
= −1,
A
=
(10)
(11)
and
−(3/4)z −1 1 − 0.25z −1 z=−.5
1.
B
=
=
(12)
(13)
A
We can now invert term-by-term. Based on the ROC of H, we associate to the term 1−0.25z
−1 the
B
ROC |z| > 0.25. To the term 1+0.5z−1 , we associate the ROC |z| > 0.5. We then obtain
h[n] = 4δ[n] − (0.25)n u[n] + (−0.5)n u[n].
(e) x[n] = u[−n − 1] has Z-transform X(z) =
Y (z) = H(z)X(z) =
−1
1−z −1 ,
(14)
|z| < 1. Thus,
−4 − 0.25z −1 + 0.5z −2
.
(1 − 0.25z −1 )(1 + 0.5z −1 )(1 − z −1 )
(15)
The region of convergence is the intersection of the region of convergence for X and that for H:
ROC(y) = {z | 1/2 < |z| < 1} .
(16)
(f ) We need to compute the inverse Z-transform of Y . We do this by partial fraction expansion.
Since the degree of the denominator is larger than the degree of the numerator, we can directly
apply partial fraction expansion, without polynomial division. We will express Y (z) as
Y (z) =
B
C
A
+
+
,
−1
−1
1 − 0.25z
1 + 0.5z
1 − z −1
(17)
with
−4 − 0.25z −1 + 0.5z −2 (1 + 0.5z −1 )(1 − z −1 ) z=0.25
= −1/3
A
=
(18)
(19)
and
−4 − 0.25z −1 + 0.5z −2 (1 − 0.25z −1 )(1 − z −1 ) z=−0.5
−1/3
B
=
=
(20)
(21)
and
−4 − 0.25z −1 + 0.5z −2 (1 − 0.25z −1 )(1 + 0.5z −1 ) z=1
= −10/3
C
=
(22)
(23)
To finish inverting the Z transform, we need to decide what ROC to associate with each of the
terms in the partial fraction expansion. Based on the ROC of y, we should associate with the term
A
B
1−0.25z −1 the ROC |z| > 1/4; with the term 1+0.5z −1 , we should associate the ROC |z| > 1/2; finally,
C
with the term 1−z
−1 , we should associate ROC |z| < 1. Inverting each term indvidually, we obtain
y[n] = −(1/3)(0.25)n u[n] − (1/3)(−0.5)n u[n] − (10/3)u[−n − 1].
(24)