 # EE 350 - Signals & Systems Sample Exam #3 - Solutions 1)

```EE 350 - Signals & Systems
Sample Exam #3 - Solutions
Spring 2005
1)
3/2
Y1(jω)
-330
(a)
-300 -270
0
1
Y2(jω)
-160
(b)
-120 -80
0
80
3/2
1
-300 -270 -160
-330
-120 -80
(c)
0
80
120 160
270 300 330
ω
1
Z(jω)
-160
-120 -80
0
80
-260 -220 -180
120 160
ω
1/2
A(jω)
(e)
ω
120 160
Y3(jω)
(d)
ω
270 300 330
-20 0 20 60 180 220
-60
ω
1/2
A(jω)
=
260
ω
-260 -220 -180
1/2
B(jω)
-20
(f)
-60
-20 0 20 60 180 220
260
ω
20
(g) From transform #9 in Table 4.2, b(t ) =
1 sin(20t )
2 πt
(h) This is frequency division multiplexing because two signals share different frequency bands.
2) (a) This is pulse amplitude modulation. See O&W Chapter 8.6
(b) C1(jω) comes from Fourier Series Table #6 with w=5, X0=1, T0 = 20 Table 3.1 #2 (time shift) and
Table 4.2#1.
a0=(5/20)e-j(0)(2π/20)(-2.5)= 5/20 = 1/4,
a1=(5/20)sinc(5(1)(2π/20/2)) e-j(1)(2π/20)(-2.5) =(1/4)sinc(π/4)ejπ/4,
a-1=(5/20)sinc(5(-1)(2π/20/2)) e-j(-1)(2π/20)(-2.5),
a2=(5/20)sinc(5(2)(2π/20/2)) e-j(2)(2π/20)(-2.5),
ak=(5/20)sinc(5(k)(2π/20/2)) e-j(k)(2π/20)(-2.5),
Y(jω)
a-1
a0
…
a1
a2
-2π/20
0
ωs
=2π/20
…
ω
4π/20
(c) This is time division multiplexing because we are modulating with a square pulse train. This will
allow another signal to fill in the gaps of time not used by this signal.
3) For the transfer function
H ( s) =
s+2
2
s + 4s + 3
Imag
X O X
-3 -2 -1
Real
a)
b) 4 possible ROCs:
Re{s}<-3, -3< Re{s} < -1, Re{s}>-1.
1/ 2 1/ 2
+
c & d) H ( s ) =
s + 3 s +1
1/ 2
Re{s}<-3 Æ h1(t)=-e-3tu(-t), Re{s}>-3 Æ h1(t)= e-3tu(t)
H 1 ( s) =
s+3
1/ 2
H 2 ( s) =
Re{s}<-1 Æ h1(t)= -e-tu(-t), Re{s} > -1 Æ h1(t)= e-tu(t)
s +1
so
h(t)= -½e-3tu(-t) -½e-tu(-t),
not stable
Re{s}<-3 Æ
h(t)= ½e-3tu(t) -½e-tu(-t)
not stable
-3< Re{s} < -1 Æ
h(t)= ½e-3tu(t)+ ½e-tu(t)
stable
Re{s}>-1 Æ
not causal
not causal
causal
4) For the transfer function
H (s) =
s 2 + s + 1.25
s+2
Imag
O
X
-2
1
-1/2
O
Real
-1
Sketch the pole-zero plot for this transfer function
|H(j ω)|
-1
1
ω
a) Sketch the magnitude of the associated Fourier Transform
b) What type of filter is this? Because of the infinite amplitudes, I would accept either Stop band
filter or high pass filter with the appropriate explanation).
c) Draw the magnitude of the straight line Bode plot for this transfer function. Does it agree with
20log10(5/8) = -4 ωn2 = 5/4
Bode Diagram
40
Magnitude (dB)
30
20
10
0
-10
-1
10
10
0
10
1
10
2
5) Given the system transfer function
H ( s) =
s +1
( s + 2)( s + 3)
(a)
Using the Initial Value Theorem, what is h(0+)?
lim
sH ( s ) = 1
s→∞
(b)
Using the Final Value Theorem, what is lim h(t ) ?
(c)
lim
sH ( s ) = 0
s→0
s +1
−1
2
H ( s) =
=
+
⇒ (−e − 2t + 2e −3t )u (t )
( s + 2)( s + 3) s + 2 s + 3
as t Æ 0 h(0)=(-1+2)=1 (checks)
as t Æ ∞ h(∞) = (0+0)=0 (checks)
t →∞
6) If x(t)=e-5tu(t) and y(t)=3e-5tu(t)+e-2tu(t), what is h(t)? What must the ROC for H(s) be?
and y(t)=3e-5tu(t)+e-2tu(t), what is h(t)
1
3
1
4s + 11
ROC: Re{s}>-5, Y ( s ) =
ROC: Re{s}>-2
X (s) =
+
= 2
s + 5 s + 2 s + 7 s + 10
s+5
3
4s + 11
H ( s) =
= 4+
ROC: Re{s}>-2
h(t)=4δ(t) + 3e-2tu(t)
s+2
s+2
7) Let x[n] = (-1)nu[n] + anu[-n-n0]. Determine the constraints on the complex number a and the integer
n0, given the ROC of X(z) is 1 < |z| <2 .
The first term transforms with equation 5 in Table 10.2 giving ROC |z|>|-1|
The second term transforms with equation 5 in Table 10.2 with ROC |z|<|a| linearity & time shift
which don’t affect the ROC so |a|=2, n0 arbitrary because time shift doesn’t affect the ROC.
8) Determine the inverse transform of
X ( z) =
z −1
1 + 89 z −1
|z|>8/9
⎛
⎞
1
⎟ eq 2 (from ROC “>”), T10.2: time shift of
By use of the transform tables X ( z ) = z −1 ⎜
⎜ 1 + 8 z −1 ⎟
9
⎝
⎠
(8/9)nu[n]
time shift of one unit Æ x[n] = (-8/9)n-1u[n-1]
b) By use of the Power Function method
Long division gives (want all negative exponents so see how many times “1” goes into the
numerator)
X(z) = z-1 – 8/9z-2 + (8/9)2z-3+…
x[n] = …0,0,0,1, -8/9, (8/9)2, … (first non-zero term at n=1). = (8/9)n-1u[n-1]
c) Sketch the pole-zero plot for this transfer function
Imag
a)
1
X
-1 -8/9
Real
-1
single pole at z=-8/9. No zeros.
d) Sketch the magnitude of the associated Fourier transform based on the pole-zero plot.
|H(ejω)|
0
e)
π
2π ω
What type of filter is this?
This is a high pass filter.
9) Given the signal with the Z-transform:
−5
8
z ( z − 2)
1 − 2 z −1
z 2 − 2z
3
3
=
=
=
+
X ( z) =
1 + 52 z −1 + z − 2 z 2 + 52 z + 1 ( z + 12 )( z + 2) 1 + 12 z −1 1 + 2 z −1
a)
Sketch the pole-zero plot for this transfer function
Imag
1
X
X O
-2 -1 -1/2
O
2 Real
-1
zeros at z={0,2}, poles at z={-1/2, -2}
b) How many and what are the possible ROCs for this transfer function?
3 possible: ROCs: |z|>2, ½<|z|<2, |z|<½
c) For each ROC in (b) above, determine the associated inverse transform.
x[n] = -5/3(- ½ )nu[n] + 8/3(-2)nu[n]
|z|>2 Æ
x[n] = -5/3 (- ½ )nu[n] + -8/3 (-2)nu[-n-1]
½<|z|<2 Æ
x[n] = 5/3 (- ½ )nu[-n-1] + -8/3 (-2)nu[-n-1]
½ >|z| Æ
d) For each ROC in (b) above, determine stability and causality of the impulse response.
not stable, yes causal
|z|>2 Æ
stable, not causal
½<|z|<2 Æ
not stable, not causal
½ >|z| Æ
e) What would the inverse transform be if you knew the signal was right sided?
x[n] = -5/3 (- ½ )nu[n] + 8/3 (-2)nu[n]
|z|>2 Æ
f) What would the inverse transform be if you knew the signal was absolutely summable?
x[n] = -5/3 (- ½ )nu[n] + -8/3 (-2)nu[-n-1]
½<|z|<2 Æ
``` # EE 179 May 7, 2014 Fourier Transform and Applications Handout #29 