1. No category

Sample Final Exam Covering Chapters 10-17 (finals00)
Sample Final Exam (finals00)
Covering Chapters 10-17 of Fundamentals of Signals & Systems
Problem 1 (10 marks)
Consider the causal FIR filter described by its impulse response
1
1
h[n] = − δ [n] + δ [n − 1] − δ [n − 2] .
2
2
(a) [3 marks] Draw a block diagram of the direct form realization of the filter. Write the transfer
function H ( z ) of the filter and specify its ROC.
(b) [5 marks] Find the frequency response H ( e
Sketch its magnitude H (e
jω
jω
) of the filter and give its magnitude and its phase.
) for ω ∈ [−π , π ] . What type of filter is it? (low-pass, band-pass or
high-pass?)
(c) [2 marks] Compute the -3dB cutoff frequency
ωc
of the filter.
Answer:
(a) [3 marks]
x[ n]
y[ n]
+
−0.5
+
z −1
+
1
+
z −1
−0.5
1
1
−0.5 z 2 + z − 0.5
H ( z ) = − + z −1 − z −2 =
,
2
2
z2
1
z >0
Sample Final Exam Covering Chapters 10-17 (finals00)
(b) [5 marks] Frequency response:
1
1
1
 1

H (e jω ) = − + e − jω − e − j 2ω = e − jω 1 − e jω − e − jω 
2
2
2
 2

= e − jω (1 − cos ω )
H (e jω ) = 1 − cos ω
∠H (e jω ) = −ω
H (e jω )
1
−π
−0.5π
0.5π
π
ω
Highpass filter
(c) [2 marks] -3dB cutoff frequency
1 − cos ω c =
1
2
ωc
of the filter:
⇐ ω c = a cos(1 −
1
2
) = 1.2735
Problem 2 (20 marks)
The system shown below processes the continuous-time signal yn (t ) = y1 (t ) + y2 (t ) + n(t )
composed of the sum of:
• Signal y1 (t ) which is a lower single-sideband amplitude modulation, suppressed carrier (lower
SSB-AM/SC) of the corresponding message signal x1 (t ) ,
•
Signal y2 (t ) which is an upper single-sideband amplitude modulation, suppressed carrier (upper
SSB-AM/SC) of the corresponding message signal x2 (t ) ,
•
a noise signal n(t ) .
Both SSB-AM/SC signals share the same carrier signal cos(ω c t ) of frequency
ω c . Also assume that
the sidebands have the same magnitude as the spectra of the message signals.
The antialiasing filter H a ( jω ) is a perfect unity-gain lowpass filter with cutoff frequency
ωa .
The
antialiased signal w(t ) is first converted to a discrete-time signal w[ n] via the CT/DT operator. Signal
w[n] is processed in parallel by two discrete-time filters: a lowpass filter H lp (e jΩ ) , and a highpass
filter H hp (e
jΩ
) . Then, the output of either filter, as selected by a switch, is the input to a discrete-time
synchronous demodulator with frequency Ω c which is used to demodulate either of the two
discretized message signals. Finally, the DT/CT operator followed by a continuous-time lowpass filter
reconstruct the desired continuous-time message signal.
2
Sample Final Exam Covering Chapters 10-17 (finals00)
wd [ n]
y (t ) +
y n (t )
+
H a ( jω )
w( t )
CT /DT
H hp (e jΩ )
n(t )
vd [ n ]
H lp (e jΩ )
T
S1
S 2 y d [ n]
qd [ n ]
×
x (t )
xd [ n]
DT /CT
cos(Ωc n)
H lp ( jω )
x (t )
T
The modulating (or message signals) x1 (t ), x2 (t ) have spectra X 1 ( jω ), X 2 ( jω ) as shown below.
The spectrum N ( jω ) of the noise signal is also shown below.
X 1 ( jω )
1
−4000π
4000π
1
−4000π
ω
X 2 ( jω )
4000π
ω
N ( jω )
0.2
−10000π
10000π
(a) [5 marks] Find the maximum carrier frequency
filter's cutoff frequency
ωa
ωc
ω
and the corresponding minimum antialiasing
that will avoid any unrepairable distortion of the modulated signals due
to the additive noise n(t ) . Sketch the spectrum W ( jω ) of the signal w(t ) for the frequencies
ωc
and
ωa
that you found. Indicate the important frequencies and magnitudes on your sketch.
Answer:
Maximum ω c = 10000π − 4000π = 6000π . Minimum
1
−10000π −6000π −2000π
ω a = 10000π .
W ( jω )
2000π 6000π 10000π ω
3
Sample Final Exam Covering Chapters 10-17 (finals00)
(b) [4 marks] Find the minimum sampling frequency
ωs =
2π
and its corresponding sampling period
T
T that will allow perfect reconstruction of the modulated signal.
Answer:
ω s = 2ω a = 20000π , T =
2π
ωs
= 10−4
(c) [5 marks] Suppose that the frequency of the discrete-time synchronous demodulator can be set to
either Ω c1 or Ω c 2 corresponding to switch positions S1 and S 2 , respectively. With the
frequencies that you found in (a) and (b), find the cutoff frequencies Ωlp and Ω hp of the ideal
discrete-time lowpass and highpass filters. Also give the frequencies Ω c1 and Ω c 2 . Sketch the
spectra Wd (e
jΩ
) , Vd (e jΩ ) and Qd (e jΩ ) over the interval [ −π , π ] , indicating the important
frequencies and magnitudes.
Answer:
Cutoff frequencies: Ωlp = 6000π T = 0.6π ,
demodulator frequencies Ω c1 = 0.6π ,
Ω hp = 6000π T = 0.6π
Ωc 2 = 0.6π
10000
−π
−0.6π
−0.2π
Wd (e jΩ )
0.6π
0.2π
π
ω
Vd (e jΩ )
10000
−π
−0.6π
−0.2π
0.2π
0.6π
π
Ω
π
Ω
Qd (e jΩ )
10000
−π
−0.6π
−0.2π
0.2π
4
0.6π
Sample Final Exam Covering Chapters 10-17 (finals00)
(d) [6 marks] Find the cutoff frequency
ωr
of the ideal continuous-time lowpass filter H lp ( jω ) and
its gain A so that the message signals can be recovered, i.e., x(t ) = x1 (t ) when the switch is at
position S1 , and x(t ) = x2 (t ) when the switch is at position S 2 . Sketch the spectra X d (e
jω
),
X ( jω ) and X ( jω ) in each case (switch at S1 and at S 2 ), indicating the important frequencies
and magnitudes.
Answer:
H lp ( jω )
A=2
−4000π
ω r = 4000π
ω
Switch at S1:
X d (e jΩ )
10000
−π
−0.4π
0.4π
π
Ω
X ( jω )
1
−20000π
−10000π
−4000π
4000π
10000π
20000π
X ( jω )
1
−4000π
4000π
5
ω
ω
Sample Final Exam Covering Chapters 10-17 (finals00)
Switch at S2:
X d (e jΩ )
10000
−π
−0.8π −0.4π
π
0.4π 0.8π
Ω
X ( jω )
1
−10000π
−8000π −4000π
4000π
10000π
8000π
Ω
X ( jω )
1
−10000π
10000π
−4000π
4000π
Ω
Problem 3 (20 marks)
Consider the discrete-time system shown below, where ↓ N represents decimation by N . This
system transmits a signal x[ n] over two low-rate channels.
x[n]
H lp1 (e jω )
x1[n]
channels
↓ N1
v1[n]
x2 [n] z2 [n]
w2 [n]
×
↓ N2
H lp 2 (e jω )
H hp (e )
v2 [n]
jω
cos(ω2 n)
Numerical values:
lowpass filters' cutoff frequencies
highpass filter's cutoff frequency
ω clp1 = ω clp 2 =
ω chp =
π
2
π
2
,
6
,
?
y[n]
Sample Final Exam Covering Chapters 10-17 (finals00)
modulation carrier frequency:
ω2 =
π
2
,
4
3π

1
ω
,
ω
−
≤
 3π
4
jω
signal's spectrum over [ −π , π ] : X (e ) = 
.
3
π
 0,
< ω <π

4
(a) [6 marks] Sketch the spectra X (e ) , X 1 (e
the important frequencies and magnitudes.
jω
jω
) , X 2 (e jω ) , Z 2 (e jω ) and W2 (e jω ) , indicating
Answer:
X (e jω )
1
−π
−0.75π
0.75π
π
Ω
π
ω
X 1 (e jω )
1
−π
−0.5π
0.5π
X 2 (e jω )
1
−π −0.75π −0.5π
0.5π 0.75π π
ω
Z 2 (e jω )
0.5
−π −0.75π −0.25π
0.25π
0.75π π
ω
W2 (e jω )
0.5
−π −0.75π −0.25π
0.25π
7
0.75π π
ω
Sample Final Exam Covering Chapters 10-17 (finals00)
(b) [6 marks] Find the maximum decimation factors N1 and N 2 avoiding aliasing and sketch the
corresponding spectra V1 (e
jω
) , V2 (e jω ) , indicating the important frequencies and magnitudes.
Answer:
Maximum decimation factors: N1 = 2 and N 2 = 4 .
1
−π
V1(e jω )
−0.5π
0.5π
π
ω
π
ω
V2 (e jω )
0.5
1/6
−π
(c) [8 marks] Design the receiver system (draw its block diagram) such that y[ n] = x[n] . You can
use upsamplers (with symbol ↑ N ), synchronous demodulators, ideal filters and summing
junctions.
Answer:
H lp (e jω )
v1[n]
v2 [n]
↑ N1
↑ N2
k1[n]
x1[n]
1
−π
−0.5π
k2 [n]
0.5π
ω
+
π
x2 [n] +
2
−π −0.75π−0.5π
0.5π 0.75π π
ω
H bp (e jω )
8
y[n]
Sample Final Exam Covering Chapters 10-17 (finals00)
Problem 4 (20 marks)
One way to discretize a continuous-time state-space system with sampling period T for simulation
purposes is to use the step-invariant transformation called "c2d".
c 2d
ud [n]
u (t )
DT /ZOH
x = Ax + Bu
yd [ n ]
y (t )
y = Cx + Du
CT/DT
The DT/ZOH operator is defined as a mapping from a discrete-time signal to an impulse train
(consisting of forming a train of impulses occurring every T seconds, with the impulse at time nT
having an area ud [ n] ), followed by a "hold" function that holds each value for a period T .
ud [ n]
conversion of
discrete-time
sequence to
impulse train
u p (t )
1
u (t )
h (t )
T t
DT/ ZOH
(a) [10 marks] Derive the mathematical relationships mapping the continuous-time state-space
matrices ( A, B, C , D ) to the discrete-time state-space matrices ( Ad , Bd , Cd , Dd ) . (Hint: start
with the general state response given by x(t ) = e
A ( t − t0 )
t
x(t0 ) + ∫ e Aτ Bu (t − τ )dτ )
t0
Answer:
The state response of the state-space system ( A, B, C , D ) to be discretized is given by the
combination of the zero-input state response and the zero-state state response.
t
x(t ) = e A(t −t0 ) x(t0 ) + ∫ e Aτ Bu (t − τ )dτ
t0
t
= e A(t −t0 ) x(t0 ) + ∫ e A( t −τ ) Bu (τ )dτ
t0
Notice that the input u (t ) is a "staircase" function as it is held constant over every sampling period.
Thus, if we take t0 = kT and t = ( k + 1)T , we can write
9
Sample Final Exam Covering Chapters 10-17 (finals00)
( n +1)T
x((n + 1)T ) = e AT x(nT ) +
∫
e A(( n +1)T −τ ) dτ Bu (nT )
nT
= e x(nT ) + e
AT
A ( n +1)T
( n +1)T
∫
e − Aτ dτ Bu ( nT )
nT
= e AT x(nT ) − e A( n +1)T A−1 e − Aτ 
( n +1)T
nT
Bu ( nT )
= e AT x(nT ) − A−1e A( n +1)T e − A( n +1)T − e − AnT  Bu ( nT )
= e AT x(nT ) − A−1  I n − e AT  Bu (nT )
AT
x(nT ) + A−1 e AT − I n  B u (nT )
= eN
Ad
Bd
A ( n +1)T
−1
where we used the fact that A commutes with e
. If we define the discrete-time state
xd [n] = x(nT ) , the last equation can be rewritten as a DT state equation as follows, while the output
equation doesn't change.
xd [n + 1] = Ad xd [n] + Bd ud [n]
yd [n] = Cd xd [n] + Dd ud [n]
,
where
Ad := e AT
Bd := A−1 e AT − I n  B
Cd := C
Dd := D
(b) [10 marks] We want to simulate the causal LTI differential system initially at rest:
2
dy
du
+y=
dt
dt
with input u (t ) = q (t ) (unit step) by discretizing it using "c2d". Find a continuous state-space
representation ( A, B, C , D ) of this system. Then, discretize it with T = 1 to get the system
xd [n + 1] = Ad xd [n] + Bd ud [n]
yd [n] = Cd xd [n] + Dd ud [n]
using your formulas obtained (a). Finally, compute and plot the first 5 values of the step response
yd [n] . Give the steady-state value lim yd [n] .
n →+∞
Answer:
The transfer function corresponding to the system is
s
1
, Re{s} > − .
2s + 1
2
10
Sample Final Exam Covering Chapters 10-17 (finals00)
Partial fraction expansion:
s
1 1 1
1
, Re{s} > − .
= −
2s + 1 2 4 2s + 1
2
Corresponding state-space system:
1
1
x = − x − u
2
4
1
y = x+ u
2
Thus, A = −
1
1
1
, B = − , C = 1, D = .
2
4
2
Discretized state-space system:
Ad = e AT = e −0.5 = 0.6065
1
1
Bd := A−1 e AT − I n  B = −2(e −0.5 − 1)(− ) = (e −0.5 − 1) = −0.1967
4
2
Cd := 1
Dd :=
1
2
Recursion:
xd [n + 1] = 0.6065 xd [n] − 0.1967ud [n]
yd [n] = 1xd [n] + 0.5ud [n]
n = 0:
xd [1] = 0.6065(0) − 0.1967(1) = −0.1967
yd [0] = 1(0) + 0.5(1) = 0.5
n = 1:
xd [2] = 0.6065(−0.1967) − 0.1967(1) = −0.3160
yd [1] = 1(−0.1967) + 0.5(1) = 0.3033
n = 2:
xd [3] = 0.6065(−0.3160) − 0.1967(1) = −0.3884
yd [2] = 1(−0.3160) + 0.5(1) = 0.1840
n = 3:
xd [4] = 0.6065(−0.3884) − 0.1967(1) = −0.4323
yd [3] = 1(−0.3884) + 0.5(1) = 0.1116
n = 4:
xd [5] = 0.6065(−0.4323) − 0.1967(1) = −0.4589
yd [4] = 1(−0.4323) + 0.5(1) = 0.0677
11
Sample Final Exam Covering Chapters 10-17 (finals00)
plot
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
−0.1967
= −0.5
Steady-state:
1 − 0.6065
y[∞] = −0.5 + 0.5 = 0
x[∞] =
Problem 5 (15 marks)
A causal recursive DLTI system is described by its (zero-state) impulse response given below:
π 
h[n] = (0.8) n sin  n  u[n] + 0.5δ [n]
4 
1
and has initial conditions y[ −1] =
, y[−2] = 0 . Calculate the zero-input response y zi [n] of the
2
system for n ≥ 0 using the unilateral z-transform.
Answer:
From the table:
π
H ( z) =
0.8sin( ) z −1
4
1
+ , z > 0.8
2
π
1 − 1.6 cos( ) z −1 + (0.8) 2 z −2
4
1
+ 0.32 z −2
2
, z > 0.8
=
π −1
2 −2
1 − 1.6 cos( ) z + (0.8) z
4
π
Difference equation: y[ n] − 1.6 cos( ) y[ n − 1] + (0.8) y[ n − 2] =
2
4
1
+ 0.32 x[n − 2]
2
Using the unilateral z-transform with zero input:
π
(
)
(
)
Y ( z ) − 1.6 cos( ) z −1Y ( z ) + y[−1] + (0.8) 2 z −2 Y ( z ) + z −1 y[−1] + y[−2] = 0
4
12
Sample Final Exam Covering Chapters 10-17 (finals00)
π
π
Y ( z ) − 1.6 cos( ) z −1Y ( z ) + (0.8) 2 z −2 Y ( z ) = 1.6 cos( ) y[−1] − (0.8) 2 y[−1]z −1 − (0.8) 2 y[−2]z −2
4
4
π −1
π 2
π −1
π −1

2 −2 
2
2
1 − 1.6 cos( 4 ) z + (0.8) z  Y ( z ) = 1.6 cos( 4 ) − (0.8) cos( 4 ) z = 0.8 − (0.8) cos( 4 ) z
Y ( z) =
π


0.8 1 − 0.8cos( ) z −1 
4


π
−1
1 − 1.6 cos( ) z + (0.8) z
4
2
−2
, z > 0.8
and from the table, we get the zero-input response:
π 
y zi [n] = (0.8) n +1 cos  n  u[n]
4 
Problem 6 (15 marks)
Design an envelope detector to demodulate the AM signal:
y (t ) = [1 + 0.5 x(t )]cos(2π 106 t )
where x(t ) is the periodic modulating signal shown below. That is, draw a circuit diagram of the
envelope detector and compute the values of the circuit components. Justify all of your approximations
and assumptions. Provide rough sketches of the carrier signal, the modulated signal and the signal at
the output of the detector. What is the modulation index m of the AM signal?
x (t )
1
−0.005
0.005
t (s)
-1
An envelope detector can be implemented with the following simple RC circuit with a diode.
+
+
C
R
-
y (t )
-
13
w( t )
Sample Final Exam Covering Chapters 10-17 (finals00)
The output voltage of the detector, when it goes from one peak at voltage v1 to the next when it
intersects the modulated carrier at voltage v 2 after approximately one period T = 1µs of the carrier,
is given by:
v 2 ≅ v1e − T / RC
Since the time constant τ = RC of the detector should be large with respect to T = 1µs , we can use
a first-order approximation of the exponential such that
v 2 ≅ v1 (1 − T RC ) .
This is a line of negative slope −
v1
between the initial voltage v1 and the final voltage v 2 so that
RC
v 2 − v1
≅ − v1 RC .
T
This slope must be more negative than the maximum negative slope of the envelope of y (t ) , which is
maximized as follows:
min
t
d (0.5 x(t ))
= −200
dt
Taking the worst-case v1 = 0.5 , we must have
0.5
< −200
RC
⇔
0.5
RC <
= 0.0025
200
−
We could take R = 2k Ω, C = 1µ F to get RC = 0.002 .
Let K be the maximum amplitude of 0.5 x(t ) , i.e., 0.5 x (t ) < 0.5 = K and let A = 1 . The
modulation index m is the ratio m = K A = 0.5 .
END OF EXAMINATION
14