DIGITAL SIGNAL PROCESSING IV
Previous Evaluation and Assessment
Module Code: EIDSV4
VUT
Vaal University of Technology
Digital Signal Processing EIDSV4
Unit 1 First Assessment 2 March 2006
Page 1
Question 1 A sequence is defined by y(k+2) + 0.5y(k) = 0 with y(0) = 1 and y(1) = 2.
Write down the first six terms, y(0) to y(5), of the sequence.
(4)
Question 2
x(k)
A signal x(k) is shown in Figure 1.
Figure 1
Sketch the following signals:
1
a) -x(k+1)
(2)
k
b) x(1-k)
(2)
0
c) x(2-k)x(k-2)
(2)
-1
Question 3 If x(k) = cosk×[u(k) – u(k-4)], with k in radians, express x(k) as the sum
of weighted and shifted impulse functions.
(4)
Question 4 Refer to the structure in Figure 2.
y(k)
x(k)
a) Determine a difference equation that
-1
z
describes the relationship between the
z-1
output y(k), and the input x(k), of the
system.
(5)
-0.5
b) Calculate the first five terms, h(0) to h(4),
Figure 2
of the system’s impulse response h(k).
(5)
+
+
k
Question 5 The impulse response of a digital filter is given by h(k) = 0.5 u(k). The
input to the system is a sinusoidal signal, x(k) = sinku(k). Use graphical convolution
methods to determine the response y(k) of the system, for k = 0, 1, 2, 3 and 4.
(10)
Question 6 Calculate the sum of the infinite sequence:
S = 1×½ + 2×(½)2 + 3×(½)3 + 4×(½)4 + 5×(½)5 ………
(6)
k
Question 7 The impulse response of a system is given by h(k) = 0.5 , for k  0,
and the input by x(k) = sink, for k  0 and k in radians. Use z transform methods
and the relationship Y(z) = H(z)X(z), to find an expression for the output y(k).
---oooOooo---
(10)
Total: 50
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
z2X(z) - z2x(0) - zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N -1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2abkcos(k + )u(k)
(a z (a z

z b z b
1
x(k) = ω
s

ωs /2
ωs /2
X()e
jkT
d , s = 2π
T
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
h(k) = c
, T= 2π
ωs
π (k -α)ωcT
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
tan (
)
2
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 1 Eerste Evaluasie 2 Maart 2006 Memorandum
Bladsy 1
Digitale Seinverwerking IV EIDSV4 Eenheid 1 Eerste Evaluasie 2 Maart 2006 Memorandum
1. y(0)=1, y(1)=2, y(2)=-½y(0)=-0.5[1], y(3)=-½y(1)=-1[1], y(4)=-½y(2)=0.25[1] en y(5)=-½y(3)=0.5[1]
[4]
{or alternatively: (z2+0.5)Y(z)=z(z+2)  Y(z)=1.5-1.231z/(z-0.7071/2+1.51.231z/(z-0.7071-/2)
y(k)=3(0.7071)kcos(k/2 – 1.231)  y(2)=-0.5, y(3)=-1, y(4)=0.25 and y(5)=0.5}
2.
a) -x(k+1)
b) x(1-k)
c) x(2-k)x(k-2)
[2]
[2]
1
-1
[6]
[2]
1
k
0
0
k
-1
0
k
-1
-1
3. x(k) = (k) [1] + 0.5403(k-1) [1] – 0.4161(k-2) [1] – 0.99(k-3) [1]
[4]
4. a)
x(k)

D
-½y(k-1)
[10]
x(k-1)-½y(k-2)
-½

b) h(k) = (k-1) + h(k-1) – ½h(k-2)
y(k)=y(k-1)+x(k-1)-½y(k-2)
h(0)=(-1)+h(-1)–½h(-2)=0+0-0 = 0 [1]
h(1)=(0)+h(0)- ½h(-1)=1+0-0 = 1 [1]
D
h(2)=(1)+h(1)- ½h(0)=0+1-0=1 [1]
h(3)=(2)+h(2)- ½h(1)=0+1-½=½ [1]
y(k-1)
h(4)=(3)+h(3)- ½h(2)=0+½-½=0 [1]
y(k) = x(k-1) + y(k-1) – ½y(k-2)
5. h(k)=0.5ku(k)
h(0)=1
h(1)=0.5
h(2)=0.25
h(3)=0.125
h(4)=0.0625
[10]
(5)
(5)
x(k)=sinku(k)
n
-4
-3
-2
-1
0
1
2
3
4
x(0)=0
x(n)
0 .8415 .9093 .1411 -.7568
x(1)=.8415
h(-n) .0625 0.125 0.25 0.5
1
y(0)=0 [2]
h(1-n)
.0625 0.125 0.25 0.5
1
y(1)=0.8415[2]
x(2)=.9093
h(2-n)
.0625 0.125 0.25 0.5
1
y(2)=1.33 [2]
x(3)=.1411
h(3-n)
.0625 0.125 0.25 0.5
1
y(3)=0.8061[2]
x(4)=-.7568
[2]
h(4-n)
.0625 0.125 0.25
0.5
1
y(4)=-0.3537
S = 1×½ + 2×(½)2 + 3×(½)3 + 4×(½)4 + 5×(½)5 …….
½S =
1×(½)2 + 2×(½)3 + 3×(½)4 + 4×(½)5 …….[1]
½S = ½ + (½)2 + (½)3 + (½)4 + (½)5 …….[1]
½(½S) =
(½)2 + (½)3 + (½)4 + (½)5 …….[1]
½S - ¼S = ½ [1]
¼S = ½
[6] S = 2 [2]
6.
h(k) = 0.5k  H(z) = z/(z-0.5) [1]
x(k) = sink  X(z) = zsin1/(z-ej)(z-e-j) [1]
Y(z) = H(z)X(z) = 0.8415z2/(z-0.5)(z-ej)(z-e-j) [1] { = 0.8415z2/(z-0.5)(z2 – 1.081z + 1)}
Y(z)/z = 0.8415z/(z-0.5)(z-ej)(z-e-j) = A/(z-0.5) + B/(z-ej) + B*/(z-e-j)
A = 0.84150.5/(0.52 – 1.0810.5 + 1) = 0.593
B = 0.841511/(11-0.5)(11-1-1) = 0.5935-2.094
Y(z)/z = 0.593/(z-0.5) + 0.5935-2.094/(z-11) + 0.59352.094/(z-1-1) [3]
Y(z) = 0.593z/(z-0.5) + 0.5935-2.094z/(z-11) + 0.59352.094z/(z-1-1)
y(k) = 0.593(0.5)k + 20.59351kcos(k-2.094)
[10]
= 0.593(0.5)k + 1.187cos(k-2.094) [4]
7.
Digital Signal Processing EIDSV4
Unit 2 First Assessment 30 March 2006
Question 1 The block diagram of a system with input x(k)
x(k) and output y(k), is shown in Figure 1.
a) Determine a difference equation that describes
z-1
the relationship between y(k) and x(k).
(2)
b) Determine an expression for the system’s
transfer function H(z) = Y(z)/X(z).
(4)
z-1
c) If the sampling period is 1 sec. (T = 1), calculate
the system’s amplitude frequency response, |H()|,
at  = 0, 1, 2 and 3 rad/sec.
(8)
Page 1
+
0.25
y(k)
0.5
Figure 1
0.25
x(k)
Question 2 A non periodic digital signal x(k), with x(-1) = 1, x(0) = 2,
2
x(1) = 1 and x(k) = 0 everywhere else, is shown in Figure 2.
1
1
a) Determine the frequency spectrum X() of this signal.
k
0
Assume a sampling period of 1 sec. (T = 1).
(4)
Figure 2
b) From X(), calculate X(0), X(/2), X() and X(3/2). (4)
Question 3 The four (N=4) sampled values x(0)=0, x(1)=1, x(2)=2 and x(3)=1, that were
obtained when a signal x(t) was sampled, are shown in Figure 3.
x(k)
Figure 3
a) For the discrete Fourier transform, X(n), of this sequence,
2
calculate the values of X(0), X(1), X(2), and X(3).
(10)
1
1
b) Assuming a sampling period of 1 sec. (T = 1), determine
k
0
the frequency of the first harmonic, X(1), in rad/sec.
(2)
Question 4 The frequency response H(), of a
s
s
s
s

high pass filter, is shown in Figure 4. Assume a
1  2  ω   4 and 4  ω  2
sampling period of T = 1 sec. (s = 2/T = 2).
H()= 


0  s  ω  s
a) Show that the impulse response h(k) for this
filter is given by h(k) =
sin k 1 sin k/2
–
.
k
2 k/2
(6)
b) Calculate in particular, the five values of h(k),
h(-2), h(-1), h(0), h(1) and h(2).
(4)
b) Sketch the structure of a FIR filter of length 5
(N = 5), that may realize this filter.
(6)
---oooOooo--Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
2
z X(z) - z2x(0) - zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N-1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
1
x(k) = ω
s
ω /2
 ωss /2 X()e
jkT
d , s = 2π
T

4
1
ω
ω
- s - s
2
4
4
H()
0 ωs
4
Figure 4
ωs
2
 [r/s]
Total: 50
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
h(k) = c
, T= 2π
ωs
π (k -α)ωcT
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
)
tan (
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
j
Euler and trig. identities: re = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 2 Eerste Evaluasie 30 Maart 2006 Memorandum
Bladsy 1
Digitale Seinverwerking IV EIDSV4 Eenheid 2 Eerste Evaluasie 30 Maart 2006 Memorandum
1.
a) y(k) = 0.25x(k) + 0.5x(k-1) + 0.25x(k-2)
b) Y(z) = 0.25X(z) + 0.5z-1X(z) + 0.25z-2X(z)
Y(z)/X(z) = H(z) = 0.25 + 0.5z-1 + 0.25z-2
{= (0.25z2 + 0.5z + 0.25)/z2 }
c) H(ejT) = [0.25(ejT)2 + 0.5(ejT) + 0.25]/(ejT)2
H(ej) = (0.25ej2 + 0.5ej + 0.25)/ej2
H() = (0.252 + 0.5 + 0.25)/12
H(0) = (0.25 + 0.5 + 0.25)/1 = 1  |H(0)| = 1
en H(1) = (0.252 + 0.51 + 0.25)/12
= 0.7702-1  |H(1)| = 0.7702
en H(2) = (0.254 + 0.52 + 0.25)/14
= 0.2919-2  |H(2)| = 0.2919
en H(3) = (0.256 + 0.53 + 0.25)/16
[14]
= 0.005-3  |H(3)| = 0.005
2.
1
a) X() =  x(k)e
(2)
x(k)
(4)
D
0.5
(2)
D
0.25
(2)
(2)
(2)
- jkT
k = -1
= 1e-j(-1) + 2 + 1e-j(1) [2] = ej + 2 + e-j
= 2 + 2[(ej +e-j)/2]
= 2 + 2cos [2]
b) X(0) = 2 + 2cos0 = 4 [1]
X(/2) = 2 + 2cos(/2) = 2 [1]
X() = 2 + 2cos() = 0 [1]
[8]
X(3/2) = 2 + 2cos(3/2) = 2 [1]
3.
N-1
a) X(n) =  x(k)e
- j(2/N)nk
k =0
3
=  x(k)e
x(k)
a) h(k) = 1
s
s /2
 s/2 H()e
(4)
x(k)
=
2
1
1
k
0
NT = 4 sec
f1 = ¼ Hz

jk
d = 1
H()e d
 


k
- j(/2)nk
 jk

jk
= 1
1e d + 1
1e d [2]
 
 
1  jk   [2]
1  jk  
e
e
+
=
jk 
jk 
 
 /2

1
0
k =0
jkT
2
1
(4)
X(0) = 0 + 1 + 2 + 1 = 4 [1]
X(1) = 0 + 1e-j(/2)×1×1 + 2e-j(/2)1*2 + e-j(/2)1×3
= 1-/2 + 2- + 1-3/2 = -2 [3]
X(2) = 0 + 1e-j(/2)×2×1 + 2e-j(/2)2×2 + 1e-j(/2)2×3
= 1- + 2-2 + 1-3 = 0 [3]
X(3) = 0 + 1e-j(/2)×3×1 + 2e-j(/2)3×2 + 1e-j(/2)3×3
= 1-3/2 + 2-3 + 1-9/2 = -2 [3]
(10)
[12] b) T1 = NT = 4×1  f1 = ¼ Hz  1 = 2×¼ = /2 r/s (2)
4.
y(k)

0.25
H()
1
ω
ω
- s - s
2
4
0
ωs
4
ωs
2
jk
- jk
jk/2
- jk/2
1  - jk/2
1  jk
e
e
e
- jk 
jk/2  e
–
e
e
+
e
e
=
jk 
jk 


jk
jk
jk
- jk
jk/2
- jk/2
1 e
e
1 e
e
1
1
sink–
sin(k/2)
–
=
k
j2
k
j2
k
k
sin k 1 sin k/2 [2]
–
=
k
2 k/2
=
(6)
Digitale Seinverwerking EIDSV4 Eenheid 2 Eerste Evaluasie 30 Maart 2006 Memorandum
b) h(-2) = sin(-2)/(-2) - ½sin(-)/(-) = 0 [½]
h(-1) = sin(-)/(-) - ½sin(-/2)/(-/2) = -0.3183 [½]
h(0) = sin0/0 - ½sin0/0 = 1 – 0.5 = 0.5 [2]
h(1) = sin/ - ½sin(/2)/(/2) = -0.3183 [½]
h(2) = sin(2)/(2) - ½sin/ = 0 [½]
Bladsy 2
h(k)
0.5
0
-0.3183
k
-0.3183
(4)
c)
x(k)

y(k)
Filter parameters
0.5
D
-.318
D
0
k
-0.3183 -0.3183
0.5
D
[16]
-.318
(6)
Digital Signal Processing EIDSV4
Unit 1 Final Assessment 25 May 2006
Page 1
Question 1 Tabulate the values of the following discrete function x(k), for –4  k  5:
(10)
x(k) = k2[u(k+1) + (k-1)] + (k-1)u(2-k).
Question 2 Express the function x(k+1) as the sum of weighted and shifted impulse
(4)
functions, if x(k) = (k+1)+(k2+2)[u(k+2)-u(k-2)].
x(k)
Question 3 Refer to the
system in Figure 1 with
input x(k) and output y(k).
Determine the first five
terms, h(0) to h(4), of
the impulse response of
the system.
(10)
+
+
y(k)
z-1
+
1
2
–
Question 4 The impulse response of a
system, h(k), is shown in Figure 2. The
input to the system is the step function,
x(k) = u(k). Use graphical convolution, to
determine the first five values, y(0) to y(5),
of the response, y(k), of the system.
(5)
1
2
z-1
Figure 1
1
2
h(k)
3
Figure 2
2
1
k
0

Question 5 Starting from the definition of the z transform, X(z)   x(k)zk , show that:
k 0
Z{x(k+1)} = zZ{x(k)} – zx(0).
(3)
Question 6 Use z transform methods to determine the k’th term, in closed form,
of the series: x(k+2) = x(k+1) + x(k), with x(0) = 1 and x(1) = 1.
(8)
Question 7 Use z transform methods to solve y(k) from the following difference equation:
y(k) + 0.5y(k-1) = sin k, for k > 0.
(10)
---oooOooo---
Total: 50
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
2
z X(z) - z2x(0) - zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N-1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
1
x(k) = ω
s

ωs /2
ωs /2
X()e
jkT
d , s = 2π
T
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
, T= 2π
h(k) = c
ωs
π (k -α)ωcT
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
)
tan (
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 1 Finale Evaluasie 25 Mei 2006 Memorandum
Bladsy 1
Digitale Seinverwerking IV EIDSV4 Eenheid 1 Finale Evaluasie 25 Mei 2006 Memorandum
1.
[10]
k
-4
-3
-2
-1
0
1
2
3
4
5
k2[u(k+1)+(k-1)]+(k-1)u(2-k)
16(0+0)+(-5)(1) = -5 [1]
9(0+0)+(-4)(1) = -4 [1]
4(0+0)+(-3)(1) = -3 [1]
1(1+0)+(-2)(1) = -1 [1]
0(1+0)+(-1)(1) = -1 [1]
1(1+1)+0(1) = 2 [1]
4(1+0)+1(1) = 5 [1]
9(1+0)+2(0) = 9 [1]
16(1+0)+3(0) = 16 [1]
25(1+0)+4(0) = 25 [1]
2. x(k) = (k+1) + (k2 + 2)[u(k+2) - u(k-2)]
= 6(k+2)+4(k+1)+2(k)+3(k-1) [3]
[4] x(k+1) = 6(k+3)+4(k+2)+2(k+1)+3(k)[1]
4.
x(n)
1 1 1 1 1 1
h(-n) 1 2 3
y(0)=3 [1]
h(1-n)
1 2 3
y(1)=5 [1]
h(2-n)
1 2 3
y(2)=6 [1]
h(3-n)
1 2 3
y(3)=6 [1]
1 2 3
y(4)=6 [1]
[5] h(4-n)
3. x(k)
w(k)
+
z
+
1
2
-
1
2
-1
z-1w(k)
z-1
z-2w(k)
1
2
w(k)=x(k)+0.5z-1w(k)-0.5z-2w(k)
w(k) = x(k)/(1-0.5z-1+0.5z-2)
y(k) = w(k)+0.5z-1w(k) = (1+0.5z-1)w(k)
= [(1+0.5z-1)/(1-0.5z-1+0.5z-2)]x(k)
y(k)-0.5z-1y(k)+0.5z-2y(k)=x(k)+0.5z-1x(k)
y(k)=x(k)+0.5x(k-1)+0.5y(k-1)-0.5y(k-2) [5]
h(0) = 1+0+0-0=1[1], h(1) = 0+0.5+0.5-0=1[1]
h(2)=0+0+0.5-0.5=0[1], h(3)=0+0+0-0.5=-0.5[1]
[10] h(4) = 0+0+0.5(-0.5)-0=-0.25[1]
5. Z{x(k)} = x(k)k=0 + z-1x(k)k=1 + z-2x(k)k=2 + ……
Z{x(k+1)} = x(1)+z-1x(2)+z-2x(3)+z-3x(4)+….[1]
= z[x(1)z-1 + x(2)z-2 + x(3)z-3 + … ]
= z[x(0) + z-1x(1) + z-2x(2) + x(3)z-3 +..]
– zx(0) [1]
[3]
= zZ{x(k)} – zx(0) [1]
6.
x(k+2) = x(k+1) + x(k)
z2X(z) – z2x(0) – zx(1) = zX(z) – zx(0) + X(z) [2]
(z2 – z –1)X(z) = z2 + z – z
X(z) = z2/(z2 –z –1)
X(z)/z = z/(z2 – z – 1) [1] = z/(z – 1.618)(z + 0.618)
= 0.7236/(z – 1.618) + 0.2764/(z + 0.618) [3]
X(z) = 0.7236z/(z – 1.618) + 0.2764z/(z + 0.618)
[8] x(k) = 0.7236(1.618)k + 0.2764(-0.618)k [2]
7.
+
y(k)
y(k) + 0.5y(k-1) = sink
Y(z) + 0.5z-1Y(z) = zsin1/(z-11)(z-1-1) [2]
zY(z) + 0.5Y(z) = z20.8415/(z-11)(z-1-1)
Y(z) = z20.8415/(z+0.5)(z-11)(z-1-1)
Y(z)/z = 0.8415z/(z+0.5)(z-11)(z-1-1)
= -0.235/(z+0.5) + 0.3737-1.251/(z-11) + 0.37371.251/(z-1-1) [4]
Y(z) = -0.235z/(z+0.5) + 0.3737-1.251z/(z-11) + 0.37371.251z/(z-1-1)
y(k) = -0.235(-0.5)k + 20.37371kcos(k – 1.251)
[10] y(k) = -0.235(-0.5)k + 0.7474cos(k – 1.251) [4]
Digital Signal Processing EIDSV4
Unit 2 Final Assessment 25 May 2006
Page 1
x(k)
y(k)
Question 1 The block diagram of a system with
input x(k) and output y(k), is shown in Figure 1.
a) Determine a difference equation that describes
z-1
z-1
the relationship between y(k) and x(k).
(2)
b) Determine an expression for the system
-1
1
function H(z) = Y(z)/X(z).
(4)
z-1
c) Plot the roots of the denominator of H(z) and indicate
the type of natural response that the system will exhibit. (4) Figure 1
-0.64
d) Using H(z), calculate the steady state response yss(k),
if the input to the system is given by x(k) = cos(/4)k. (5)
Question 2 A non periodic digital signal x(k), with x(-1) = -1,
x(k)
x(1) = 1 and x(k) = 0 everywhere else, is shown in Figure 2.
1
k
a) Determine the frequency spectrum X() of this signal.
0
-1
Assume a sampling period of 1 sec. (T = 1).
(4)
Figure 2
b) From X(), calculate X(0), X(/2), X() and X(3/2). (4)
Question 3 The four (N=4) sampled values x(0)=0, x(1)=-1, x(2)=0 and x(3)=1, that were
obtained when a signal x(t) was sampled, are shown in Figure 3.
x(k)
a) For the discrete Fourier transform, X(n), of this sequence,
1
calculate the values of X(0), X(1), X(2), and X(3).
(10)
k
0
b) Assuming a sampling period of 1 sec. (T = 1), determine
-1
Figure 3
the frequency of the first harmonic, X(1), in rad/sec.
(2)
Question 4 The frequency response H(), of a
s
s

low pass filter, is shown in Figure 4. Assume a
1  4  ω  4
H()= 
sampling period of T = 1 sec. (s = 2/T = 2).




0  s  ω   s and s  ω  s
a) Show that the impulse response h(k) for this
2
4
4
2

+
filter is given by h(k) =
1 sin k/2
.
2 k/2
(5)
b) Calculate in particular, the five values of h(k),
h(-2), h(-1), h(0), h(1) and h(2).
(4)
b) Sketch the structure of a FIR filter of length 5
(N = 5), that may realize this filter.
(6)
---oooOooo--Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
z2X(z) - z2x(0) - zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N-1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2abkcos(k + )u(k)
(a z (a z

z b z b
1
x(k) = ω
s
ω /2
 ωss /2 X()e
jkT
d , s = 2π
T
1
ω
ω
- s - s
2
4
H()
0 ωs
4
Figure 4
ωs
2
 [r/s]
Total: 50
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
h(k) = c
, T= 2π
ωs
π (k -α)ωcT
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
)
tan (
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 2 Finale Evaluasie 25 Mei 2006 Memorandum
Bladsy 1
Digitale Seinverwerking IV EIDSV4 Eenheid 2 Finale Evaluasie 25 Mei 2006 Memorandum
1. a) y(k)=x(k)-x(k-1)+y(k-1)–0.64y(k-2)
b) Y(z) = X(z) – z-1X(z) + z-1Y(z) – 0.64z-2Y(z) [2]
z2Y(z) = z2X(z) – zX(z) + zY(z) – 0.64Y(z)
Y(z)(z2 – z + 0.64) = z(z – 1)X(z)
H(z) = Y(z)/X(z) = z(z – 1)/(z2 - z + 0.64) [2]
c) z2 - z + 0.64 = 0
z = 0.8±0.8957 [2]
(2)
x(k)
y(k)

D
(4)
[2]
D
1
-1
D
(4)
-0.64
j/4
d) H(/4) = H(e ) = H(10.7854)
= 10.7854(10.7854– 1)/[(10.7854)2 – 10.7854 + 0.64) = 2.5470.9529 [4]
[15] yss(k) = 2.547cos[(/4)k + 0.9529) [1]
1
2. a) X() =  x(k)e
- jkT
= -1e-j(-1) + 1e-j(1) [2]
x(k)
k = -1
= -ej + e-j = -j2[(ej - e-j)/j2]
= -j2sin [2]
b) X(0) = -j2sin0 = 0 [1]
X(/2) = -j2sin(/2) = -j2 [1] (= 2-/2)
X() = -j2sin() = 0 [1]
[8]
X(3/2) = -j2sin(3/2) = j2 [1] (=2/2)
3.
N-1
a) X(n) =  x(k)e
k =0
- j(2/N)nk
3
=  x(k)e
(4)
s /2

jk
jkT
a) h(k) = 1
H()e d
H()e
d = 1





/2
s
s
1  jk   [1]

jk
= 1
1e d [1] =
e
 
jk 
 
x(k)
1
0
[1]
=
e
jk/2
- jk/2
e
jk
jk/2
- jk/2
1 e
e
1
sin(k/2)
=
k
k
j2
1 sin k/2 [2]
=
2 k/2
k
-1
NT = 4 sec
f1 = ¼ Hz


1  jk/2
- jk/2 
=
e
e
jk 

0
-1
- j(/2)nk
k =0

k
(4)
X(0) = 0 - 1 + 0 + 1 = 0 [1]
X(1) = 0 - 1e-j(/2)×1×1 + 0e-j(/2)1*2 + e-j(/2)1×3
= -1-/2 + 1-3/2 = 21.571 [3]
X(2) = 0 - 1e-j(/2)×2×1 + 0e-j(/2)2×2 + 1e-j(/2)2×3
= -1- + 1-3 = 0 [3]
X(3) = 0 - 1e-j(/2)×3×1 + 0e-j(/2)3×2 + 1e-j(/2)3×3
= -1-3/2 + 1-9/2 = 2-1.571 [3]
(10)
[12] b) T1 = NT = 4×1  f1 = ¼ Hz  1 = 2×¼ = /2 r/s (2)
4.
1
(5)
1
ω
ω
- s - s
2
4
H()
0
ωs
4
ωs
2
=
(5)
Digitale Seinverwerking EIDSV4 Eenheid 2 Finale Evaluasie 25 Mei 2006 Memorandum
b) h(-2) = ½sin(-)/(-) = 0 [½]
h(-1) = ½sin(-/2)/(-/2) = 0.3183 [½]
h(0) = ½sin0/0 = 0.5 [2]
h(1) = ½sin(/2)/(/2) = 0.3183 [½]
h(2) = ½sin/ = 0 [½]
Bladsy 2
h(k)
0.3183
0.5
0.3183
k
0
(4)
c)
x(k)

y(k)
Filter parameters
0.3183 0.5
D
.318
0
0.3183
k
D
0.5
D
[15]
.318
(6)
Digital Signal Processing EIDSV4
Unit 1 First Assessment 8 March 2007
Page 1
Question 1 Write down the first seven values y(0) to y(6), of the sequence:
y(k + 2) = y(k + 1) – 0.5y(k) + u(k), with y(0) = 0 and y(1) = 3.
Question 2
A signal x(k) = [1 2 1] is shown in Figure 1.
a) Express x(k) as the sum of weighted
and shifted impulse functions.
(3)
(2)
b) Sketch the function x(2 – k)u(k – 2).
Question 3 Refer to the structure in Figure 2.
a) Determine a difference equation that
describes the relationship between the
output y(k), and the input x(k), of the
system.
(6)
b) Calculate the first five terms, h(0) to h(4),
of the system’s impulse response h(k). (5)
(5)
x(k)
2
1
Figure 1
1
k
0
x(k)
Question 4 The input to a system, x(k) = [1 2 3],
and the output of the system, y(k) = [3 8 14 8 3],
is shown in Figure 3.
a) Determine the impulse response, h(k),
of the system.
(7)
b) Use long division, and the relation
Y(z)
, to verify your answer
H(z) =
X(z)
for h(k), in question 4 a).
(6)
+
+
3
y(k)
z-1
2
7
Figure 2
x(k)
2
1
3
k
y(k)
14
8
3
Figure 3
8
3
k
Question 5 Calculate the sum of the infinite sequence:
(4)
S = ¼ + (¼)2 + (¼)3 + (¼)4 + (¼)5 ………
Question 6 Use z transform methods to determine x(k) in closed form, from the
difference equation: x(k+2) – x(k+1) + 0.5x(k) = u(k), with x(0) = 0 and x(1) = 3.
(12)
---oooOooo---
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
z2X(z) - z2x(0) - zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N-1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
1
x(k) = ω
s

ωs /2
ωs /2
X()e
jkT
d , s = 2π
T
Total: 50
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
, T= 2π
h(k) = c
π (k -α)ωcT
ωs
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
tan (
)
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 1 Eerste Evaluasie 8 Maart 2007 Memorandum
Bladsy 1
1. y(0) = 0, y(1) = 3  y(2) = y(1) – 0.5y(0) + u(0) = 3 – 0 + 1 = 4 [1]
y(3) = y(2) – 0.5y(1) + u(1) = 4 – 1.5 + 1 = 3.5 [1] y(4) = y(3) – 0.5y(2) + u(2) = 3.5 – 2 + 1 = 2.5 [1]
[5] y(5) = y(4) – 0.5y(3) + u(3) = 2.5 – 1.75 + 1 = 1.75 [1] y(6) = y(5) – 0.5y(4) + u(4) = 1.75 – 1.25 + 1 = 1.5 [1]
2. a) x(k) = (k + 1) [1] + 2(k) [1] + (k – 1) [1]
[5]
3. a) y(k) = 3a(k) + 7z-1a(k) [2] …… (1)
and a(k) = x(k) + 2z-1a(k) [2]
a(k)×(1 – 2z-1) = x(k)
a(k) = x(k)/(1 – 2z-1) ……… (2)
(2) in (1):
y(k) = 3[x(k)/(1 – 2z-1)] + 7z-1[x(k)/(1 – 2z-1)]
y(k)×(1 – 2z-1) = 3x(k) + 7z-1x(k)
y(k) – 2y(k – 1) = 3x(k) + 7x(k – 1) [2]
2 [1]
1 [1]
b) x(2-k)u(k-2)
(3)
(2)
k
0
x(k)
a(k)
+
+
3
y(k)
z-1
2
(6)
7
z-1a(k)
[11] b) h(0) = 3 [1] h(1) = 6 + 7 = 13 [1] h(2) = 26 [1] h(3) = 52 [1] h(4) = 104 [1]
(5)
y(0) = x(0)h(0)  3 = 1×h(0)  h(0) = 3 [2]
y(1) = x(0)h(1)+x(1)h(0)  8 = 1×h(1)+23  h(1) = 2 [2]
y(2) = x(0)h(2)+x(1)h(1)+x(2)h(0)  14 = 1×h(2)+22+33  h(2) = 1 [2]
y(3)=x(0)h(3)+x(1)h(2)+x(2)h(1)+x(3)h(0)  8 = 1×h(3)+21+32+03  h(3) = 0
h(k) = 0 vir k >=3 [1]
(7)
14
y(k)
x(k)
3
b) X(z) = 1 + 2z-1 + 3z-2 [2]
8
8
-1
-2
-3
-4 [2]
2
Y(z) = 3 + 8z + 14z + 8z + 3z
1
3
3
k
H(z) = Y(z)/X(z)
k
3+2z-1+ z-2 [2]
1+2z-1+3z-2 | 3+8z-1+14z-2+8z-3+3z-4
3+6z-1+ 9z-2
n
-1 0 1 2 3
4 5
-1
-2
-3
x(n)
1
2
3
2z + 5z +8z
h(-n) ? 3
y(0)=3
2z-1 + 4z-2 +6z-3
-2
-3
-4
h(1-n) ? 2 3
y(1)=8
z +2z +3z
h(2-n) ? 1 2 3
y(2)=14
-2
-3
-4
z +2z +3z
4) a)
[13]
5.
[4]
6.
H(z) = 3 + 2z-1 + z-2
h(0) = 3, h(1) = 2 en h(2) = 1
(6)
h(3-n) ?
h(4-n) ?
h(5-n) 0
0
0
0
1
0
0
2
1
0
3
2
1
3
2
3
y(3)=8
y(4)=3
y(5)=0
S = ¼ + (¼)2 + (¼)3 + (¼)4 + (¼)5 + …….
¼S =
(¼)2 + (¼)3 + (¼)4 + (¼)5 + …….[1]
¾S = ¼ [1]  S = 1/3 [2]
x(k+2) – x(k+1) + 0.5x(k) = u(k)
z2X(z) – z2x(0) – zx(1) – [zX(z) – x(0)] + 0.5X(z) = z/(z – 1) [2]
z2X(z) – zX(z) + 0.5X(z) – z2×0 – z×3 + 0 = z/(z – 1)
(z2 – z + 0.5)X(z) = 3z + z/(z – 1) = [3z(z – 1) + z]/(z – 1) = (3z2 – 3z + z)/(z – 1) = (3z2 – 2z)/(z – 1)
X(z) = (3z2 – 2z)/(z – 1)(z2 – z + 0.5)
X(z) = (3z2 – 2z)/(z – 1)(z – 0.70710.7854)(z – 0.7071– 0.7854) [2]
X(z)/z = (3z – 2)/(z – 1)(z – 0.70710.7854)(z – 0.7071– 0.7854) [1]
X(z)/z = 2/(z – 1) + 2.236– 2.034/(z – 0.70710.7854) + 2.2362.034/(z – 0.7071– 0.7854) [3]
X(z) = 2z/(z – 1) + 2.236– 2.034z/(z – 0.70710.7854) + 2.2362.034z/(z – 0.7071– 0.7854
x(k) = 2u(k) + 2×2.236×0.7071kcos(0.7854k – 2.034)u(k) [4]
x(0)=0,x(1)=3,x(2)=4,x(3)=3.5,x(4)=2.5,
[12]
= 2u(k) + 4.472×0.7071kcos(0.7854k – 2.034)u(k)
x(5)=1.75 and x(6)=1.5 as per Question 1
Digital Signal Processing EIDSV4
Unit 2 First Assessment 12 April 2007
Page 1
Question 1 The difference equation that describes the relationship between the
output y(k) and the input x(k), of a system, is given by:
y(k) = x(k) + 0.6y(k – 1) – 0.25y(k – 2)
a) Determine an expression for the system’s transfer function H(z) = Y(z)/X(z).
(4)
b) Comment on the basic character of the natural (transient) response of the system.
(3)
c) Calculate the steady state output response, yss(k), of the system, if the input is
given by x(k) = sin2k.
(6)
Question 2 The transfer function of a digital filter, H(), that aims to combat high
frequency sinx/x attenuation caused by first order hold, is proposed in Figure 1. In


the fundamental interval,  s    s , H() is defined as,
H()
Figure 1
2
2
1½

1.5
H() = 
1





when  s  ω   s and s  ω  s
2
4
4
2
s
s
when
ω
4
4
1
ω
ω
- s - s
2
4

0
ωs
4
ωs
2
where s is the sampling frequency. Assuming a sampling period of T = 1 sec,
determine an expression for the impulse response, h(k), required for this digital filter. (10)
Question 3 The four (N = 4) sampled values x(0) = 1,
x(k)
Figure 2
2 2
x(1) = 2, x(2) = 2 and x(3) = 1, shown in Figure 2,
1
1
were obtained when a signal x(t) was sampled.
k
a) For the discrete Fourier transform, X(n), of this,
0
sequence, calculate the values of X(0), X(1), X(2), and X(3).
(10)
b) Calculate the average value of x(t).
(1)
c) Calculate the amplitude of the fundamental frequency (X(1)), contained in x(t).
(2)
d) If the sampling period was 0.5 sec (T = 0.5), calculate the frequency of the
fundamental component (X(1)), contained in x(t).
(2)
Question 4 Design a finite impulse response low pass filter, of length seven (N=7),
with cut off frequency c = ¼ s, where s is the sampling frequency. Draw a
(12)
structure for the filter.
---oooOooo--Total: 50
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
2
z X(z) - z2x(0) - zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N -1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
(z e j )(z e j )
z(sin)
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2abkcos(k + )u(k)
(a z (a z

z b z b
1
x(k) = ω
s

ωs /2
ωs /2
X()e
jkT
d , s = 2π
T
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
, T= 2π
h(k) = c
ωs
π (k -α)ωcT
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
)
tan (
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 2 Eerste Evaluasie 12 April 2007 Memorandum
Bladsy 1
1. a) y(k) = x(k) + 0.6y(k – 1) – 0.25y(k – 2).
Y(z) = X(z) + 0.6z-1Y(z) – 0.25z-2Y(z) [2]
Y(z)[1 – 0.6z-1 + 0.25z-2] = X(z)  Y(z)/X(z) = 1/(1 – 0.6z-1 + 0.25z-2) [2]
H(z) = z2/(z2 – 0.6z + 0.25)
b) z2 – 0.6z + 0.25 = 0  z = [0.6  (0.62 – 4×0.25)]/2 = [0.6  (– 0.64)]/2
z = [0.6  j(0.64)]/2 = [0.6  j0.8]/2 = 0.3  j0.4 = 0.50.9273r [2]
Natural response: decaying sinusoid [1]
c) H(ej2) = (12)2/[(12)2 – 0.6×12 + 0.25] = 14/(14 – 0.62 + 0.25)
= 0.7625–0.5947 [4]
[13]
yss(k) = 0.7625sin(2k – 0.5947) [2]
2. s = 2/T = 2: h(k) = 1
s
s /2
 s/2 H()e
(6)

jk
H()e d
d = 1
 

N-1

3. a) X(n) =  x(k)e
- j(2/N)nk
k =0
[15]
(3)
H()
1½



jk
jk
jk
1.5e d + 1
1e d + 1
1.5e d [3]
= 1
1

 
 
 
1.5  jk  
1  jk  
1.5  jk   [3]
0 
- - 

e
e
e
+
+
=
2
2
jk 
jk 
 
 /2 jk 
 /2
1.5  - jk/2
1  jk/2
1.5  jk
- jk 
- jk/2 
jk/2 
=
e
e
+
e
e
+
e
e
jk 
jk 
jk 



jk
- jk
jk/2
- jk/2
jk
- jk
jk/2
- jk/2
e
e
e
e
1.5 e
e
0.5 e
e
– 0.5×
=
–
= 1.5×
jk
jk
k
j2
k
j2
1.5
0.5
sin k
sin k/2
sink –
sin(k/2) [4] = 1.5
– 0.25
=
k
k
k
k/2

[10]
jkT
(4)

3
=  x(k)e
- j(/2)nk
x(k)
k =0
[1]
X(0) = 1 + 2 + 2 + 1 = 6
X(1) = 1e-j(/2)×1×0 + 2e-j(/2)×1×1 + 2e-j(/2)1×2 + 1e-j(/2)1×3
= 10 + 2– /2 + 2–  + 1– 3/2 = 1.414– 2.356 [3]
X(2) = 1e-j(/2)×2×0 + 2e-j(/2)×2×1 + 2e-j(/2)2×2 + 1e-j(/2)2×3
= 10 + 2–  + 2– 2 + 1– 3 = 0 [3]
X(3) = 1e-j(/2)×3×0 + 2e-j(/2)×3×1 + 2e-j(/2)3×2 + 1e-j(/2)3×3
= 10 + 2– 3/2 + 2– 3 + 1– 9/2 = 1.4142.356 [3]
b) avg[x(t)] = X(0)/N = 6/4 = 1.5
c) Amplitude of first harmonic = |X(1)|/(N/2) = 1.414/2 = 0.7071
d) T1 = NT = 4×0.5 = 2  f1 = ½ Hz [2]  1 = 2×½ =  r/s
2
2
1
1
k
0
(10)
(1)
(2)
(2)
4. h(k) = (cT/)sin[(k - )cT]/[(k - )cT].
With T = 2/s and c = s/4: h(k) = [(s/4)(2/s)/]sin[(k - 3)(s/4)(2/s)]/[(k - 3)(s/4)(2/s)]
h(k) = (1/2)sin[(k - 3)/2]/[(k - 3)/2] [1]
h(0) = (1/2)sin(-3/2)/(-3/2) = -0.1061 [1]
h(1) = (1/2)sin(-2/2)/(-2/2) = 0 [1]
[1]
h(2) = (1/2)sin(-/2)/(-/2) = 0.3183
h(3) = (1/2)sin0/0 = 0.5 [2]
h(5) = h(1) = 0 [1]
h(4) = h(2) = 0.3183 [1]
[1]
h(6) = h(0) = -0.1061
x(k)
z-1
– 0.1061
[12]
z-1
z-1
z-1
z-1
z-1
[3]
0
0.3183
0.5
0.3183
0
– 0.1061
+
+
+
+
+
+
y(k)
Digital Signal Processing EIDSV4
Unit 1 Final Assessment 24 May 2007
Page 1
Question 1 Write down the first seven values y(0) to y(6), of the sequence:
y(k + 2) = u(k) + 0.6y(k + 1) – 0.25y(k), with y(0) = 0 and y(1) = 1.
(5)
Question 2 Tabulate the values of the function x(k) = (k + e – k)[u(k + 1) – u(k – 2)]
and express x(k) as the sum of a series of weighted and shifted impulse functions.
(6)
Question 3 Refer to the structure in Figure 1.
a) Determine a difference equation that
describes the relationship between the
output y(k), and the input x(k), of
the system.
(6)
b) Calculate the first five terms, h(0) to h(4),
of the system’s impulse response h(k).
(5)
x(k)
+
0.5
-1
+
y(k)
z
z-1
Figure 1
k
Question 4 The impulse response of a digital filter is given by h(k) = 0.8 u(k). The
k
input to the system is the signal, x(k) = 0.5 u(k).
a) Use graphical convolution methods to determine the response y(k) of the system,
for k = 0, 1, 2 and 3.
b) Verify your results in part a), by using z transform methods to obtain an expression
for y(k) from the inverse z transform of Y(z) = H(z)X(z), and then calculating
y(k) for k = 0, 1, 2 and 3.
(8)
(8)
Question 5 Use z transform methods to determine x(k) in closed form, from the
difference equation: x(k+2) – 0.6x(k+1) + 0.25x(k) = u(k), with x(0) = 0 and x(1) = 1. (12)
---oooOooo---
Total: 50
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
2
z X(z) - z2x(0) - zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N-1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
1
x(k) = ω
s

ωs /2
ωs /2
X()e
jkT
d , s = 2π
T
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
, T= 2π
h(k) = c
ωs
π (k -α)ωcT
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
)
tan (
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 1 Finale Evaluasie 24 Mei 2007 Memorandum
Bladsy 1
1. y(0) = 0, y(1) = 1  y(2) = u(0) + 0.6y(1) – 0.25y(0) = 1 + 0.6 – 0 = 1.6 [1]
y(3) = u(1) + 0.6y(2) – 0.25y(1) = 1 + 0.6×1.6 – 0.25×1 = 1 + 0.96 – 0.25 = 1.71 [1]
y(4) = u(2) + 0.6y(3) – 0.25y(2) = 1 + 0.6×1.71 – 0.25×1.6 = 1 + 1.026 – 0.4 = 1.626 [1]
y(5) = u(3) + 0.6y(4) – 0.25y(3) = 1 + 0.6×1.626 – 0.25×1.71 = 1 + 0.9756 – 0.4275 = 1.548 [1]
[5]
y(6) = u(1) + 0.6y(5) – 0.25y(4) = 1 + 0.6×1.548 – 0.25×1.626 = 1 + 0.9288 – 0.4065 = 1.522 [1]
2.
[6]
k
-1
0
1
k + e–k
1.718 [1]
1 [1]
1.368 [1]
x(k) = 1.718(k+1)+(k)+1.368(k-1) [3]
3. a) y(k) = a(k) + z-1a(k) [2] …………………..… (1)
x(k)
and a(k) = 0.5(x(k) + z-1a(k) + z-2a(k) [2]
a(k)×(1 – 0.5z-1 – 0.5z-2) = 0.5x(k)
a(k) = 0.5x(k)/(1 – 0.5z-1 – 0.5z-2) ….…… (2)
(2) in (1):
y(k) = 0.5x(k)/(1 – 0.5z-1 – 0.5z-2)
+ 0.5z-1x(k)/(1 – 0.5z-1 – 0.5z-2)
y(k)×(1 – 0.5z-1 – 0.5z-2) = 0.5x(k) + 0.5z-1x(k)
y(k) – 0.5y(k – 1) – 0.5y(k – 2) = 0.5x(k) + 0.5x(k – 1) [2]
0.5
+
4. a) h(k)=0.8ku(k) x(k)=0.5ku(k)
h(0)=1
x(0)=1
h(1)=0.8
x(1)=0.5
h(2)=0.64
x(2)=0.25
h(3)=0.512 x(3)=0.125
[1]
z-1
n
-3
-2
-1
0
1
2
3
x(n)
1
0.5 0.25 0.125
h(-n) 0.512 0.64 0.8
1
y(0)=1 [2]
h(1-n)
0.512 0.64 0.8
1
y(1)=1.3 [2]
h(2-n)
0.512 0.64 0.8
1
y(2)=1.29 [2]
h(3-n)
0.512 0.64 0.8
1
y(3)=1.157 [2]
b) H(z) = z/(z-0.8)
X(z) = z/(z-0.5) [1]
Y(z) = z2/(z-0.8)(z-0.5)  Y(z)/z = z/(z-0.8)(z-0.5) = 2.667/(z-0.8) – 1.667/(z-0.5)
Y(z) = 2.667z/(z-0.8) – 1.667z/(z-0.5)  y(k) = 2.667×0.8k – 1.667×0.5k [2]
y(0) = 1 [1] , y(1) = 1.3 [1] , y(2) = 1.29 [1] & y(3) = 1.157 [1]
[16]
5.
y(k)
z-1
b) h(0) = 0.5 [1] h(1) = 0.75 [1] h(2) = 0.625 [1] h(3) = 0.6875 [1] h(4) = 0.6563 [1]
[11]
[12]
+
a(k)
(6)
(5)
(8)
(8)
x(k+2) – 0.6x(k+1) + 0.25x(k) = u(k)
z2X(z) – z2x(0) – zx(1) – 0.6[zX(z) – x(0)] + 0.25X(z) = z/(z – 1) [2]
z2X(z) – 0.6zX(z) + 0.25X(z) – z2×0 – z×1 + 0.6×0 = z/(z – 1)
(z2 – 0.6z + 0.25)X(z) = z + z/(z – 1) = [z(z – 1) + z]/(z – 1) = (z2 – z + z)/(z – 1) = z2/(z – 1)
X(z) = z2/(z – 1)(z2 – 0.6z + 0.25)
X(z) = z2/(z – 1)(z – 0.50.9273)(z – 0.5– 0.9273) [2]
X(z)/z = z/(z – 1)(z – 0.50.9273)(z – 0.5– 0.9273) [1]
X(z)/z = 1.5385/(z – 1) + 0.775213.0172/(z – 0.50.9273) + 0.77521–3.0172/(z – 0.5–0.9273) [3]
X(z) = 1.5385z/(z – 1) + 0.775213.0172z/(z – 0.50.9273) + 0.77521–3.0172z/(z – 0.5–0.92732)
x(k) = 1.5385u(k) + 2×0.77521×0.5kcos(0.9273k + 3.0172)u(k) [4]
x(0)=0,x(1)=1,x(2)=1.6,x(3)=1.71,x(4)=1.626
x(5)=1.548 and x(6)=1.522 as per Question 1
= 1.5385u(k) + 1.5504×0.5kcos(0.9273k + 3.0172)u(k)
Digital Signal Processing EIDSV4
Unit 2 Final Assessment 24 May 2007
Page 1
Question 1 The block diagram of a system with
x(k)
0.25
input x(k) and output y(k), is shown in Figure 1.
z-1
y(k)
a) Determine a difference equation that describes
0.25
the relationship between y(k) and x(k).
(2)
z-1
b) Determine an expression for the system’s
Figure 1
0.25
transfer function H(z) = Y(z)/X(z).
(4)
c) Calculate the frequency response, H() = H(ejT), of the system, for a sampling
period T = 0.6283 sec, and frequency  = 1 rad/sec. (that is at T = 0.6283 rad).
(4)
d) A sinusoidal signal, sint, with frequency  = 1 rad/sec, is sampled with a
sampling period T = 0.6283 sec. The resulting signal, sin0.6283k, is applied
(4)
to the input of the filter in Figure 1. Calculate the steady state filter output, yss(k).
Question 2
H()
Figure 2
1
The proposed transfer function of a
narrow digital band pass filter with

centre frequency 0.1s and cut off
0
ωs
ω
frequencies 0.09s and 0.11s, is
9ωs ωs 11ωs
11ωs - ωs 9ωs
- s
2
2
10 100
100
100 10 100
shown in Figure 2.
In the fundamental interval, – 0.5ωs    0.5ωs , H() is defined as,
+
0
H() = 
 1
when  0.5ωs  ω   0.11ωs or  0.09ωs  ω  0.09ωs or 0.11ωs  ω  0.5ωs
when  0.11ωs  ω   0.09ωs or 0.09ωs  ω  0.11ωs
where s is the sampling frequency. Assuming a sampling period of T = 1 sec,
determine an expression for the impulse response, h(k), required for this digital filter. (12)
Question 3 Calculate the discrete Fourier transform X(0), X(1) and X(2) of the
sequence x(0) = – 1, x(1) = 0 and x(2) = 1.
(6)
Question 4 Design a finite impulse response low pass filter, of length five (N=5),
with cut off frequency c = s/8, where s is the sampling frequency. Draw a
structure for the filter.
(10)
Question 5 The transfer function of a first order high pass filter with cutoff frequency
0 = 1 rad/sec, is given by: F(s) = s/(s + 1). Use the bilinear transformation to determine
the transfer function (in standard form) of the corresponding digital IIR high pass filter,
with cut off frequency f0 = 1000 Hz, assuming a sampling rate of 8000 samples/sec.
(8)
---oooOooo--Total: 50
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
2
z X(z) - z2x(0) - zx(1)
x(k)
(1/a)kaku(k)
sinku(k)
X(z)
z/(z - a)2
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2abkcos(k + )u(k)
(a z (a z

z b z b
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
h(k) = c
, T= 2π
ωs
π (k -α)ωcT
Fourier transforms:
Bilinear Transformation:

- jkT
jkT
1 ω /2
Ωo
x(k) = ω  ωs /2 X()e
X() =  x(k)e
d , s = 2π
z-1
T
s= 
s
s
k = -
ωoT z +1
tan (
)
N -1
2
- j(2 /N)nk
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
X(n) =  x(k)e
n =0
k =0
j -j
j -j
j
Euler and trig. identities: re = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 2 Finale Evaluasie 24 Mei 2007 Memorandum
a) y(k) = 0.25x(k) + 0.25x(k-1) + 0.25x(k-2)
(2)
-1
-2
b) Y(z) = 0.25X(z) + 0.25z X(z) + 0.25z X(z)
x(k)
0.25
(3)
Y(z)/X(z) = H(z) = 0.25 + 0.25z-1 + 0.25z-2
2
2
{= (0.25z + 0.25z + 0.25)/z }
D
c) H() = H(ejT)
= [0.25(ejT)2 + 0.25(ejT) + 0.25]/(ejT)2
0.25
H(1) = H(ej0.6238)
D
= [0.25(ej0.6238)2 + 0.25(ej0.6238) + 0.25]/(ej0.6238)2
= (0.251.2476 + 0.250.6238 + 0.25)/11.2476
0.25
= 0.65580.6238/11.2477 = 0.6558– 0.6239
(5)
jT
jT
d) If x(k) = Asin(kT) = then yss(k) = A|H(e )|sin[kT + H(e )]
yss(k) = 1×|H(1)|sin[1×kT + H(1)] = 1×0.6558sin[1×k×0.6283 + (– 0.6239)]
[14]
yss(k) = 0.6558sin(0.6283k – 0.6239)
H()
2. s = 2/T = 2 [1]
1
s /2
jkT
h(k) = 1
H()e
d


/2
s
s
0

jk
π 11π
11π - π - 9π
9π
-π
H()e d
= 1
 
5
5
50
50
50
50
Bladsy 1
1.
y(k)

(4)


= 1

=
=
=
[12]
=
3. X(n) =
9

π
11
 e jk 
 e j k 
[2]


(
1)e d + 1
(
1)e d [3] = 
+
 9
11


jk
jk





 11 
 9
- j9k/50
- j11k/50
j11k/50
j9k/50
e
e
e
e
–
+
–
jk
jk
jk
jk
j11k/50
- j11k/50
j9k/50
- j9k/50
j0.22 k
- j0.22 k
j0.18 k
- j0.18 k
e
e
e
e
e
e
e
e
–
–
+
)
–
–
+
(=
jk
jk
jk
jk
jk
jk
jk
jk
j11k/50
- j11k/50
j9k/50
- j9k/50
e
e
1 e
1 e
[5]
–
j2
j2
k
k
11 sin 11k/50 9 sin 9k/50
1
1
sin 0.22k
sin 0.18k
–
(=.22
-.18
)
sin(11k/50) –
sin(9k/50) [1] =
k
k
50 k/50
50 k/50
k
k
N-1
- j(2/N)nk
 x(k)e

9 
jk

11
j k
k =0
X(0) = -1 + 0 + 1 = 0 [2] X(1) = -1 + 0 + 1e-j(2/3)2 = -1 + 1-4/3 = 1.7322.618 [2] (-1.5+j0.866)
[6]
X(2) = -1 + 0 + 1e-j(2/3)4 = -1 + 1-8/3 = 1.732-2.618 [2] (-1.5-j0.866)
4. h(k) = (cT/)sin[(k - )cT]/[(k - )cT]
With T = 2/s and c = s/8: h(k) = [(s/8)(2/s)/]sin[(k-2)(s/8)(2/s)]/[(k-2)(s/8)(2/s)]
x(k)
h(k) = (1/4)sin[(k-2)/4]/[(k-2)/4] [1]
[3]
z-1
z-1
z-1
z-1
[1]
h(0) = (1/4)sin(-/2)/(-/2) = 0.1592
h(1) = (1/4)sin(-/4)/(-/4) = 0.2251 [1]
0.1592
0.2251
0.25
0.2251
0.1592
h(2) = (1/4)sin0/0 = 0.25 [2]
h(3) = h(1) = 0.2251 [1]
y(k)
[1]
[10]
+
+
+
+
h(4) = h(0) = 0.1592

z  1 [3]

z -1
z -1
0
5. s
= 
= 2.414
z1
  T  z + 1 tan (2  1000)  (1/8000)  z + 1


tan 0 


2
 2 
[8]
F(s) = s/(s+1) = 2.414[(z - 1)/(z + 1)]/{2.414[(z - 1)/(z + 1)] + 1} = 2.414(z - 1)/[2.414(z - 1) + (z + 1)]
= (2.414z - 2.414)/(2.414z - 2.414 + z + 1) = (2.414z - 2.414)/(3.414z - 1.414) [3]
=(0.7071 - 0.7071z-1)/(1 - 0.414z-1) [2]
Digital Signal Processing EIDSV4
Unit 1 First Assessment 28 February 2008
Page 1
Question 1 Calculate the value of x(1) for each of the following functions x(k):
(2)
a) x(k) = (k + e – k)[u(k + 1) – u(k – 2)]
–k
2
(2)
b) x(k) = 2 ×(k + 1) + (k + 1)×[u(k + 4) – u(k – 3)]
k
2
(2)
c) x(k) = 2 ×(k – 1) + (k – k – 1)×[u(k + 2) – u(k – 2)]
d) x(k) = –k + u(k + 4) – (k + 3) + 2×[u(–k + 2) – u(k + 2)
(2)
Question 2
x(k)
Figure 1
A signal x(k) = [1 2 1 2] is shown in Figure 1.
2
2
a) Express x(k) as the sum of weighted
1
1
and shifted impulse functions.
(2)
k
(2)
b) Sketch the function x(1 – k)u(k – 2).
0
Question 3 Refer to the structure in Figure 2.
a) Determine a difference equation that
x(k)
y(k)
-1
describes the relationship between the
z
output y(k), and the input x(k), of the
z-1
system.
(5)
– 0.25
b) Calculate the first five terms, h(0) to h(4),
Figure 2
of the system’s impulse response h(k).
(5)
k
Question 4 The impulse response, h(k), of a system is given by h(k) = (0.5) u(k).
The input signal, x(k), to this system, is given by:
h(k)
 0 when k  0
1
x(k) = (–1)ku(k) =  1 when k  0, 2, 4, 6, .......
½
……
¼
 1 when k  1, 3, 5, 7, .......
k
0
The functions h(k) and x(k) are depicted in Figure 3.
Figure 3
a) Use graphical convolution methods to
x(k)
determine the response y(k) of the system,
1
1
1
for k = 0, 1, 2 and 3.
(8)
……
k
b) Use z transform methods to verify your
0
results in Question 4 a), by finding y(k) for
–1
–1
k = 0, 1, 2 and 3, from Y(z) = H(z)X(z).
(8)
Question 5 Use z transform methods to determine x(k) in closed form, from the
difference equation: x(k+2) – x(k+1) + 0.25x(k) = u(k), with x(0) = 0 and x(1) = 0.
(12)
---oooOooo--Total: 50
Appendix: z transforms
+
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z – 1)
z/(z – a)
z/(z – 1)2
z-1X(z)
z-2X(z)
zX(z) – zx(0)
2
z X(z) – z2x(0) – zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N-1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z – a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
1
x(k) = ω
s

ωs /2
ωs /2
X()e
jkT
d , s = 2π
T
+
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
, T= 2π
h(k) = c
π (k -α)ωcT
ωs
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
tan (
)
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 1 Eerste Evaluasie 28 Februarie 2008 Memorandum Bladsy 1
x(k) = (k + e – k)[u(k + 1) – u(k – 2)]  x(1) = (1 + e-1)[u(2) – u(-1)] = 1.368
x(k) = 2 – k×(k + 1) + (k2 + 1)×[u(k + 4) – u(k – 3)]  x(1) = 2-1(2) + (1+1)[u(5) – u(-2)] = 2
x(k) = 2k×(k – 1) + (k2 – k – 1)×[u(k + 2) – u(k – 2)]  x(1) = 21(0) + (1-1-1)[u(3)-u(-1)] = 1
x(k) = –k + u(k + 4) – (k + 3) + 2×[u(–k + 2) – u(k+2)  x(1) = -1+u(5)-(4)+2[u(1)-u(3)] = 0
1. a)
b)
c)
d)
[8]
[½]
[½]
[½]
[1]
(2)
[4] 2. a) x(k) = (k+1) + 2(k) + (k–1) + 2(k-2)
3. a)
x(k)

D
-¼y(k-1)
x(k-1)-¼y(k-2)
-¼
4. a) h(k)=(½)ku(k)
h(0)=1
h(1)=½
h(2)=¼
h(3)=1/8
x(k)=(-1)ku(k)
x(0)=1
x(1)=-1
x(2)=1
x(3)=-1
x(1-k)u(k-2)
1
(2)
k
0
b) h(k) = (k-1) + h(k-1) – ¼h(k-2)
y(k)=y(k-1)+x(k-1)-¼y(k-2) h(0)=(-1)+h(-1)–¼h(-2)=0+0-0 = 0 [1]
h(1)=(0)+h(0)- ¼h(-1)=1+0-0 = 1 [1]
D
h(2)=(1)+h(1)- ¼h(0)=0+1-0=1 [1]
h(3)=(2)+h(2)- ¼h(1)=0+1-¼=¾ [1]
y(k-1)
h(4)=(3)+h(3)- ¼h(2)=0+¾-¼=½ [1]
y(k) = x(k-1) + y(k-1) – ¼y(k-2)
[10]
n
-3
x(n)
0
h(-n) 1/8
h(1-n)
h(2-n)
h(3-n)
(5)
-2
0
¼
1/8
(5)
-1
0
½
¼
1/8
0
1
1
½
¼
1/8
1
-1
1
½
¼
2
1
1
½
3
-1
1
y(0)=1 [2]
y(1)= -½ [2]
y(2)= 3/4 [2]
y(3)=-5/8 [2]
b) H(z) = z/(z-0.5) [1]
(8)
[1]
X(z) = z/(z-(-1))=z/(z+1)
Y(z) = H(z)X(z) = z2/(z-0.5)(z+1)  Y(z)/z = z/(z-0.5)(z+1) = 0.3333/(z-0.5) + 0.6667/(z+1)
Y(z) = 0.3333z/(z-0.5) + 0.6667z/(z-(-1))  y(k) = 0.3333×0.5k + 0.6667×(-1)k [2]
y(0) = 1 [1] , y(1) = -0.5 [1] , y(2) = 0.75 [1] & y(3) = -0.625 [1]
(8)
[16]
5.
[12]

b)
(2)
(2)
(2)
(2)
x(k+2) – x(k+1) + 0.25x(k) = u(k)
z2X(z) – z2x(0) – zx(1) – [zX(z) – zx(0)] + 0.25X(z) = z/(z – 1) [2]
z2X(z) – zX(z) + 0.25X(z) – z2×0 – z×0 + 0 = z/(z – 1)
(z2 – z + 0.25)X(z) = z/(z – 1)
X(z) = z/(z – 1)(z2 – z + 0.25)
X(z)/z = 1/(z – 1)(z2 – z + 0.25) = 1/(z – 1)(z – 0.5)(z – 0.5) [2]
X(z)/z = 4 [1]/(z – 1) – 2 [1]/(z – 0.5)2 – 4 [2]/(z – 0.5)
X(z) = 4z/(z – 1) – 2z/(z – 0.5)2 – 4z/(z – 0.5)
x(k) = 4u(k) [1] – 2×[(1/0.5)×k×(0.5)k]u(k) [2] – 4(0.5)ku(k) [1]
= 4 – 4k(0.5)k – 4(0.5)k = 4[1 – (0.5)k(k + 1)] k > 0
Digital Signal Processing EIDSV4
Unit 2 First Assessment 10 April 2008
Page 1
Question 1 The block diagram of a system with
x(k)
y(k)
input x(k) and output y(k), is shown in Figure 1.
a) Determine a difference equation that describes
z -1
the relationship between y(k) and x(k).
(2)
0.5
b) Determine an expression for the system’s
Figure 1
transfer function H(z) = Y(z)/X(z).
(2)
c) Assuming that the sampling period is one second (T = 1), calculate the
frequency response, H(ejT), of the system at the following frequencies:
i)  = –  (2) ii)  = – /2 (2) iii)  = 0 (2) iv)  = /2 (2) v)  =  (2)
d) Using the results in Question 1 c) above, sketch a rough graph of |H(ejT)|
(2)
against , for variations of  between –  and  radians/second.
Question 2 The transfer function H(), of an all pass digital filter, is defined in the


fundamental interval,  s    s , by: H() = e jω , where s is the sampling
2
2
frequency (s = 2/T). Assuming a sampling period of one second (T = 1), determine
an expression for the impulse response, h(k), required for this digital filter.
(10)
Question 3 A signal x(t) was sampled with a sampling
x(k)
1
k
period of one second (T = 1) to obtain the sequence of
0
three (N = 3) sampled values, x(0) = – 2, x(1) = – 1 and
–1
Figure 2
–2
x(2) = 1, shown in Figure 2.
+
a) For the discrete Fourier
b) Using the inverse Fourier transform x(k) of
transform, X(n), of this
X(n), verify your answer in Question 3 a),
sequence, calculate the values
by calculating:
i) x(0)
(3)
of X(0), X(1) and X(2).
(5)
ii) x(1)
(4)
Question 4 The transfer function of a low pass filter with cutoff frequency
0 = 1 rad/sec, is given by F(s) = 1/(s+1). The bilinear transformation is used
to construct a corresponding digital infinite impulse response (IIR) low pass filter
with cutoff frequency f0 = 100 Hz. and a sample rate of 1000 samples/sec.
a) Determine the transfer function of the IIR filter.
(8)
b) Draw a block diagram structure for this filter.
(4)
Appendix: z transforms
---oooOooo--Total: 50
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
2
z X(z) - z2x(0) - zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N-1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
1
x(k) = ω
s
ω /2
 ωss /2 X()e
jkT
d , s = 2π
T
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
h(k) = c
, T= 2π
π (k -α)ωcT
ωs
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
tan (
)
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 2 Eerste Evaluasie 10 April 2008 Memorandum
1. a) y(k) = x(k) + 0.5y(k – 1)  y(k) – 0.5y(k–1) = x(k)
b) Y(z) – 0.5z-1Y(z) = X(z)  Y(z)[1 – 0.5z-1 ] = X(z)
Y(z)/X(z) = 1/(1 – 0.5z-1)  H(z) = z/(z – 0.5)
c) i) H(e-j) = (1-)/[(1-)–0.5] = 0.66670
ii) H(e-j/2) = (1-/2)/[(1-/2)–0.5] = 0.89440.4636
iii) H(ej0) = (10)/[(10)–0.5] = 20
iv) H(ej/2) = (1/2)/[(1/2)–0.5] = 0.8944-0.4636
v) H(ej) = (1)/[(1) – 0.5] = 0.66670
d)
|H()|
2
(2)
x(k)
(2)
(2)
(2)
(2)
(2)
(2)
Bladsy 1
y(k)

0.5
D
1

2. h(k) = 1
s
0
-
2
-
[16]

s /2
jkT
H()e
d
s /2

2

(2)


jk
j jk
= 1
H()e d = 1
e e
d
 
 



1
1

j(k  1)
e j(k 1)  [3] =
= 1
e
d =
 
j(k 1)
j(k 1) 
 
j(k  1)
- j(k  1)
1
e
1
e
[2]
=
sin(k  1) =
=

1)
(k
(k 1)
j2

[10]
N-1
3. a) X(n) =  x(k)e
- j(2/N)nk
k =0
[12]
2
=  x(k)e
- j(2/3)nk
k =0
e j(k 1)  e - j(k 1) 


sin(k  1)
(k 1)
x(k)
0
s = 2/T = 2
[3]
1
–1
[2]
k
–2
X(0) = – 2 – 1 + 1 = – 2 [1]
-j(2/3)×1×0
-j(2/3)×1×1
-j(2/3)1×2
X(1) = –2e
–1e
+e
= –20–1–2/3+1–4/3 = 2.6462.428 [2]
X(2) = –2e-j(2/3)×2×0–1e-j(2/3)×2×1+e-j(2/3)2×2 = –20–1–4/3+1–8/3 = 2.646-2.428 [2]
1 N-1
j(2/N)nk 1 2
j(2/3)nk
=  X(n)e
b) x(k) =
 X(n)e
N n =0
3 n =0
i) x(0) = 1/3(– 2 + 2.6462.428 + 2.646-2.428) = – 2 [3]
ii) x(1) = 1/3[– 2ej(2/3)×0×1 + 2.6462.428×ej(2/3)×1×1 + 2.646-2.428×ej(2/3)×2×1]
= 1/3[–20+2.6462.428×12/3+2.646-2.428×14/3
= (1/3)[–20 +2.6464.522 + 2.6461.761] = (1/3)[33.142] = – 1 [4]
4. a) s

z1

z -1
z -1
0
= 
= 3.0777
z1
  T  z + 1 tan (2  100)  (1/1000)  z + 1


tan 0 


2
 2 
(5)
(7)
[4]
H(z) = 1/(s+1)|s=3.0777(z-1)/(z+1) = 1/{3.0777[(z - 1)/(z + 1)] + 1} = (z + 1)/[3.0777(z - 1) + (z + 1)]
= (z + 1)/(3.0777z –3.0777 + z + 1) = (z + 1)/(4.0777z – 2.0777) [3]
(8)
= (0.2452 + 0.2452z-1)/(1 – 0.5095z-1) [1]
{y(k)/x(k) = (0.2452+0.2452D)/(1-0.5095D)  y(k)(1-0.5095D) = x(k)(0.2452+0.2452D)
y(k) - 0.5095y(k-1) = 0.2452x(k)+0.2452x(k-1)}
b)
x(k)
0.2452
+
y(k)=0.5095y(k-1)+0.2452x(k)+0.2452x(k-1)
+
z-1
[12]
z-1
0.2452
0.5095
(4)
Digital Signal Processing EIDSV4
Unit 1 Final Assessment 22 May 2008
Page 1
Question 1 State Shannon’s sampling theorem.
(2)
Question 2 A sequence y(k) is defined by the difference equation:
y(k+2) + y(k) = sin k
with y(0) = –1 and y(1) = 1. Write down the values of y(2) and y(3).
Question 3 A signal x(k) is shown in Figure 1.
Sketch the following signals:
a) x(k + 1)u(k)
(2)
b) x(2 – k)
(2)
c) x(1 – k)(k + 2)
(2)
(4)
x(k)
Figure 1
2
1
1
0
k
–1
Question 4 The difference equation that describes the relationship between the
output y(k) and the input x(k), of a system, is given by:
y(k) = x(k) + 0.5x(k – 1)
a) Draw a block diagram structure representing this system.
b) Calculate the two terms, h(0) and h(1), of the system’s impulse response.
c) Express the impulse response of the system, h(k), as the sum of weighted and
shifted impulse functions.
d) Confirm your answer for h(k) in Question 4 c), by calculating H(z) = Y(z)/X(z)
from the given difference equation.
e) Use graphical convolution methods to determine the response y(k) of the system
for k = 0, 1, 2 and 3, if the input to the system is the signal, x(k) = sink u(k).
f) Use z transform methods to verify your results in Question 4 e), by finding y(k)
for k = 0, 1, 2 and 3, from Y(z) = H(z)X(z).
Question 5 Use z transform methods to determine x(k) in closed form, from
the sequence: x(k+2) = x(k+1) + 2x(k), with x(0) = 0 and x(1) = 1.
---oooOooo---
(2)
(2)
(2)
(4)
(8)
(8)
(12)
Total: 50
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
z2X(z) - z2x(0) - zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N -1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2abkcos(k + )u(k)
(a z (a z

z b z b
1
x(k) = ω
s

ωs /2
ωs /2
X()e
jkT
d , s = 2π
T
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
h(k) = c
, T= 2π
π (k -α)ωcT
ωs
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
)
tan (
2
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 1 Finale Evaluasie 22 Mei 2008 Memorandum
Bladsy 1
1. ‘n Sein x(t), wat gemonster word teen ‘n frekwensie fm, kan teoreties volkome herwin word uit
[2] die gemonsterde weergawe van x(t), indien x(t) aanvanklik bandbeperk was tot fm/2.
(2)
2. y(k+2) + y(k) = sin k y(0) = -1, y(1) = 1
set k = 0: y(2) + y(0) = sin 0  y(2) + (-1) = 0  y(2) = 1 [2]
[2]
[4] set k = 1: y(3) + y(1) = sin 1  y(3) + 1 = 0.8415  y(3) = -0.1585
3. a)
b)
x(k+1)u(k)
1
x(2-k)
k
2
1
c)
1
x(1-k)(k+2)
k
k
–1
[6]
4. a)
(2)
x(k)
+
–1
(2)
y(k)
z-1
0.5
c) h(k) = (k) + 0.5(k-1)
e) x(k) = sin k u(k)
x(0) = 0
x(1) = 0.8415
x(2) = 0.9093
x(3) = 0.1411
[26]
–1
(2)
b) y(k) = x(k) + 0.5x (k-1)
With x(k) = (k):
h(0) = 1 [1]
h(1) = 0.5 [1]
(2)
d) Y(z) = X(z) + 0.5z-1X(z)
H(z) = Y(z)/X(z) = 1 + 0.5z-1 [2]
h(k) = (k) + 0.5(k-1) [2]
{or h(0) = 1 [1] and h(1) = 0.5 [1]}
(4)
(2)
(2)
n
-1
x(n)
h(-n) 0.5
h(1-n)
h(2-n)
h(3-n)
0
1
2
3
0 0.8415 0.9093 0.1411
1
y(0)=0 [2]
0.5
1
y(1)=0.8415[2]
0.5
1
y(2)=1.33 [2]
0.5
1
y(3)=0.5958[2]
(8)
f) H(z) = 1 + 0.5z-1 = (z + 0.5)/z
X(z) = zsin1/(z-ej)(z-e-j) = 0.8415z/(z-11)(z-1-1) [1]
Y(z) = H(z)X(z) = [(z+0.5)/z]×[0.8415z/(z-11)(z-1-1)] = 0.8415(z+0.5)/[(z-11)(z-1-1)]
Y(z)/z = 0.8415(z+0.5)/[z(z-11)(z-1-1)] = A/z + B/(z-11) + B*/(z-1-1)
A = 0.4208 & B = 0.8415×(11+0.5)/(11)(11-1-1) = 0.669-1.891
Y(z)/z = 0.4208/z + 0.669-1.891/(z-11) + 0.6691.891/(z-1-1)
Y(z) = 0.4208 + (0.669-1.891)z/(z-11) + (0.6691.891)z/(z-1-1)
y(k) = 0.4208(k) + 2×0.669×1k×cos(k-1.891) = 0.4208(k) + 1.338cos(k-1.891) [5]
y(0) = 0.4208 + 1.338cos(-1.891) = 0.0003  0, y(1) = 1.338cos(1-1.891) = 0.8411
[2]
y(2) = 1.338cos(2-1.891) = 1.33 en y(3) = 1.338cos(3-1.891) = 0.5962
(8)
5. x(k+2) = x(k+1) + 2x(k)
z2X(z) – z2x(0) – zx(1) = zX(z) – zx(0) + 2X(z) [2]
z2X(z) – z = zX(z) + 2X(z)
(z2 – z – 2)X(z) = z
X(z) = z/(z2 – z – 2) [2]
X(z)/z = 1/(z - 2)(z + 1) [2]
= 0.3333/(z-2) – 0.3333/(z+1) [2]
X(z) = 0.3333z/(z-2) – 0.3333z/(z+1)
k
k [4]
[12] x(k) = 0.33332 – 0.3333(-1)
Digital Signal Processing EIDSV4
Unit 2 Final Assessment 22 May 2008
Page 1
Question 1 The block diagram of a system with
x(k)
y(k)
input x(k) and output y(k), is shown in Figure 1.
a) Determine a difference equation that describes
z -1
the relationship between y(k) and x(k).
(2)
0.5
b) Determine an expression for the system’s
Figure 1
transfer function H(z) = Y(z)/X(z).
(2)
c) Use z transform methods to determine the response y(k), if a signal
x(k) = sink u(k), is applied at the input of the system.
(7)
j
d) Calculate H(e ) and confirm the steady state part, yss(k), of the expression for
y(k), found in Question 1 c).
(5)
x(k)
Question 2 A signal x(k), with x(0) = 1, x(1) = 1, and
Figure 2
x(k) = 0 for all other k, is shown in Figure 1. The samples
1
1
k
were obtained with a sampling period of T = 1 second.
0
a) Determine the frequency spectrum, X(), of x(k).
(2)
-1
1
2
b) Calculate the magnitude |H(1.5)|, of X() at  = 1.5 rad/sec.
(4)
c) Calculate the angle X(2) of X() at  = 2 rad/sec.
(4)
Question 3 The transfer function, H(), of a digital
H()
1
digital low pass filter with sampling frequency s
Figure 3
and a cut off frequency of s/4, is shown in Figure 3.

H() = 0 for s/4  || < s/2 and H() = 1 for
0
ω
ω
ωs
ωs
|| < s/4. Assuming a sampling period of T = 1 sec,
– s – s
2
4
4
2
determine an expression for the impulse response,
h(k), required for this digital filter.
(8)
Question 4 A signal is sampled every 1 millisecond to produce a batch of 250 samples.
The discrete Fourier frequency transform (DFT) of the samples is then calculated.
a) Determine the increment in Hz, between successive frequency components.
(2)
b) Determine the repetition period (frequency length), of the spectrum, in Hz.
(2)
c) Determine the highest frequency in Hz, permitted in the spectrum of the sampled
signal, to avoid interspectra interference (aliasing).
(2)
Question 5 The ideal impulse response of a certain digital low pass filter, is given
1 sin(k/2)
by h(k) =
. Design a corresponding FIR filter of length five (N=5), and
2 k/2
draw a structure for this filter.
(10)
---oooOooo--Total:
50
Appendix: z transforms
+
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
2
z X(z) - z2x(0) - zx(1)
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
, T= 2π
h(k) = c
ωs
π (k -α)ωcT
Fourier transforms:
Bilinear Transformation:

jkT
- jkT
1 ωs /2
2π
Ωo
x(k) = ω  ω /2 X()e
d , s =
X() =  x(k)e
z-1
T
s= 
s
s
k = -
ωoT z +1
tan (
)
N -1
2
- j(2 /N)nk
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
X(n) =  x(k)e
n =0
k =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 2 Finale Evaluasie 22 Mei 2008 Memorandum
Bladsy 1
1. a) y(k) = x(k) + 0.5y(k – 1)  y(k) – 0.5y(k–1) = x(k)
(2)
-1
-1
x(k)
y(k)
b) Y(z) – 0.5z Y(z) = X(z)  Y(z)[1 – 0.5z ] = X(z)

-1
Y(z)/X(z) = 1/(1 – 0.5z )  H(z) = z/(z – 0.5)
(2)
c) x(k) = sinku(k)
0.5
D
X(z) = 0.815z/(z - 11)(z - 1-1) [2]
Y(z)=H(z)X(z)
= [z/(z – 0.5)]×[0.815z/(z – 11)(z – 1-1)] = 0.8415z2/(z – 0.5)(z – 11)(z – 1-1)
Y(z)/z = 0.8415z/(z – 0.5)(z – 11)(z – 1-1)
= 0.5929/(z – 0.5) + 0.5935–2.094/(z – 11) + 0.59352.094/(z – 1–1) [3]
y(k) = 0.5929(0.5)k + 2×0.5935×1k×cos(k – 2.094) [2]
= 0.5929(0.5)k + 1.187sin(k – 0.5232)
(7)
k
{As k   then 0.5929(0.5)  0 and yss(k) = 1.187sin(k – 0.5232)}
[16] d) H(ej1) = 11/(11 – 0.5) [2] = 1.187-0.5229  yss(k) = 1.187sin(k – 0.5229) [2]
(5)

2. a) X() =  x(k)e
- jkT
k =- 
[10]
1
=  x(k)e
- j k
k =0
= 1 + e-j
(2)
b) X(1.5) = 1 + 1-1.5 = 1.463-0.75 [2]  |X(1.5)| = 1.463 [2]
c) X(2) = 1 + 1-2 = 1.081-1 [2]  arg[X(2)] = -1 rad [2]
3. s = 2/T = 2

s /2
jk
jkT
H()e d
h(k) = 1
H()e
d = 1


s s /2



 e jk 

jk
1
[2]
[2]

=
1e d = 
 
 jk 

 
jk/2
- jk/2
jk/2
- jk/2
1 e
e
e
e
=
–
=
jk
jk
j2
k

[8]
1
=
H()
ω
– s
2
–
[2]
(4)
(4)
ωs
2
–
2
0

2
1 sin k/2
1
sin(k/2) =
2 k/2
k


[2]
4. N=250 en T = 1 ms
a) f0 = 1/NT = 1/250 ms = 4 Hz.
b) Frequency repetition length (period) = fs = 1/T = 1/0.001 = 1000 Hz.
[6]
c) fmaks = 0.51000 = 500 Hz
(2)
(2)
(2)
1 sin(k/2)
1 sin[(k -2)/2)]
therefore FIR filter of length 5: h(k) =
k = 0, 1, 2, 3, 4 [2]
k

/2
[(k
2)

/2]
2
2
[1]
[1]
[1]
[1]
h(1)
=
0.3183
h(2)
=
0.5
h(3)
=
0.3183
h(4)
= 0 [1]
h0) = 0
5. h(k) =
x(k)
z-1
z-1
0.3183
[10]
[3]
z-1
0.5
0.3183
+
+
y(k)=0x(k)+0.3183x(k-1)+0.5x(k-2)+0.3183x(k-3)+0x(k-4)
Digital Signal Processing EIDSV4
Unit 1 First Assessment 5 March 2009
Page 1
Question 1 A signal x(t), is given by x(t) = 10cos200t.
a) Determine the theoretical minimum frequency at which x(t) should be sampled,
so that x(t) could again be recovered from the samples.
(2)
b) Draw a representation of the frequency spectrum of the sampled version of x(t),
if x(t) is sampled at 500 Hz by an ideal sampler.
(2)
Question 2
x(k)
A signal x(k) = [2 1 –1 1 2] is
2
2
shown in Figure 1.
1
1
a) Sketch the following signals:
k
i) x(k + 1)u(k – 1).
(2)
(2)
ii) – x(k – 1).
–1
Figure 1
iii) x(–k – 2).
(2)
b) Express x(k – 1) as the sum of weighted and shifted impulse functions,
(2)
Question 3 The difference equation of a digital filter, with input x(k) and output
y(k), is given by: 0.5y(k – 2) – y(k – 1) + y(k) = x(k).
a) Draw a blockdiagram for the system.
(3)
b) Determine the first four terms h(0) to h(3), of the impulse response of the system.
(3)
Question 4
The impulse response of a system, h(k) = [3 2 1], as well as the input to the system,
x(k) = [2 1 0 –1], is shown in Figure 2.
a) Use graphical convolution
h(k)
x(k)
Figure 2
methods, to determine the
3
response, y(k), of the system. (6)
2
2
b) Use z transform methods to
1
1
k
k
calculate Y(z) and from Y(z),
the response y(k).
(8)
–1
Question 5 Use z transform methods to determine y(k) in closed form, from the
following difference equations:
a) y(k+2) = y(k+1) + y(k), with y(0) = 0 and y(1) = 1.
(8)
b) y(k) – 0.6y(k – 1) + 0.25y(k – 2) = u(k).
(10)
---oooOooo---
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z – 1)
z/(z – a)
z/(z – 1)2
z-1X(z)
z-2X(z)
zX(z) – zx(0)
z2X(z) – z2x(0) – zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N-1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z – a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
1
x(k) = ω
s

ωs /2
ωs /2
X()e
jkT
d , s = 2π
T
Total: 50
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
, T= 2π
h(k) = c
π (k -α)ωcT
ωs
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
tan (
)
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 1 Eerste Evaluasie 5 March 2009 Memorandum
1. a) f = 200/2 = 100 Hz  fs = 200 Hz.
b)
[4]
-600
2. a)
i) x(k+1)u(k-1)
-500
-400
(2)
ii) -x(k-1)
2
[8]
0
100
-1
400
500
iii)
(2)
1
600
f
k
-1
(2)
(2)
x(-k-2)
2
2
1
k
-1
b) x(k)=2(k+2)+(k+1)–(k)+(k–1)+2(k–2)  x(k–1)=2(k+1)+(k)–(k–1)+(k–2)+2(k–3) (2)
-2
-2
3. a) 0.5y(k–2)–y(k–1)+y(k)=x(k)
y(k) = y(k-1)-0.5y(k-2)+x(k)
y(k
x(k)
D
D
b) y(k) = y(k - 1) - 0.5y(k - 2) + x(k)
h(k) = h(k - 1) - 0.5h(k - 2) + (k) [1]
h(0)=h(-1)-0.5h(-2)+(0)=0-0+1 = 1 [½]
h(1)=h(0)-0.5h(-1)+(1)=1-0+0 = 1 [½]
h(2)=h(1)-0.5h(0)+(2)=1-0.51+0=0.5 [½]
h(3)=h(2)-0.5h(1)+(3)=0.5-0.51+0=0 [½]

– 0.5
[6]
[14]
(2)
X*(f)
-100
k
Bladsy 1
4. a)
n
x(n)
h(-n)
h(1-n)
h(2-n)
h(3-n)
h(4-n)
h(5-n)
-2
0
1
0
0
0
0
0
-1
0
2
1
0
0
0
0
0
2
3
2
1
0
0
0
1
1
0
3
2
1
0
0
2
0
0
0
3
2
1
0
3
-1
0
0
0
3
2
1
4
0
0
0
0
0
3
2
(3)
5
0
0
0
0
0
y(0)=6 [1]
y(1)=7 [1]
y(2)=4 [1]
y(3)=-2 [1]
y(4)=-2 [1]
3 y(5)=-1 [1]
(3)
b) X(z) = 2 + z-1 + 0z-2 – z-3 [1]
H(z) = 3 + 2z-1 + z-2 [1]
Y(z) = H(z)X(z)
= (3+2z-1+z-2)(2+z-1–z-3)
= 6 + 3z-1 – 3z-3
+ 4z-1 + 2z-2 – 2z-4
+ 2z-2 + z-3 – z-5
Y(z)=6+7z-1+4z-2–2z-3–2z-4–z-5 [4]
y(0)=6 y(1)=7 y(2)=4 y(3)=-2
(6)
(8)
y(4)=-2 y(5)=-1 [2]
5. a) z2Y(z) – [z2y(0) + zy(1)] = zY(z) – zy(0) + Y(z) [2]
(z2 – z – 1)Y(z) = z20 + z1 – z0  (z2 – z – 1)Y(z) = z
Y(z)/z = 1/(z2 – z – 1) = A/[z – (1 + 5)/2] + B/[z – (1 - 5)/2]
A = 1/[z – (1 – 5)/2]|z=(1+5)/2 = 1/[(1 + 5)2 – (1 – 5)/2] = 1/5
B = 1/[z – (1 + 5)/2]|z=(1–5)/2 = 1/[(1 – 5)2 – (1 + 5)/2] = –1/5
Y(z) = (1/5)z/[z – (1 + 5)/2] – (1/5)z/[z – (1 – 5)/2] [2]
y(k) = (1/5)[(1 + 5)/2]k – (1/5)[(1 – 5)/2]k
y(k) = (1/5){[(1 + 5)/2]k – [(1 – 5)/2]k } [4] (=0.4472[1.618k – (-0.618)k])
(8)
b) y(k) – 0.6y(k – 1) + 0.25y(k – 2) = u(k)
Y(z) – 0.6z-1Y(z) + 0.25z-2Y(z) = z/(z-1) [2]
z2Y(z) – 0.6zY(z) – 0.25Y(z) = z3/(z-1)
(z2 – 0.6z + 0.25)Y(z) = z3/(z-1)
Y(z)/z = z2/(z – 1)(z2 – 0.6z + 0.25)
Y(z)/z = z2/(z – 1)(z – 0.50.9273)(z – 0.5–0.9273) [2]
Y(z)/z = 1.538/(z-1)+0.3876–2.339/(z–0.50.9273)+0.38762.339/(z–0.5–0.9273) [3]
Y(z) = 1.538z/(z-1)+(0.3876–2.339)z/(z–0.50.9273)+(0.38762.339)z/(z–0.5–0.9273)
y(k) = 1.538 + 20.3876(0.5)kcos(k0.9273 – 2.339)
[18]
y(k) = 1.538 + 0.7752(0.5)kcos(k0.9273 – 2.339) [3]
(10)
Digital Signal Processing EIDSV4
Unit 2 First Assessment 16 April 2009
Page 1
2z 2  2.5z
.
Question 1 The transfer function of a system is given by H(z) =
z 2  2.5z  1
Draw the poles of H(z) on the z plane, and judge whether the system is stable or not.
(5)
z
Question 2 The transfer function of a system is given by H(z) =
. The input
z  0.4
to the system, is given by x(k) = sin(0.8k)u(k).
a) Determine the response y(k) of the system, using the relation Y(z) = H(z)X(z).
(10)
b) Confirm the steady state part of your answer to Question 2 a), by calculating
H() at T = 0.8
(4)
H()
Question 3 The transfer function H(),
Figure 1
1
of an ideal band pass digital filter, is
shown in Figure 1. Assuming a sampling
frequency s = 1 (T = 2), determine an
expression for the impulse response h(k),
required for this digital filter.
(8)

-0.5 -0.3 -0.2
(-s/2)
0
0.2
0.3
0.5
(s/2)
Question 4 The four (N = 4) sampled values x(0) = 2,
x(k)
Figure 2
x(1) = –1, x(2) = –1 and x(3) = 2, shown in Figure 2,
2
2
were obtained when a signal x(t) was sampled.
k
a) For the discrete Fourier transform, X(n), of this,
0
sequence, calculate the values of X(0), X(1), X(2)
–1 –1
and X(3).
(10)
b) Calculate the average value of x(t).
(1)
c) Calculate the amplitude of the fundamental frequency (X(1)), contained in x(t).
(1)
d) If the sampling period was 0.5 sec (T = 0.5), calculate the frequency of the
fundamental component (X(1)), contained in x(t).
(1)
Question 5 Design and draw the structure of a digital finite impulse response (FIR)
ω
low pass filter of order five (N = 5), with cutoff frequency, c = s , where s is the
8
sampling frequency.
(10)
---oooOooo--Total: 50
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
2
z X(z) - z2x(0) - zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N-1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
1
x(k) = ω
s
ω /2
 ωss /2 X()e
jkT
d , s = 2π
T
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
h(k) = c
, T= 2π
π (k -α)ωcT
ωs
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
tan (
)
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 2 Eerste Evaluasie 16 April 2009 Memorandum
2
1. z – 2.5z + 1 = (z – 0.5)(z – 2)
[5]
Bladsy 1
[2]
[1]
The system is unstable [2]
2. a) Y(z) = H(z)X(z); H(z) = z/(z = 0.4) & x(k) = sin(0.8k)u(k)
= [z/(z – 0.4)]×[zsin0.8/(z – 10.8)(z – 1–0.8)] = 0.7174z2/[(z – 0.4)(z – 10.8)(z – 1–0.8)]
Y(z)/z = 0.7174z/[(z – 0.4)(z – 10.8)(z – 1–0.8)]
= 0.4762/(z – 0.4) [2] + 0.6441–1.949/(z – 10.8) [2] + 0.64411.949/(z – 1–0.8) [2]
Y(z) = 0.4762z/(z – 0.4) + (0.6441–1.949)z/(z – 10.8) + (0.64411.949)z/(z – 1–0.8)
y(k) = 0.4762(0.4)k [2] + 2×0.6441×(1)kcos(0.8k – 1.949) [2]
= 0.4762(0.4)k + 1.288cos(0.8k – 1.949)
(10)
b) H(ej0.8) = 10.8/(10.8 – 0.4) = 1.288-0.3786 [2]
x(k) = sin(0.8k)  yss(k) = 1.288sin(0.8k – 0.3786) [1] = 1.288cos(0.8k – 0.3786 – /2)
[14]
yss(k) = 1.288cos(0.8k – 1.949) [1]
(4)

0 j2k
03 j2k
s /2
jk2
jkT
H()e
d =
e
d +
e
d
3. h(k) = 1
H()e
d = 1
 0.5
0.2
0.3
s s /2
= 1/j2k[ej2k]-0.3-0.2 + 1/j2k[ej2k]0.20.3 [2]
= 1/j2k[ej2(-0.2)k – ej2(-0.3)k] + 1/j2k[ej2(0.3)k – ej2(0.2)k] [2]
= 1/j2k[ej0.6k – e-j0.6k] - 1/j2k[ej0.4k – e-j0.4k]
= 1/k[ej0.6k – e-j0.6k]/j2 - 1/k[ej0.4k – e-j0.4k]/j2
= sin(0.6k)/k – sin(0.4k)/k [2]
[8]


N-1
4. a) X(n) =  x(k)e
k =0
[13]
- j(2/N)nk
[1]
3
=  x(k)e


[2]
- j(/2)nk
k =0
X(0) = 2 – 1 – 1 + 2 = 2
X(1) = 2e-j(/2)×1×0 – 1e-j(/2)×1×1 – 1e-j(/2)1×2 + 2e-j(/2)1×3
= 20 – 1– /2 – 1–  + 2– 3/2 = 4.2430.7854 [3]
X(2) = 2e-j(/2)×2×0 – 1e-j(/2)×2×1 – 1e-j(/2)2×2 + 2e-j(/2)2×3
= 20 – 1–  – 1– 2 + 2– 3 = 0 [3]
X(3) = 2e-j(/2)×3×0 – 1e-j(/2)×3×1 – 1e-j(/2)3×2 + 2e-j(/2)3×3
= 20 – 1– 3/2 – 1– 3 + 2– 9/2 = 4.243–0.7854 [3]
b) avg[x(t)] = X(0)/N = 2/4 = 0.5
c) Amplitude of first harmonic = |X(1)|/(N/2) = 4.243/2 = 2.122
d) T1 = NT = 4×0.5 = 2  f1 = ½ Hz [1]  1 = 2×½ =  r/s
5) h(k) = (cT/)sin[(k - )cT]/[(k - )cT]
Met T = 2/s, c = s/8 en  = (5-1)/2 = 2:
h(k) = [(s/8)(2/s)/]sin[(k - 2)(s/8)(2/s)]
/[(k - 2)(s/8)(2/s)]
h(k) = (1/4)sin[(k - 2)/4]/[(k - 2)/4] [2]
h(0) = (1/4)sin(-/2)/(-/2) = 0.1592 [1]
h(1) = (1/4)sin(-/4)/(-/4) = 0.2251 [1]
h(2) = (1/4)sin0/0 = 0.25 [2]
h(3) = h(1) = 0.2251
h(4) = h(0) = 0.1592
[10]
x(k)
x(k)
2
2
k
0
-1 -1
(10)
(1)
(1)
(1)
0.1592
+
0.2251
+
0.25
+
0.2251
+
-1
z
y(k)
-1
z
-1
z
-1
z
0.1592
[4]
Digital Signal Processing EIDSV4
Unit 1 Final Assessment 21 May 2009
Page 1
Question 1 Sketch the following signal x(k), for –2  k  3:
x(k) = k[u(1-k) – u(1+k)] + 2ku(k+1).
(6)
Question 2 A sequence y(k) is defined by the difference equation:
y(k+2) + y(k+1) + y(k) = sin k
with y(0) = –1 and y(1) = 1. Write down the values of y(2), y(3) and y(4).
Question 3 Refer to the system
x(k)
in Figure 1. Determine the
impulse response, h(k), of the
system and calculate the first
four terms, h(0) to h(3), of the
system’s impulse response, h(k)
(6)
+
(6)
y(k)
+
z–1

1
2

1
4
Figure 1
Question 4 The difference equation that describes the relationship between the
output y(k) and the input x(k), of a system, is given by:
y(k) = x(k) + 0.5y(k – 1)
z
and
a) Show that the transfer function of the system is given by: H(z) =
z  0.5
therefore that the system’s impulse response, is given by: h(k) = (0.5)ku(k) .
b) Use graphical convolution methods to determine the response y(k) of the system
for k = 0, 1, 2 and 3, if the input to the system is the signal, x(k) = sink u(k).
c) Use z transform methods to verify your results in Question 4 b), by finding y(k)
for k = 0, 1, 2 and 3, from Y(z) = H(z)X(z).
(10)
Question 5 Use z transform methods to determine x(k) in closed form, from
the sequence: x(k+1) = x(k) + 2, with x(0) = 0.
(10)
---oooOooo---
(4)
(8)
Total: 50
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
z2X(z) - z2x(0) - zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N-1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
1
x(k) = ω
s

ωs /2
ωs /2
X()e
jkT
d , s = 2π
T
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
, T= 2π
h(k) = c
ωs
π (k -α)ωcT
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
tan (
)
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 1 Finale Evaluasie 21 Mei 2009 Memorandum
1.
[6]
k
-2
-1
0
1
2
3
x(k)=k[u(1-k)–u(1+k)]+2ku(k+1)
-2(1-0) + (-4)0 = -2 [1]
-1(1-1) + (-2)1 = -2 [1]
0(1-1) + (0)21 = 0 [1]
1(1-1) + (2)1 = 2 [1]
2(0-1) + (4)1 = 2 [1]
3(0-1) + (6)1 = 3 [1]
Bladsy 1
x(k)
2
2
3
k
0
-2 -2
2. y(k+2) + y(k+1) + y(k) = sin k y(0) = -1, y(1) = 1
set k = 0: y(2) + y(1) + y(0) = sin 0  y(2) + (1) + (-1) = 0  y(2) = 0 [2]
set k = 1: y(3) + y(2) + y(1) = sin 1  y(3) + (0) + 1 = 0.8415  y(3) = -0.1585 [2]
[6]
set k = 2: y(4) + y(3) + y(2) = sin 2  y(4) + (-0.1585) + 0 = 0.9093  y(4) = 1.068 [2]
3. a(k) = x(k) – 0.5z-1a(k)  a(k) = x(k)/(1+0.5z-1)
a(k)
x(k)
y(k) = a(k)-0.25z-1a(k) = (1-0.25z-1)a(k)
= [(1-0.25z-1)/(1+0.5z-1)]x(k)
z–1
-1
-1
y(k) + 0.5z y(k) = x(k) – 0.25z x(k)
1
1
y(k) = x(k) – 0.25x(k-1) – 0.5y(k-1) [2]
4
2
h(0) = 1 [1] h(1) = -0.75 [1] h(2) = 0.375 [1] h(3) = -0.1875 [1]
[6]
4. a) y(k) = x(k) + 0.5y(k-1)  Y(z) = X(z) + 0.5z-1Y(z) [1]  Y(z) – 0.5z-1Y(z) = X(z)
H(z) = Y(z)/X(z) = 1/(1 – 0.5z-1) = z/(z – 0.5) [2]
h(k) = Z-1{z/(z – 0.5)} = (0.5)k u(k) [1]
b) x(k) = sin k u(k)  x(0) = 0, x(1) = 0.8415, x(2) = 0.9093, x(3) = 0.1411
h(k) = (0.5)ku(k)  h(0) = 1, h(1) = 0.5, h(2) = 0.25, h(3) = 0.125
+
n
-3
-2
-1
0
1
2
3
x(n)
0
0.8415 0.9093 0.1411
h(-n) 0.125 0.25
0.5
1
y(0)=0 [2]
h(1-n)
0.125 0.25
0.5
1
y(1)=0.8415[2]
h(2-n)
0.125 0.25
0.5
1
y(2)=1.33 [2]
h(3-n)
0.125 0.25
0.5
1
y(3)=0.8061 [2]
+
y(k)
(4)
(8)
c) H(z) = z/(z–0.5)
X(z) = zsin1/(z-ej)(z-e-j) = 0.8415z/(z-11)(z-1-1) [1]
Y(z) = H(z)X(z) = [z/(z-0.5)]×[0.8415z/(z-11)(z-1-1)] = 0.8415z2/(z-0.5)/[(z-11)(z-1-1)]
Y(z)/z = 0.8415z/[(z-0.5)(z-11)(z-1-1)] = A/(z-0.5) + B/(z-11) + B*/(z-1-1)
A = 0.8415×0.5/[0.5-11)(0.5-1-1)] = 0.59286
B = 0.8415×11/[(11-0.5)(11-1-1) = 0.59354-2.0937
Y(z)/z = 0.59286/(z-0.5) + 0.59354-2.0937/(z-11) + 0.593542.0937/(z-1-1)
Y(z) = 0.59286z/(z-0.5) + (0.59354-2.0937)z/(z-11) + (0.593542.0937)z/(z-1-1)
y(k) = 0.59286(0.5)k + 2×0.59354×1k×cos(k-2.0937)
= 0.59286(0.5)k + 1.1871cos(k-2.0937) [5]
y(0) = 0.59286 + 1.1871cos(-2.0937) = 0.000025  0 [1]
y(1) = 0.59286×0.5 + 1.1871cos(1-2.0937) = 0.8415 [1]
y(2) = 0.59286×0.25 + 1.1871cos(2-2.0937) = 1.33 [1]
[22]
(10)
y(3) = 0.59286×0.125 + 1.1871cos(3-2.0937) = 0.8061 [1]
5. x(k+1) = x(k) + 2 x(0) = 0
zX(z) – zx(0) = X(z) + 2z/(z-1) [3]
zX(z) – 0 = X(z) + 2z/(z-1)
(z – 1)X(z) = 2z/(z-1)
X(z) = 2z/(z-1)(z-1) [3]
X(z) = 2z/(z-1)2
x(k) = 2ku(k) [4]
[10]
Digital Signal Processing EIDSV4
Unit 2 Final Assessment 21 May 2009
Page 1
Question 1
The impulse response of a system is given by h(k) = (0.8) k u(k), and the input
signal, x(k), to the system, by x(k) = (cos k)u(k).
a) Determine the output y(k) of the system, using z transform methods and the
relation Y(z) = H(z)X(z).
b) Verify the steady state component of your answer in Question 1a), by
calculating the frequency response of the system, H(ejT), at T = 1.
Question 2
Determine an expression
for the impulse response,
h(k), of an ideal high
pass filter, with frequency
characteristic, shown in
Figure 1. The sampling
(8)
frequency s = 1.
H()
(12)
(4)
Figure 1
1
 (r/s)
-0.5
-0.25
(-s/2)
0
0.25
0.5
(s/2)
Question 3
Calculate the frequency spectrum X() of the signal x(k) = (k). Assume T = 1.
(3)
Question 4
Calculate the discrete Fourier transform X(0), X(1), X(2) and X(3) of the four
(N = 4) samples x(0) = 2, x(1) = 3, x(2) = 2 and x(3) = 1. Verify your answer by
finding the inverse discrete Fourier transform of X(0), X(1), X(2) and X(3).
(12)
Question 5
Design and draw the structure of a digital finite impulse response (FIR) low pass
ω
filter of order seven (N = 7), with cutoff frequency, c = s , where s is the
5
sampling frequency.
(11)
---oooOooo---
Total: 50
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
2
z X(z) - z2x(0) - zx(1)
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
, T= 2π
h(k) = c
π (k -α)ωcT
ωs
Fourier transforms:
Bilinear Transformation:

- jkT
jkT
1 ωs /2
2π
Ωo
X() =  x(k)e
x(k) = ω  ω /2 X()e
d , s =
z-1
T
s= 
s
s
k = -
ωoT z +1
tan (
)
N-1
2
- j(2 /N)nk
j(2 /N)nk
1 N-1
X(n) =  x(k)e
x(k) = N  X(n)e
n =0
k =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 2 Finale Evaluasie 21 Mei 2009 Memorandum
Bladsy 1
1. a) H(z) = z/(z-0.8) en X(z) = z2 - zcos1/(z-11)(z-1-1)
Y(z) = H(z)X(z) = [z/(z-0.8)]×[(z2–0.5403z)/(z-11)(z-1-1)] [3]
Y(z)/z = (z2–0.5403z)/(z-0.8)(z-11)(z-1-1)
= 0.2679/(z-0.8) + 0.56777-0.87015/(z-11) + 0.567770.87015/(z-1-1) [5]
Y(z) = 0.2679z/(z-0.8) + (0.56777-0.87015)z/(z-11) + (0.567770.87015)z/(z-1-1)
y(k) = 0.2679(0.8)k + 20.567771kcos(k-0.87015)
= 0.2679(0.8)k + 1.1355cos(k-0.87015) [4]
b) H(z) = z/(z-0.8)  H(ej) = ej/(ej – 0.8) = 1.1355-0.8701 [2]
Therefore with an input cosk, the steady state output will be 1.1355cos(k-0.8701) [1]
yss = 1.1355cos(k-0.8701) [1]
[16]
s /2

 j2k
 j2k
jk2
jkT
2. h(k) = 1
H()e
d = 1
H()e
d =
e
d +
e
d
 0.5
0.5
0.25
s s /2


[8]

(4)
[3]
= 1/j2k[ej2k]-0.5-0.25+1/j2k[ej2k]0.250.5 = 1/j2k[ej2(-0.25)k-ej2(-0.5)k] + 1/j2k[ej2(0.5)k-ej2(0.25)k] [1]
= 1/j2k[ejk – e-jk] – 1/j2k[ej0.5k – e-j0.5k] = 1/k[ejk – e-jk]/j2 – 1/k[ej0.5k – e-j0.5k]/j2
= sin(k)/k – sin(0.5k)/k [4]

3. X() =  x(k)e
k =- 
[3]
N-1
- jkT
4. a) X(n) =  x(k)e
k =0
[12]

(12)
= 1e-j0 = 1
- j(2/N)nk
[2]
3
=  x(k)e - j(/2)nk
k =0
X(0) = 2 + 3 + 2 + 1= 8
X(1) = 2 + 3e-j/2 + 2e-j(/2)2 + 1e-j(/2)3 = 2 + 3-/2 + 2- + 1-3/2 = 2-1.571 [2]
X(2) = 2 + 3e-j(/2)2 + 2e-j(/2)4 + 1e-j(/2)6 = 2 + 3- + 2-2 + 1-3 = 0 [2]
X(3) = 2 + 3e-j(/2)3 + 2e-j(/2)6 + 1e-j(/2)9 = 2 + 3-3/2 + 2-3 + 1-9/2 = 21.571[2]
N-1
j(2/N)nk 1 3
b) x(k) = 1  X(n)e
=
 X(n)e j(/2)nk
4 n =0
N n =0
x(0) = ¼[8 + 2-1.571 +0 + 21.571] = ¼(8) = 2 [1]
x(1) = ¼[8 + 2-1.571ej/2 + 0ej(/2)2 + 21.571ej(/2)3]
= ¼[8 + 2(-1.571+/2) + 0 + 2(1.571+3/2)] = ¼(12) = 3 [1]
x(2) = ¼[8 + 2-1.571ej(/2)2 + 0ej(/2)4 + 21.571ej(/2)6]
= ¼[8 + 2(-1.571+) + 0 + 2(1.571+3)] = ¼(8) = 2 [1]
x(3) = ¼[8 + 2-1.571ej(/2)3 + 0ej(/2)6 + 21.571ej(/2)9]
= ¼[8 + 2(-1.571+3/2) + 0 + 2(1.571+9/2)] = ¼(4) = 1 [1]
5. h(k) = (cT/)sin[(k - )cT]/[(k - )cT]
N = 7   = (N-1)/2 = 3
With T = 2/s en c = s/5, h(k) becomes:
h(k) = [(s/5)(2/s)/]sin[(k - 3)(s/5)(2/s)]/
[(k - 3)(s/5)(2/s)]
h(k) = (2/5)sin[(k - 3)2/5]/[(k - 3)2/5] [2]
h(0) = (2/5)sin(-6/5)/(-6/5) = -0.06237 [1]
h(1) = (2/5)sin(-4/5)/(-4/5) = 0.09355 [1]
h(2) = (2/5)sin(-2/5)/(-2/5) = 0.3027 [1]
h(3) = (2/5)sin 0/0 = 0.4 [2]
h(4) = h(2) = 0.3027
h(5) = h(1) = 0.09355
[11] h(6) = h(0) = -0.06237
x(k)
-0.06237
+
0.09355
+
0.3027
+
0.4
+
0.3027
+
0.09355
+
y(k)
z-1
x(k-1)
-1
z
x(k-2)
-1
z
x(k-3)
-1
z
x(k-4)
-1
z
x(k-5)
-1
z
-0.06237
[4]
Digital Signal Processing EIDSV4
Unit 1 First Assessment 25 February 2010
Page 1
Question 1 State Shannon’s sampling theorem.
(3)
x(k)
Question 2 A signal x(k) = [ –1 2 1] is shown in Figure 1.
2
a) Express x(k+1) as the sum of
weighted and shifted impulses.
(3)
1
k
b) Sketch the following signals:
i) x(k + 1)u(k).
(3)
–1
Figure 1
ii) x(3 – k).
(3)
iii) x(k) – (k).
(3)
x(k)
y(k)
Question 3 Refer to the system in Figure 2.

a) Determine a difference equation that
z-1
describes the relationship between the
output y(k) and the input x(k) of the
z-1
system.
(2)
b) Calculate the first five terms, h(0) to
– 0.5
Figure 2
h(4), of the impulse response of the
system.
(5)
h(k)
Question 4 The impulse response of a system,
3
h(k) = [2 3 1], as well as the input to the system,
2
1
x(k) = [1 2 –1], is shown in Figure 3.
k
a) Use graphical convolution methods, to determine
the response, y(k), of the system.
(5) x(k)
Figure 3
b) From your answer for y(k) in Question 4 a),
2
calculate Y(z), using z transform methods. From
1
k
this, calculate H(z) = Y(z)/X(z), and confirm the
impulse response h(k), given in Figure 3.
(8)
–1
Question 5 Use z transform methods to determine y(k) in closed form, from the
following difference equations:
a) y(k) = 0.25y(k – 2) + cos k
b) y(k) = u(k) + y(k – 1) – 0.25y(k – 2)
---oooOooo---
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z – 1)
z/(z – a)
z/(z – 1)2
z-1X(z)
z-2X(z)
zX(z) – zx(0)
z2X(z) – z2x(0) – zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N -1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z – a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
1
x(k) = ω
s

ωs /2
ωs /2
X()e
jkT
d , s = 2π
T
(8)
(7)
Total: 50
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
h(k) = c
, T= 2π
ωs
π (k -α)ωcT
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
tan (
)
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 1 Eerste Evaluasie 25 February 2010 Memorandum Bladsy 1
[3]
1.
‘n Sein x(t), wat gemonster word teen ‘n frekwensie fm, kan teoreties volkome herwin word
uit die gemonsterde weergawe van x(t), indien x(t) aanvanklik bandbeperk was tot fm/2.
2. a) x(k) = -(k+1) + 2(k) + (k-1)
x(k+1) = -(k+2) + 2(k+1) + (k)
b) i) x(k+1)u(k)
(3)
ii) x(3-k)
1
1
k
[12]
iii) x(k)-(k)
2
1
k
(3)
-1
(3)
-1
3. a) y(k) = x(k) + y(k–1) – 0.5y(k–2)
b) h(k) = (k) + h(k-1) – 0.5h(k-2)
[7]
h(0)=1[1], h(1)=1[1], h(2)=1-0.5=0.5[1], h(3)=0.5-0.5=0[1], h(4)=0-0.25=-0.25 [1]
4. a)
n
x(n)
h(-n)
h(1-n)
h(2-n)
h(3-n)
h(4-n)
-2
0
1
0
0
0
0
-1
0
3
1
0
0
0
0
1
2
3
1
0
0
1
2
0
2
3
1
0
2
-1
0
0
2
3
1
3
0
0
0
0
2
3
4
0
0
0
0
0
2
y(0)=2 [1]
y(1)=7 [1]
y(2)=5 [1]
y(3)=-1 [1]
y(4)=-1 [1]
(5)
[13]
1
k
(3)
(2)
(5)
b)
X(z) = 1+2z-1–z-2 [1]
Y(z) = 2+7z-1+5z-2–z-3–z-4 [1]
H(z) = Y(z)/X(z)
=(2+7z-1+5z-2–z-3–z-4)/(1+2z-1-z-2)
[6]
2+3z-1+z-2
-1 -2
-1
-2 -3 -4
1+2z –z | 2+7z +5z –z –z
2+4z-1–2z-2
3z-1+7z-2
3z-1+6z-2–3z-3
z-2+2z-3
z-2+2z-3–z-4
H(z) = 2+3z-1+z-2
0
(8)
h(0)=2, h(1)=3 en h(2)=1
5. a) y(k) = 0.25y(k - 2) + cos k
Y(z) = 0.25z-2Y(z) + (z2 – zcos1)/(z-11)(z-1-1) [2]
z2Y(z) – 0.25Y(z) = z2(z2 – 0.5403z)/(z-11)(z-1-1)
Y(z)(z-0.5)(z+0.5) = z2(z2 – 0.5403z)/(z-11)(z-1-1)
Y(z)/z = z(z2 – 0.5403z)/(z-0.5)(z+0.5)(z-11)(z-1-1)
Y(z)/z = -0.0142/(z-0.5)+0.1453/(z+0.5)+0.4436-0.2031/(z-11)+0.44360.2031/(z-1-1)[4]
Y(z)=-0.0142z/(z-0.5)+0.1453z/(z+0.5)+0.4436-0.2031z/(z-11)+0.44360.2031z/(z-1-1)
y(k) = -0.0142(0.5)k + 0.1453(-0.5)k + 0.8872cos(k-0.2031) [2]
(8)
b) y(k) = 1 + y(k - 1) - 0.25y(k - 2)
Y(z) = z/(z-1) + z-1Y(z) - 0.25z-2Y(z) [2]
z2Y(z) - zY(z) + 0.25Y(z) = z3/(z-1)
(z2 – z + 0.25)Y(z) = z3/(z-1)
(z – 0.5)(z – 0.5)Y(z) = z3/(z-1)
Y(z)/z = z2/(z-1)(z-0.5)2
Y(z)/z = 4/(z-1) – 0.5/(z-0.5)2 - 3/(z-0.5) [3]
Y(z) = 4z/(z-1) – 0.5z/(z-0.5)2 – 3z/(z-0.5)
y(k) = 4 – (0.5)(1/0.5)k(0.5)k - 3(0.5)k
[15]
y(k) = 4 – 3(0.5)k - k(0.5)k [2] = 4 – (3 + k)(0.5)k , k>=0
(7)
Digital Signal Processing EIDSV4
Unit 2 First Assessment 25 March 2010
Page 1
Question 1 Determine whether the following systems, are stable or unstable:
1
1
(2)
(2)
b) H(z) =
a) H(z) =
2
2
z  0.64
z  2z  4
Question 2 The impulse response of a system,
h(k)
Figure 1
h(k) = (k) + (k – 1), is shown in Figure 1.
1
1
a) Determine H(z).
(2)
jT
k
b) Calculate the frequency response, H(e ),
of the system at the following values of T:
i) T = 0
ii) T = /4 iii) T = /2 iv) T = 3/4 v) T = 
(10)
c) Sketch a rough graph of |H()| versus T.
(2)
d) If the input to this particular system is x(k) = sin(/4)k, determine an expression
(2)
for the steady state output signal yss(k).
Question 3 The transfer function, H(), of a digital filter (Figure 2), is given by:
for    /2
Figure 2
 0
H()

H() =  2 cos ω for  /2    /2
2cos
 0






for
/2


Assume a sampling period of T = 1 sec.
0 /2
(s = 2/T = 2), and calculate the impulse
–/2

–
response h(k), required for this digital filter. (10)
(s/2)
(–s/2)
Question 4 The eight (N = 8) sampled values x(0) = 1, x(1) = 1, x(2) = 1, x(3) = 1,
x(4) = –1, x(5) = –1, x(6) = –1 and x(7) = –1, shown in Figure 3, were obtained when
a square wave signal x(t) was sampled.
x(k)
Figure 3
a) For the discrete Fourier transform, X(n), of
1 1 1 1
k
this sequence, calculate the value of X(1). (8)
b) Calculate the amplitude of the fundamental
0
–1 –1 –1 –1
component, X(1), contained in x(t).
(2)
Question 5 The ideal impulse response of a certain digital low pass filter, is given
1 sin(k/2)
. Design a corresponding FIR filter of length five (N=5), and
by h(k) =
2 k/2
draw a structure for this filter.
(10)
---oooOooo--Total: 50
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
2
z X(z) - z2x(0) - zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N-1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
(z e j )(z e j )
z(sin)
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2abkcos(k + )u(k)
(a z (a z

z b z b
1
x(k) = ω
s

ωs /2
ωs /2
X()e
jkT
d , s = 2π
T
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
h(k) = c
, T= 2π
ωs
π (k -α)ωcT
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
tan (
)
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 2 Eerste Evaluasie 25 Maart 2010 Memorandum
1. a) H(z) = 1/(z2-0.64) = 1/(z-0.8)(z+0.8)  Stelsel stabiel.
[4]
b) H(z) = 1/(z2 – 2z + 4) = 1/(z-21.0471)(z-2-1.0471)  Stelsel onstabiel.
Bladsy 1
(2)
(2)
2. a) h(k) = (k) + (k-1)  H(z) = 1 + z-1
b) i) H(ej0) = 1 + (ej0)-1 = 1 + 1 = 20
ii) H(ej/4) = 1 + e-j/4 = 1 + 1-/4 = 1.848-0.3927
iii) H(ej/2) = 1 + e-j/2 = 1 + 1-/2 = 1.414-0.7854
iv) H(ej3/4) = 1 + e-j3/4 = 1 + 1-3/4 = 0.7654-1.178
v) H(ej) = 1 + e-j = 1 + 1- = 00
c)
|H()|
d) yss(k) = 1.848sin(k/4 – 0.3927)
2
(2)
(2)
(2)
(2)
(2)
(2)
(2)
1
0
[16]

(2)
/4 /2 3/4 
H()
s /2
1 
jk
jkT
3. h(k) = 1
H()e
dω
H()e
d =
2 
s s /2
2cos
1 /2
jk [2] 1 /2  e j  e  j  jk
2cos  e
d =
d [2]
=
2
e
2
2 /2 
2 /2

0

– –/2
/2

1 /2
j  j jk
1 /2
j(k  1)
j(k  1)
[e  e
]e
d =
[e
e
]d
=
2 /2
2 /2
/2
/2
1 /2 j(k  1)
1 /2 j(k  1)
1 e j(k  1)
1 e j(k  1)
[4]
e
d +
e
d =
=
+
2 /2
2 /2
2 j(k  1)
2 j(k  1)
 /2
 /2
 j(k 1)/2





j
(k
1)/2
j
(k
1)/2
j
(k
1)/2


1 e
1 e
e
e
+
=
j(k  1)
j(k  1)
2
2
[2]
j(k 1)/2  j(k 1)/2
j(k 1)/2  j(k 1)/2 sin(k  1)/2 sin(k  1)/2
1
1
e
e
e
e
+
=
=
+
j2
j2


 k  1
 k  1



k
1
(k
1)
[10]








N-1
4. X(n) =  x(k)e
k =0
-j(/4)×1×0
7
=  x(k)e
k =0
-j(/4)×1×1
- j(2/8)nk
-j(/4)×1×2
7
=  x(k)e
- j(/4)nk
k =0
-j(/4)×1×3
x(k)
+ 1e
+ 1e
1 1
-j(/4)1×6
-j(/4)1×7
– 1e
– 1e
– 1e
– 1e
= 10 + 1– /4 + 1– /2 + 1– 3/4
0
– 1– – 1–5/4 – 1–3/2 – 1–7/4 [6]
[2]
(8)
= 5.2262-1.1781
b) A = |X(1)|/(N/2) [1] = 5.1258/4 = 1.2815 [1]
(2)
a) X(1) = 1e
[10]
- j(2/N)nk
+ 1e
-j(/4)1×4
-j(/4)1×5
1 1
k
–1 –1 –1 –1
1 sin(k/2)
1 sin[(k -2)/2)]
k = 0, 1, 2, 3, 4 [2]
therefore FIR filter of length 5: h(k) =
2 k/2
2 [(k-2)/2]
h0) = 0 [1] h(1) = 0.3183 [1] h(2) = 0.5 [1] h(3) = 0.3183 [1] h(4) = 0 [1]
5. h(k) =
x(k)
z-1
z-1
0.3183
[10]
[3]
z-1
0.5
0.3183
+
+
y(k)=0x(k)+0.3183x(k-1)+0.5x(k-2)+0.3183x(k-3)+0x(k-4)
Digital Signal Processing EIDSV4
Unit 1 Final Assessment 13 May 2010
Question 1
A signal x(k) = [2 1 –1 1 2] is shown in
Figure 1. Sketch the following signals:
a) x(k + 1)u(k – 1)
(4)
b) –x(k – 1)
(4)
c) x(–k – 2)
(4)
x(k)
2
1
-1
Page 1
2
Figure 1
1
k
0
Question 2 If x(k) is given by x(k) = (k + 1)[u(k+2) – u(k–3)], express x(k-1) as the
sum of weighted and shifted impulse functions.
(6)
z -1
Question 3
Refer to the system in Figure 2 and
determine the first four terms,
h(0) to h(3), of the impulse
response of the system.
(6)
x(k)
Figure 2
+
1
4
y(k)
z -1
Question 4 The
impulse
h(k)
response of a system,
4
Figure 3
h(k) = [4 3 2 1], as well as the
3
x(k)
2
input x(k) =[0 1 2 1] to the
2
1
system, is shown in Figure 3.
1
1
a) Use graphical convolution
k
k
summation, to determine
the response, y(k), of the system.
(7)
b) Confirm your answer in Question 4 a), by calculating y(k) from Y(z) = H(z)X(z),
using z transform methods.
(6)
Question 5
a) Show from first principles that Z{x(k+2)} = z2X(z) - z2x(0) - zx(1).
b) Use z transform methods to determine the k’th term, in closed form, of the series:
x(k+2) = x(k+1) + 2x(k), x(0) = 0 and x(1) = 1.
---oooOooo---
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
2
z X(z) - z2x(0) - zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N -1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
1
x(k) = ω
s

ωs /2
ωs /2
X()e
jkT
d , s = 2π
T
(5)
(8)
Total: 50
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
h(k) = c
, T= 2π
ωs
π (k -α)ωcT
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
tan (
)
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 1 Finale Evaluasie 13 Mei 2010 Memorandum
1. a) x(k+1)u(k-1)
b) -x(k-1)
2
k
[12]
2
k
-1
1
2
1
k
-1
-2
(4)
x(-k-2)
c)
1
Bladsy 1
-2
-1
(4)
(4)
2. x(k) = -(k+2) +0(k+1) + (k) + 2(k-1) + 3(k-2) [4]
x(k-1) = - (k+1) + (k-1) + 2(k-2) + 3(k-3) [2]
[6]
3. y(k)=.25[x(k)+x(k-1)+y(k-1)] [2]
[6]
h(2) = 0.25(0.3125) = 0.07813 [1]
4. a) y(0) = 0 [1]
y(1) = 4 [1]
y(2) = 11 [1]
y(3) = 12 [1]
y(4) = 8 [1]
y(5) = 4 [1]
y(6) = 1 [1]
(7)
[13]
1
2
1
3
2
1
0
4
3
2
1
1
2
4
3
2
1
4
3
2
1
1
4
3
2
1
4
3
2
4
3
4
y(0)=0
y(1)=4
y(2)=11
y(3)=12
y(4)=8
y(5)=4
y(6)=1
b) X(z) = z-1 + 2z-2 + z-3 [1] and H(z) = 4 + 3z-1 + 2z-2 + z-3 [1]
Y(z) = H(z)X(z) = (4 + 3z-1 + 2z-2 + z-3)(z-1 + 2z-2 + z-3)
= 4(z-1 + 2z-2 + z-3) + 3z-1(z-1 + 2z-2 + z-3) + 2z-2(z-1 + 2z-2 + z-3) + z-3(z-1 + 2z-2 + z-3)
= 4z-1 + 8z-2 + 4z-3 + 3z-2 + 6z-3 + 3z-4 + 2z-3 + 4z-4 + 2z-5 + z-4 + 2z-5 + z-6
= 4z-1 + 11z-2 + 12z-3 + 8z-4 + 4z-5 + z-6 [3]
y(0) = 0, y(1) = 4, y(2) = 11, y(3) = 12, y(4) = 8, y(5) = 4, y(6) = 1 [1]
5. a)
b)
[13]
x(n)
h(-n)
h(1-n)
h(2-n)
h(3-n)
h(4-n)
h(5-n)
h(6-n)
h(0) = 0.25(1) = 0.25 [1] h(1) = 0.25(1+0.25) = 0.3125 [1]
h(3) = 0.25(0.07813) = 0.01953 [1]

Z{x(k)} =  x(k)z
-k

 Z{x(k+2)} =  [x(k  2)]z
k =0
-1
Z{x(k+2)} = x(2) + x(3)z
k =0
(6)
-k
+ x(4)z-2 + x(5)z-3.+ .....[2]
= z2[x(2)z-2 + x(3)z-3 + x(4)z-4 + x(5)z-5.+ .........]
= z2[x(2)z-2 + x(3)z-3 + x(4)z-4 + x(5)z-5.+ .........] + z2[x(0) + z-1x(1)] – z2[x(0) + z-1x(1)] [2]
= z2[x(0) + x(1)z-1 + x(2)z-2 + x(3)z-3 + x(4)z-4 + x(5)z-5.+ .........] – z2[x(0) + z-1x(1)]
(5)
= z2X(z) – z2x(0) – zx(1) [1]
x(k+2)=x(k+1)+2x(k)
z2X(z) – z2x(0) – zx(1) = zX(z) – zx(0) + 2X(z) [2]
z2X(z) – z = zX(z) + 2X(z)
(z2 – z – 2)X(z) = z [1]
X(z) = z/(z2 – z – 2)
X(z)/z = 1/(z2 – z – 2)
X(z)/z = 1/(z-2)(z+1) [1]
X(z)/z = 0.3333/(z-2) – 0.3333/(z+1) [2]
X(z) = 0.3333z/(z-2) – 0.3333z/(z+1)
x(k) = 0.33332k – 0.3333(-1)k [2]
(8)
Digital Signal Processing EIDSV4
Unit 2 Final Assessment 13 May 2010
Page 1
Question 1 The block diagram of a system
x(k)
y(k)
with input x(k) and output y(k), is shown in
0.318
Figure 1.
z-1
a) Determine a difference equation that
describes the relationship between
0.5
y(k) and x(k).
(2)
z-1
b) Determine an expression for the
system function H(z) = Y(z)/X(z).
(4)
0.318
Figure 1
c) Calculate the frequency response,
jT
H(e ), of the system, at the following values of T:
i) T = 0
ii) T = /4 iii) T = /2 iv) T = 3/4 v) T = 
(10)
d) If the input to this particular system is x(k) = sin(/2)k, determine an
(2)
expression for the steady state output signal yss(k).
Question 2 The transfer function, H(), of a digital filter (Figure 2), is given by:
Figure 2
H()
for    /2
 0

H() =  1  cosω for  /2    /2
1 – cos
 0
for /2   


Assume a sampling period of T = 1 sec.
0 /2

–
–/2
(s = 2/T = 2), and calculate the impulse
(s/2)
(–s/2)
response h(k), required for this digital filter. (10)
Question 3 The eight (N = 8) sampled values x(0) = 1, x(1) = 1, x(2) = 1, x(3) = 1,
x(4) = –1, x(5) = –1, x(6) = –1 and x(7) = –1, shown in Figure 3, were obtained when
a square wave signal x(t) was sampled.
x(k)
Figure 3
a) For the discrete Fourier transform, X(n), of
1 1 1 1
this sequence, calculate the value of X(3). (8)
k
b) Calculate the amplitude of the third harmonic
0
component, X(3), contained in x(t).
(2)
–1 –1 –1 –1
Question 4 The ideal impulse response of a certain digital low pass filter, is given
1 sin(k/2)
by h(k) =
. Design a corresponding FIR filter of length seven (N=7), and
2 k/2
draw a structure for this filter.
(12)
---oooOooo--Total: 50
Appendix: z transforms
+
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
2
z X(z) - z2x(0) - zx(1)
x(k)
(1/a)kaku(k)
sinku(k)
X(z)
z/(z - a)2
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2abkcos(k + )u(k)
(a z (a z

z b z b
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
, T= 2π
h(k) = c
ωs
π (k -α)ωcT
Fourier transforms:
Bilinear Transformation:

- jkT
jkT
1 ω /2
Ωo
X() =  x(k)e
x(k) = ω  ωs /2 X()e
d , s = 2π
z-1
T
s= 
s
s
k = -
ωoT z +1
tan (
)
N-1
2
- j(2 /N)nk
j(2 /N)nk
1 N-1
X(n) =  x(k)e
x(k) = N  X(n)e
n =0
k =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 2 Finale Evaluasie 13 Mei 2010 Memorandum
Bladsy 1
1. a) y(k) = 0.318x(k) + 0.5x(k – 1) + 0.318x(k – 2)
(2)
x(k)
y(k)
-1
-2
0.318
b) Y(z) = 0.318X(z) + 0.5z X(z) + 0.318z X(z)
Y(z) = (0.318 + 0.5z-1 + 0.318z-2)X(z)
z-1
-1
-2
Y(z)/X(z) = H(z) = (0.318 + 0.5z + 0.318z )
(4)
0.5
c) i) H(ej0) = 0.318 + 0.5(ej0)-1 + 0.318(ej0)-2
= 0.318 + 0.5 + 0.318 = 1.136
(2)
z-1
ii) H(ej/4) = 0.318 + 0.5(ej/4)-1 + 0.318(ej/4)-2
0.318
= 0.318 + 0.5-/4 + 0.318-/2
= 0.9497-0.7854
(2)
iii) H(ej/2) = 0.318+0.5(ej/2)-1+0.318(ej/2)-2 = 0.318+0.5-/2+0.318- = 0.5-1.571
(2)
j3/4
j3/4 -1
j3/4 -2
iv) H(e )=0.318+0.5(e ) +0.318(e ) =.318+.5-3/4+.318-3/2=0.0528-2.356 (2)
(2)
v) H(ej) = 0.318 + 0.5(ej)-1 + 0.318(ej)-2 = 0.318 + 0.5- + 0.318-2 = 0.1360
[18] d) T = /2: yss(k) = 0.5sin(k/2 – 1.571)
(2)
s /2
1 /2
1 
jk
jk [2]
jkT
2. h(k) = 1
H()e
dω =
(
1

cos  e
d
H()e
d =
2


2 /2
s s /2
1 /2
jk
1 /2
j  j jk
1 /2  e j  e  j  jk
e
d –
[e  e
]e
d
=
d [2] =
1 
e
/2
/2






/2
2
2
4
2


1 /2
jk
1 /2
j(k  1)
j(k  1)
=
e
d –
[e

e
]d
H() 1–cos
2 /2
4 /2
1 /2
jk
1 /2 j(k  1)
1 /2 j(k  1)
=
e
d –
e
d –
e
d
2 /2
4 /2
4 /2
0
/2
/2
/2
– –/2
/2 
1 e jk
1 e j(k  1)
1 e j(k  1)
[4]
=
–
–
2 jk
4 j(k  1)
4 j(k  1)
 /2
 /2
 /2
 jk/2
 j(k 1)/2


j

k/2
j
(k
1)/2
1 e
1 e
1 e j(k 1)/2  e  j(k1)/2
e
e
–
–
=
jk
j(k  1)
j(k  1)
2
4
4
[2]
j(k 1)/2  j(k 1)/2
j(k 1)/2  j(k 1)/2
1
1
1 e jk/2  e  jk/2
e
e
e
e
–
=
–
j2
j2
j2
 k  1
(k  1)
k
+




[10]
=







sink/2 sin(k  1)/2 sin(k  1)/2
sink/2 1 sin(k  1)/2 1 sin(k  1)/2
–
–
–
–
= 1
2 k/2
4  k  1
4  k  1
k
 k  1
 k  1
N-1
3. X(n) =  x(k)e
- j(2/N)nk
k =0
7
=  x(k)e
- j(2/8)nk
k =0
7
=  x(k)e
- j(/4)nk
k =0
x(k)
a) X(3) = 1e-j(/4)×3×0 + 1e-j(/4)×3×1 + 1e-j(/4)×3×2 + 1e-j(/4)×3×3
1 1 1 1
-j(/4)3×4
-j(/4)3×5
-j(/4)3×6
-j(/4)3×7
– 1e
– 1e
– 1e
– 1e
k
= 10 + 1–3 /4 + 1–3 /2 + 1– 9/4
0
– 1–3 – 1–15/4 – 1–9/2 – 1–21/4 [6]
–1 –1 –1 –1
= 2.164784-0.392699 [2]
(8)
[1]
[1]
[10] b) A = |X(3)|/(N/2) = 2.164784/4 = 0.5412
(2)
1 sin(k/2)
1 sin[(k-3)/2)]
4. h(k) =
k = 0, 1, 2, 3, 4, 5, 6 [2]
therefore FIR filter of length 7: h(k) =
2 [(k-3)/2]
2 k/2
h0) = -0.1061[1] h(1) = 0[1] h(2) = 0.3180[1] h(3) = 0.5[1] h(4) = 0.13831[1] h(5) = 0[1] h(6) = -0 10610[1]
x(k)
[3]
z-1
z-1
z-1
z-1
z-1
z-1
-0.1061
[12]
0.3183
0.5
0.3183
-0.1061
+
+
+
+
y(k)=-0.1061x(k)+0.3183x(k-2)+0.5x(k-3)+0.3183x(k-4)-0.1061x(k-6)
Digital Signal Processing EIDSV4
Unit 1 First Assessment 24 February 2011
Page 1
Question 1 Write down the first seven values x(0) to x(6), of the sequence:
x(k + 2) = x(k + 1) – 0.8x(k) + (k), with x(0) = 0 and x(1) = 1.
(5)
Question 2 Calculate the value of x(2) for each of the following functions x(k):
a) x(k) = (k2 + 1)[u(k + 1) – u(k – 4)]
b) x(k) = 2 – k×(k – 2) + (k2 + 1)×[u(k + 4) – u(k – 2)]
c) x(k) = 2k×u(k – 1) + (k2 – k – 1)×u(k – 2)
d) x(k) = –k + u(k + 3)
(2)
(2)
(2)
(2)
Question 3 Refer to the system in Figure 1.
a) Determine a difference equation that
describes the relationship between the
output y(k) and the input x(k) of the
system.
(3)
b) Calculate the first four terms, h(0) to
h(3), of the impulse response h(k),
of the system.
(4)
x(k)
y(k)

z-1
z-1
– 0.8
Figure 1
Question 4 The difference equation that describes the relationship between the
output y(k) and the input x(k), of a system, is given by:
y(k) = x(k) + y(k – 1) – 0.8y(k – 2)
a) Show that the transfer function of the system, H(z) = Y(z)/X(z), is given by:
z2
H(z) =
(2)
z 2  z  0.8
b) From H(z) in Question 4 a), show that the system’s impulse response, is given by:
h(k) = 1.206 (0.8944)k cos(0.9776k – 0.5932).
(4)
c) From h(k) in Question 4 b), calculate the first four terms, h(0) to h(3) of the
impulse response of the system.
(4)
d) Use graphical convolution methods to determine the response y(k) of the system
for k = 0, 1, 2 and 3, if the input to the system is the signal, x(k) = (0.5)k u(k).
(8)
e) Use z transform methods to verify your results in Question 4 d), by finding y(k)
for k = 0, 1, 2 and 3, from Y(z) = H(z)X(z).
(12)
---oooOooo---
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z – 1)
z/(z – a)
z/(z – 1)2
z-1X(z)
z-2X(z)
zX(z) – zx(0)
2
z X(z) – z2x(0) – zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N -1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z – a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
1
x(k) = ω
s

ωs /2
ωs /2
X()e
jkT
d , s = 2π
T
Total: 50
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
h(k) = c
, T= 2π
π (k -α)ωcT
ωs
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
)
tan (
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
j
Euler and trig. identities: re = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 1 Eerste Evaluasie 24 February 2011 Memorandum Bladsy 1
1.
x(0) = 0, x(1) = 1  x(2) = x(1) – 0.8x(0) + (0) = 1 – 0 + 1 = 2 [1]
x(3) = x(2) – 0.8x(1) + (1) = 2–0.8+0 = 1.2 [1] x(4) = x(3) – 0.8x(2) + (2) = 1.2–1.6+0 = –0.4 [1]
x(5)=x(4)–0.8x(3)+(3)=–0.4–0.96 + 0=–1.36[1] x(6)=x(5)–0.8x(4)+(4)=–1.36–(–0.32)+0=–1.04 [1]
2.
a)
b)
c)
d)
[5]
[8]
x(k)=(k2+1)[u(k+1)–u(k–4)]  x(2)=(22+1)[u(3)–u(–2)]=5[1–0] = 5
x(k)=2–k×(k–2)+(k2+1)×[u(k+4)–u(k–2)]  x(2)=2–2×(0)+5[u(6)-u(0)]=¼+5(1-1)=0.25
x(k) = 2k×u(k – 1) + (k2 – k – 1)×u(k – 2)  x(2) = 22u(1)+(22-2-1)u(0) = 41+11 = 5
x(k) = –k + u(k + 3)  x(2) = –2 + u(5) = –2 + 1 = –1
3. a) y(k) = x(k) + y(k–1) – 0.8y(k–2)
b) h(k) = (k) + h(k–1) – 0.8h(k–2)
[7]
h(0) = 1[1], h(1) = 1[1], h(2) = 1 – 0.8 = 0.2[1], h(3) = 0.2 – 0.8 = –0.6[1],
-1
-2
(3)
(4)
[1]
4. a) y(k) = x(k) + y(k – 1) – 0.8y(k – 2)  Y(z) = X(z) + z Y(z) – 0.8z Y(z)
Y(z) – z-1Y(z) + 0.8z-2Y(z) = X(z)  z2Y(z) – zY(z) + 0.8Y(z) = z2X(z)
(z2 – z + 0.8)Y(z) = z2X(z)  H(z) = Y(z)/X(z) = z2/(z2 – z + 0.8) [1]
b) H(z)/z = z/(z2 – z + 0.8)
= 0.6030–0.5932/(z – 0.89440.9776) + 0.60300.5932/(z – 0.8944–0.9776) [2]
H(z) = (0.6030–0.5932)z/(z–0.89440.9776)+(0.60300.5932)z/(z–0.8944–0.9776)
h(k) = 1.206(0.8944k)cos(0.9776k – 0.5932) [2]
c) h(0) = 1.206(1)cos(–0.5932) = 0.99996  1 [1]
h(1) = 1.206(0.8944)cos(0.9776 – 0.5932) = 0.99993  1 [1]
h(2) = 1.206(0.89442)cos(20.9776 – 0.5932) = 0.19997  0.2 [1]
h(3) = 1.206(0.89443)cos(30.9776 – 0.5932) = –0.59993  –0.6 [1]
d) x(k) = (0.5)k u(k)  x(0) = 1, x(1) = 0.5, x(2) = 0.25, x(3) = 0.125
h(k) = 1.206(0.8944k)cos(0.9776k – 0.5932)  h(0) = 1, h(1) = 1, h(2) = 0.2, h(3) = –0.6
n
-3
x(n)
h(-n) -0.6
h(1-n)
h(2-n)
h(3-n)
[30]
-2
-1
0.2
-0.6
1
0.2
-0.6
0
1
1
1
0.2
-0.6
1
0.5
1
1
0.2
2
0.25
1
1
(2)
(2)
(2)
(2)
(2)
(4)
(4)
3
0.125
1
y(0)=1 [2]
y(1)=1.5[2]
y(2)=0.95 [2]
y(3)=-0.125 [2]
(8)
e) H(z) = z2/(z2 – z + 0.8) =z2/(z–0.89440.9776)(z–0.8944–0.9776)
X(z) = z/(z – 0.5) [2]
Y(z) = H(z)X(z) = [z2/(z–0.89440.9776)(z–0.8944–0.9776)][z/(z – 0.5)]
Y(z)/z = z2/(z – 0.5)(z – 0.89440.9776)(z – 0.8944–0.9776)
= A/(z – 0.5) + B/(z – 0.89440.9776) + B*/(z – 0.8944–0.9776)
A = 0.52/[0.5–0.89440.9776)(0.5–0.8944–0.9776)] = 0.4545
B = (0.89440.9776)2/[(0.89440.9776 – 0.5)(0.89440.9776 – 0.8944–0.9776)]
= 0.7273-1.186
Y(z)/z=0.4545/(z–0.5)+0.7273–1.186/(z–0.89440.9776)+0.72731.186/(z–0.8944–0.9776)
Y(z)=0.4545z/(z–0.5)+(0.7273–1.186)z/(z–0.89440.9776)+(0.72731.186)z/(z–0.8944–0.9776)
y(k) = 0.4545(0.5)k + 2×0.7273×(0.8944k)cos(0.9776k – 1.186)
= 0.4545(0.5)k + 1.455(0.8944k)cos(0.9776k – 1.186) [6]
y(0) = 0.4545 + 1.455(1)cos(– 1.186) = 1.0006  1 [1]
y(1) = 0.4545(0.5) + 1.455(0.8944)cos(0.9776 – 1.186) = 1.5004  1.5 [1]
y(2) = 0.4545(0.5)2 + 1.455(0.89442)cos(0.97762 – 1.186) = 0.94987  0.95 [1]
y(3) = 0.4545(0.5)3 + 1.455(0.89443)cos(0.97763 – 1.186) = –0.12547  –0.1255 [1]
(12)
Digital Signal Processing EIDSV4
Unit 2 First Assessment 24 March 2011
Page 1
Question 1
x(k)
y(k)
The block diagram of a system with input
0.5
x(k) and output y(k), is shown in Figure 1.
z-1
z-1
a) Determine a difference equation that
0.25
0.5
describes the relationship between
Figure 1
y(k) and x(k).
(2)
b) Determine an expression for the system function H(z) = Y(z)/X(z).
(4)
jT
c) Calculate the frequency response, H(e ), of the system, at the following values
of T:
i) T = 0
ii) T = 
(5)
d) Determine an expression for the steady state output signal yss(k), if the input x(k)
to the system in Figure 1, is: i) x(k) = u(k) and ii) x(k) = sink.
(4)
Question 2 The transfer function of a digital
H()
Figure 2
1
filter, H(), shown in Figure 2, is given by:
/2

 0 for       /2

0
0
– –/2

H() =  1 for  /2   
1
for
0
/2







–1
(s/2)
(–s/2)
/2   

 0 for
Assume a sampling period of T = 1 sec. (s = 2/T = 2), and calculate the impulse
response h(k), required for this digital filter.
(10)
+
Question 3 The eight (N = 8) sampled values x(0) = 0, x(1) = 0, x(2) = 1, x(3) = 1,
x(4) = 0, x(5) = 0, x(6) = 0 and x(7) = 0, shown in Figure 3, were obtained when
a signal x(t) was sampled.
x(k)
a) For the discrete Fourier transform X(n), of
1 1
k
this sequence, calculate the values of X(0),
X(1) and X(2).
(10)
0
Figure 3
b) Calculate the average value of x(t).
(1)
c) Calculate the amplitude A1, of the fundamental frequency X(1), and the
amplitude A2, of the second harmonic frequency X(2), contained in x(t).
(2)
Question 4 The ideal impulse response required for a digital filter, is given
by h(k) = cosk. Design a corresponding FIR filter of length five (N=5), and
draw a structure for this filter.
(12)
---oooOooo--Total: 50
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
2
z X(z) - z2x(0) - zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N-1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
(z e j )(z e j )
z(sin)
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2abkcos(k + )u(k)
(a z (a z

z b z b
1
x(k) = ω
s
ω /2
 ωss /2 X()e
jkT
d , s = 2π
T
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
h(k) = c
, T= 2π
ωs
π (k -α)ωcT
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
tan (
)
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 2 Eerste Evaluasie 24 Maart 2011 Memorandum
1. a) y(k) = 0.5x(k) + 0.25x(k – 1) + 0.5y(k – 1)
(2)
x(k)
-1
-1
b) Y(z) = 0.5X(z) + 0.25z X(z) + 0.5z Y(z)
0.5
Y(z) – 0.5z-1Y(z) = 0.5X(z) + 0.25z-1X(z)
-1
z
H(z) = (0.5 + 0.25z-1 )/(1 – 0.5z-1)
(4)
j0
j0 -1
j0 -1
c) i) H(e ) = [0.5 + 0.25(e ) ]/[1 – 0.5(e ) ]
0.25
0.5
= [0.5 + 0.25]/[1 – 0.5] = 1.5
(2)
ii) H(ej) = [0.5 + 0.25(ej)-1]/[1 – 0.5(ej)-1] = (0.5 + 0.25–)/(1 – 0.5–) = 0.16670
[15]
d) i) yss(k) = 1.5u(k)
(2)
ii) yss(k) = 0.1667sink

Bladsy 1
y(k)
z-1
(3)
(2)
s /2
1 0
1 /2
1 
jk
jkT
2. h(k) = 1
H()e
d =
H()e
dω =
1
e jk d +
 1 e jk d
2

/2
0






/2
2

2

s
s
0
/2
H()
1  e j k 
1  e j k 
[3]
=
–
1
2  jk 
2  jk 
/2
 /2
0

1  e j/2k 1 
1  1 e  j/2k 
0
=
–

 
– –/2



jk 
jk  2  jk
2  jk
–1
j/2k  e  j/2k
j/2k  e  j/2k
1
1
1
 1 e
=
=
e
=
 1 cos  k , [4]
[10]
2
jk

2

jk

jk
jk
jk jk


N-1
3. X(n) =  x(k)e
- j(2/N)nk
- j(2/8)nk
k =0
k =0
[13]
7
=  x(k)e

7
=  x(k)e

[3]
- j(/4)nk
k =0
a) X(0) = 0 + 0 + 1 + 1 + 0 + 0 + 0 + 0 = 2 [2]
x(k)
X(1) = 0e-j(/4)×1×0 + 0e-j(/4)×1×1 + 1e-j(/4)×1×2 + 1e-j(/4)×1×3
1 1
-j(/4)1×4
-j(/4)1×5
-j(/4)1×6
-j(/4)1×7
+ 0e
+ 0e
+ 0e
+ 0e
k
= 00 + 0– /4 + 1– /2 + 1– 3/4
0
+ 0– + 0–5/4 + 0–3/2 + 0–7/4
[4]
= 1.848-1.963
X(2) = 0e-j(/4)×2×0+0e-j(/4)×2×1+1e-j(/4)×2×2+1e-j(/4)×2×3+0e-j(/4)2×4+0e-j(/4)2×5+0e-j(/4)2×6+0e-j(/4)2×7
= 00 + 0– /2 + 1–  + 1– 3/2 + 0–2 + 0–5/2 + 0–3 + 0–7/2
= 1.4142.356 [4]
(10)
(1)
b) xav = X(0)/N = 2/8 = 0.25
[1]
[1]
A2 = |X(2)|/(N/2) = 1.414/4 = 0.3535
(2)
c) A1 = |X(1)|/(N/2) = 1.848/4 = 0.462
4. h(k) = cosk, therefore FIR filter of length 5: h(k) = cos(k – 2), k = 0, 1, 2, 3, 4 [2]
h0) = –0.4161 [1] h(1) = 0.5403 [1] h(2) = 1 [1] h(3) = 0.5403 [1] h(4) = –0.4161 [1]
x(k)
z-1
x(k)
–0.4161
[12]
z-1
x(k-1)
z-1
x(k-2)
[5]
z-1
x(k-3)
x(k-4)
0.5403
1
0.5403
–0.4161
+
+
+
+
y(k) = –0.4161 x(k)+ 0.5403x(k-1)
+x(k-2)+ 0.540x(k-3)
–0.4161x(k-4)
Digital Signal Processing EIDSV4
Unit 1 Final Assessment 5 May 2011
Page 1
Question 1 Write down the first seven values x(0) to x(6), of the sequence:
x(k + 2) = x(k + 1) + x(k), with x(0) = 0 and x(1) = 1.
Question 2
A signal x(k) = [ 0 1 1 2 3 5 ] is shown
in Figure 1. Sketch the following signals:
a) x(k) + x(–k)
(3)
b) x(k + 3)u(k)
(3)
c) x(k + 3)  x(3 – k)
(4)
x(k)
5
Figure 1
1
(5)
1
2
3
k
0
x(k)
Question 3
Refer to the system in Figure 2 and
determine the first six terms,
h(0) to h(5), of the impulse
response of the system.
(10)

y(k)
0.5
z-1
z-1
0.5
z-1
Figure 2
– 0.5
Question 4 The difference equation that describes the relationship between the
output y(k) and the input x(k), of a system, is given by:
y(k) = 0.5y(k–1) – 0.25y(k–2) + 0.5x(k) + 0.25x(k–1)
a) Determine the system’s transfer function H(z) and from H(z), show that the
(5)
impulse response h(k) = Z–1{H(z)} = (0.5)k cos(1.0472k – 1.0472).
b) From h(k) in Question 4 a), calculate the first four terms, h(0) to h(3) of the
impulse response of the system.
(4)
c) Use graphical convolution methods to determine the response y(k) of the system
(6)
for k = 0, 1, 2 and 3, if the input to the system is the signal, x(k) = (0.5)k u(k).
d) Use z transform methods to verify your results in Question 4 c), by finding y(k)
for k = 0, 1, 2 and 3, from Y(z) = H(z)X(z).
(10)
---oooOooo---
Total: 50
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
z2X(z) - z2x(0) - zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N-1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2abkcos(k + )u(k)
(a z (a z

z b z b
1
x(k) = ω
s

ωs /2
ωs /2
X()e
jkT
d , s = 2π
T
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
, T= 2π
h(k) = c
ωs
π (k -α)ωcT
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
tan (
)
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 1 Finale Evaluasie 5 Mei 2011 Memorandum
1.
[5]
2.
x(0) = 0, x(1) = 1  x(2) = x(1) + x(0) = 1 + 0 = 1 [1]
x(3) = x(2) + x(1) = 1 + 1 = 2 [1] x(4) = x(3) + x(2) = 2 + 1 = 3 [1]
x(5)=x(4) + x(3) = 3 + 2 = 5 [1] x(6) = x(5) + x(4) = 5 + 3 = 8 [1]
a) 5
b) x(k+3)u(k)
x(k)+x(–k) 5
3
2
1 1
2
1 1
2
0
c)
x(k+3)
1 1
[10
3
2
3
k
0
2
k
(3)
x(k+3)x(3–k)
x(3–k)
5
3
0
(3)
5
5
3
k
Bladsy 1
5
3
1 1
4
k
5
3
k
0
0
[4]
(4)
3. y(k) = 0.5[y(k–1) – 0.5y(k–2) + x(k) + 0.5x(k–1)]
h(0) = 0.5[0 – 0 + 1 + 0] = 0.5 [1]
h(1) = 0.5[0.5 – 0 + 0 + 0.5] = 0.5 [1]
h(2) = 0.5[[0.5 – 0.50.5 + 0 + 0] = 0.125 [1]
h(3) = 0.5[0.125 – 0.50.5] = –0.0625 [1]
h(4) = 0.5[–0.0625 – 0.50.125] = –0.0625 [1]
h(5) = 0.5[–0.0625 – 0.5–0.0625] = -0.015625 [1]
[10]
4. a) y(k)=0.5y(k–1)–0.25y(k–2)+0.5x(k)+0.25x(k–1)
Y(z)=0.5z-1Y(z)–0.25z-2Y(z)+0.5X(z)+0.25z-1X(z)
Y(z)–0.5z-1Y(z)+0.25z-2Y(z)=0.5X(z)+0.25z-1X(z)
z2Y(z)–0.5zY(z)+0.25Y(z)=0.5z2X(z)+0.25zX(z)  (z2–0.5z+0.25)Y(z) = (0.5z2+0.25z)X(z)
H(z) = Y(z)/X(z) = (0.5z2+0.25z)/(z2–0.5z+0.25) [1]
H(z)/z = (0.5z + 0.25)/(z2 – 0.5z + 0.25)
= 0.5–1.0472/(z – 0.51.0472) + 0.51.0472/(z – 0.5–1.0472) [2]
H(z) = (0.5–1.0472)z/(z – 0.51.0472) + (0.51.0472)z/(z – 0.5–1.0472)
h(k) = (0.5k)cos(1.0472k – 1.0472) [2]
(5)
[1]
[1]
h(1) = (0.5)cos(1.0472 – 1.0472) = 0.5
b) h(0) = (1)cos(–1.0472) = 0.499998  0.5
2
[1]
h(2) = (0.5 )cos(21.0472–1.0472) = 0.125
h(3) = 0.53cos(31.0472–1.0472)=–0.0625 [1] (4)
c) x(k) = (0.5)k u(k)  x(0) = 1 [½] , x(1) = 0.5 [½] , x(2) = 0.25 [½] , x(3) = 0.125 [½]
h(k) = (0.5k)cos(1.0472k – 1.0472)  h(0) = 0.5, h(1) = 0.5, h(2) = 0.125, h(3) = –0.0625
n
-3
-2
-1
0
x(n)
1
h(-n) -0.0625 0.125
0.5
0.5
h(1-n)
-0.0625 0.125
0.5
h(2-n)
-0.0625 0.125
h(3-n)
-0.0625
2
[25]
2
1
0.5
0.5
0.5
0.125
2
0.25
0.5
0.5
3
0.125
0.5
y(0)=0.5 [1]
y(1)=0.75 [1]
y(2)=0.5 [1]
y(3)=-0.1875 [1]
(6)
d) H(z) = (0.5z + 0.25z)/(z – 0.5z + 0.25) = [0.5z(z + 0.5)]/[(z – 0.51.0472)(z – 0.5–1.0472)]
X(z) = z/(z – 0.5) [2]
Y(z) = H(z)X(z) = [0.5z(z + 0.5)]/[(z – 0.51.0472)(z – 0.5–1.0472)][z/(z – 0.5)]
Y(z)/z = [0.5z(z + 0.5)]/[(z – 0.5)(z – 0.51.0472)(z – 0.5–1.0472)]
= A/(z – 0.5) + B/(z – 0.51.0472) + B*/(z – 0.5–1.0472)
A = [0.50.5(0.5 + 0.5)]/[(0.5)2 – 0.5(0.5) + 0.25)] = 1
B = [0.50.51.0472(0.51.0472+0.5)]/(0.51.0472–0.5)(0.51.0472–0.5–1.0472)
= 0.5–2.0944
Y(z) = z/(z – 0.5) + 0.5–2.0944z/(z – 0.51.0472) + 0.52.0944z/(z – 0.5–1.0472)
y(k) = (0.5)k + 20.5(0.5)kcos(1.0472k + 0) = (0.5)k + (0.5)kcos(1.0472k – 2.0944) [6]
y(0) = 1 + cos(–2.0944) = 0.5 [1]
y(1) = (0.5) + (0.5)cos(1.0472 – 2.0944) = 0.75 [1]
y(2) = (0.5)2 + (0.5)2cos(2.0944 – 2.0944) = 0.5 [1]
y(3) = (0.5)3 + (0.5)3cos(3.1416 – 2.0944) = 0.1875 [1]
(10)
Digital Signal Processing EIDSV4
Unit 2 Final Assessment 5 May 2011
Page 1
Question 1 The block diagram of a system with x(k)
y(k)
input x(k) and output y(k), is shown in Figure 1.

0.5
a) Determine a difference equation that describes
z-1
z-1
the relationship between y(k) and x(k).
(2)
0.5
b) Determine an expression for the system
z-1
function H(z) = Y(z)/X(z).
(4)
– 0.5
jT
Figure 1
c) Calculate the frequency response, H(e ), of
the system, at the following values of T: i) T = 0
ii) T = /2
(5)
d) Determine an expression for the steady state output signal yss(k), if the input x(k)
to the system in Figure 1, is: i) x(k) = u(k) and ii) x(k) = sin(/2)k.
(4)
Question 2 The transfer function of a digital
H()
Figure 2
filter, H(), shown in Figure 2, is given by:

1
e
for       /2
 0
/2

 
for  /2    0
0
H() =  e
– –/2



–
for 0    /2
 e
–
e
–1
(–s/2)
(s/2)
for /2    
 0
Assume a sampling period of T = 1 sec. (s = 2/T = 2), and calculate the impulse
response h(k), required for this digital filter. Do not attempt to simplify your answer.
(10)
Question 3 The sixteen (N=16) sampled values x(0) = 0, x(1) = 0, x(2) = 0, x(3) = 0,
x(4)=1, x(5)=1, x(6)=1. x(7)=1, x(8)=0, x(9)=0, x(10)=0, x(11)=0, x(12)=0, x(13) = 0,
x(14)=0 and x(15)=0, shown in Figure 3, were obtained when a signal x(t) was sampled.
a) For the discrete Fourier transform X(n), of
Figure 3
x(k)
1111
this sequence, calculate the values of X(0),
k
X(1) and X(2).
(10)
0
b) Calculate the average value of x(t).
(1)
c) Calculate the amplitude A1, of the fundamental frequency X(1), and the
amplitude A2, of the second harmonic frequency X(2), contained in x(t).
(2)
Question 4 The ideal impulse response h(k), required for a digital filter,
h(k)
 k for k  0
ek
e–k
is shown in Figure 4 and given by: h(k) = e-|k| =  e
k
 e  k for k  0
0
Figure 4
Design and draw the structure of a corresponding FIR filter (h(k) shifted
twice to the right and then truncated) of length five (N=5).
(12)
---oooOooo--Total: 50
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
z2X(z) - z2x(0) - zx(1)
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2abkcos(k + )u(k)
(a z (a z

z b z b
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
, T= 2π
h(k) = c
ωs
π (k -α)ωcT
Fourier transforms:
Bilinear Transformation:

- jkT
jkT
1 ω /2
Ωo
X() =  x(k)e
x(k) = ω  ωs /2 X()e
d , s = 2π
z-1
T
s= 
s
s
k = -
ωoT z +1
tan (
)
N-1
2
- j(2 /N)nk
j(2 /N)nk
1 N-1
X(n) =  x(k)e
x(k) = N  X(n)e
n =0
k =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 2 Finale Evaluasie 5 Mei 2011 Memorandum
Bladsy 1
1. a) y(k) = 0.5y(k–1)–0.25y(k–2)+0.5x(k)+0.25x(k–1)
(2)
x(k)
-1
-2
-1
b) Y(z) = 0.5z Y(z)–0.25z Y(z)+0.5X(z)+0.25z X(z)
0.5
Y(z)–0.5z-1Y(z)+0.25z-2Y(z) = 0.5X(z)+0.25z-1X(z)
z-1
z2Y(z)–0.5zY(z)+0.25Y(z) = 0.5z2X(z)+0.25zX(z)
0.5
H(z)=Y(z)/X(z)=(0.5z2+0.25z)/(z2–0.5z+0.25)
(4)
c) i) H(ej0) = [0.5(ej0)2+0.25(ej0)]/[(ej0)2–0.5(ej0)+0.25]
= [0.5+0.25]/[1– 0.5+0.25] = 1
(2)
–0.5
j/2
j/2 2
j/2
j/2 2
j/2
ii) H(e ) = [0.5(e ) +0.25(e )]/[(e ) –0.5(e )+0.25]
=(0.5+0.25/2)/(1–0.5/2+0.25) = 0.62017–1.0517
[15] d) i) yss(k) = u(k)
(2)
ii) yss(k) = 0.62017sin[(/2)k–1.0517]
y(k)

2. h(k)= 1
s
= 1
2
s /2
 s/2 H()e
jkT
0
d =
1

/2
/2 e(jk  1) d – 2 0

/2
 ( jk  1)/2 

= 1  1 e

jk  1
2  jk  1


  /2
= 1  1  1 –e
2  jk  1 jk  1
2
3. X(n) =  x(k)e
k =0
[13]
- j(2/N)nk
(3)
(2)

e (jk  1) d
 (jk  1) 
 (jk  1) 
– 1 e
= 1 e

2  jk  1   /2 2  jk  1  0
N-1
z-1
0
/2
1 
jk
  j k
e
H()e
dω = 1
e  e jk d + 1
e
d [3]
2 
2 /2
2 0
0
[10]
z-1
H()
e
1
[3]
 (jk  1)/2

 1 
– 1 e
jk  1
2  jk  1
 e  jk(/2) e jk(/2)  [4]
 jk  1  jk  1 


15
15
k =0
k =0
–
=  x(k)e- j(2/16)nk =  x(k)e- j(/8)nk
–/2
/2
0
–1 –e–


x(k) 1 1 1 1
k
0
a) X(0) = 0+0+0+0+1+1+1+1+0+0+0+0+0+0+0+0 = 4
-j(/8)×1×0
-j(/8)×1×1
-j(/8)×1×2
-j(/8)×1×3
-j(/8)1×4
-j(/8)1×5
+0e
+0e
+0e
+1e
+1e
+1e-j(/8)1×6+1e-j(/8)1×7
X(1) = 0e
-j(/8)1×8
-j(/8)1×9
-j(/8)1×10
-j(/8)1×11
-j(/8)1×12
-j(/8)1×13
+0e
+0e
+0e
+0e
+0e
+0e-j(/8)1×14+0e-j(/8)1×15
+0e
= 00 + 0– /8 + 0– /4 + 0– 3/8 +1–/2 + 1–5/8 + 1–3/4 + 1–7/8
+ 0– + 0–9 /8 + 0–5 /4 + 0–11/8 + 0–3/2 + 0–13/8 + 0–7/4 + 0–15/8
= 3.6245–2.1598 [4]
X(2) = 0e-j(/8)×2×0+0e-j(/8)×2×1+0e-j(/8)×2×2+0e-j(/8)×2×3+1e-j(/8)2×4+1e-j(/8)2×5+1e-j(/8)2×6+1e-j(/8)2×7
+0e-j(/8)2×8+0e-j(/8)2×9+0e-j(/8)2×10+0e-j(/8)2×11+0e-j(/8)2×12+0e-j(/8)2×13+0e-j(/8)2×14+0e-j(/8)2×15
= 00 + 0– /4 + 0– /2 + 0– 3/4 +1– + 1–5/4 + 1–3/2 + 1–7/4
+ 0–2 + 0–9 /4 + 0–5 /2 + 0–11/4 + 0–3 + 0–13/4 + 0–7/2 + 0–15/4
(10)
= 2.61311.9635 [4]
b) xav = X(0)/N = 4/16 = 0.25
(1)
[1]
[1]
c) A1 = |X(1)|/(N/2) = 3.6245/8 = 0.453
A2 = |X(2)|/(N/2) = 2.6131/8 = 0.32664
(2)
[2]
4. hideal(k) = e–|k|, therefore FIR filter of length 5: h(k) = e–|k–2|, k = 0, 1, 2, 3, 4 [2]
h0) = e-|0-2|= 0.1353 [1] h(1) = e-|1-2| = 0.3679 [1] h(2) = e-|2-2| =1 [1] h(3) = 0.3679 [1] h(4) = 0.1353 [1]
[5]
h(k)
x(k)
Ideal and perfect but
z-1
z-1
z-1
z-1
k
e
e–k
non-causal, infinite
x(k-4)
x(k)
x(k-1)
x(k-2)
x(k-3)
k
and unrealizable
0
0.1353
0.3679
1
0.3679
0.1353
Not perfect and
hpractcal(k)
y(k)
only approxima+
+
+
+
ting ideal filter
y(k) = 0.1353x(k)+ 0.3679x(k-1)
behaviour, but
k
[12]
+x(k-2)+ 0.3679x(k-3)
causal, finite
0
+0.1353x(k-4)
and realizable
Digital Signal Processing EIDSV4
Unit 1 First Assessment 23 February 2012
Page 1
Question 1 A signal x(t), is given by x(t) = 10cos100t. Determine the theoretical
minimum frequency at which x(t) should be sampled, so that x(t) could again be
recovered from the samples.
(3)
x(k)
2
Question 2 A signal x(k) = [ –2 –1 0 1 2 ] is
1
shown in Figure 1. Sketch the following signals:
k
a) x(k + 1)u(k – 1).
(3)
0
b) x(–k + 2)(k – 3).
(3)
–1
Figure 1
–2
c) –x(k – 1) – (k – 1).
(3)
Question 3 Refer to the structure in Figure 2.
a) Determine a difference equation that
describes the relationship between the
output y(k) and the input x(k) of the
system.
(4)
b) Calculate the first four terms, h(0) to
h(3), of the impulse response h(k), of the
system.
(4)
z-1
x(k)

z-1
0.4

0.6
y(k)
Figure 2
Question 4 The difference equation that describes the relationship between the
output y(k) and the input x(k), of a system, is given by:
y(k) = x(k) + y(k–1) – 0.24y(k–2).
a) Determine the system’s transfer function H(z) = Y(z)/X(z), and from H(z), show
(4)
that the impulse response h(k) = Z–1{H(z)} = 3(0.6)k – 2(0.4)k, for k  0.
b) From h(k) in Question 4 a), calculate the first four terms, h(0) to h(3), of the
impulse response of the system.
(2)
c) Use graphical convolution methods to determine the response y(k) of the system
for k = 0, 1, 2 and 3, if the input to the system is the signal, x(k) = cos k u(k)
(calculating cosk for k = 0, 1, 2 and 3, radians must be used for k).
(12)
d) Use z transform methods to verify your results in Question 4 c), by finding y(k)
(12)
for k = 0, 1, 2 and 3, from Y(z) = H(z)X(z) and y(k) = Z-1{Y(z)}.
---oooOooo---
Total: 50
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z – 1)
z/(z – a)
z/(z – 1)2
z-1X(z)
z-2X(z)
zX(z) – zx(0)
2
z X(z) – z2x(0) – zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N-1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z – a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
1
x(k) = ω
s

ωs /2
ωs /2
X()e
jkT
d , s = 2π
T
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
, T= 2π
h(k) = c
π (k -α)ωcT
ωs
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
tan (
)
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
j
Euler and trig. identities: re = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 1 Eerste Evaluasie 23 Februarie 2012 Memorandum Bladsy 1
[3]1. fx(t) = 100/2 = 50 Hz  fs=100 Hz.
2.
a)
2
[9]
[8]
–x(k–1)–(k–1)
c)
x(–k+2)(k–3)
2
k
k
0
3.
b)
x(k+1)u(k–1)
0
(3)
–1
0
–1 –1
(3)
a(k) = x(k) + 0.6Da(k) [1]
a(k) = x(k)/(1 – 0.6D) ........... (1)
y(k) = a(k) + 0.4Dy(k) [1]
y(k) = a(k)/(1 – 0.4D) ........... (2)
x(k)
(1) in (2):
y(k) = x(k)/(1 – 0.6D)(1 – 0.4D)
(1 – 0.6D)(1 – 0.4D)y(k) = x(k)
y(k) – Dy(k) + 0.24D2y(k) = x(k)
y(k) = x(k) + y(k–1) – 0.24y(k–2) [2]
b) h(k) = (k) + h(k–1) – 0.24h(k–2)
h(0) = 1 [1] , h(1) = 1 [1] , h(2) = 0.76 [1] , h(3) = 0.52 [1]
1
(3)
–2
a)
D

D
a(k)

n
-3
x(n)
h(-n) 0.52
h(1-n)
h(2-n)
h(3-n)
2
2
-2
-1
0.76
0.52
1
0.76
0.52
0
1
1
1
0.76
0.52
2
1
0.5403
1
1
0.76
2
-0.4161
1
1
0.4
y(k)
0.6
(4)
(4)
4. a) y(k) = x(k) + y(k–1) – 0.24y(k–2)  Y(z) = X(z) + z-1Y(z) – 0.24z-2Y(z) [1]
Y(z) – z-1Y(z) + 0.24z-2Y(z) = X(z)  z2Y(z) – zY(z) + 0.24Y(z) = z2X(z)
(z2 – z + 0.24)Y(z) = z2X(z)  H(z) = Y(z)/X(z) = z2/(z2 – z + 0.24)
H(z)/z = z/(z2 – z + 0.24) = z/(z – 0.6)(z – 0.4) = 3/(z – 0.6) – 2/(z – 0.4) [1]
H(z) = 3z/(z – 0.6) – 2z/(z – 0.4)  h(k) = 3(0.6)k – 2(0.4)k, k  0. [2]
b) h(0) = 3(0.6)0 – 2(0.4)0 = 3 – 2 = 1 [½]
h(1) = 3(0.6)1 – 2(0.4)1 = 1.8 – 0.8 = 1 [½]
h(2) = 3(0.6)2 – 2(0.4)2 = 1.08 – 0.32 = 0.76 [½]
h(3) = 3(0.6)3 – 2(0.4)3 = 0.648 – 0.128 = 0.52 [½]
c) x(k) = cosk  x(0) = 1 [½] , x(1) = 0.5403 [½] , x(2) = –0.4161 [½] , x(3) = –0.9900 [½]
h(0) = 1, h(1) = 1, h(2) = 0.76, h(3) = 0.52
[30]
k
(4)
(2)
3
-0.99
1
y(0)=1 [2]
y(1)=1.5403 [2]
y(2)=0.8842 [3]
y(3)= –0.4755 [3]
(12)
d) H(z) = z /(z – z + 0.24) = z /(z – 0.6)(z – 0.4)
X(z) = (z2 – 0.5403z)/(z – 11)(z – 1–1) [2]
Y(z) = H(z)X(z) = [z2(z2 – 0.5403z)]/[(z – 0.6)(z – 0.4)(z – 11)(z – 1–1)]
Y(z)/z = [z(z2 – 0.5403z)]/[(z – 0.6)(z – 0.4)(z – 11)(z – 1–1)]
= A/(z – 0.6) + B/(z – 0.4) + C/(z – 11) + C*/(z – 1–1)
A = 0.6(0.62 – 0.54030.6)/(0.6 – 0.4) (0.6 – 11)(0.6 – 1–1) = 0.1510
B = 0.4(0.42 – 0.54030.4)/(0.4 – 0.6)(0.4 – 11)(0.4 – 1–1) = 0.1542
C = [11((11)2–0.540311)]/(11–1–1)(11–0.6)(11–0.4) = 0.6948–1.047
Y(z)/z=0.151/(z–0.6)+ 0.1542/(z–0.4)+0.6948–1.047/(z–11)+0.69481.047/(z–1–1)
Y(z)=0.151z/(z–0.6)+0.1542z/(z–0.4)+0.6948–1.047z/(z–11)+0.69481.047z/(z–1–1) [4]
y(k) = 0.151(0.6)k + 0.1542(0.4)k + 2×0.6948×(1k)cos(k – 1.047)
= 0.151(0.6)k + 0.1542(0.4)k + 1.3896cos(k – 1.047) [2]
y(0) = 0.151(0.6)0 + 0.1542(0.4)0 + 1.3896cos(0 – 1.047) = 1.000 [1]
y(1) = 0.151(0.6)1 + 0.1542(0.4)1 + 1.3896cos(1 – 1.047) = 1.5403 [1]
y(2) = 0.151(0.6)2 + 0.1542(0.4)2 + 1.3896cos(2 – 1.047) = 0.8839 [1]
y(3) = 0.151(0.6)3 + 0.1542(0.4)3 + 1.3896cos(3 – 1.047) = –0.4758[1]
(12)
Digital Signal Processing EIDSV4
Unit 2 First Assessment 22 March 2012
Page 1
Question 1 The block diagram of a system with
z–1
0.4
input x(k) and output y(k), is shown in Figure 1.
a) Determine a difference equation that describes
x(k)


the relationship between y(k) and x(k).
(4)
y(k)
b) Determine an expression for the system
function H(z) = Y(z)/X(z).
(4)
0.6
Figure 1
z–1
jT
c) Calculate the frequency response, H(e ), of
the system, at the following values of T:
i) T = 0
ii) T = 1
(5)
d) Determine an expression for the steady state output signal yss(k), if the input x(k)
to the system in Figure 1, is: i) x(k) = u(k) and ii) x(k) = sink.
(2)
Question 2 The transfer function
of a digital filter is given by: H() =
0
e j
for

    /2
  2   
for
0
for
/2   
The sampling period for the filter is T = 1 sec. (s = 2/T = 2).
a) Calculate the impulse response h(k), required for this digital filter.
b) From h(k), obtain values for h(–1), h(0) and h(1).
/2

(10)
(3)
Question 3 The eight (N = 8) sampled values x(0) = 0, x(1) = 1.25, x(2) = 2.5,
x(3) = 3.75, x(4) = 5.0, x(5) = 6.25, x(6) = 7.5
x(k)
7.50 8.75
and x(7) = 8.75, shown in Figure 2, were
6.25
5.0
obtained when a signal x(t) was sampled.
3.75
2.50
a) For the discrete Fourier transform X(n), of
1.25
k
this sequence, calculate the values of X(0)
and X(1).
(10)
0
Figure 2
b) Calculate the average value of x(t).
(1)
c) Calculate the amplitude A1, of the fundamental frequency X(1) contained in x(t).
(1)
Question 4 The ideal impulse response required for a digital filter, is given
sin[(1  k)  (]
. Design and draw the structure of a corresponding FIR
by h(k) = 1
2 [(1  k)  (
filter of length five (N=5). (Shift h(k) twice to the right and then truncate).
(10)
---oooOooo---
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
2
z X(z) - z2x(0) - zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N-1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
1
x(k) = ω
s
ω /2
 ωss /2 X()e
jkT
d , s = 2π
T
Total: 50
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
h(k) = c
, T= 2π
ωs
π (k -α)ωcT
Bilinear Transformation:
Ωo
z-1
s= 
ω
T
N
1
z
o
j(2

/N)nk
1
tan (
) +1
x(k) = N  X(n)e
2
n =0
j -j
j -j
j
Euler and trig. identities: re = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 2 Eerste Evaluasie 22 Maart 2012 Memorandum
Bladsy 1
1. a) a(k) = x(k)+0.6Da(k)  a(k) = x(k)/(1–0.6D) & y(k) = a(k)+0.4Dy(k)  y(k) = a(k)/(1–0.4D)
y(k) = x(k)/(1 – 0.6D)(1 – 0.4D)
0.4
D
(1 – 0.6D)(1 – 0.4D)y(k) = x(k)
2
y(k) – Dy(k) + 0.24D y(k) = x(k)
a(k)
x(k)
y(k)
y(k) = x(k) + y(k–1) – 0.24y(k–2) (4)


b) y(k) = x(k) + y(k–1) – 0.24y(k–2)
Y(z) = X(z) + z-1Y(z) – 0.24z-2Y(z) [1]
0.6
D
Y(z) – z-1Y(z) + 0.24z-2Y(z) = X(z)
2
2
z Y(z) – zY(z) + 0.24Y(z) = z X(z)
(z2 – z + 0.24)Y(z) = z2X(z)  H(z) = Y(z)/X(z) = z2/(z2 – z + 0.24)
c) i) H(ej0) = [(ej0)2]/[(ej0)2 – (ej0) + 0.24] = 1/(1 – 1 + 0.24) = 4.167
ii) H(ej1) = [(ej1)2]/[(ej1)2 – (ej1) + 0.24] = (11)2/[(11)2 – (11) + 0.24) = 1.3896–1.047
d) i) yss(k) = 4.167u(k)
(1)
ii) yss(k) = 1.3896sin(k – 1.047)
[15]
(4)
(2)
(3)
(1)
s /2
1 
jk
1 /2 j jk
jkT
2. a) h(k) = 1
H()e
d =
H()e
dω =
e e
d
2 
2 /2
s s /2
/2
1 /2 j  k)
1  e j1  k) 
1  e j(1  k)/2 e  j(1  k)/2  [4]
[3]
[3]

e
d = 
=
=



j(1  k) 
2  j(1  k)
2 /2
2  j(1  k) 


  /2

j(1  k)/2
 j(1  k)/2 
1 sin[(1  k) 2]
e
1  e
1
 =
(10)
sin[(1  k)/2 =
=
j2
2 (1  k) 2
(1  k) 
 (1  k)


b) i) h(–1) = 0.5[sin0/0] = 0.5 {or (1/2)(1/0 – 1/0) = ?}
(1)
ii) h(0) = 0.5[sin(/2)]/(/2) = 0.3183
(1)
{or (1/2)[(1/2/1/2) – (1–/2/1/2)] = (1/2)(1 – 1–) = 0.3183}
iii) h(1) = 0.5[sin(2/2)]/(2/2) = 0
(1)
{or (1/2)[(1/2/2) – (1–/2/2)] = (1/2)(1 – 1–) = (1/2)(–1+1) = 0}
[13]




N-1
3. X(n) =  x(k)e
- j(2/N)nk
- j(2/8)nk
k =0
k =0
[12]
7
=  x(k)e
7
=  x(k)e
- j(/4)nk
k =0
a) X(0) = 0 + 1.25 + 2.5 + 3.75 + 5 + 6.25 + 7.5 + 8.75 = 35 [3]
X(1) = 0e–j(/4)×1×0 + 1.25e–j(/4)×1×1 + 2.5e–j(/4)×1×2
+ 3.75e–j(/4)×1×3 + 5.0e–j(/4)1×4 + 6.25e–j(/4)1×5
+ 7.5e–j(/4)1×6 + 8.75e–1 j(/4)1×7
= 00 + 1.25– /4 + 2.5– /2 + 3.75– 3/4
+ 5– + 6.25–5/4 + 7.5–3/2 + 8.75–7/4
(10)
= 13.0661.9635 [7]
b) xav = X(0)/N = 35/8 = 4.375
c) A1 = |X(1)|/(N/2) = 13.066/4 = 3.267
x(k)
1.25
2.50
3.75
5.0
6.25
7.50
8.75
k
0
(1)
(1)
4. h(k)=½sin[(1+k)/2]/[(1+k)/2], therefore FIR filter of length 5: h(k)=½sin[(1+k–2)/2]/[(1+k–2)/2]
h(k) = ½sin[(k–1)/2]/[(k–1)/2], k = 0, 1, 2, 3, 4 [1]
h0) = 0.3183 [1] h(1) = 0.5 [1] h(2) = 0.3183 [1] h(3) = 0 [1] h(4) = –0.1061 [1]
x(k)
z-1
x(k)
0.3138
[10]
z-1
x(k-1)
0.5
+
z-1
x(k-2)
0.3183
+
[4]
z-1
x(k-3)
0
+
x(k-4)
–0.1061
+
y(k) = –0.3138 x(k)+ 0.5x(k–1)
+0.3183x(k–2) –0.1061x(k–4)
Digital Signal Processing EIDSV4
Unit 1 Final Assessment 3 May 2012
Question 1 A signal x(k) = [ –1 –2 0 2 1 ] is
shown in Figure 1.
a) Express x(k – 2) as the sum of weighted
and shifted impulse functions.
(3)
b) Sketch the following signals:
i) x(k + 1)u(k – 1).
(3)
ii) x(–k + 2)(k – 3).
(3)
iii) –x(k – 1) – (k – 1).
(3)
Question 2 Refer to the structure in Figure 2.
a) Determine a difference equation that
describes the relationship between the
output y(k) and the input x(k) of the
system.
(4)
b) Calculate the first four terms, h(0) to
h(3), of the impulse response h(k), of the
system.
(4)
Page 1
x(k)
2
–1
1
k
0
Figure 1
–2
z-1
x(k)

z-1
0.7

0.3
y(k)
Figure 2
Question 3 The difference equation that describes the relationship between the
output y(k) and the input x(k), of a system, is given by:
y(k) = x(k) + y(k–1) – 0.21y(k–2).
a) Determine the system’s transfer function H(z) = Y(z)/X(z), and from H(z), show
(4)
that the impulse response h(k) = Z–1{H(z)} = 1.75(0.7)k – 0.75(0.3)k, for k  0.
b) From h(k) in Question 3 a), calculate the first four terms, h(0) to h(3), of the
impulse response of the system.
(2)
c) Use graphical convolution methods to determine the response y(k) of the system
for k = 0, 1, 2 and 3, if the input to the system is the signal, x(k) = cos k u(k)
(calculating cosk for k = 0, 1, 2 and 3, radians must be used for k).
(12)
d) Use z transform methods to verify your results in Question 3 c), by finding y(k)
(12)
for k = 0, 1, 2 and 3, from Y(z) = H(z)X(z) and y(k) = Z-1{Y(z)}.
---oooOooo---
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
2
z X(z) - z2x(0) - zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N-1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
1
x(k) = ω
s

ωs /2
ωs /2
X()e
jkT
d , s = 2π
T
Total: 50
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
, T= 2π
h(k) = c
ωs
π (k -α)ωcT
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
)
tan (
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 1 Finale Evaluasie 3 Mei 2012 Memorandum
1.
a)
x(k) = –(k+2) – 2(k+1) + 2(k–1) + (k–2)
x(k–2) = –(k) – 2(k-1) + 2(k–3) + (k–4)
b) i)
[8]
k
0
2.
ii)
x(k+1)u(k–1)
1
[12]
Bladsy 1
(3)
x(–k+2)(k–3)
k
1
0
–2
(3)
–x(k–1)–(k–1)
iii)
k
0
–1
(3)
a(k) = x(k) + 0.3Da(k) [1]
a(k) = x(k)/(1 – 0.3D) ........... (1)
y(k) = a(k) + 0.7Dy(k) [1]
y(k) = a(k)/(1 – 0.7D) ........... (2)
(1) in (2):
x(k)
y(k) = x(k)/(1 – 0.3D)(1 – 0.7D)
(1 – 0.3D)(1 – 0.7D)y(k) = x(k)
y(k) – Dy(k) + 0.21D2y(k) = x(k)
y(k) = x(k) + y(k–1) – 0.21y(k–2) [2]
b) h(k) = (k) + h(k–1) – 0.21h(k–2)
h(0) = 1 [1] , h(1) = 1 [1] , h(2) = 0.79 [1] , h(3) = 0.58 [1]
2
a)
–2
D

D
a(k)

0.3
2
[30]
2
-2
-1
0.79
0.58
1
0.79
0.58
0
1
1
1
0.79
0.58
2
1
0.5403
1
1
0.79
2
-0.4161
1
1
0.7
y(k)
(4)
(4)
3. a) y(k) = x(k) + y(k–1) – 0.21y(k–2)  Y(z) = X(z) + z-1Y(z) – 0.21z-2Y(z) [1]
Y(z) – z-1Y(z) + 0.21z-2Y(z) = X(z)  z2Y(z) – zY(z) + 0.21Y(z) = z2X(z)
(z2 – z + 0.21)Y(z) = z2X(z)  H(z) = Y(z)/X(z) = z2/(z2 – z + 0.21)
H(z)/z = z/(z2 – z + 0.21) = z/(z – 0.7)(z – 0.3) = 1.75/(z – 0.7) – 0.75/(z – 0.3) [1]
H(z) = 1.75z/(z – 0.7) – 0.75z/(z – 0.3)  h(k) = 1.75(0.7)k – 0.75(0.3)k, k  0. [2]
b) h(0) = 1.75(0.7)0 – 0.75(0.3)0 = 1.75 – 0.75 = 1 [½]
h(1) = 1.75(0.7)1 – 0.75(0.3)1 = 1.225 – 0.225 = 1 [½]
h(2) = 1.75(0.7)2 – 0.75(0.3)2 = 0.8575 – 0.0675 = 0.79 [½]
h(3) = 1.75(0.7)3 – 0.75(0.3)3 = 0.60025 – 0.02025 = 0.58 [½]
c) x(k) = cosk  x(0) = 1 [½] , x(1) = 0.5403 [½] , x(2) = –0.4161 [½] , x(3) = –0.9900 [½]
h(0) = 1, h(1) = 1, h(2) = 0.79, h(3) = 0.58
n
-3
x(n)
h(-n) 0.58
h(1-n)
h(2-n)
h(3-n)
(3)
–1
(4)
(2)
3
-0.99
1
y(0)=1 [2]
y(1)=1.5403 [2]
y(2)=0.9142 [3]
y(3)= –0.399263 [3]
(12)
d) H(z) = z /(z – z + 0.21) = z /(z – 0.7)(z – 0.3)
X(z) = (z2 – 0.5403z)/(z – 11)(z – 1–1) [2]
Y(z) = H(z)X(z) = [z2(z2 – 0.5403z)]/[(z – 0.7)(z – 0.3)(z – 11)(z – 1–1)]
Y(z)/z = [z(z2 – 0.5403z)]/[(z – 0.7)(z – 0.3)(z – 11)(z – 1–1)]
= A/(z – 0.7) + B/(z – 0.3) + C/(z – 11) + C*/(z – 1–1)
A = 0.7(0.72 – 0.54030.7)/(0.7 – 0.3)(0.7 – 11)(0.7 – 1–1) = 0.266683
B = 0.3(0.32 – 0.54030.3)/(0.3 – 0.7)(0.3 – 11)(0.3 – 1–1) = 0.070601
C = [11((11)2–0.540311)]/(11–1–1)(11–0.7)(11–0.3) = 0.66709–1.051
Y(z)/z=0.266683/(z–0.7)+ 0.070601/(z–0.3)+0.66709–1.051/(z–11)+0.667091.051/(z–1–1)
Y(z)=0.266683z/(z–0.7)+ 0.070601z/(z–0.3)+(0.66709–1.051)z/(z–11)+(0.667091.051)z/(z–1–1) [4]
y(k) = 0.26683(0.7)k + 0.070601(0.3)k + 2×0.66709×(1k)cos(k – 1.051)
= 0.26683(0.7)k + 0.070601(0.3)k + 1.33418cos(k – 1.051) [2]
y(0) = 0.26683(0.7)0 + 0.070601(0.3)0 + 1.33418cos(0 – 1.051) = 1.0001 [1]
y(1) = 0.26683(0.7)1 + 0.070601(0.3)1 + 1.33418cos(1 – 1.051) = 1.5404 [1]
y(2) = 0.26683(0.7)2 + 0.070601(0.3)2 + 1.33418cos(2 – 1.051) = 0.91426 [1]
y(3) = 0.26683(0.7)3 + 0.070601(0.3)3 + 1.33418cos(3 – 1.051) = –0.39919[1]
(12)
Digital Signal Processing EIDSV4
Unit 2 Final Assessment 3 May 2012
Page 1
Question 1 The block diagram of a system with
x(k)
y(k)

input x(k) and output y(k), is shown in Figure 1.
a) Determine a difference equation that describes
z-1
the relationship between y(k) and x(k).
(2)
b) Determine an expression for the system
function H(z) = Y(z)/X(z).
(4)
z-1
– 0.24
Figure
1
c) Calculate the frequency response, H(ejT), of
the system, at the following values of T: i) T = 0
ii) T = 1
(4)
d) Determine an expression for the steady state output signal yss(k), if the input x(k)
to the system in Figure 1, is: i) x(k) = u(k) and ii) x(k) = sin k.
(2)
e) Plot the poles of H(z) on the z-plane, and illustrate the system’s transient response.
(3)
Question 2 The transfer function
of a digital filter is given by: H() =
0
e  j
for

  2
for
0
for
/2
The sampling period for the filter is T = 1 sec. (s = 2/T = 2).
a) Calculate the impulse response h(k), required for this digital filter.
b) From h(k), obtain values for h(–1), h(0) and h(1).
    /2
  
  
/2

(10)
(3)
Question 3 The eight (N = 8) sampled values x(0) = 8.75, x(1) = 7.5, x(2) = 6.25,
x(3) = 5.0, x(4) = 3.75, x(5) = 2.5, x(6) = 1.25
x(k)
and x(7) = 0, shown in Figure 2, were 8.75
7.50
obtained when a signal x(t) was sampled.
6.25
5.0
3.75
a) For the discrete Fourier transform X(n), of
2.50
1.25
this sequence, calculate the values of X(0)
0 k
and X(1).
(10)
0
Figure 2
b) Calculate the average value of x(t).
(1)
c) Calculate the amplitude A1, of the fundamental frequency X(1), contained in x(t).
(1)
Question 4 The ideal impulse response required for a digital filter, is given
sin[(k 1)  (]
. Design and draw the structure of a corresponding FIR
by h(k) = 1
2 [(k 1)  (
filter of length five (N=5). (Shift h(k) twice to the right and then truncate).
(10)
---oooOooo--Total: 50
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
2
z X(z) - z2x(0) - zx(1)
x(k)
(1/a)kaku(k)
sinku(k)
X(z)
z/(z - a)2
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2abkcos(k + )u(k)
(a z (a z

z b z b
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
, T= 2π
h(k) = c
ωs
π (k -α)ωcT
Fourier transforms:
Bilinear Transformation:

- jkT
jkT
1 ω /2
Ωo
X() =  x(k)e
x(k) = ω  ωs /2 X()e
d , s = 2π
z-1
T
s= 
s
s
k = -
ωoT z +1
tan (
)
N-1
2
- j(2 /N)nk
j(2 /N)nk
1 N-1
X(n) =  x(k)e
x(k) = N  X(n)e
n =0
k =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking EIDSV4 Eenheid 2 Finale Evaluasie 3 Mei 2012 Memorandum
Bladsy 1
1. a) y(k) = x(k) + z-1y(k) – 0.24z-2y(k)
x(k)
y(k)
y(k) = x(k) + y(k – 1) – 0.24y(k – 2)
(2)
b) y(k) = x(k) + y(k–1) – 0.24y(k–2)
z-1
Y(z) = X(z) + z-1Y(z) – 0.24z-2Y(z) [1]
Y(z) – z-1Y(z) + 0.24z-2Y(z) = X(z)
z2Y(z) – zY(z) + 0.24Y(z) = z2X(z)
z-1
– 0.24
(z2 – z + 0.24)Y(z) = z2X(z)
(4)
H(z) = Y(z)/X(z) = z2/(z2 – z + 0.24)
c) i) H(ej0) = [(ej0)2]/[(ej0)2 – (ej0) + 0.24] = 1/(1 – 1 + 0.24) = 4.167
(2)
j1
j1 2
j1 2
j1
2
2
ii) H(e ) = [(e ) ]/[(e ) – (e ) + 0.24] = (11) /[(11) – (11) + 0.24) = 1.3896–1.047 (2)
d) i) yss(k) = 4.167u(k) (1)
ii) yss(k) = 1.3896sin(k – 1.047)
(1)
2
e) H(z)= z /(z-0.4)(z-0.6)
0.6
0.4k
0.6k

0.4
[15]
(3)
s /2
/2  j jk
1 
jk
jkT
H()e
dω = 1
e
e
d
2. a) h(k) = 1
H()e
d =
2




2 /2
s s /2
/2
 j(k  1)/2 e  j(k  1)/2  [4]
 e j k  1) 
/2 j k  1)
[3]
[3]
1
1
=
e
d = 
= 1 e



j(k  1) 
2  j(k  1) 
2 /2
2  j(k  1)

 /2
 e j(k  1)/2  e  j(k  1)/2 
1
1 sin[(k  1)/2 = 1 sin[(k  1) 2]

 =
(10)
=
j2
2 (k  1) 2
(k  1) 
 (k  1)


b) i) h(–1) = 0.5[sin(–2/2)/( –2/2)] = 0.5[sin(–)/(–)] = 0
(1)
{or [1/(–22)][(1(–2/2)/1/2 – (1– (–2/2)/1/2)]
= (–1/4)[(1–/1/2) – (1/1/2)] = (–1/4)[1(–3/2) – 1(/2)] = 0 }
ii) h(0) = 0.5[sin(–/2)]/( –/2) = 0.50.63662 = 0.31831
(1)
{or (1/–2)[(1–/2/1/2) – (1– (–/2)/1/2)] = (–1/2)(1– – 10) = 0.3183}
[13]
iii) h(1) = 0.5[sin(0)/(0)] = 0.5
{or [1/(0)][(1/1/2) – (1/1/2)] = ?}
(1)
N-1
7
7
- j(2/N)nk
- j(2/8)nk
- j(/4)nk
3. X(n) =  x(k)e
=  x(k)e
=  x(k)e
k =0
k =0
k =0
x(k)
a) X(0) = 8.75 + 7.5 + 6.25 + 5 + 3.75 + 2.5 + 1.25 + 0 = 35 [3]
8.75
X(1) = 8.75e–j(/4)×1×0 + 7.5e–j(/4)×1×1 + 6.25e–j(/4)×1×2
7.50
–j(/4)×1×3
–j(/4)1×4
–j(/4)1×5
6.25
+ 3.75e
+ 2.5e
+ 5.0e
5.0
–j(/4)1×6
–1 j(/4)1×7
3.75
+ 0e
+ 1.25e
2.50
1.25
= 8.750 + 7.5– /4 + 6.25– /2 + 5– 3/4
0
k
+ 3.75– + 2.5–5/4 + 1.25–3/2 + 0–7/4
0
(10)
= 13.066–1.1781 [7]
b) xav = X(0)/N = 35/8 = 4.375
(1)
(1)
[12] c) A1 = |X(1)|/(N/2) = 13.066/4 = 3.267
4. h(k)=½sin[(k–1)/2]/[(k–1)/2], therefore FIR filter of length 5: h(k)=½sin[(k–2–1)/2]/[(k–2–1)/2]
h(k) = ½sin[(k–3)/2]/[(k–3)/2], k = 0, 1, 2, 3, 4 [1]
h0) = –0.1061 [1] h(1) = 0 [1] h(2) = 0.3183 [1] h(3) = 0.5 [1] h(4) = 0.3183 [1]
x(k)
[4]
z-1
z-1
z-1
z-1




x(k)
–0.1061
[10]
x(k-1)
0
+
x(k-2)
0.3183
+
x(k-3)
0.5
+
x(k-4)
0.3183
+
y(k) = –0.3138 x(k)+ 0.5x(k–1)
+0.3183x(k–2) –0.1061x(k–4)
Digital Signal Processing IV EIDSV4A Unit 1 First Assessment 28 February 2013
Page 1
Question 1 State Shannon’s sampling theorem.
Question 2 A signal x(k) = [ –1 2 1 ] is shown
in Figure 1. Sketch the following signals:
a) x(k + 1)u(k).
(3)
b) x(3 – k).
(3)
c) x(k) – (k).
(3)
(2)
x(k)
2
–1
1
0
k
Figure 1
Question 3 Calculate the values of the function x(k) = e j2k , for k = 0, 1, 2 and 3.
Sketch x(k) in the complex plane, for these values of k.
(5)
Question 4 The difference equation of a digital filter, with input x(k) and output y(k),
is given by:
y(k) = y(k–1) – 0.5y(k–2) + x(k)
a) Draw a block diagram for the system.
b) Determine the first four terms h(0) to h(3), of the impulse response of the system.
(4)
(2)
Question 5 The difference equation that describes the relationship between the
output y(k) and the input x(k), of a digital system, is given by:
y(k) = x(k) + y(k–1) – 0.5y(k–2).
a)
b)
c)
d)
Determine the system’s transfer function H(z) = Y(z)/X(z).
(2)
–1
(4)
From H(z), determine the impulse response, h(k) = Z {H(z)}, of the system.
From the impulse response h(k), calculate the first four terms, h(0) to h(3).
(4)
Use graphical convolution methods to determine the response y(k) of the system
for k = 0, 1, 2 and 3, if the input to the system is the signal, x(k) = u(k).
(8)
e) Use z transform methods to verify your results in Question 4 d), by finding y(k)
(10)
for k = 0, 1, 2 and 3, from Y(z) = H(z)X(z) and y(k) = Z –1{Y(z)}.
---oooOooo---
Total: 50
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z – 1)
z/(z – a)
z/(z – 1)2
z-1X(z)
z-2X(z)
zX(z) – zx(0)
2
z X(z) – z2x(0) – zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N -1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z – a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
1
x(k) = ω
s

ωs /2
ωs /2
X()e
jkT
d , s = 2π
T
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
h(k) = c
, T= 2π
π (k -α)ωcT
ωs
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
)
tan (
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
j
Euler and trig. identities: re = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking IV EIDSV4A Eenheid 1 Eerste Evaluasie 28 Februarie 2013 Memorandum
Bladsy 1
1. An analog signal containing components up to some maximum frequency, f1 hertz, may be completely
[2]
represented by regularly-spaced samples, provided the sampling rate is at least 2f1 samples per second.
2. a)
b)
c)
x(k)–(k)
2
x(k+1)u(k) 1
x(3–k)
1
1
1
k
k
k
(3)
3. x(k) = ej2k = 12k [1]
k = 0: x(0) = 10 [1]
k = 1: x(1) = 12r (2 rad  114) [1]
k = 2: x(2) = 14r (4 rad  229) [1]
[5]
k = 3: x(3) = 16r (6 rad  344) [1]
x(k)
y(k)
4. a)

[9]
[6]
5. a)
b)
c)
d)
(3)
–1
k=1
k=0
k=3
k=2
b) y(k) = y(k–1) – 0.5y(k – 2) + x(k)
h((k) = h(k–1) – 0.5h(k–2) + (k)
D
h(0) = h(–1) – 0.5h(–2) + (0) = 0 – 0 + 1 = 1 [½]
h(1) = h(0) – 0.5h(–1) + (1) = 1 – 0 + 0 = 1 [½]
h(2) = h(1) – 0.5h(0) + (2) = 1 – 0.5 + 0 = 0.5 [½]
D
–0.5
(4)
h(3) = h(2) – 0.5h(1) + (3) = 0.5 – 0.51 + 0 = 0 [½]
y(k) = x(k) + y(k–1) – 0.5y(k–2)  Y(z) = X(z) + z-1Y(z) – 0.5z-2Y(z) [1]
Y(z) – z-1Y(z) + 0.5z-2Y(z) = X(z)  z2Y(z) – zY(z) + 0.5Y(z) = z2X(z)
(z2 – z + 0.5)Y(z) = z2X(z)  H(z) = Y(z)/X(z) = z2/(z2 – z + 0.5) [1]
H(z)/z = z/(z2 – z + 0.5) = z/(z – 0.7071/4)(z – 0.7071–/4) [1]
= [0.7071–/4/(z – 0.7071/4)] + [0.7071–/4/(z – 0.7071–/4)] [2]
H(z) = [(0.7071–/4)z/(z – 0.7071/4)] + [(0.7071/4)z/(z – 0.7071–/4)]
h(k) = 20.7071(0.7071)kcos[(/4)k – /4] = 1.414(0.7071)kcos[(k – 1)/4], k  0. [1]
h(0) = 1.414(0.7070)0cos[(0 – 1)/4] = 1.4141cos(–/4) = 1.4140.7071 = 1 [1]
h(1) = 1.414(0.7070)1cos[(1 – 1)/4] = 1.4140.7071cos(0) = 11 = 1 [1]
h(2) = 1.414(0.7070)2cos[(2 – 1)/4] = 1.4140.5cos(/4) = 1.4140.50.7071 = 0.5 [1]
h(3) = 1.414(0.7070)3cos[(3 – 1)/4] = 1.4140.3536cos(/2) = 0.50 = 0 [1]
x(k) = u(k)  x(0) = 1, x(1) = 1, x(2) = 1, x(3) = 1 and h(0) = 1, h(1) = 1, h(2) = 0.5, h(3) = 0
n
x(n)
h(-n)
h(1-n)
h(2-n)
h(3-n)
2
[28]
(3)
–1
-3
-2
-1
0
0.5
0
1
0.5
0
0
1
1
1
0.5
0
1
1
1
1
0.5
2
1
1
1
(2)
(2)
(4)
(4)
3
1
1
y(0)=1 [2]
y(1)=2 [2]
y(2)=2.5 [2]
y(3)=2.5 [2]
[1]
(8)
e) H(z) = z /(z – 0.7071/4)(z – 0.7071–/4) (Q 5 b) and X(z) = Z{u(k)} = z/(z – 1)
Y(z)=H(z)X(z)=[z2/(z–0.7071/4)(z–0.7071–/4)][z/(z–1)]=z3/(z–1)(z–0.7071/4)(z–0.7071–/4)
Y(z)/z = z2/(z–1)(z–0.7071/4)(z–0.7071–/4) = A/(z–1)+B/(z–0.7071/4)+B*/(z–0.7071–/4)
A = z2/(z2 – z + 0.5)|z = 1 = 1/(1 – 1 + 0.5) = 2 [1]
B = z2/(z – 1)(z – 0.7071–/4)|z = 0.7071/4 = 0.7071–2.356 [1]
Y(z)/z = 2/(z – 1) + 0.7071–2.356/(z – 0.7071/4) + 0.70712.356/(z – 0.7071–/4)
Y(z) = [2z/(z–1)] + [(0.7071–2.356)z/(z–0.7071/4)] + [0.70712.356/(z–0.7071–/4)] [1]
y(k) = 2u(k)+2×0.7071×(0.7071k)cos[(/4)k–2.356)[2] = 2u(k)+1.414(0.7071k)cos[(/4)k–2.356)
y(0)=2+1.414(0.70710)cos[(/4)0–2.356]=2+1.4141cos(–2.356)=2+1.414(–0.707)=2-1=1 [1]
and y(1)=2+1.414(0.70711)cos[(/4)1–2.356]=2+1.4140.7071cos(0.7854–2.356)
=2+cos(–1.571)=2+0=2 [1]
and y(2)=2+1.414(0.70712)cos[(/4)2–2.356]=2+1.4140.5cos(1.571–2.356)
=2+0.7071cos(–0.785)=2+0.707107074=2.5[1]
and y(3)=2+1.414(0.70713)cos[(/4)3–2.356]=2+1.4140.3535cos(2.356–2.356)
=2+0.5cos(0)=2+0.51=2.5[1]
(10)
Digital Signal Processing IV EIDSV4A Unit 2 First Assessment 11April 2013
Page 1
x(k)
y(k)
Question 1 The block diagram of a system with
0.31
input x(k) and output y(k), is shown in Figure 1.
-1
z
a) Determine a difference equation that describes
the relationship between y(k) and x(k).
(2)
0.43
b) Determine an expression for the system’s
z-1
transfer function H(z) = Y(z)/X(z).
(4)
0.31
Figure 1
jT
c) Calculate the frequency response, H(e ), of the system, at
the following values of T: i) T = 0 ii) T = 0.4 iii) T = 0.5 iv) T =  (10)
d) If the input to this particular system is x(k) = sin(0.4k), determine an expression
(2)
for the steady state output signal yss(k).
Question 2 The transfer function of a low pass digital filter, H(), that aims to
extract the audio part from a mixed audio/data signal, is proposed in Figure 2. In the


H()
fundamental interval,  s    s , H() is defined as,
1
2
2
Figure 2
 0.213  s  ω  0.213  s
1
H() = 
s
s
0
ωs 
ω
 ω   0.213  s or 0.213  s  ω 
– s
0 
–0.213s
0.213s


2
2
2
2
where s is the sampling frequency. Assume a sampling period of T = 1 sec. (s = 2).
a) Determine an expression for the impulse response, h(k), required for this digital filter. (6)
(4)
b) Calculate h(–1), h(0) and h(1).
Question 3 The ten (N = 10) sampled values x(0) = 1.0, x(1) = 0.9, x(2) = 0.8, x(3) = 0.7,
x(k)
x(4) = 0.6, x(5) = 0.5, x(6) = 0.4, x(7) = 0.3
Figure 3
1.0 0.9
x(8) = 0.2 and x(9) = 0.1, shown in Figure 3,
0.8
0.7
were obtained when a signal x(t) was sampled,
0.6
0.5
with a sampling period of 0.1 sec.
0.4
0.3
a) For the discrete Fourier transform X(n), of this
0.2
0.1
(8)
sequence, calculate the values of X(0) and X(1).
b) Calculate the average value of x(t).
(2)
k
0
c) Calculate the amplitude A1 and the frequency of the component X(1), contained in x(t). (2)
Question 4 The ideal impulse response required for a digital filter, is given
by h(k) = 0.426
sin (0.426  k)
. Design and draw the structure of a corresponding FIR filter of
(0.426  k)
length five (N=5). (Shift h(k) twice to the right and then truncate, that is, let  = 2).
(10)
Total: 50
Appendix: Fourier and z transforms ---oooOooo--x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
2
z X(z) - z2x(0) - zx(1)
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
h(k) = c
, T= 2π
2ab cos(k + )u(k)
ωs
π (k -α)ωcT

- jkT
jkT
1 ω /2
Bilinear Transformation:
x(k) = ω  ωs /2 X()e
d , s = 2π
X() =  x(k)e
T
Ωo
s
s
z-1
k = -
s= 
ω
T
N-1
z
- j(2 /N)nk
j(2 /N)nk
1 N-1
tan ( o ) +1
x(k) = N  X(n)e
X(n) =  x(k)e
2
n =0
k =0
j e-j
j e-j
e
e
j
Euler and trig. identities: re = rcos + jrsin = r, cos =
, sin =
, cos = sin(+/2), sin = cos(-/2)
2
j2
k
(a z (a z

z b z b
Digitale Seinverwerking IV EIDSV4A Eenheid 2 Eerste Evaluasie 11 April 2013 Memorandum
Bladsy 1
1. a) y(k) = 0.31x(k) + 0.43x(k – 1) + 0.31x(k – 2)
(2)
x(k)
b) Y(z) = 0.31X(z) + 0.43z-1X(z) + 0.31z-2X(z)
0.31
-1
-2
Y(z) = (0.31 + 0.43z + 0.31z )X(z)
z-1
Y(z)/X(z) = H(z) = 0.31 + 0.43z-1 + 0.31z-2
(4)
c) i) H(ej0) = 0.31 + 0.43(ej0)-1 + 0.31(ej0)-2
0.43
= 0.31 + 0.43 + 0.31 = 1.05
(1)
z-1
ii) H(ej0.4) = 0.31 + 0.43(ej0.4)-1 + 0.31(ej0.4)-2
= 0.31 + 0.43–0.4 + 0.31–0.8
0.31
= 0.6216–1.257
(3)
iii) H(ej0.5)=0.31+0.43(ej0.5)-1+0.31(ej0.5)-2 = 0.31+0.43–0.5+0.31– = 0.43–1.571
iv) H(ej) = 0.31 + 0.43(ej)-1 + 0.31(ej)-2 = 0.31 + 0.43– + 0.31–2 = 0.190
[18]
d) T = 0.4: yss(k) = 0.6216sin(0.4k – 1.257)
+
y(k)
(3)
(3)
(2)
s /2
1 0.426
1 
jk
j k
jkT
2. a) h(k) = 1
H()e
dω =
1 e
d
H()e
d =
2





0
.
426


/2
2
s
s




j0.426k
 j0.426k 
e
1  e
1  e j0.426k e  j0.426k 
 [2]

 =

j2
jk
jk

k
2 






sin(0.426k)
1
=
(= 0.31831sin1.3383k)
(6)
sin(0.426k) [2] = 0.426
k
0.426k
b) h(–1) = 0.426[sin(-0.426)]/(-0.426) = 0.30975 [1] , h(0) = 0.426 [2]
(4)
and h(1)=0.426[sin(0.426)]/(0.426)=0.30975 [1]
1  e jk 
=


2  jk 

  0.426
[10]
N-1
3. X(n) =  x(k)e
- j(2/N)nk
=
9
=  x(k)e
- j(2/10)nk
k =0
k =0
[12]
[2]
9
=  x(k)e
- j(/5)nk
k =0
a) X(0) = 1.0 + 0.9 + 0.8 + 0.7 + 0.6 + 0.5 + 0.4 + 0.3 + 0.2 + 0.1 = 5.5 [2]
X(1) = 1.0e–j(/5)×1×0+0.9e–j(/5)×1×1+0.8e–j(/5)×1×2+0.7e–j(/5)×1×3
x(k)
+0.6e–j(/5)1×4+0.5e–j(/5)1×5+0.4e–1 j(/5)1×6+0.3e–j(/5)1×7
1.0 0.9
0.2e–1 j(/5)1×8+0.1e–1 j(/5)1×9 = 1.00+0.9– /5
0.8
+0.8–2/5+0.7–3/5 +0.6–4/5+0.5–5/5
+0.4–6/5+0.3–7/5+0.2–8/5+0.1–9/5
(8)
= 1.6180-1.256637 [6]
b) xav = X(0)/N = 5.5/10 = 0.55
(2)
c) A1 = |X(1)|/(N/2) = 1.6180/5 = 0.3236 [1]
0
(2)
and f1 = 1/10T = 1 Hz [1] (or 2 rad/sec)
0.7
0.6
0.5
0.4
0.3
0.2 0.1
k
sin (0.426  k)
sin [0.426  k - 2)]
, therefore FIR filter of length 5: h(k) = 0.426
, k = 0,..,4 [1]
(0.426  k)
[0.426  k - 2)
h0) = 0.071362 [1] h(1) = 0.309747 [1] h(2) = 0.426 [1] h(3) = 0.309747 [1] h(4) = 0.071362 [1]
4. h(k) = 0.426
x(k)
z-1
x(k)
0.07136
[10]
z-1
x(k-1)
z-1
x(k-2)
[4]
z-1
x(k-3)
x(k-4)
0.30975
0.426
0.30975
0.07136
+
+
+
+
y(k) = 0.07136x(k)+0.30975x(k–1)
+0.426x(k–2)+0.30975x(k–3)
+0.07136x(k–4)
Digital Signal Processing IV EIDSV4A
Unit 1 Final Assessment 9 May 2013
Page 1
Question 1 State the basic requirement that a linear system must satisfy.
Question 2 A signal x(k) = [ 1 2 1 ] is shown
in Figure 1. Sketch the following signals:
a) x(k)u(k) + x(k + 1)u(k)
(3)
b) x(–k + 2)
(3)
c) x(k) (k – 1)
(3)
(2)
x(k)
Figure 1
2
1
1
k
0
Question 3 Tabulate the values of the function x(k) = k e –k  [u(k) – u(k – 4)] , and
express x(k) as the sum of a series of weighted and shifted impulse functions.
(6)
Question 4 The difference equation of a digital filter, with input x(k) and output y(k),
is given by:
y(k) = 0.5y(k–1) – 0.25y(k–2) + x(k)
a) Draw a block diagram for the system.
b) Determine the first four terms h(0) to h(3), of the impulse response of the system.
(4)
(2)
Question 5 The difference equation that describes the relationship between the
output y(k) and the input x(k), of a digital system, is given by:
y(k) = x(k) + 0.5y(k–1) – 0.25y(k–2).
a)
b)
c)
d)
Determine the system’s transfer function H(z) = Y(z)/X(z).
From H(z), determine the impulse response, h(k) = Z –1{H(z)}, of the system.
From the impulse response h(k), calculate the first four terms, h(0) to h(3).
Use graphical convolution methods to determine the response y(k) of the system
for k = 0, 1, 2 and 3, if the input to the system is the signal, x(k) = u(k).
e) Use z transform methods to verify your results in Question 4 d), by finding y(k)
for k = 0, 1, 2 and 3, from Y(z) = H(z)X(z) and y(k) = Z –1{Y(z)}.
---oooOooo---
(2)
(4)
(4)
(8)
(9)
Total: 50
Appendix: z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
z2X(z) - z2x(0) - zx(1)
Fourier transforms:

- jkT
X() =  x(k)e
k = -
N-1
- j(2 /N)nk
X(n) =  x(k)e
k =0
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
1
x(k) = ω
s

ωs /2
ωs /2
X()e
jkT
d , s = 2π
T
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
, T= 2π
h(k) = c
π (k -α)ωcT
ωs
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
tan (
)
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking IV EIDSV4A Eenheid 1 Finale Evaluasie 9 Mei 2013 Memorandum
[2]
Bladsy 1
1. A linear system must obey the principle of superposition. If the system responds with y1(k) to an input x1(k)
(2)
and with y2(k) to an input x2(k), then y1(k)+y2(k) must be the response to the input x1(k)+x2(k).
2. a)
b)
c)
x(k)(k–1)
x(–k+2)
x(k)u(k)+x(k+1)u(k) 3
2
1
1
1
1
k
k
k
[9]
(3)
3. a)
[6]
4. a)
x(k)
[6]
5. a)
b)
c)
d)
k
x(k)
0
0 [1]
1
0.3679 [1]
2
0.2707 [1]
(3)
3
0.1494 [1]
(3)
b) x(k) = 0.3679(k–1) + 0.2707(k–2)
+0.149(k–3)
(2)
(4)
b) y(k) = 0.5y(k–1) – 0.25y(k – 2) + x(k)
h((k) = 0.5h(k–1) – 0.25h(k–2) + (k)
h(0) = 0.5h(–1) – 0.25h(–2) + (0) = 0 – 0 + 1 = 1 [½]
D
h(1) = 0.5h(0) – 0.25h(–1) + (1) = 0.51 – 0 + 0 = 0.5 [½]
0.5
h(2) = 0.5h(1) – 0.25h(0) + (2) = 0.50.5 – 0.251 + 0 = 0 [½]
h(3) = 0.5h(2) – 0.25h(1) + (3) = 0.50 – 0.250.5 + 0 = –0.125 [½]
D
–0.25
(4)
(2)
-1
-2
[1]
y(k) = x(k) + 0.5y(k–1) – 0.25y(k–2)  Y(z) = X(z) + 0.5z Y(z) – 0.25z Y(z)
Y(z) – 0.5z-1Y(z) + 0.25z-2Y(z) = X(z)  z2Y(z) – 0.5zY(z) + 0.25Y(z) = z2X(z)
(z2 – 0.5z + 0.25)Y(z) = z2X(z)  H(z) = Y(z)/X(z) = z2/(z2 – 0.5z + 0.25) [1]
(2)
2
[1]
H(z)/z = z/(z – 0.5z + 0.25) = z/(z – 0.5/3)(z – 0.5–/3)
= [0.577351–/6/(z – 0.5/3)] + [0.577351–/6/(z – 0.5–/3)] [2]
H(z) = [(0.577351–/6)z/(z – 0.5/3)] + [(0.577351/6)z/(z – 0.5–/3)]
h(k) = 20.577351(0.5)kcos[(/3)k – /6] = 1.1547(0.5)kcos[1.0472k – 0.5236], k  0. [1]
(4)
0
[1]
h(0) = 1.1547(0.5) cos[0 – 0.5236] = 1.15471cos(–0.5236) = 1.15470.86602  1
h(1) = 1.1547(0.5)1cos[1.0472 – 0.5236] = 1.15470.5cos(0.5236) = 0.577350.86602  0.5 [1]
h(2) = 1.1547(0.5)2cos[2.0944 – 0.5236] = 1.15470.25cos(1.5708)  0.288680 = 0 [1]
h(3) = 1.1547(0.5)3cos[3.1416 – 0.5236] = 1.15470.125cos(2.618)  –0.125 [1]
(4)
x(k) = u(k)  x(0) = 1, x(1) = 1, x(2) = 1, x(3) = 1 and h(0) = 1, h(1) = 0.5, h(2) = 0, h(3) = –0.125
y(k)

n
-3
x(n)
h(-n) –0.125
h(1-n)
h(2-n)
h(3-n)
2
-2
-1
0
–0.125
0.5
0
–0.125
0
1
1
0.5
0
–0.125
1
1
1
0.5
0
2
1
1
0.5
3
1
1
y(0)=1 [2]
y(1)=1.5 [2]
y(2)=1.5 [2]
y(3)=1.375 [2]
[1]
(8)
e) H(z) = z /(z – 0.5/3)(z – 0.5–/3) (Q 5 b) and X(z) = Z{u(k)} = z/(z – 1)
Y(z)=H(z)X(z)=[z2/(z – 0.5/3)(z – 0.5–/3)][z/(z–1)] = z3/(z–1)(z–0.5/3)(z–0.5–/3)
Y(z)/z = z2/(z–1)(z–0.5/3)(z–0.5–/3) = A/(z–1)+B/(z–0.5/3)+B*/(z–0.5–/3)
A = z2/(z2 – 0.5z + 0.25)|z = 1 = 1/(1 – 0.5 + 0.25) = 1.333 [1]
B = z2/(z – 1)(z – 0.5–/3)|z = 0.5/3 = 0.3333–2.0944 [1]
Y(z)/z = 1.3333/(z – 1) + 0.3333–2.0944/(z – 0.5/3) + 0.33332.0944/(z – 0.5–/3)
Y(z) = [1.3333z/(z–1)] + [(0.3333–2.0944)z/(z–0.5/3)] + [0.33332.0944/(z–0.5–/3)] [1]
y(k) = 1.3333u(k)+2×0.3333×(0.5k)cos[(/3)k–2.0944)[1] =1.3333u(k)+0.6667(0.5k)cos[(/3)k–2.0944)
[27]
y(0) = 1.3333 + 0.6667(0.50)cos[(/3)0–2.0944] = 1.3333 + 0.66671cos(–2.0944)
= 1.3333 + 0.6667(–0.5) = 1.3333 – 0.3333 = 1 [1]
and y(1) = 1.3333 + 0.6667(0.51)cos[(/3)1–2.0944] = 1.3333 + 0.66670.5cos(1.0473–2.0944)
= 1.3333 + 0.3333cos(–1.0471) = 1.3333 + 0.16667 = 1.49997  1.5 [1]
and y(2) = 1.3333 + 0.6667(0.52)cos[(/3)2–2.0944] = 1.3333+0.66670.25cos(2.0944–2.0944)
=1.3333 + 0.166675cos(0) = 1.3333 + 0.1666751 = 1.499975  1.5 [1]
and y(3) = 1.3333+0.6667(0.53)cos[(/3)3–2.0944] = 1.3333+0.66670.125cos(3.1416–2.0944)
= 1.3333 + 0.083338cos(1.0472) = 1.3333 + 0.0833380.49999=1.3749681.375[1]
(9)
Digital Signal Processing IV EIDSV4A Unit 2 Final Assessment 9 May 2013
Page 1
Question 1 The output y(k) of a digital filter, with input x(k), is given by:
y(k) = 0.07x(k) + 0.31x(k–1) + 0.43x(k–2) + 0.31x(k–3) + 0.07x(k–4)
a) Determine an expression for the system’s transfer function H(z) = Y(z)/X(z).
(3)
jT
b) Calculate the frequency response, H(e ), of the system, at the following values of T:
ii) T = 0.4
iii) T = 0.43
iv) T = 0.5 v) T = 
(10)
i) T = 0
c) If the input to this particular system is x(k) = sin(0.43k), determine an expression
(2)
for the steady state output signal yss(k).
Figure 1
H()
Question 2 The transfer function of a
1

band pass filter H(), that aims to
extract the data section from a mixed
0
ωs
ωs
–0.275

0.225

0.275

–
s –0.225s
s
s
audio/data signal, is proposed in Figure 1.
2
2
In the fundamental interval, – 0.5ωs    0.5ωs , H() is defined as,
0
H() = 
 1
when  0.5ω  ω   0.275ω or  0.225ω  ω  0.225ω or 0.275ω  ω  0.5ω
s
s
s
s
s
s
when  0.275ω  ω   0.225ω or 0.225ω  ω  0.275ω
s
s
s
s
where s is the sampling frequency. Assuming a sampling period of T = 1 sec (s=2),
determine an expression for the impulse response, h(k), required for this digital filter.
(10)
Question 3 The sixteen (N = 16) sampled values, x(0) = 1, x(1) = 1, x(2) = 1, x(3) = 1
x(4) = 1, x(5) = 1, x(6) = 1, x(7) = 1, x(8) = 0, x(9) = 0, x(10) = 0, x(11) = 0 x(12) = 0
x(13) = 0, x(14) = 0 and x(15) = 0, as shown in
x(k)
Figure 2
Figure 2, were obtained when a signal x(t) was
1
1
1
1
1
1
1
1
sampled with a sampling period of 0.0625 sec.
a) For the discrete Fourier transform X(n), of this
k
sequence, calculate the values of X(0) and X(1). (12)
0
b) Calculate the average value of x(t), the amplitude
A1, as well as the frequency of the fundamental component X(1), contained in x(t).
(3)
Question 4 The ideal impulse response required for a digital filter, is given by,
h(k) = 0.426
sin (0.426  k)
. Design and draw the structure of a corresponding FIR filter of length
(0.426  k)
seven (N=7). (Shift h(k) three positions to the right and then truncate, that is, let =3).
---oooOooo---
(10)
Total: 50
Appendix: Fourier and z transforms
Convolution:
k
k
(k)
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
u(k)
n 0
n 0
sinku(k)
for x(k) = 0, k < 0
aku(k)

ku(k)
y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
2 -z(cos
z
x(k-1)
cosku(k)
n 0
x(k-2)
(z e j )(z e j )
FIR low-pass filter:
x(k+1)
ω T sin(k -α)ωcT
(a z (a z
k
, T= 2π
h(k) = c

2ab cos(k + )u(k)
x(k+2)
ωs
π (k -α)ωcT
z b z b

- jkT
jkT
1 ω /2
Bilinear Transformation:
X() =  x(k)e
x(k) = ω  ωs /2 X()e
d , s = 2π
T
Ωo
s
s
z-1
k = -
s= 
ω
T
N-1
z
N
1
- j(2 /N)nk
j(2 /N)nk
1
tan ( o ) +1
X(n) =  x(k)e
x(k) = N  X(n)e
2
n =0
k =0
j e-j
j e-j
e
e
j
Euler and trig. identities: re = rcos + jrsin = r, cos =
, sin =
, cos = sin(+/2), sin = cos(-/2)
2
j2
x(k)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
z2X(z) - z2x(0) - zx(1)
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
z(sin)
j
(z e  )(z e j )
Digitale Seinverwerking IV EIDSV4A Eenheid 2 Finale Evaluasie 9 Mei 2013 Memorandum
Bladsy 1
1. a) y(k) = 0.07x(k) + 0.31x(k–1) + 0.43x(k–2) + 0.31x(k–3) + 0.07x(k–4)
Y(z) = 0.07X(z) + 0.31z-1X(z) + 0.43z-2X(z) + 0.31z-3X(z) + 0.07z-4X(z) [1]
Y(z) = (0.07 + 0.31z-1 + 0.43z-2 + 0.31z-3 + 0.07z-4)X(z)
H(z) = Y(z)/X(z) = 0.07 + 0.31z-1 + 0.43z-2 + 0.31z-3 + 0.07z-4 [1]
(3)
j0
–j0
–j0
–j0
–j0
b) i) H(e ) = 0.07+ 0.31e + 0.43e + 0.31e + 0.07e = 0.07 + 0.31 + 0.43 + 0.31 + 0.07 = 1.19 (2)
ii) H(ej0.4) = 0.07 + 0.31(ej0.4)–1 + 0.43(ej0.4)–2 + 0.31(ej0.4)–3 + 0.07(ej0.4)–4
(2)
= 0.07 + 0.31–0.4 + 0.43–0.8 + 0.31–1.2 + 0.07–1.6 = 0.50833–2.5133
j0.43
j0.43 –1
j0.43 –2
j0.43 –3
j0.43 –4
iii) H(e
) = 0.07 + 0.31(e
) + 0.43(e
) + 0.31(e
) + 0.07(e
)
= 0.07 + 0.31–0.43 + 0.43–0.86 + 0.31–1.29 + 0.07–1.72 = 0.43857–2.7018 (2)
iv) H(ej0.5) = 0.07 + 0.31(ej0.5)–1 + 0.43(ej0.5)–2 + 0.31(ej0.5)–3 + 0.07(ej0.5)–4
= 0.07 + 0.31–0.5 + 0.43– + 0.31–1.5 + 0.07–2 = 0.29– (= –0.29)
(2)
v) H(ej) = 0.07 + 0.31(ej)–1 + 0.43(ej)–2 + 0.31(ej)–3 + 0.07(ej)–4
= 0.07 + 0.31– + 0.43–2 + 0.31–3 + 0.07–4 = –0.05
(2)
[15] c) yss(k) = 0.43857sin(k – 2.7018)
(2)
s /2
jkT
1 H()
2. s=2/T=2 [1] & h(k)= 1
H()e
d

s s /2
0

ωs
ωs
jk
h(k) = 1
H()e d
–0.275s –0.225s
0.225s 0.275s
–
 
2
2


0.2252 
0.2722
j k
(
1)e d + 1
(1)e jk d [3]
= 1
 0.2752
 0.2252


0.45
0.55
 j0.45k
 j0.55k
j0.55k
j0.45k
 e jk 
 e jk 
e
e
e
e
[3]


=
+
=
–
+
–
jk
jk
jk
jk
 jk 
 jk 
 0.55 
 0.45

j0.55k
j0.45k
- j0.45k
 j0.55k
e
e
1 e
1 e
1
1
=
–
=
sin0.55k –
sin0.45k
k
[10]
k
j2
= 0.55
sin0.55k
0.55k
N-1
– 0.45
3. a) X(n) =  x(k)e
sin0.45k
0.45k
- j(2/N)nk
(= 0.55
15
sin1.72788k
=  x(k)e
k =0
k =0
k
j2
172788k
– 0.45
- j(2/16)nk
[2]
k
sin1.41371k
141371k
15
=  x(k)e
) [1]
- j(/8)nk
k =0
X(0) = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 = 8 [2]
X(1) = 1e–j(/8)×1×0+1e–j(/8)×1×1+1e–j(/8)×1×2+1e–j(/8)×1×3+1e–j(/8)×1×4+1e–j(/8)×1×5+1e–j(/8)1×6+1e–j(/8)1×7
x(k)
+ 0e–j(/8)1×8 + 0e–j(/8)1×9 + 0e–j(/8)1×10 + 0e–j(/8)1×11
–j(/8)1×12
–1 j(/8)1×13
–1 j(/8)1×14
–1 j(/8)1×15
1
1
1 1 1 1 11
+ 0e
+ 0e
+ 0e
+ 0e
[15]
= 10+1–/8+1–/4+1–3/8+1–/2+1–5/8+1–3/4+1–7/8+0–
+0–9/8+0–5/8+0–11/8+0–3/2+0–13/8+0–7/4+0–15/8
= 5.125831–1.374447
[1]
[10]
[1]
(12)
k
0
b) xav=X(0)/N=8/16=0.5 , A1=|X(1)|/(N/2)=5.125831/8=0.640729 , f1=1/NT=1/(160.0625)=1Hz [1]
(3)
sin (0.426  k)
sin [0.426  k  3)]
, therefore FIR filter of length 7: h(k) = 0.426
, k = 0,..,6
4. h(k) = 0.426
(0.426  k)
[0.426  k  3)
h0) = –0.08133 [1] h(1) = 0.07136 [1] h(2) = 0.30975 [1] h(3) = 0.426 [1] h(4) = 0.30975 [1]
h(5) = 0.07136 [1] h(6) = –0.08133 [1]
x(k)
[3]
y(k) = 0.08133x(k)
z-1
z-1
z-1
z-1
z-1
z-1
+0.07136x(k–1)
x(k-5)
x(k-6)
x(k)
x(k-1)
x(k-2)
x(k-3)
x(k-4)
+0.30975x(k–2)
+0.426x(k–3)
-0.08133 0.07136 0.30975
0.426
0.30975 0.07136 -0.08133
+0.30975x(k–4)
+0.07136x(k–5)
[10]
+
+
+
+
+
+
+0.08133x(k–6)
Digital Signal Processing IV EIDSV4A Unit 1 First Assessment 27 February 2014
Question 1 A signal x(k) = [ 1 2 –1 1 1 ] is
shown in Figure 1. Sketch the following signals:
a) x(k – 1)u(k).
(3)
b) x(k – 1) – (k – 1).
(3)
c) x(3 – k)u(3 – k).
(4)
Page 1
x(k)
1
2
0
1
1
–1
k
Figure 1
Question 2 Sketch the values of the function x(k) = e(k jk) in the complex plane,
for k = 0, 1 and 2.
Question 3 Refer to the structure in Figure 2. x(k)
a) Determine a difference equation that
describes the relationship between the
output y(k), and the input x(k), of the
system.
(4)
b) Calculate the first four terms, h(0) to h(3),
of the system’s impulse response h(k).
(4)
+
z –1
(4)
y(k)
+
z–1
– 0.25
Figure 2
Question 4 The difference equation that describes the relationship between the
output y(k) and the input x(k), of a digital system, is given by:
y(k) = x(k–1) + y(k–1) – 0.25y(k–2).
a) Determine the system’s transfer function H(z) = Y(z)/X(z).
(4)
–1
(4)
b) From H(z), determine the impulse response, h(k) = Z {H(z)}, of the system.
c) From the impulse response h(k), obtained in Question 4 b), calculate and confirm
that h(0) = 0, h(1) = 1, h(2) = 1 and h(3) = 0.75
(2)
d) Use graphical convolution methods to determine the response y(k) of the system
for k = 0, 1, 2 and 3, if the input to the system is the signal, x(k) = (0.8)ku(k),
with the impulse response h(k) as was calculated in Question 4 c).
(8)
e) Use z transform methods to verify your results in Question 4 d), by finding y(k)
(10)
for k = 0, 1, 2 and 3, from Y(z) = H(z)X(z) and y(k) = Z –1{Y(z)}.
Appendix: Fourier and z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
2
z X(z) - z2x(0) - zx(1)

- jkT
 x(k)e
k = -
N-1
- j(2 /N)nk
X(n) =  x(k)e
k =0
X() =
---oooOooo---
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
1
x(k) = ω
s
ω /2
 ωss /2 X()e
jkT
d , s = 2π
T
Total: 50
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
, T= 2π
h(k) = c
ωs
π (k -α)ωcT
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
tan (
)
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
j
Euler and trig. identities: re = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking IV EIDSV4A Eenheid 1 Eerste Evaluasie 27 Februarie 2014 Memorandum
Bladsy 1
1. a)
b)
c)
x(k–1)u(k)
x(3–k)u(3–k)
x(k–1) – (k–1)
2
1
2
1
1
k
1
1
1
k
1
k
–1
–1
(3)
[10]
(3)
–2
2. x(k) = e(k+jk) = ekejk = ekk [1]
k = 0: x(0) = e00 = 10 [1]
k = 1: x(1) = e11 = 2.7181r (1 rad  57) [1]
[4]
k = 2: x(2) = e22 = 7.3892r (2 rad  114) [1]
3.
a)
x(k)

x(k)-¼y(k-1)
D

k=2
2.718
7.389
y(k)=y(k-1)+x(k-1)-¼y(k-2)
2
k=1
1
1 k=0
Re
b) h(k) = (k-1) + h(k-1) – ¼h(k-2)
h(0)=(-1)+h(-1)–¼h(-2)=0+0-0 = 0 [1]
h(1)=(0)+h(0)- ¼h(-1)=1+0-0 = 1 [1]
y(k-1)
h(2)=(1)+h(1)- ¼h(0)=0+1-0=1 [1]
(4)
h(3)=(2)+h(2)- ¼h(1)=0+1-¼=¾ [1] (4)
-¼
y(k) = x(k-1) + y(k-1) – ¼y(k-2)
4. a) y(k) = x(k–1) + y(k–1) – 0.25y(k–2)  Y(z) = z–1X(z) + z-1Y(z) – 0.25z-2Y(z) [2]
Y(z) – z-1Y(z) + 0.25z-2Y(z) = z-1X(z)  z2Y(z) – zY(z) + 0.25Y(z) = zX(z)
(z2 – z + 0.25)Y(z) = zX(z)  H(z) = Y(z)/X(z) = z/(z2 – z + 0.25) [2]
b) H(z) = z/(z2 – z + 0.25) = z/(z – 0.5)(z – 0.5) [2] = z/(z – 0.5)2
h(k) = (1/0.5)k(0.5)k = 2k(0.5)k, k  0. [2]
c) h(0) = 20(0.5)0 = 201 = 0 [½]
h(1) = 21(0.5)1 = 210.5 = 1 [½]
h(2) = 22(0.5)2 = 22¼ = 1 [½]
h(3) = 23(0.5)3 = 23(1/8) = 6/8 = ¾ [½]
d) x(k) = (0.8)ku(k)  x(0) = 1, x(1) = 0.8, x(2) = 0.64, x(3) = 0.512
and h(0) = 0, h(1) = 1, h(2) = 1, h(3) = 0.75
n
-3
x(n)
h(-n) 0.75
h(1-n)
h(2-n)
h(3-n)
[28]
Im
D
-¼y(k-1)
[8]
x(k-1)-¼y(k-2)
(4)
-2
-1
1
0.75
1
1
0.75
0
1
0
1
1
0.75
1
0.8
0
1
1
2
0.64
0
1
(4)
(4)
(2)
3
0.512
0
y(0)=0 [2]
y(1)=1 [2]
y(2)=1.8 [2]
y(3)=2.19 [2]
(8)
e) H(z) = z/(z – 0.5)2 (Q 4 b) and X(z) = Z{(0.8)ku(k)} = z/(z – 0.8) [1]
Y(z)=H(z)X(z)=[z/(z – 0.5)2][z/(z–0.8)] = z2/(z – 0.5)2(z – 0.8)
Y(z)/z = z/(z – 0.5)2(z – 0.8) = –1.6667/(z – 0.5)2 – 8.8889/(z – 0.5) + 8.8889/(z – 0.8) [3]
Y(z) = –1.6667z/(z – 0.5)2 – 8.8889z/(z – 0.5) + 8.8889z/(z – 0.8)
y(k) = –1.6667(1/0.5)k(0.5)k – 8.8889(0.5)k + 8.8889(0.8)k [2]
= –3.3334k(0.5)k – 8.8889(0.5)k + 8.8889(0.8)k k  0
y(0) = –3.33340(0.5)0 – 8.8889(0.5)0 + 8.8889(0.8)0
= –3.333401 – 8.88891 + 8.88891 = 0 [1]
and y(1) = –3.33341(0.5)1 – 8.8889(0.5)1 + 8.8889(0.8)1
= –3.33340.5 – 8.88890.5 + 8.88890.8 = –1.6667 – 4.4445 + 7.1111 = 0.9999  1 [1]
and y(2) = –3.33342(0.5)2 – 8.8889(0.5)2 + 8.8889(0.8)2
= –6.66680.25 – 8.88890.25 + 8.88890.64 = –1.6667 – 2.2222 + 5.6889 = 1.8 [1]
and y(3) = –3.33343(0.5)3 – 8.8889(0.5)3 + 8.8889(0.8)3
= –100.125 – 8.88890.125 + 8.88890.512 = –1.25 – 1.1111 + 4.5511 = 2.19 [1]
(10)
Digital Signal Processing IV EIDSV4A
Unit 2 First Assessment 27 March 2014
Page 1
Question 1 The block diagram of a system with
x(k)
y(k)
input x(k) and output y(k), is shown in Figure 1.
a) Determine a difference equation that describes
z-1
z-1
the relationship between y(k) and x(k).
(3)
b) Determine and confirm that the system function
–1
1
H(z) = Y(z)/X(z) = z(z – 1)/(z2 – z + 0.64). (3)
z-1
c) Plot the poles of H(z) on the z plane and
determine the system’s impulse response h(k). (5)
Figure 1
– 0.64
d) Using H(z), calculate the steady state response
yss(k), if the input to the system is given by x(k) = sin(0.8957k). (5)
Question 2 A device that will delay the phase of all incoming frequencies by 90, must
have a frequency transfer function, as shown in Figure 2, that is defined by:
+

 j
H() = 
j


s
2
 ω 
0  ω 
0
s
H()
where j =
-1
2
ω
– s
2
–j
j
0
ωs
2
Figure 2

Assuming that the sampling frequency s = 1, determine an expression for the
impulse response, h(k), of such a 90 phase shifter (also called a Hilbert transformer). (10)
Question 3 The ten (N = 10) sampled values x(0) = 0.1, x(1) = 0.2, x(2) = 0.3, x(3) = 0.4,
x(4) = 0.5, x(5) = 0.6, x(6) = 0.7, x(7) = 0.8
Figure 3
x(k)
1.0
0.9
x(8) = 0.9 and x(9) = 1.0, shown in Figure 3,
0.8
0.7
were obtained when a signal x(t) was sampled,
0.6
0.5
with a sampling period of T = 0.1 sec.
0.4
0.3
a) For the discrete Fourier transform X(n), of this
0.2
sequence, calculate the values of X(0) and X(1). (8) 0.1
k
0
b) Calculate the average value of x(t).
(2)
c) Calculate the amplitude A1 and the frequency f1 of the component X(1), contained in x(t). (2)
Question 4 The ideal impulse response required for a certain digital high-pass filter, is
sin (  k)
sin (0.45  k)
given by:
h(k) =
– 0.45
( k)
(0.45  k)
Design and draw the structure of a corresponding FIR filter of length five (N = 5).
(Shift h(k) twice to the right by calculating h(k–2) and then truncate the sequence.)
(12)
Appendix: Fourier and z transforms ---oooOooo--Total: 50
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
z2X(z) - z2x(0) - zx(1)
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2abkcos(k + )u(k)
(a z (a z

z b z b
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
h(k) = c
, T= 2π
ωs
π (k -α)ωcT
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
tan (
)
2

- jkT
jkT
1 ω /2
x(k) = ω  ωs /2 X()e
d , s = 2π
 x(k)e
T
s
s
k = -
N-1
- j(2 /N)nk
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
X(n) =  x(k)e
n =0
k =0
j -j
j -j
j
Euler and trig. identities: re = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
X() =
Digitale Seinverwerking IV EIDSV4A Eenheid 2 Eerste Evaluasie 27 March 2014 Memorandum
Bladsy 1
x(k)
1. a) y(k) = x(k) – x(k–1) + y(k–1) – 0.64y(k–2)
(3)

b) Y(z) = X(z) – z-1X(z) + z-1Y(z) – 0.64z-2Y(z) [1]
z2Y(z) = z2X(z) – zX(z) + zY(z) – 0.64Y(z)
D
Y(z)(z2 – z + 0.64) = z(z – 1)X(z)
H(z) = Y(z)/X(z) = z(z – 1)/(z2 - z + 0.64) [2]
(3)
1
-1
2
[1]
c) z – z + 0.64 = 0  z = 0.8±0.8957
H(z)/z = (z–1)/(z–0.80.8957)(z–0.8-0.8957)
= 0.640520.67514/(z–0.80.8957)
-0.64
[1]
+0.64052–0.67514/(z–0.8–0.8957)
h(k) = 1.281(0.8)kcos(0.8957k+0.67514) [3]
(5)
d) H(0.8957) = H(ej0.8957) = H(10.8957)
= 10.8957(10.8957– 1)/(10.8957)2 – 10.8957 + 0.64) = 3.06960.536415 [4]
[16]
yss(k) = 3.0696sin(0.8957k + 0.53642) [1]
1 s /2
1 /2
2. h(k) = 
H()e jkT d =
H()e jk2 d
s s /2
1 1/2


h(k) =

 1/2
je jk2d +
/2
0
 je jk2d
[2]
D
D
(5)
, with s = 1 and T = 2/s = 2.
= (j/j2k)[ej2k] 0½ – (j/j2k)[ej2k] ½
[2]
y(k)
[2]
0
= (1/2k)[e0 – ej(–1/2)2k] – [(1/2k)[ej(1/2)2k – e0] = (1/2k)[1 – e–jk – ejk + 1] [2]
= 2/2k – [(ejk + e–jk)/2k] = 1/k – (1/k)[(ejk + e–jk)/2]
= (1/k)[1 – cosk] [2]
[10]
N-1
9
9
k =0
k =0
k =0
3. X(n) =  x(k)e j(2/N)nk =  x(k)e j(2/10)nk =  x(k)e j(/5)nk
a) X(0) = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 + 0.8 + 0.9 + 1.0 = 5.5 [2]
X(1) = 0.1e–j(/5)×1×0+0.2e–j(/5)×1×1+0.3e–j(/5)×1×2+0.4e–j(/5)×1×3
+0.5e–j(/5)1×4+0.6e–j(/5)1×5+0.7e–1 j(/5)1×6+0.8e–j(/5)1×7
x(k)
1.0
–1 j(/5)1×8
–1 j(/5)1×9
0.9
0.9e
+1.0e
= 0.10+0.2– /5
0.8
0.7
+0.3–2/5+0.4–3/5 +0.5–4/5+0.6–5/5
0.6
0.5
+0.7–6/5+0.8–7/5+0.9–8/5+1.0–9/5
0.4
[6]
0.3
= 1.6180341.884956
(8)
0.2
b) xav = X(0)/N = 5.5/10 = 0.55
(2)
0.1
c) A1 = |X(1)|/(N/2) = 1.618034/5 = 0.3236068 [1]
k
0
and f1 = 1/10T = 1 Hz [1] (or 2 rad/sec)
(2)
[12]
hideal(k)
sin (  k)
sin (0.45  k)
– 0.45
4. h Ideal HPF (k) =
.
( k)
(0.45  k)
Therefore FIR filter approximation of length 5:
h(k) =
1
0.55
sin [ (k  2)]
sin [0.45  k  2)]
, k = 0,..,4 [2]
– 0.45
[ (k  2)]
[0.45  k  2)]
-0.04918
-0.31439
h(k)
[4]
-1
-1
-1
z
z
x(k-1)
x(k-2)
x(k-3)
–0.04918 –0.31439
+
0.55
+
0.55
k
x(k-4)
–0.31439 –0.04918
+
k
-0.31439
-1
z
z
x(k)
[12]
0.45
-0.04918
h(0) = –004918 [1] h(1) = –0.31439 [1] h(2) = 0.55 [2]
h(3) = –0.31439 [1] h(4) = –004918 [1]
x(k)
(1/k)sink
0.45(1/0.45k)sin0.45k
+
-0.04918
-0.04918
-0.31439
-0.31439
y(k) = –0.04918x(k)–0.31439x(k–1)
+0.55x(k–2)–0.31439x(k–3)
–0.04918x(k–4)
Digital Signal Processing IV EIDSV4A
Unit 1 Final Assessment 8 May 2014
Page 1
x(k)
Question 1 A signal x(k) = [ 1 2 –1 1 1 ] is
shown in Figure 1. Sketch the following signals:
a) x(k – 1)u(k + 1).
(3)
b) x(k + 1)(k – 1).
(3)
c) x(2 – k)u(k – 2).
(4)
1
2
1
–1
1
k
Figure 1
Question 2 A sequence y(k) is defined by the difference equation:
y(k+2) + 0.5y(k) = sin k
with y(0) = 1 and y(1) = –1.
Write down the values of y(2) and y(3).
Question 3 Refer to the system
in Figure 2. Determine the
impulse response, h(k), of the
system and calculate the first
four terms, h(0) to h(3), of the
system’s impulse response, h(k)
x(k)
(4)
+
+
y(k)
z–1
(7)

1
2

1
4
Figure 2
k
Question 4 The impulse response of a digital filter is given by h(k) = 0.8 u(k). The
k
input to the system is the signal, x(k) = 0.5 u(k).
a) Use graphical convolution methods to determine the response y(k) = x(k) * h(k) of
the system, for k = 0, 1, 2 and 3.
b) Use z transform methods to verify your results in Question 4 a), by finding y(k)
for k = 0, 1, 2 and 3, from Y(z) = H(z)X(z) and y(k) = Z –1{Y(z)}. (Confirmation
must be independent and without assuming any answers from Question 4 a)
Question 5 Use z transform methods to determine expressions for y(k) from the
following difference equations:
a) y(k) = 0.25y(k – 2) + (k)
b) y(k) = 0.25y(k – 2) + cos k
Appendix: Fourier and z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
2
z X(z) - z2x(0) - zx(1)
---oooOooo---
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
(7)
(7)
(6)
(9)
Total: 50
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
, T= 2π
h(k) = c
ωs
π (k -α)ωcT
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z+1
tan (
)
2

- jkT
jkT
1 ω /2
x(k) = ω  ωs /2 X()e
d , s = 2π
 x(k)e
T
s
s
k = -
N-1
- j(2 /N)nk
j(2 /N)nk
1 N-1
X(n) =  x(k)e
x(k) = N  X(n)e
n =0
k =0
j -j
j -j
j
Euler and trig. identities: re = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
X() =
Digitale Seinverwerking IV EIDSV4A Eenheid 1 Finale Evaluasie 8 Mei 2014 Memorandum
1. a)
1
b)
x(k–1)u(k+1)
2
1 1
c)
x(k+1)(k–1)
x(2–k)u(k–2)
2
1
k
1
k
–1
[10]
Bladsy 1
k
–1
(3)
(3)
(4)
2. y(k+2) + 0.5y(k) = sin k y(0) = 1, y(1) = –1
set k = 0: y(2) + 0.5y(0) = sin 0  y(2) + 0.5 = 0  y(2) = –0.5 [2]
set k = 1: y(3) + 0.5y(1) = sin 1  y(3) – 0.5 = 0.8415  y(3) = 1.3415 [2]
[4]
3. a(k) = x(k) – 0.5z-1a(k)  a(k) = x(k)/(1+0.5z-1)
x(k)
y(k) = a(k) – 0.25z-1a(k) = (1 – 0.25z-1)a(k)
-1
-1
= [(1 – 0.25z )/(1 + 0.5z )]x(k)
y(k) + 0.5z-1y(k) = x(k) – 0.25z-1x(k)
y(k) = x(k) – 0.25x(k – 1) – 0.5y(k – 1) [3]
h(0) = 1 [1] h(1) = –0.75 [1] h(2) = 0.375 [1] h(3) = –0.1875 [1]
[7]
4. a) h(k)=0.8ku(k) x(k)=0.5ku(k)
h(0)=1
x(0)=1
h(1)=0.8
x(1)=0.5
h(2)=0.64
x(2)=0.25
h(3)=0.512 x(3)=0.125
[14]
a(k)
+
+
z–1
1
2
y(k)
1
4
n
-3
-2
-1
0
1
2
3
x(n)
1
0.5 0.25 0.125
h(-n) 0.512 0.64 0.8
1
y(0)=1 [1]
h(1-n)
0.512 0.64 0.8
1
y(1)=1.3 [2]
h(2-n)
0.512 0.64 0.8
1
y(2)=1.29 [2]
h(3-n)
0.512 0.64 0.8
1
y(3)=1.157 [2]
b) H(z) = z/(z–0.8) [½] X(z) = z/(z–0.5) [½]
Y(z) = z2/(z–0.8)(z–0.5)  Y(z)/z = z/(z–0.8)(z–0.5) = 2.667/(z0.8) – 1.667/(z–0.5)
Y(z) = 2.667z/(z–0.8) – 1.667z/(z–0.5)  y(k) = 2.667×0.8k – 1.667×0.5k [2]
y(0) = 1 [1] , y(1) = 1.3 [1] , y(2) = 1.29 [1] & y(3) = 1.157 [1]
or: H(z) = 1+0.8z-1+0.64z-2+0.512z-3+ ... and X(z) = 1+0.5z-1+0.25z-2+0.125z-3+ ...
Y(z) = H(z)X(z) = (1+0.8z-1+0.64z-2+0.512z-3)(1+0.5z-1+0.25z-2+0.125z-3)
(7)
(7)
=1+0.5z-1+0.25z-2+0.125z-3 + 0.8z-1+0.4z-2+0.2z-3+0.1z-4 + 0.64z-2+0.32z-3+0.16z-4+0.08z-5 + 0.512z-3+0.256z-4+0.128z-5+0.064z6
= 1 + 1.3z-1 + 1.29z-2 + 1.157z-3 + .......  y(0) = 1 , y(1) = 1.3 , y(2) = 1.29 & y(3) = 1.157
5. a) y(k) = 0.25y(k – 2) + (k)  Y(z) = 0.25z-2Y(z) + 1 [2]  z2Y(z) – 0.25Y(z) = z2
Y(z)(z–0.5)(z+0.5) = z2  Y(z)/z = z/(z–0.5)(z+0.5)  Y(z)/z = 0.5/(z–0.5) + 0.5/(z+0.5) [2]
Y(z) = 0.5z/(z – 0.5) + 0.5z/(z+0.5)  y(k) = 0.5(0.5)k + 0.5(–0.5)k [2]
(6)
-2
2
[2]
b) y(k) = 0.25y(k – 2) + cos k  Y(z) = 0.25z Y(z) + (z – zcos1)/(z–11)(z–1–1)
z2Y(z) – 0.25Y(z) = z2(z2 – 0.5403z)/(z–11)(z–1–1)
Y(z)(z–0.5)(z+0.5) = z2(z2 – 0.5403z)/(z-11)(z–1–1)
Y(z)/z = z(z2 – 0.5403z)/(z–0.5)(z+0.5)(z–11)(z–1–1)
Y(z)/z = –0.0142/(z–0.5)+0.1453/(z+0.5)+0.4436–0.2031/(z–11)+0.44360.2031/(z–1–1)[4]
Y(z)= –0.0142z/(z–0.5)+0.1453z/(z+0.5)+0.4436–0.2031z/(z–11)+0.44360.2031z/(z–1–1)
[15]
y(k) = –0.0142(0.5)k + 0.1453(–0.5)k + 0.8872cos(k – 0.2031) [3]
(9)
Digital Signal Processing IV EIDSV4A Unit 2 Final Assessment 8 May 2014
Page 1
Question 1 A digital system with input x(k) and output y(k), is shown in Figure 1.
a) Determine a difference equation that describes
x(k)
y(k)
the relationship between y(k) and x(k).
(3)
b) Determine H(z) = Y(z)/X(z).
(3)
z-1
c) Plot the poles of H(z) on the z plane and
1
determine the system’s impulse response h(k).
(5)
Figure
1
d) Using H(z), calculate the steady state response
z-1
yss(k), if the input to the system is given by
– 0.64
x(k) = sin(0.8957k).
(5)
+
Question 2 The
a digital band
shown in Figure
interval – 0.5ωs
defined as:
0
H() = 
1
transfer function of
pass filter, H(), is
2. In the fundamental
   0.5ωs , H() is
Figure 2
H()
1
–0.5s –0.3s –0.231s

0
0.231s
0.3s
0.5s
when  0.5ωs  ω   0.3ωs or  0.231ωs  ω  0.231ωs or 0.3ωs  ω  0.5ωs
when  0.3ωs  ω   0.231ωs or 0.231ωs  ω  0.3ωs
where s is the sampling frequency. Assuming a sampling period of T = 1 sec (s=2),
determine an expression for the impulse response, h(k), required for this digital filter. (10)
Question 3 The twelve (N = 12) sampled values x(0) = 1.0, x(1) = 1.366, x(2) = 1.366,
x(3) = 1.0, x(4) = 0.366, x(5) = –0.366,
x(k)
x(6) = –1.0, x(7) = –1.366, x(8) = –1.366,
Figure 3
1.366 1.366
x(9) = –1.0, x(10) = –0.366 and x(11) = 0.366, 1.0
1.0
0.366
shown in Figure 3, were obtained when a
0.366
k
signal x(t) was sampled, with a sampling
–0.366
0
–0.366
period of T = 0.1 sec. For the discrete
–1.0
–1.0
Fourier transform X(n), of this sequence,
–1.366 –1.366
(10)
a) Calculate the value of X(1)
b) Calculate the amplitude A1 and the frequency f1 of the component X(1), contained in x(t) (2)
Question 4 The ideal impulse response required for a digital band pass filter, is
given by, h(k) = 0.6
sin(0.6k)
sin(0.462k)
– 0.462
. Design and draw the structure
(0.6k)
(0.462k)
of a corresponding FIR filter of length seven (N=7).
Appendix: Fourier and z transforms ---oooOooo--x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
z2X(z) - z2x(0) - zx(1)

- jkT
 x(k)e
k = -
N-1
- j(2 /N)nk
X(n) =  x(k)e
k =0
X() =
x(k)
(1/a)kaku(k)
sinku(k)
X(z)
z/(z - a)2
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2ab cos(k + )u(k)
(a z (a z

z b z b
k
1
x(k) = ω
s
ω /2
 ωss /2 X()e
jkT
d , s = 2π
T
(12)
Total: 50
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0

y(k)=  h(n)x(k  n) , for x(k)  0, k < 0
n 0
FIR low-pass filter:
ω T sin(k -α)ωcT
, T= 2π
h(k) = c
π (k -α)ωcT
ωs
Bilinear Transformation:
Ωo
z-1
s= 
ωoT z +1
)
tan (
2
j(2 /N)nk
1 N-1
x(k) = N  X(n)e
n =0
j -j
j -j
j
Euler and trig. identities: re = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking IV EIDSV4A Eenheid 2 Finale Evaluasie 8 Mei 2014 Memorandum
1. a) y(k) = x(k) + y(k–1) – 0.64y(k–2)
(3)
x(k)
-1
-2
[1]
b) Y(z) = X(z) + z Y(z) – 0.64z Y(z)

z2Y(z) = z2X(z) + zY(z) – 0.64Y(z)
Y(z)(z2 – z + 0.64) = z2X(z)
H(z) = Y(z)/X(z) = z2/(z2 - z + 0.64) [2]
(3)
c) z2 – z + 0.64 = 0  z = 0.8±0.8957 [1]
H(z)/z = z/(z–0.80.8957)(z–0.8-0.8957)
= 0.64049–0.6751/(z–0.80.8957)
+0.640490.6751/(z–0.8–0.8957)
[1]
h(k) = 1.281(0.8)kcos(0.8957k–0.6751) [3]
(5)
d) H(0.8957) = H(ej0.8957) = H(10.8957)
= (10.8957)2/(10.8957)2 – 10.8957 + 0.64) = 3.5444-0.58653 [4]
[16]
yss(k) = 3.5333sin(0.8957k – 0.58653) [1]
s /2
jkT
2. s=2/T=2 [1] & h(k)= 1
H()e
d
s s /2

jk
ω
h(k) = 1
H()e d
– s
 
2
–0.3s
0.2312 
0.32 
jk
jk
= 1
(
1)e
d + 1
(
1)e
d
 0.32 
 0.2312 


0.462 
y(k)
D
1
D
-0.64
–0.231s
0
0.231s
[10]
= 0.6
k
0.6 k
N-1
– 0.462
3. a) X(n) =  x(k)e
k =0
11
sin0.462 k
ωs
2
j2
k
[2]
k
[1]
0.462 k
 j(2/N)nk
=  x(k)e
0.3s
0.6 
j2
sin0.6 k

[3]
 j0.462 k
 j0.6 k
j0.6 k
j0.462 k
 e jk 
 e jk 
e
e
e
e
[3]


=
+
=
–
+
–
jk
jk
jk
jk
 jk 
 jk 
0.6 
0.462 
j0.6 k
j0.462 k
- j0.462 k
 j0.6 k
1 e
1 e
1
1
e
e
–
=
sin0.6k –
sin0.462k
=
k
(5)
1 H()


Bladsy 1
11
=  x(k)e
 j(2/12)nk
k =0
 j(/6)nk
x(k)
1.366 1.366
1.0
1.0
k =0
X(1) = 1e–j(/6)×1×0+1.366e–j(/6)×1×1+1.366e–j(/6)×1×2
+1.0e–j(/6)×1×3+0.366e–j(/6)×1×4–0.366e–j(/6)×1×5
–1.0e–j(/6)1×6–1.366e–j(/6)1×7–1.366e–j(/6)1×8
–1.0e–j(/6)1×9–0.366e–j(/10)1×10 + 0.366e–j(/6)1×11
0
0.366
–0.366
–1.0
–1.366
0.366
k
–0.366
–1.0
–1.366
= 1.00+1.366–/6+1.366–2/6+1.0–3/6+0.366–4/6–0.366–5/6–1.0–6/6–1.366–7/6
(10)
–1.366–8/6–1.0–9/6–0.366–10/6+0.366–11/6 = 8.485157–0.785398163
[12] b) A1=|X(1)|/(N/2)= 8.485157/6=1.408595 [1], f1=1/NT=1/(120.1)=0.8333 Hz [1]
(2)
sin0.6 k
sin0.462 k
sin0.6  k - 3)
sin0.462  k - 3)
4. h(k)=0.6
–0.462
. Realizable filter of length 7: h(k)=0.6
–0.462
0.6  k - 3)
0.462  k - 3)
0.6 k
0.462 k
[10]
k = 0,..,6  h0) = 0.03701 [1] h(1) = –0.13119 [1] h(2) = –0.01331 [1] h(3) = 0.138 [1]
h(4) = –0.01331 [1] h(5) = –0.13119 [1] h(6) = 0.03701 [1]
y(k) = 0.03701x(k)
[3]
z-1
z-1
z-1
z-1
z-1
z-1
–0.13119x(k–1)
–0.01331x(k–2)
x(k-5)
x(k-6)
x(k)
x(k-1)
x(k-2)
x(k-3)
x(k-4)
+0.138x(k–3)
0.03701 -0.13119 -0.01331
0.138
-0.01331 -0.13119 0.03701
–0.01331x(k–4)
–0.13119x(k–5)
y(k)
+0.03701x(k–6)
+
+
+
+
+
+
Digital Signal Processing IV EIDSV4A Unit 1 First Assessment 26 February 2015
Question 1 A signal x(k) = [ 1 1 0 1 2 3 ]
is shown in Figure 1.
a) Express x(k) in terms of weighted and
shifted impulse functions.
(3)
b) Express x(1–k) in terms of weighted and
shifted impulse functions.
(3)
c) Sketch the function x(3–k)[u(k–2) – u(k–5)]. (4)
x(k)
1
1
1
0
2
Page 1
3
k
Figure 1
Question 2 Sketch the values of the function x(k) = e  jk in the complex plane,
for k = 0, 1 and 2.
(3)
Question 3 The difference equation that describes the relationship between the
output y(k) and the input x(k), of a digital system, is given by:
1
5
y(k) = x(k) + y(k–1) – y(k–2)
6
6
a) Draw a block diagram of the system.
b) Determine the system’s transfer function H(z).
c) Determine the system’s impulse response h(k).
d) Calculate h(0) and h(1).
(3)
(4)
(4)
(2)
Question 4 The impulse response of a system is given by h(k) = 0.8k , for k  0, and
the input by x(k) = cosk, for k  0 and k in radians. Use graphical convolution
methods to determine the response, y(k), of the system for k = 0, 1, 2 and 3.
(12)
Question 5 The impulse response of a system is given by h(k) = 0.8k , for k  0, and
the input by x(k) = cosk, for k  0 and k in radians. Use z transform methods and the
relationship Y(z) = H(z)X(z), to find an expression for the output y(k), for k  0.
(12)
---oooOooo---
Total: 50
Appendix: Fourier and z transforms
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
z2X(z) - z2x(0) - zx(1)
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2abkcos(k + )u(k)
(a z (a z

z b z b

jk
- jkT
1
1 
X() =  x(k)e
x(k) = 2   X()e
d = ω
s
k = -
N-1
N
1
- j(2 /N)nk
j(2

/N)nk
1
X(n) =  x(k)e
x(k) = N  X(n)e
n =0
k =0
ω /2
ωss/2 X()e
jkT
d
2π
s =
T
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0
FIR Low pass filter:
ω T sin(k -α)ωcT
,T= 2π
h(k)= c
ωs
π (k -α)ωcT
Ωo
z-1
Bilinear Transform: s = 
ωoT z +1
tan (
)
2
Diverse:
ax
ax
ax
 x e ax dx xea  ea 2  e a x 1a


Lim sin x
1
x 0 x


Lim cos x  1
 ½
x 0
x2
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
Digitale Seinverwerking IV EIDSV4A Eenheid 1 Eerste Evaluasie 26 Februarie 2015 Memorandum
Bladsy 1
[½]
[½]
[½]
1. a) x(k) = (k+2) +(k+1) +0(k)
x(k)
x(1–k)
3
3
+(k–1)[½]+2(k–2)[½]+3(k–3)[½]
(3)
2
2
1 1
1
k
1
1 1 k
b) x(1–k) = 3(k+2)[½] +2(k+1)[½]
[½]
[½]
[½]
+(k) +(k–2) +(k–3)
(3)
0
0
{or x(1–k) = (3–k)+(2–k)+(–k)+2(–1–k)+3(–2–k) = (k–3)+(k–2)+(k)+2(k+1)+3(k+2)
as we may conclude from the definition of the  function, that (x) = (–x)}
c)
u(k–2)–u(k–5)
x(3–k)
x(3–k)[u(k–2)–u(k–5]
3
2
k
k
1
1 1 k
1 1 1
1
1
0
0
0
[10]
(4)
Im
2. x(k) = e– jk = 1–k
k = 0: x(0) = 10 [1]
k = 1: x(1) = 1–1r (1 rad  57) [1]
[3]
k = 2: x(2) = 1–2r (2 rad  114) [1]
3. a)
x(k)

[1]
[½]
[½]
5/6
(5/6)y(k-1)
[½]
[½]
D
k=0
k=2
y(k)=x(k)
+(5/6)y(k–1)
–(1/6)y(k–2)
D
-1/6
(3)
–(1/6)y(k–2)
[13]
4. x(k) = cosk
h(k) = 0.8k
n
-3
x(n)
h(-n) 0.512
h(1-n)
h(2-n)
h(3-n)
-2
-1
0.64
0.512
0.8
0.64
0.512
Re
k=1
b) Y(z) = X(z)+(5/6)z-1Y(z)–(1/6)z-2Y(z) [2]
z2Y(z) = z2X(z)+ (5/6)zY(z)–(1/6)Y(z)
[z2–(5/6)z+(1/6)]Y(z) = z2X(z)
H(z)=Y(z)/X(z)=z2/[z2–(5/6)z+(1/6)] [2] (4)
c) H(z)/z = z/[z2 – (5/6)z + (1/6)]
= z/[z – (1/2)][z – (1/3)]
= 3/[z – (1/2)] – 2/[z – (1/3)]
H(z) = 3{z/[z – (1/2)]} – 2{z/[z – (1/3)]}
h(k) = 3(1/2)k – 2(1/3)k , k  0
(4)
[1]
[1]
h(1) = 5/6 { = 0.83333}
(2)
d) h(0) = 1
{or h(0) = (0) + (5/6)h(–1) – (1/6)h(–2) = 1
& h(1) = (1) + (5/6)h(0) – (1/6)h(–1) = 5/6}
0
1
2
3
1
0.5403 –0.4161 -0.99
1
0.8
1
0.64
0.8
1
0.512 0.64
0.8
1
y(0)=1 [3]
y(1)=1.340 [3]
y(2)=0.6561 [3]
y(3)= –0.4651[3]
[12]
5. h(k) = 0.8k  H(z) = z/(z–0.8) [1]
x(k) = cosk  X(z) = (z2–zcos1)/(z–ej)(z–e-j) [1]
Y(z) = H(z)X(z) = z(z2 – 0.5403z)/(z–0.8)(z–ej)(z–e-j) [2] { = z(z2-0.5403z)/(z-0.8)(z2 – 1.081z + 1)}
Y(z)/z = (z2 – 0.5403z)/(z–0.8)(z–ej)(z–e-j) = A/(z–0.8) + B/(z–ej) + B*/(z–e-j)
A = (0.82 – 0.54030.8)/(0.82 – 1.0810.8 + 1) = 0.268
B = ((11)2 – 0.540311)/(11–0.8)(11–1–1) = 0.5678–0.8701
Y(z)/z = 0.268/(z–0.8) + 0.5678–0.8701/(z–11) + 0.56780.8701/(z–1–1) [4]
Y(z) = 0.268z/(z–0.8) + 0.5678–0.8701z/(z–11) + 0.56780.8701z/(z–1–1)
y(k) = 0.268(0.8)k + 20.56781kcos(k–0.8701)
[12]
= 0.268(0.8)k + 1.136cos(k–0.8701) [4]
{just as a check: y(0) = 0.268(0.8)0 + 1.136cos(0 – 0.8701) = 1
y(1) = 0.268(0.8)1 + 1.136cos(1–0.8701) = 1.341
y(2) = 0.268(0.8)2 + 1.136cos(2–0.8701) = 0.6563
y(3) = 0.268(0.8)3 + 1.136cos(3–0.8701) = –0.4653}
Digital Signal Processing IV EIDSV4A Unit 2 First Assessment 26 March 2015 Page 1
Question 1 The relation between the input x(k) and the output y(k) of a certain
“moving average exponential” filter, is described x(k)
y(k)
0.5
by the block diagram in Figure 1, as well as by the

difference equation: y(k) = 0.5x(k) + 0.5y(k – 1)
z-1
a) Determine the frequency response H(ejT) of
0.5
Figure 1
this filter, at the following frequencies:
3

i) T = 0 ii) T = 
(12)
4 iii) T = 2 iv) T = 4 and v) T = 
b) Draw a rough graph of |H(ejT)| versus T, and indicate the type of filter
action provided by this moving average filter.
(2)
Question 2 A digital filter is required to have a frequency characteristic, shown
(10)
H()
in Figure 2. The transfer function H() is defined by:
Figure 2
–
     1
 0
1 
 
H() = 



0
1    0
0    1
1    
(–s/2)
(s/2)

0
–

–1
1
Assuming a sampling period of T = 1 (s = 2), calculate the impulse response h(k), required for
this filter. You may use the result from integral calculus:

 e

a  d e a  e a  e a 1
a
a
a
a2



Question 3 The twelve (N = 12) sampled values x(0) = 1.2, x(1) = 1.1, x(2) = 1.0, x(3) = 0.9,
x(k)
x(4) = 0.8, x(5) = 0.7, x(6) = 0.6, x(7) = 0.5, x(8) = 0.4
1.1
1.2
Figure 3
x(9) = 0.3, x(10) = 0.2 and x(11) = 0.1, shown in
1.0
0.9
0.8
Figure 3, were obtained when a signal x(t) was
0.7
0.6
sampled, with a sampling period of T = 0.08333 sec.
0.5
0.4
a) For the discrete Fourier transform X(n), of this
0.3
0.2
0.1
sequence, calculate the values of X(0) and X(1). (10)
b) Estimate the average value of x(t).
(2)
11 k
0
c) Calculate the approximate r.m.s value of the first harmonic contained in x(t).
(2)
Question 4 The ideal impulse response required for a certain digital filter, is given by: (12)
(cos k)1
Design and draw the structure of a corresponding FIR filter of
h(k) = 1 sin k  1
 k
 k2
Lim 
 1 = – 0.5
length seven (N = 7). You may use the result from the theory of limits: x  0  (cos x)

 x2

---oooOooo--Appendix: Fourier and z transforms
Total: 50
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
z2X(z) - z2x(0) - zx(1)
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2abkcos(k + )u(k)
(a z (a z

z b z b
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0
FIR Low pass filter:
ω T sin(k -α)ωcT
,T= 2π
h(k)= c
ωs
π (k -α)ωcT
Ωo
z-1
Bilinear Transform: s = 
ωoT z +1
tan (
)
2
Diverse:
ax
ax
ax
 x e ax dx xea  ea 2  e a x 1a



jk
jkT
- jkT
1 ω /2
1 
X() =  x(k)e
x(k) = 2   X()e
d
d = ω ωs /2 X()e
s
s
k = -
2π
Lim sin x
Lim cos x  1
s =
N-1
1
 ½
T
- j(2/N)nk
j(2/N)nk
1 N-1
x 0 x
x 0
X(n) =  x(k)e
x(k) = N  X(n)e
x2
n =0
k =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2


Digitale Seinverwerking IV EIDSV4A Eenheid 2 Eerste Evaluasie 26 March 2015 Memorandum
Bladsy 1
1. a) y(k) = 0.5x(k)+0.5y(k–1)  Y(z)=0.5X(z)+0.5z –1Y(z)
b)
|H()|
–1
–1
Y(z)–0.5z Y(z) = 0.5X(z)  Y(z)(1–0.5z ) = 0.5X(z)
1
–1
[2]
Y(z)/X(z) = 0.5/(1–0.5z )  H(z) = 0.5z/(z–0.5)
0.5
i) H(e-j0) = 0.5ej0/(ej0–0.5) = 0.5/(1–0.5) = 0.5/0.5 = 10 [2]
ii) H(ej/4)=0.5ej/4/(ej/4–0.5)=0.5/4/(1/4–0.5)=0.6786–0.5005 [2]
0 /4 /2 3/4  T
iii) H(ej/2)=0.5ej/2/(ej/2–0.5)=0.5/2/(1/2–0.5)=0.4472–0.4636 [2]
[14]
Low pass filter
iv) H(ej3/4)=0.5ej3/4/(ej3/4–0.5)=0.53/4/(13/4–0.5) = 0.3574–0.2555[2]
j
j
j
[2]
v) H(e )=0.5e /(e –0.51)=0.5/(1–0.5) = 0.33330
(12)
(2)


/2
H()
jk
jkT
s
2. h(k) = 1
H()e d
H()e
d = 1
1

=
2




/2
s
2
s
– 
s
and
T
=
1
1
0

jk
j k
h(k) = 1
  e d + 1
 e d [2].
2 0
2 1
0 1

–
–1




0
[10]
1
0
1
jk
1 
1  e jk
1  [2]
1 e jk (  1 )  + 1 e jk (  1 )
)
+ 2
(

 )
=
–
h(k) = – 21  e jk (  jk
j2k 
jk 
jk 
j2k 
jk 

  1
 jk
0
1
0
1 e 0 (0  1 )  e  jk ( 1  1 )  + 1 e jk (1  1 )  e 0 (0  1 )  = – 1 (  1 )  e  jk (1  1 ) 
h(k) = – j2k
 
jk
jk  j2k 
jk
jk 
jk 
j2k  jk
 jk
 jk
jk
jk
jk
1
+ 1 e (1  1 )  ( 1 ) =
e
 e
 e  e
 1
j2k 
jk
jk 
j2 2k 2 j2k j2 2k 2 j2k j2 2k 2 j2 2k 2
jk
jk
jk  jk
jk  jk
 jk e jk
 1 [4] =  1  1 e  e
+ e  e
 1 e e
h(k) =  1  e

2
2k 2 j2k 2k 2
j2k 2k 2 2k 2
k 2 k
j2
k 2
x(k)
h(k) =  1  1 sin k  1 cos k = 1 sin k  1 cos k  1 [2]
k
1.2 1.1
k 2 k
k 2
k 2
1.0
N-1
 j(2/N)nk 11
 j(2/12)nk 11
 j(/6)nk
0.9
3. a) X(n) =  x(k)e
=  x(k)e
=  x(k)e
0.8
k =0
k =0
k =0
0.7
0.6
0.5
0.4
0.3
0.2
0.1
[2]
X(0) = 1.2+1.1+1.0+0.9+0.8+0.7+0.6+0.5+0.4+0.3+ 0.2+0.1=7.8
X(1) = 1.2e–j(/6)×1×0+1.1e–j(/6)×1×1+1.0e–j(/6)×1×2+0.9e–j(/6)×1×3
+0.8e–j(/6)1×4+0.7e–j(/6)1×5+0.6e–1 j(/6)1×6+0.5e–j(/6)1×7
+0.4e–j(/6)1×8+0.3e–1 j(/6)1×9+0.2e– j(/6)1×10 +0.1e– j(/6)1×11
0
= 1.20+1.1– /6+1.0–/3+0.9–/2 +0.8–2/3+0.7–5/6+0.6–+0.5–7/6+0.4–4/3+0.3–3/2+0.2–5/3+0.1–11/6
[8]
[14]
k
= 2.31822198–1.308996939
(10)
b) xav = X(0)/N = 7.8/12 = 0.65
(2)
c) A1 = |X(1)|/(N/2) = 2.31822198/6 = 0.38637 [1]  rms(A1) = 070710.38637 = 0.2732 [1]
(2)
(cos k)1
4. hideal(k) = 1 sin k  1
Therefore FIR filter approximation of length 7:
 k
 k2
hideal(k)
0.318
sin
(k

3)
cos
(k

3)

1
[1]
sink/k
h(k) = 1
k = 0,..,6
1
 (k 3)
 (k 3)2
3
[12]
[1]
[1]
[1]
1
2
h(0) = –0.055408 h(1) = 0.032026 h(2) = 0.121522
k
2
[2]
[1]
[1]
(cosk
–1)/
k
–0.159
h(4) = 0.121522
h(5) = 0.032026
h(3) = 0.159155
h(6) = –0.055408 [1]
[3]
h(k)
x(k)
-1
-1
-1
-1
-1
-1
0.159155
z
z
z
z
z
z
x(k)
x(k-1)
x(k-2)
x(k-3)
x(k-4)
x(k-5)
x(k-6)
–0.05541 0.032026 0.121522 0.159155 0.121522 0.032026 –0.0554
+
+
+
+
+
+
0.121522
0.032026
-0.0554
0.121522
0.032026
k
-0.0554
y(k)=–0.05541x(k)
+0.032026x(k–1)+0.121522x(k–2)
+0.159155x(k–3)+0.121522x(k–4)
+0.032026x(k–5)–0.05541x(k–6)
Digital Signal Processing IV EIDSV4A
Unit 1 Final Assessment 7 May 2015
Question 1 A signal x(k) = [ 1 2 3 2 1 ] is
shown in Figure 1. Sketch the following signals:
a) x(k – 1)u(k + 1).
(3)
b) x(1 – k)(1 – k).
(3)
c) x(2 – k)u(2 – k).
(4)
x(k)
1
2
3
Page 1
Figure 1
2
1
k
Question 2 Express the function x(k) = (k2 – k + 2)[u(k+2) – u(k–3)], as the sum of
weighted and shifted impulse functions.
Question 3 Refer to the system in Figure 2.
a) Determine a difference equation that
z –1
describes the relationship between the
x(k)
output y(k) and the input x(k) of the
+
system.
(2)
b) Calculate the first four terms, h(0) to
h(3), of the impulse response h(k) of
the system.
(4)
c) Determine an expression for the system transfer function H(z)
d) From H(z), determine an expression for the impulse response h(k)
(5)
Figure 2
y(k)
1
2
z –1
(4)
(5)
Question 4 Show that the z transform of the function x(k) = u(k) – u(k–2) is given by
X(z) = (z + 1)/z.
(5)
k
Question 5 The impulse response of a digital filter is given by h(k) = 0.5 u(k). The
input to the system is the pulse signal, x(k) = u(k) – u(k–2).
a) Use graphical convolution methods to determine the response y(k) = x(k) * h(k) of
the system, for k = 0, 1, 2 and 3.
b) Use z transform methods to verify your results in Question 5 a), by finding y(k)
for k = 0, 1, 2 and 3, from Y(z) = H(z)X(z) and y(k) = Z –1{Y(z)}. (Confirmation
must be independent and without assuming any answers from Question 5 a)
---oooOooo---
(7)
(8)
Total: 50
Appendix: Fourier and z transforms
Convolution:
k
k
(k)
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
u(k)
z(sin)
for
x(k)
=
0,
k
<
0
sinku(k)
k
a u(k)
(z e j )(z e j )
FIR Low pass filter:
ku(k)
2 -z(cos
ω T sin(k -α)ωcT
z
x(k-1)
,T= 2π
h(k)= c
cosku(k)
ωs
π (k -α)ωcT
j


j

x(k-2)
(z e )(z e
)
Ωo
z-1
x(k+1)
Bilinear Transform: s = 
(a z (a z
ω
T
k
z
o

2ab cos(k + )u(k)
tan (
) +1
z b z b
x(k+2)
2
ax
xe ax e ax e ax
1

jk
jkT
- jkT
1 ω /2
1 
dx a  2  a (x a )
X() =  x(k)e
x(k) = 2   X()e
d  xe
d = ω ωs /2 X()e
a
s
s
k = -
s = 2π
Lim 1  cos x
Lim sin x
T
N-1

1
½
- j(2/N)nk
j(2/N)nk
1 N-1
jT
x 0
x 0 x
X(n) =  x(k)e
x(k) = N  X(n)e
H() = H(e )
x2
n =0
k =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2
x(k)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
z2X(z) - z2x(0) - zx(1)
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2

Digitale Seinverwerking IV EIDSV4A Eenheid 1 Finale Evaluasie 7 Mei 2015 Memorandum
1. a)
1
x(k–1)u(k+1)
3
2
2
1
b)
x(1–k)(1–k)
3
c)
Bladsy 1
x(2–k)u(2–k)
3
2
1
k
k
(3)
(3)
(4)
2
2. x(k) = (k – k + 2)[u(k+2) – u(k–3)]
{u(k+2) – u(k–3) = 1 for k = –2 to 2 and zero elsewhere
[5] x(k) = 8(k+2) + 4(k+1) + 2(k) + 2(k–1) + 4(k–2)
[10]
k
x(k–1)
3. a) y(k) = ½[x(k) + x(k–1) + y(k–1)]
(2)
z– 1
x(k)
b) h(k) = ½(k) + ½(k–1) + ½h(k–1)
1 y(k)
h(0)=½ [1], h(1)=¾ [1], h(2)=⅜ [1], h(3)=3/16 [1] (4)
x(k)
x(k–1)+x(k)+y(k–1)
2
c) Y(z) = ½X(z) + ½z–1X(z) + ½z–1Y(z) [2]
(1–½z–1)Y(z) = ½(1+z–1)X(z)
z– 1
y(k–1)
–1
–1 [2]
Y(z)/X(z) = ½(1+z )/(1–½z )  H(z) = ½(z+1)/(z–½)
(4)
[2]
d) H(z)/z = ½(z+1)/z(z–½) = –1/z + (3/2)/(z–½)
[15]
H(z) = –1 + (3/2)z/(z–½)  h(k) = –(k) + (3/2)(½)ku(k) [3]
(5)
–1
{or H(z)=½(z+1)/(z–½)=[½z/(z–½)]+[½/(z–½)]=[½z/(z–½)]+z [½z/(z–½)]
h(k)=½(½)ku(k)+½(½)k–1u(k–1) check: h(0)=½+0=½ h(1)=¼+½=¾ h(2)=⅛+¼=⅜ h(3)=1/16+⅛=3/16}
+
4. Z{u(k) – u(k–2)} = z/(z–1) – z –2[z/(z–1)] [2] = z/(z–1) – z –1/(z–1) = z/(z–1) – 1/z(z–1)
X(z) = (z2–1)/[z(z–1)] = [(z–1)(z+1)]/[z(z–1)] = (z+1)/z [3]
[5] {or x(0) = 1 and x(1) = 1  X(z) = x(0) + z–1x(1) = 1 + z–1 = (z+1)/z}
5. a) h(k)=0.5ku(k) x(k)=u(k)–u(k–2)
n
-3
-2
-1
0
1
2
3
h(0)=1
x(0)=1
x(n)
1
1
0
0
h(1)=0.5
x(1)=1
h(-n) 0.125 0.25 0.5
1
y(0)=1 [1]
h(2)=0.25
h(1-n)
0.125 0.25 0.5
1
y(1)=1.5 [2]
h(2-n)
0.125 0.25 0.5
1
y(2)=0.75 [2]
h(3)=0.125
h(3-n)
0.125 0.25 0.5
1
y(3)=0.375 [2]
b) H(z) = z/(z–½) [½]
X(z) = (z+1)/z [½]
Y(z) = z/(z–½)(z+1)/z  Y(z)/z = (z+1)/[z(z–½)] = –2/z + 3/(z–½)
Y(z) = –2 + 3z/(z–½)  y(k) = –2(k) + 3(½)ku(k) [3]
[15]
y(0) = –2(0) + 3(½)0u(0) = –2 + 3 = 1 [1]
y(1) = –2(1) + 3(½)1u(1) = 0 + (3/2) = 1.5 [1]
y(2) = –2(2) + 3(½)2u(2) = 0 + (3/4) = 0.75 [1]
y(3) = –2(3) + 3(½)3u(3) = 0 + (3/8) = 0.375 [1]
{or: H(z) = 1 + 0.5z-1 + 0.25z-2 + 0.125z-3 + ...... and X(z) = 1 + z-1
Y(z) = H(z)X(z) = (1 + 0.5z-1 + 0.25z-2 + 0.125z-3 + .... )(1 + z-1)
= (1 + 0.5z-1 + 0.25z-2 + 0.125z-3 + ......... ) + (z-1 + 0.5z-2 + 0.25z-3 + 0.125z-4 + ........ )
= 1 + 1.5z-1 + 0.75z-2 + 0.375z-3 + ...........
y(0) = 1 , y(1) = 1.5 , y(2) = 0.75 & y(3) = 0.375 }
(7)
(8)
Digital Signal Processing IV EIDSV4A Unit 2 Final Assessment 7 May 2015
Page 1
Question 1 The relationship between the output y(k) and the input x(k) of the digital filter
shown in Figure 1, is given by: y(k) = 0.070382x(k) + 0.112693x(k–1) + 0.146326x(k–2)
+ 0.159155x(k–3) + 0.146326x(k–4) + 0.112693x(k–5) + 0.070382x(k–6).
Figure 1
x(k)
-1
-1
-1
-1
-1
-1
z
z
z
z
z
z
0.070382
0.112693
0.146326
0.159155
0.146326
0.112693
0.070382
+
+
+
+
+
+
+
y(k)
a) Determine the system transfer function H(z) = Y(z)/X(z).
(4)
jT
b) Determine the frequency response H(e ) of this filter, at the following frequencies:


i) T = 0
ii) T = 8 iii) T = 4 and iv) T = 
(12)
Question 2 A digital filter is required to have a frequency characteristic, shown
Figure 2
H()
in Figure 2. The transfer function H() is defined by:
1
     
 0
1+
1–

    
(–
/2)
(s/2)
s
H() = 
    
 
0
–
 
1
 0
    
–1
Assuming a sampling period of T = 1 (s = 2), calculate the impulse response h(k),
(10)
a
required for this filter. You may find the following integral useful:  e ad  e a   1a )
Question 3 The sixteen (N = 16) sampled values x(0) = 0.159155, x(1) = 0.146326,
x(2) = 0.112693, x(3) = 0.070382, x(4) = 0, x(5) = 0, x(6) = 0, x(7) = 0, x(8) = 0, x(9) = 0, x(10) = 0,
x(11) = 0, x(12) = 0, x(13) = 0.070382, x(14) = 0.112693 and x(15) = 0.146326, shown in Figure 3,
were obtained when a signal x(t) was sampled, with a sampling period of T = 1 sec.
Figure 3
x(k)
a) For the discrete Fourier transform X(n), of this
0.159155
sequence, calculate the values of X(0) and X(1). (10)
0.146326
0.112693
b) Estimate the average value of x(t).
(2)
0.070382
c) Calculate the amplitude A1 and the frequency of
k
(2)
the first harmonic X(1), contained in x(t).
0
15
Question 4 The ideal impulse response required for a certain digital filter, is given by:
[1  (cos k)]
Design and draw the structure of a corresponding FIR filter of length
h(k) = 1

k2
Lim 1  (cos x)  = 0.5
seven (N = 7). You may find the following limit useful:
(10)

x  0  x 2
---oooOooo--Appendix: Fourier and z transforms
Total: 50
x(k)
(k)
u(k)
aku(k)
ku(k)
x(k-1)
x(k-2)
x(k+1)
x(k+2)
X(z)
1
z/(z - 1)
z/(z - a)
z/(z - 1)2
z-1X(z)
z-2X(z)
zX(z) - zx(0)
z2X(z) - z2x(0) - zx(1)
x(k)
(1/a)kaku(k)
X(z)
z/(z - a)2
sinku(k)
z(sin)
j
(z e  )(z e j )
cosku(k)
z 2 -z(cos
(z e j )(z e j )
2abkcos(k + )u(k)
(a z (a z

z b z b
Convolution:
k
k
y(k)=  h(n)x(k  n) =  x(n)h(k  n) ,
n 0
n 0
for x(k) = 0, k < 0
FIR Low pass filter:
ω T sin(k -α)ωcT
,T= 2π
h(k)= c
π (k -α)ωcT
ωs
Ωo
z-1
Bilinear Transform: s = 
ωoT z+1
tan (
)
2
ax
xe ax e ax e ax
1
 xe dx a  a 2  a (x a )

jk
jkT
- jkT
1 ω /2
1 
X() =  x(k)e
x(k) = 2   X()e
d
d = ω ωs /2 X()e
s
s
k = -
s = 2π
Lim sin x
Lim 1  cos x
T
N-1
1
½
- j(2/N)nk
j(2/N)nk
1 N-1
x 0 x
x 0
x(k) = N  X(n)e
H() = H(ejT)
X(n) =  x(k)e
x2
n =0
k =0
j -j
j -j
Euler and trig. identities: rej = rcos + jrsin = r, cos = e e , sin = e e , cos = sin(+/2), sin = cos(-/2)
2
j2

Digitale Seinverwerking IV EIDSV4A Eenheid 2 Finale Evaluasie 7 Mei 2015 Memorandum
Bladsy 1
1. a) Y(z) = 0.070382X(z)
z-1
z-1
z-1
z-1
z-1
z-1
-1
+0.112693z X(z)
x(k-5)
x(k-6)
x(k)
x(k-1)
x(k-2)
x(k-3)
x(k-4)
+0.146326z-2X(z)
-3
+0.159155z X(z) 0.070382 0.112693 0.146326 0.159155 0.146326 0.112693 0.070382
+0.146326z-4X(z)
y(k)
+0.112693z-5X(z)
-6
[2]
+
+
+
+
+
+
+0.070382z X(z)
H(z) = 0.070382+0.112693z-1+0.146326z-2+0.159155z-3+0.146326z-4+0.112693z-5+0.070382z-6 [2] (4)
b) i) H(ej0)=0.070382+0.112693e-j0+0.146326e-j0+0.159155e-j0+0.146326e-j0+0.112693e-j0+0.070382e-j0=0.817957[3]
ii) H(ej/8)=0.070382+0.112693e-j/8+0.146326e-j2/8+0.159155e-j3/8+0.146326e-j4/8+0.112693e-j5/8+0.070382e-j6/8=0.6427702-1.178097 [3]
iii) H(ej/4)=0.070382+0.112693e-j/4+0.146326e-j2/4+0.159155e-j3/4+0.146326e-j4/4+0.112693e-j5/4+0.070382e-j6/4=0.266556-2.356194 [3]
[16]
(12)
iv) H(ej)=0.070382+0.112693e-j+0.146326e-j2+0.159155e-j3+0.146326e-j4+0.112693e-j5+0.070382e-j6=0.048869 [3]


/2
jk
jkT
s
H()
2. h(k) = 1
H()e d
H()e
d = 1
s = 2
1
2  
s s/2
and T = 1
+1
–+1
0

jk
jk
[2]
1
1

[  1] e d +
[  1] e d
=
2 1
2 0
0 1
–

–1
0
0




j
k
j
k
j
k
j
k


= 1
 e d + 1
e d – 1
 e d + 1
e d
a

e
a
1
 1
2 1
 0
2 0
Using   e
d a  
a
0

0

jk
jk 
1  jk (  1 ) + 1 e jk  [2]
(  1 ) + 1 e
= 1 e
 – j2k e
 0
jk  0 j2k 
j2k 
jk    j2k 

1   1    jk
1 
1 
  jk   – 1 e jk (1  1 )    1   + 1 e jk   1
( 1  )   +
=
1  e


     e
 

  j2k 


j2 k   jk  
jk   j2 k 
jk   jk   j2 k 






=
[10]
1
2 k 2
=

+
1
j2 k
e
 jk
–
1
2 k 2

e  jk +
1
j2 k
–

1
j2 k
e
 jk
–
1
j2 k
e
jk
–
1
2 k 2
1 1  cos k
1
1
1
1 e jk  e  jk
–
–
cos k =
=
2
k 2
k 2
k 2
k 2

k2
N-1
3. a) X(n) =  x(k)e
 j(2/N)nk
k =0
15
15
k =0
k =0
1
e jk +
2 k 2
+
1
j2 k
e
jk
–
1
[4]
j2k
[2]
=  x(k)e  j(2/16)nk =  x(k)e  j(/8)nk
x(k)
0.159155
0.146326
0.112693
0.070382
X(0) =0.159155+0.146326+0.112693+0.070382+0+0+0+0+0
+0+0+0+0+0.070382+0.112693+0.146326=0.817957 [2]
X(1) =0.159155e–j(/8)×1×0+0.146326e–j(/8)×1×1 +0.112693e–j(/8)×1×2
+0.070382e–j(/8)×1×3+0e–j(/8)1×4+0e–j(/8)1×5+0e–j(/8)1×6+0e–j(/8)1×7
k
+0e–j(/8)1×8+0e–j(/8)1×9+0e–j(/8)1×10+0e–j(/8)1×11+0e–j(/8)1×12
–j(/8)1×13
–j(/8)1×14
–j(/8)1×15
0
15
+0.112693e
+0.146326e
+0.070382e
=0.1591550+0.146326–/8+0.112693–/4+0.070382–3/8+0–/2+0–5/8+0–3/4+0–7/8
+0–+0–9/8+0–5/4+0–11/8+0–3/2+0.070382–13/8+0.112693–7/4+0.146326–15/8
= 0.642770 [8]
(10)
b)
x
=
X(0)/N
=
0.817957/16
=
0.0511223
(2)
[14]
av
[1]
[1]
c) A1 = |X(1)|/(N/2) = 0.64277/8 = 0.080346 f0 = 1/NT = 1/(161) = 0.0625 Hz
(2)
1
cos
(k
3)
1

(cos
k)


1
1
4. hideal(k) = 
Therefore FIR filter approximation of length 7: h(k) = 
k = 0,.,6 [1]
k2
(k 3)2
h(0)=0.070382[1] h(1)=0.112693[1] h(2)=0.146326[1] h(3)=0.159155[2] h(4)=0.146326[1] h(5)=0.112693[1] h(6)=0.070382[1]
[1]
x(k)
z
x(k)
-1
z
-1
x(k-1)
z
-1
x(k-2)
z
-1
x(k-3)
z
-1
x(k-4)
z
-1
x(k-5)
x(k-6)
[10]
+
+
+
+
0.146326
0.112693
0.070382 0.112693 0.146326 0.159155 0.146326 0.112693 0.070382
+
h(k)
+
0.159155
0.146326
0.112693
k
0
0.070382
0.070382
y(k) = 0.070382x(k)
+ 0.112693x(k–1)+0.146326x(k–2)
+ 0.159155x(k–3)+0.146326x(k–4)
+ 0.112693x(k–5)+0.070382x(k–6)