DSP MARKING SCHEME : Main exam 2013
Section A: Compulsory
Question 1: a) Distinguish between energy and power signals. Test whether the discrete-time signal
x[n] = (1/2)n u[n] is a power signal or energy signal.
1. 𝑥[𝑛] is said to be an energy signal if and only if 0 < 𝐸 < ∞, and so 𝑃 = 0.
2. 𝑥[𝑛] Is said to be a power signal if and only if 0 < 𝑃 < ∞, and so 𝐸 = ∞.
1. For a discrete-time signal 𝑥[𝑛], the normalized energy content 𝐸 of 𝑥[𝑛] is defined as:
The normalized average power 𝑃 of 𝑥[𝑛] is defined as:
Given a signal 𝑥[𝑛] = (1/2)𝑛 𝑢[𝑛]
Energy of 𝑥[𝑛] is given by:
2
1 𝑛
1 𝑛
∞
∞
2
𝐸 = ∑∞
𝑛=−∞|𝑥[𝑛]| = ∑𝑛=0 [(2) ] = ∑𝑛=0 (4)
𝐸=
1
1
1−4
=
1
4
=
4−1 3
4
2. Power of 𝑥[𝑛] is given by:
𝑁
𝑁
𝑛=−𝑁
𝑛=0
1
1
1 𝑛
𝑃 = lim
∑ |𝑥[𝑛]|2 = lim
∑( )
𝑁→∞ 2𝑁 + 1
𝑁→∞ 2𝑁 + 1
4
1 𝑁+1
1−( )
1
4
= lim
[
]=0
1
𝑁→∞ 2𝑁 + 1
1−4
The energy is finite 0 < 𝐸 < ∞, and 𝑃 = 0 , therefore 𝑥[𝑛] is an energy signal.
[5 Marks]
b) Using the property of z-transform, determine the z-transforms of the following signals
and the corresponding pole-zero patterns.
x (n) = (1 + n) u (n)
Or in terms of positive powers
The pole-zero patterns are as follows:
Double pole at z = 1 and a zero at z = 0.
(5 Marks)
c) Compare finite-impulse response (FIR) filters with infinite-impulse response (IIR) filters
in terms phase, flexibility, stability and noise. Mathematical expressions are required to
clarify your answers.
Finite-impulse response (FIR) filters
Infinite-impulse response (IIR) filters
1. Have precisely linear phase
2. Greater flexibility
3. Are always stable while IIR
4. Less severe errors due to round off noise are
5. Restricted to finite number of samples
1. Do not have linear phase
2. Less flexibility
3. Are not always stable
4. More susceptible to errors due to round off noise.
5. Extend over an infinite duration.

y[n]  k 0 h[k ] x[n  k ]
N
y[n]  k  h[k ] x[n  k ]
(5 Marks)
d) Compute 4-Point DFT of a sequence x[n] = {0, 1, 2, 3} using DIT algorithm and sketch its
butterfly diagram.
[5 Marks]
DIT Algorithm
Twiddle factors associated with the butterflies are:
W40 = e-j(2π/4)0 = e0 = cos (0)
– j sin (0) = 1
W41 = e-j(2π/4)1 = e-j(π/2) = cos (π/2) – j sin (π/2) = - j
W42 = e-j(2π/4)2 = e-j(π) = cos (π) – j sin(π) = - 1
W43 = e-j(2π/4)3 = e-j(3π/2) = cos (3π/2) – j sin(3π/2) = j
Input
x(0)= 0
x(2)= 2
x(1)= 1
x(3)= 3
S1
0+2=2
0 – 2 = -2
1+ 3 = 4
1 – 3 = -2
Output X(k)
X(0) = 2 + 4 = 6
X(1) = -2 + (-j)(-2) = -2 +2j
X(2) = 2- 4 = -2
X(3) = -2 + (-2) (j) = -2 - 2j
X (k) = {6, -2 +2j, -2, -2-2j}
SECTION B: Choose any two questions
Question 2
a) A discrete-time signal x [n] is defined as
i)
ii)
iii)
i)
Determine its values and sketch the signal x[n]
Give the sequential representations of x[-n], x[-n +4], x[-n -4]
Can you express the signal x[n] in terms of δ[n] and u[n]? ( in one expression)
ii)
Yes
(6 Marks)
b) Determine the causal signal x[n] if its z-transform is:
Eliminate the negative powers by multiplying both numerator and denominator by z2, we get
Dividing X (z) by z and by using partial fraction, we obtain:
(6 Marks)
c) Make a comparison between analogue and digital filters.
(4 Marks)
Analog filters
Digital filters
1.Work on analog signals
1. Operate on the digital samples of the signals.
2.Defined by linear differential equations
2. Defined by linear differential equations.
3. Made in electrical components like
resistors, capacitors and inductors are used.
3. Made in digital logic components such as adders,
delays, multipliers, subtractors, etc…
4. For stability and causality, the poles
should lie on the left half of s-plane.
4. For stability and causality, the poles should lie inside
the unit circle in z-plane.
c) What is the speed improvement factor in calculating 64-point DFT of a sequence using
direct computation DFT and FFT algorithms? Hints: use only number of multiplications!
[4 Marks]
For N point sequence (DFT:
FFT)
Total number of addition N (N-1):
N log2N.
Total number of multiplication N2: (N/2) log2N.
The number of multiplications required using direct computation DFT is N2 = 642 = 4096
The number of multiplications required using FFT is:
(N/2) log2N = (64/2)log2 64 =32 x(ln 64/ ln2) = 32 x 6 =192
Speed improvement factor= 4096/ 192 = 21.33
[4 marks]
Question 3
a) List 4 disadvantages of digital filters compared to analog filters. [4 marks]
1.
2.
3.
4.
Speed limitation
Long design and development time
Finite word length effects: quantization noise
Finite word length effects: round off noise
b) State and prove the differentiation property in z-transform.
[5 marks]
If
ZT
x[n] 
X ( z ),
with ROC  R
Then
We know that:
X ( z) 

n
x
[
n
]
z

n  
Differentiating both sides with respect to z, we have
By multiplying both sides by – z, we get:
Thus, we conclude that
dX ( z )
nx[n]   z
, with ROC  R
dz
ZT
b) What are the differences and similarities between Decimation-in-time (DIT) and Decimationin-frequency (DIF) algorithms?
[5 Marks]
Differences
 In DIT, the input is bit–reversed while the output is in natural order. For DIF, the
input is normal order, while output is
bit–reversed.
 Butterfly diagram of DIF slightly different from DIT butterfly. The DIF FFT is
the transpose of the DIT FFT
 Considering the butterfly diagram, in DIF, the complex multiplication takes place
after the add–subtract operation.
Similarities
 Both algorithms require same number of operations to compute the DFT.
 Both algorithms can be done in place and both require bit–reversed at same place
during computation.
d) Compute the Inverse DFT of:
[6 Marks]
j
1 N 1
x(n)   X (k )e
N n 0
2nk
N
1 N 1
  X (k )WNnk
N n 0
n  0,1, N  1
Question 4
a) List any 5 advantages of digital filters over analog filters.
(5 marks)
Any five advantages of digital filters of the following:
1.
2.
3.
4.
5.
6.
7.
Linear phase response
Performance does not vary with environment
Automatically adjustable frequency response (adaptive filters)
Several input signals can be filtered by one digital filter
Both filtered and unfiltered data can be saved for future use
Reproducibility & Can work at very low frequencies
Fast re-designing or modifications
b) Compute 4-Point DFT of a sequence x[n] = {0, 1, 2, 3} using DIF algorithm and sketch its
butterfly diagram.
[6 Marks]
DIF Algorithms
Twiddle factors associated with the butterflies are:
W40 = e-j(2π/4)0 = e0 = cos (0)
– j sin (0) = 1
W41 = e-j(2π/4)1 = e-j(π/2) = cos (π/2) – j sin (π/2) = - j
W42 = e-j(2π/4)2 = e-j(π) = cos (π) – j sin(π) = - 1
W43 = e-j(2π/4)3 = e-j(3π/2) = cos (3π/2) – j sin(3π/2) = j
Input
x(0) = 0
x(1) = 1
x(2) = 2
x(3) = 3
S1
0+2=2
1+ 3 = 4
0–2=-2
(-j) + 3j = 2j
Output X(k)
X(0) = 2 + 4 = 6
X(2) = 2 + 4 (-1) = -2
X(1) = -2 +2j
X(3) = -2 + (2j)(-1)= - 2 -2j
X (k) = {6, -2 +2j, -2, -2-2j}
d) Nowadays, there is a need of other very fast processing architectures; briefly explain and
sketch a not detailed Super Harvard Architecture (SHARC® DSPs)
[4 Marks]
The Super Harvard Architecture is a digital signal processor with higher speed that is
improved upon the Harvard design which uses separate memories for data and
instructions by adding an instruction cache and a dedicated I/O controller.
d) Express the following discrete-time sequence:
as a sum of weighted impulse sequences and as a sum of minimum of scaled and shifted unit
steps.
[5 Marks]
x[n] = 2 δ [n +1] + 4 δ [n] + δ [n -1] + 3 δ [n -2]
as
 [n]  u[n]  u[n  1]
x[n] = 2{u[n +1]-u[n]} + 4{ u[n]-u[n -1]} + { u[n- 1] -u[n- 2]}+ 3{u[n-2]- u[n-3]}
x[n] = 2u [n+1] + 2u [n]- 3u [n-1] + 2u [n -2]- 3u[n – 3]
Marking Scheme : Supplementary DSP Exam 2013
SECTION A: Compulsory
Question 1
a) Determine the z-transforms of the following signals, the corresponding pole-zero patterns and
sketch the pole-zero plot with the ROC of the following sequence.
[5 Marks]
Or X
(z) = 3z/(z -2) – 4z/(z -3)
X (z) = (3 z2 – 9 z – 4 z2 + 8 z)/(z -2)(z -3) = (- z2 – z)/(z -2)(z -3)
The pole-zero patterns are as follows:
(a) Poles at z = 2 and z = 3 and
And zeros at z = 0 and z = - 1 and
POLE-ZERO PLOT
b) State and prove the differentiation property in z-transform.
[5 Marks]
If
ZT
x[n] 
X ( z ),
with ROC  R
Then
We know that:
X ( z) 

 x[n]z
n
n  
Differentiating both sides with respect to z, we have
By multiplying both sides by – z, we get:
Thus, we conclude that
ZT
nx[n] 
z
dX ( z )
, with ROC  R
dz
c) Find the circular convolution of the two sequences:
x1 (n) = {1, 2, 2, 1} and x2 (n) = { 1, 2, 3, 1} using circle method or matrix method
Circle method
[5 Marks]
The circular convolution is x3[n] = {11, 9, 10, 12}
Or Matrix method
d) Using tabular method, determine the convolution sum of two sequences: [5 Marks]
SECTION B: Choose any two questions
Question 2
a) Any 5 advantages of digital signal processing over analog signal processing. [5 Marks]
1. Noise-resistance and high accuracy
2. Great flexibility: DSP operations can be changed by changing the program in digital
programmable system.
3. Stability and repeatability: easily duplicated and independent of temperature, ageing
and other external parameters.
4. Simplicity: easy to build any digital system
5. Easy upgradations: because of use of software
6. Compatibility: all applications need standard hardware. Thus operations of DSP system
are mainly dependent on software.
7. Remote processing: Easy information storage on magnetic media without loss of quality
of reproduction of signal and easily transported, digital signals can be processed off line.
8. In some cases, Cheaper, digital hardware cheaper and digital circuits can be reproduced
easily in large quantities or by time-sharing of given processor among a number of
signals.
9. Sophisticated signal processing algorithms can be implemented by DSP method
b) Find the z-transform of each of the following sequences: In all cases assume that n ≥ 0
[6 Marks]
or
c) Make a comparison between fixed-point and floating-point digital signal processors.
[9 Marks]
Fixed versus floating point
Fixed-point DSPs
1. A minimum of 16 bits
2. Up to 65,536 possible bit
patterns (216).
3. Represent and manipulate
integers : whole numbers
4. Generally cheaper
5. Worst precision
6. More quantization ‘noise’
7. Lower dynamic range
8. Longer development cycle
9. For high-volume, general
purpose applications
Floating-point DSPs
1. A minimum of 32 bits
2. Up to 4,294,967,296
possible bit patterns (232).
3. Represent and manipulate
rational numbers
4. Generally more expensive
5. Better precision
6. Less quantization ‘noise’
7. Higher dynamic range,
8. Shorter development cycle.
9. For specialized,
computationally intensive
applications
Question 3
a) A causal discrete-time LTI system is described by:
Where x[n] and y[n] are the input and output of the system, respectively.
i) Determine the system function H (z).
ii) Find the impulse response h[n] of the system.
Solution
[6 Marks]
b)Sketch and describe the working principle of the Harvard architecture processor compared
with the traditional processor von Neumann machine.
[6 Marks]
The Harvard architecture is based on separate memories for data and program instructions
with separate buses for each. Since the buses operate independently, program instructions and
data can be fetched at the same time, improving the speed over the traditional processor von
Neumann Machine with its single bus design.
Having two separate buses for the program and data memory; content of program memory and
data memory can be accessed in parallel. The instruction code can be fed from the program
memory to the control unit while the operand is fed to the processing unit from the data
memory.
c)
[4 Marks]
d) What are the limitations of digital signal processing compared with the analog signal
processing?
[4 Marks]
1. Bandwidth limitations, limited speed of operation of AD converters and digital signal
processors
2. System complexity: usage of ADC or DAC, time required for this conversion is more
3. In some cases, DSP systems are expensive as compared to analog system
4. More power consumption
Question 4
a) List any five main applications of digital signal processing.
[5 Marks]
Any five from the following applications
1. Audio and speech signal processing
2. SONAR (Sound Navigation And Ranging) signal processing
3. RADAR (Radio Detection And Ranging) signal processing
4. Digital image processing
5. Signal processing for communications
6. Control of systems
7. Biomedical signal processing
8. Seismic data processing, etc…
b) State and prove the time-shifting property in z-transform.
[5 Marks]
ZT
x[n] 
X ( z ), with ROC  R
If
Then
ZT
x[n  n0 ] 
z  n0 X ( z ),
Proof
with ROC  R
Verification of the time-shifting property
By the change of variables m = n – no & n = m +no
c) Sketch the butterflies diagram and compute 4-Point DFT of a sequence x[n] = {1, 2, 3, 0} using
DIT algorithm and DIF algorithm.
[10 Marks]
i)DIT Algorithm
Twiddle factors associated with the butterflies are:
W40 = e-j(2π/4)0 = e0 = cos (0)
– j sin (0) = 1
W41 = e-j(2π/4)1 = e-j(π/2) = cos (π/2) – j sin (π/2) = - j
W42 = e-j(2π/4)2 = e-j(π) = cos (π) – j sin(π) = - 1
W43 = e-j(2π/4)3 = e-j(3π/2) = cos (3π/2) – j sin(3π/2) = j
Input
x(0) = 1
x(2) = 3
x(1) = 2
x(3) = 0
S1
1+3=4
1 - 3=-2
2+0=2
2 -0=2
Output X(k)
X(0) = 4 + 2= 6
X(1) = -2 + (-j) (2)= -2 - 2j
X(2) = 4 + 2 (-1) = 2
X(3) = (-2) + 2 (j) + = – 2 + 2j
X (k) = {6, - 2- 2j, 2, -2 + 2j}
ii)DIF Algorithm
Input
x(0) = 1
x(1) = 2
x(2) = 3
S1
1+3=4
2+0=2
1– 3 = - 2
Output X(k)
4 + 2 = 6 = X(0)
4 – 2 = 2 = X(2)
-2 -2j= X(1)
x(3) = 0
(-j) (2) + 0 (j)= - 2j
X (k) = {6, -2 - 2j, 2, -2 + 2j}
-2 + (-2j)(-1)= -2 +2j= X(3)