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SJcIT Pere 1
ai
[ Internal Test Question paper format
Fame of he staf: Ravi M V/ Manjula K
Date:_02/12/2020
Cc Reviewer's Signature:
NOTE: Only the following information's to be given tothe students.
S.J.C. Institute of Technology
Department:-Electronics & Communication Engg.
Test: 10 .
Semester: V Section: A
Subject Name & Code: Digital Signal Processing (18EC52)
Duration:1.5 Hour Max Mark
Note: Answer 5 Questions choosing Ifull question each part
cate Marks | co
umber
Mlustrate & prove the following properties of DFT
i)Symmetry property ii)Parseval's theorem. 10M | COI
OR
TFx(n)=(1,2,0,3,-2,4,7,5). Interpret the following
19K (0) i1)XG) ii)E" Co CK) iv) Deo [XG |? tomo!
a. Compare the differences and similarities between DIT
and DIF FFT algorithms? am | coz
3 | b. Summarize and prove the two properties of phase factor
Wr that are exploited in Fast Fourier Transform
algorithms? om | coz
1
OR
3 Explicate with necessary diagrams and equations the concept]
of overlap-save method for linear filtering.
A long sequence x(n) is filtered with a filter with
impulse response h(n) to produce output y(n). If
5 x(n)= (1,4,3,0,7,4,-7,-7,-1,3,4,3} and h(n)={1,2}.Infer y(n) | 10M | coz
using overlap add method by using only 5 point circular
convolution.
10M | coz
OR
Interpret 8-point DFT of the sequence, x(n)=(I,1,1,1,0,0,0,0)
6 |using Redix-2 DIT-FFT algorithm by showing’ the
intermediate results.
10M co2 SJCIT
064Form#02b - Rev.?
Page: 12
Devélop DIT-FFT algorithm and write Siggal Flow graph
for N=8.
10M
cc
OR
Solve for 4-point Circular Convolution of x[nJ={1,1,1,1}
and_h{n]=(1,0,1,0) using radix -2 DIF-FFT algorithm,
10M
_L
co
A filter is to be designed with the following desired
frequency response
w .
ei, as" Xk) = Yoxrywyi wi" (since wi" = 1)
aad
N-1
yxy"
aad ~
Hence, X*(k) = X(N-k)
‘The above equation conveys the message that the DFT of a real sequence possesses conjugate
symmetry about the midpoint.
= X*(k)
i] IN is odd, the conjugate symmetry is about ¥. The index, k = $ is called the folding inde,
This aspect is illustrated in Fig. 3.7
Conjugate symmetry: X*(k) = AWW = A) or A(R) = PUN ~
Conjugate symmetry: X(k) = X(N ~ k)
we (LEE _
- #
0 2 3 5 eee)
Not
.
120. CHAPTERS © Digital Signal Processing
Fig. 2.7 Symmetry of Xk) for x(n) being real.
Conjugate symmetry implies that we need to compute only half of the DET values to find the
entire DFT sequences ~ a great labor saving help! A similar result holds good for IDFT also.
—
siete Yori Serced ale q 3.7.10 Inner product (Parsoval)
Proof:
From the definition of IDFT, we have
W-1
Lermym => P earay
nel) ind
N-I
| -t
rma Dy xaowy'
def
‘Taking conjugates on both the sides, we get
Hence,
Corollary:
If y(n) = x(n), we get
W-I
Yr'mrn
nw
Wet
atin) = 0 Na
dad
NeI
YL ayn =
n=l
Not yy Not
> (3 Ds awe) y(n)
120
nw
1 “2! N-1
ww (= voy)
imo ned
NI
I
7 LX ra)
dad
w-1
DietmpP
n=0 The DFT andFFT =—-145,
N-I
1
= VL wxe
k=O
1 N-1
= FL ikwer
We
Thus, we have proved that
N-I 1 We
2 2
Leo =y Lor
n=O k=O slays eal
sole
3 Re Ree
BALERS TS
4,7,5),N=8,
1) To obtain x0)
1
DFT, XW) = Sawin
Fn}
With k= 0 and N = 8, x00) = Fxq =1424+043-24+44+745=20
: 0 :
Above equation shows that X(Q is summation of x(x). Hence if x(n) is real, X(0 is also real.
Il) To obtain x{4)
=
DFT, XW) = Sane
m0
N
With k= =4 and N=8, X(4) = Secs )
n=O = .
= Fejy ction an
= Baeye we Sere - Sayer
ned
= Seay since €F** = (4)"
0
= #(0)~*(0) +2(2)-2(3) +2(4) -2(5) + 2(6) 0)
= 1-24+0-3-2-447-52-8
Ill) To obtain Fxn) :
0
141
IDFT, ata) = Ole
1
5x
With n = 0 and N=8, (0) Digital Signe! Processing Properties of DFT and its Ap
Sw =
8x(0) =8x1=8
iv) To obtain Sx?
ke
Parseval's theorem,
be
Nin
'
= Deer
N= 8, and real x(n) ,
= 83240) =9[2+20)+2 Ort?)
vat
B[l+4+0+9+4+16+49+25] = 864.
EE ue ne
Finesse eteneinnmncnenen te caesarean Differences between FFT algorithms
_ Ae DET at the output jain bit reversed order.
Decimates the sequence step by step to 2-point
ce nnd cae DER pes
i Properties of Phase Factor (Ww)
© Now let us consider the properties of Wy. These properties are used by FFT algorithms to redu
of calculations.
* We know the N-point DFT of sequence x(n) is given as,
=
X= D xQwe, k=0,1,..N-1
n=0
© Here Wy is called twiddle factor and it is defined as,
* This twiddle factor exhibit symmetry and periodicity properties,
Periodicity Property of Wy
x TECHNICAL PUBLICATIONS” An up thrust for knowledge We know that Ww =
2
Ween we TRON tite
w= (254)
~j2%%
eet NE ein
Here eF2% = cos 2n-jsin2n =1-j0= 1 always
Hence equation (2.5.4) becomes, 2
Win =
‘This shows that Wi, is periodic with period ‘N’.
This property was illustrated earlier in Fig, 24.1. Observe that for N= 8
wpe = Wh
ww (255)
Proof :
We know that
~ (256)
Here” cit a cosn~jsin ts ajon tna
Hence equation (2.5.6) becomes,
ah
Wy ? 2
~ 257)
(258) yeh Is fo fuel
aa NM UL
hea N - 2
um — le
ES Bi | 0 for clock
gl ca ofa
fo be divded Orr ton)
—
fucerker> MMe.
—a_ —* jae
vin wl eect
> -b >
a L= Ntaiel
c
[sl Toe tk? iM. sy ae
ee: > a
—->
Petes io zee | een
Giep 3 tome Ma Pret i Ge Vid
Shp on =
BG). sate E Ken.
-- so Sod mor CAD) Semyes from JOH), Feed
np 5 Tee. dak Ceawtivhan Gip Is Chien cl bee Contaiting
je “Cy ter ce) > br ceoe nel
nepared by Bh Deeps a ve Saot Hn) = (1, ie. M=2 and
x(t) = {1,4,3,0,7,4,-7,-7,-1,3,4,3}
Since 5 - point circular convolution is to be used, N = 5. Since N= M+L-1,
Now,
And
x
~ Now let us calculate,
Last M1 Le. 1 sample of eich yi(1) should be added to frst M~1 Le, 1 sample of succeeding iC)"
5 =2+L-1
Hn) = {1,2,0,0,0}
x(n) = £,4,3,0,0}
xo(n) = {7,4,-7,-7,0}
xy(n) = £1,3,4,3,0}
vie) = 1) He)
vale) = x20) 0)
ws) = 3@®© +0)
n ae
vi) ey
wQ2)| =}3 4
(| |o 3
n(4) 90
v()| [7
o77 4
y)| [4 7 07 7
nQ)=\7 4 7 0 -7Vol=
=L=4
w(3)| |7 7 4 7
wi] loz7s
ys(0)] [20 3 4
y(l)| |3 -1 0 3
w(2)|=|4 3 20
w(| [0 3 43 4
so
ou ne
y(3)| |3 4 3-1
TECHNICAL PUBLICATIONS" An up trust for knowledge a Processing
pose ; 2107 Proportss of DFT and its Applications
‘This is shown below
‘Thus the output sequence is,
0
{1,6,11,6,7,18,1,-21,-15,1,10,11,6 } uel
(EIIEER) Develop an’ 8-point Drr-FFT algoritn: Draw te’ signal oi graph Debora the DET ofthe Flat Beueee NY
x) = {11,1,1,0,0,0,g a
using the signal flow graph. Shoo all the intermediate results on the signal flow graph.
:
I
We have already discussed the theory of DIT-FFT algorithm. The signal flow graph for 8-point DIT-FFT
Soluti
algorithm is given in Fig. 3.2.7. Let us use this signal flow graph to obtain the DFT of given sequence, The samples
of x(n) are:
x@=1 x(4)=0
x()=1 x @)=0
TECHNICAL PUBLICATIONS”. An up trust for knowedye x(Qal x(=0
x(3)"1 x@)=0
The phase factors are calculated in example 3.2.1 and given in Table 3.2.3. These factors are reproduce
convenience.
weed
Ww} = 0707 ~/ 0707
We = -j
We = -0707-j 0707
Computation of 2-point DFTs :
Observe the signal flow graph of Fig. 327. The 2-point DFTs are first computed. Table 324 thon
computation of these 2-point DFTs.
nal flow graphs aa
tee computaion f-pont DFTs Cokcisalions of 2 6ehk OEE
4x0) = x(0) + Wy" xc) 4
yee em |
= x(0)—Wy'x(4)
= 1-10
=a
°
= x(2) + Wy x(6)
= 1410
4 ie.
= x(2)-W,"x6)
= x(t) + Wy x5)
= 1+ 4x0
=
x(1) =W4"x5)
pe te Fam=t]
= (3) + Wear)
t+ so
Le.
= x(3)—W4"x¢7)
1-10
1
Table 3.24 Computation of 2polnt DFTa
Combining 2-point DFTs :
‘The next stage is to combine the above 2-point DFTs to 4 ll
Fig. 327 and its calculations are shown in Table 3.2.5 = “Point DFTs, This part of the signal flow 47" inal Processing 3-10
I flow for
toes tty Sy on
tame} FO) = Vs(O)eWNg V4 210)
ret | fer)
oe} Fy Vy (neg Vat
a! aa
) Wg t F\(2) = Vyy(OWNy Vj 210)
ett
oe
ray F (8) = Vy tN Vat
= lepxt
et
PT F20) = V2y(0}+\%y Vo2(0)
F(t) = Vay 0g Vag 1)
Combining 4-point DFTe :
eat
i Gy
F2) = Vay(O-We V22(0)
sion
iy Ea)
F (3) = V2y(1}-W¥g Vo9(1)
sitet
=f
Table 3.2.6 Combining 2-point DFTs:
The next stage is to combine the two 4-point DFTs to get &point DFT. This part of the signal flow graph of
F327 and its calculations are shown below in Table 32.6.
Refer Table 32.6 on next page)
Tho complete signal flow graph :
The complete signal flow graph with all the values calculated is reproduced below in Fig. 32.14.
(Refer Fig. 32.14 on page 3-21.) Afar. tyferuite? eer=n-’
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ase
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Pret acer
Hfe) = Ferns) : Pare
Gr fic fo eT]
lave wen, TECHNICAL PUBLICATIONS”. An up thrust for knowedg gta! Signal Processing
an .
For n=3, fy(t) = =B fo
de
‘Thus
a(n)
forns3
) Step 2 : To perform windowing and obtain h(n), Hore hamming window Is given,
y(n) = ost nbof 1)
For M= 7, wy(n) = ost a46ens( 3
for n= 04,
for M1 = 0,1 pornnenne 6
Following table shows uC toy (2) and hn)
sin .75 n=O)" fore 3°
iyi fore
for es
An) = {e006 00:929,017325,075,017325,004925 0006}
Stop 3: To obtain frequency response
M3
wagon)
n=O
peo EF
0) 22 SM)20(0-3) = 0.75 + 2[0005 cos 310-001929 cos2ea+ 017325 cos
. nad
