Question

A discrete time waveform x(n) is composed of three sinusoidal components: 

30 Hz with magnitude 2.0 

50 Hz with magnitude 3.0 

80 Hz with magnitude 1.0 

The time between samples is ts= 0.005 sec. (fs= 200 samples/sec .). Sketch the magnitude of the N point DFT X(m)|= |DFT{x(n)} | versus m over the entire range shown, assuming that the DFT is performed on 20 consecutive samples of x(n) (i.e. N = 20). (By the way, the magnitude of a DFT element corresponding to a sinusoidal input of magnitude A, is A*N/2) 

Answer 1

From the given information we can compose time varying signal x(t) as some of three sinusoidal signals whose frequencies and magnituds are given below

30Hz magnitude 2

50Hz magnitude 3

80Hz magnitude 1

x(t)=2sin(60$\pi$t)+3sin(100$\large \pi$t)+sin(160$\large \pi$t) ................(1)

now to obtian the descrete time signal we need to keep t=nt_s where t_s is sampling time period given as 0.005sec whereas n is descrete time variable

so x(n)= 2sin(60$\pi$*n*0.005)+3sin(100$\large \pi$*n*0.005)+sin(160$\large \pi$*n*0.005)

x(n)=2sin(0.3$\pi$n)+3sin(0.5$\large \pi$n)+sin(0.8$\large \pi$n) ....................(2)

now we have to obtain 20 point DFT of the above expression hence we need to find

x(0),x(1),x(2),x(3),x(4), .........................upto x(19) .

so by keeping n=1,2,3,4,5,6,7,8,9, .........................19 we will get below values

x(0)= 0 x(10)=0

x(1)=5.2058 x(11)=-4.03

x(2)=0.95 x(12)=-2.853

x(3)=-1.43 x(13)=3.333

x(4)=-1.763 x(14)=0.5877

x(5)=1 x(15)=-1

x(6)=-0.587 x(16)=1.7633

x(7)=-3.333 x(17)=1.4309

x(8)=2.8531 x(18)=-0.951

x(9)=4.0302 x(19)=-5.205

for finding N point DFT we have the formula as given below

X(m)=   .............................(3)

n 19 Σ χ(n) x ewm e-wmn n 0

where $\large \omega$ =
2T N
=
10

so by keeping value of N=20 and m=0, 1, 2, 3, ...................19   in equation (3) we get

DFT of x(n)

X(0)=
n 19 Σ Χ1) n-0
=75.49*10^-16
20

X(1)=
n 19 Σ Χ(n) x erun n-0
=84.89*10^-16
46,98

X(2)=
n 19 Σ Χ1) x e 102n n-0 eju2n
=35.544*10^-16
a39.8

similarly

X(3)=
n-19 Σ χ(n) x eu3η n 0
=20
a36.93

X(4)=
n-19 Σ χ(n) x n=0 e "u 4n
=8.01*10^-16
Z89.95

X(5)=
n 19 Σχ(n) x eu 5η e 5n n 0
=30
09.43

X(6)=
n 19 Σ χ(n) x eu 6n n 0
=119*10^-16
44.95

X(7)=
n 19 Σ Χ1) x efu 7η n 0
=69.12*10^-16$\large \angle-175.66$

X(8)=
n 19 Σ Χ(n) x eu 8n η-0
=10
a5.16

X(9)=
n 19 Σ χ() x elu 9n n-0
=116*10^-16
a235

X(10)=
n 19 Σ χ(n) x eu10n η-0
=595*10^-16$\large \angle174.75$

X(11)=
n 19 x(n) x eu11n n 0
=173*10^-16$\large \angle-95.56$

X(12)=
n 19 Σ Χn) x eiu1 2η n-0 elu12η
=10
Z3349

X(13)=
n-19 Σχ(n) x e13n n-0 e-w13n
=243*10^-16$\large \angle-162.14$

X(14)=
n 19 Σ Χ(n) x eiu14n eiu14η n 0
=442.52*10^-16$\large \angle-124.9$

X(15)=
n=19 Σ χ(n) x eju15η n 0
=30$\large \angle143.64$

X(16)=
n 19 Σ Χn) n-0 eiu 16η
=147.74*10^-16$\large \angle-97.4$

X(17)=
n 19 Σ Χ1) eiu 17η Χ n 0
=20
Zi40 37

X(18)=
n 19 Σ χ(n) x eiu18η n 0
=615*10^-16$\large \angle-121.46$

X(19)=
n 19 Σ Χn) n-0 e ju 19n
=572.111*10^-16$\large \angle-74.48$

from the above results considering the magnitude of those whose value is in powers of -16 is equel to zero

we have

magnitudes of

X(3)=X(17)=20

X(5)=X(15)=30

X(8)=X(12)=10 and remaining all others are zeros

by drawing the magnitude spectrum of the above values we will get below figure

X(m) 30 20 20 20 10 10 17 15 12 10 5 3 0