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0 1 2 3 4 5 6 7 8 9 100
0.5
1
n
ampl
itude
discrete unit step sequence
0 1 2 3 4 5 6 7 8 9 100
0.5
1
1.5
2
n
ampl
itude
continuous unit step signal
-5 -4 -3 -2 -1 0 1 2 3 4 50
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1
n
ampl
itude
unit impulse sequence
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-0.5
0
0.5
1
time
ampl
itude
sine signal
0 1 2 3 4 5 6 7 8 9 10-1
-0.5
0
0.5
1
time
ampl
itude
cosine signal
0 2 4 6 8 10 12-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
time
Am
plitu
desquare signal
0 1 2 3 4 5 6 7 8 9 10-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
time
Am
plitu
de
rectangular signal
0 5 100
5
10
time
Am
plitu
de
positive ramp signal
0 5 100
5
10
time
Am
plitu
de
positive ramp signal
0 5 100
5
10
time
Am
plitu
de
negative ramp
0 5 100
5
10
time
Am
plitu
de
negative ramp
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1-1
-0.5
0
0.5
1
Time
Am
plitu
de
saw tooth wave
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1-1
-0.5
0
0.5
1
time
ampl
itude
OUTPUT
0 5 10 15 20 25 30-1
0
1
Time Index t (sec.)
x1(
t)
Signal 1 : Sine Wave of Frequency 1/3 Hz
0 5 10 15 20 25 30-1
0
1
Time Index t (sec.)
x2(
t)
Signal 2 : Sine Wave of Frequency 1/5 Hz
0 5 10 15 20 25 30-2
0
2
Time Index t (sec.) y(t
) =
x1(
t) +
x2(
t) Resultant Signal : Signal 1 + Signal 2
OUTPUT
0 5 10 15 20 25 30-1
0
1
Time Index t (sec.)
x1(
t)
Signal 1 : Sine Wave of Frequency 1/3 Hz
0 5 10 15 20 25 30-1
0
1
Time Index t (sec.)
x2(
t)
Signal 2 : Sine Wave of Frequency 1/5 Hz
0 5 10 15 20 25 30-1
0
1
Time Index t (sec.) y(t
) =
x1(
t) .
* x2
(t) Resultant Signal : Dot Product of Signal 1 and Signal 2
OUTPUT
0 1 2 3 4 5 6 7 8 9 10-2
-1
0
1
2
Time Index t (sec.)
x(t
)
Signal 1 : Sine Wave of Frequency 1/5 Hz
0 1 2 3 4 5 6 7 8 9 10-2
-1
0
1
2
Time Index t (sec.)
y(t
)
Signal 2 : Scaled Version of Signal 1
OUTPUT
0 100 200 300 400 500 600 700 800 900 10000
0.2
0.4
0.6
0.8
nx
x(n
x)
Original Signal
-1000 -900 -800 -700 -600 -500 -400 -300 -200 -100 00
0.2
0.4
0.6
0.8
nf
xf(
nf)
Folded Signal
OUTPUT
1 1.5 2 2.5 3 3.5 40
2
4
samplesam
plitu
de
original signal
1 1.5 2 2.5 3 3.5 40
2
4
samples
ampl
itude
even part of sequence
1 1.5 2 2.5 3 3.5 4-2
0
2
samples
ampl
itude
odd part of sequence
OUTPUT
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0.2
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0.5
0.6
0.7
0.8
0.9
1
time n
ampl
itude
Energy of the signal (1/2).n
Output:
Type a value for N 1The calculated power of the signal
P =
0.3750
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10
0.1
0.2
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0.5
0.6
0.7
0.8
0.9
1
time n
ampl
itude
input signal
Result: Hence the energy and power of the signal are calculated.
Output:Enter x[n]: [1 2 3 4]Enter h[n]: [1 2]Z= 1 4 7 10 8
0 0.5 1 1.5 2 2.5 30
2
4
Time
Am
plitu
de
Input sequence x[n]
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
1
2
Time
Am
plitu
de
Impulse response of the system h[n]
0 0.5 1 1.5 2 2.5 3 3.5 40
5
10
Time
Am
plitu
de
Linear Convolution
Output:enter x[n]: [1 2 3 4]Enter h[n]: [1 2]
xnew =
1 2 3 4 0
hnew =
1 2 0 0 0
xf =
10.0000 -4.0451 - 1.3143i 1.5451 - 2.1266i 1.5451 + 2.1266i -4.0451 + 1.3143i
hf =
3.0000 1.6180 - 1.9021i -0.6180 - 1.1756i -0.6180 + 1.1756i 1.6180 + 1.9021i
zf =
30.0000 -9.0451 + 5.5676i -3.4549 - 0.5020i -3.4549 + 0.5020i -9.0451 - 5.5676i
z = 1 4 7 10
0 0.5 1 1.5 2 2.5 30
2
4
time
Am
plitu
de
First sequence x(n)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
1
2
time
Am
plitu
de
Second sequence h(n)
0 0.5 1 1.5 2 2.5 3 3.5 40
5
10
time
Am
plitu
de
Convolution
Result:The convolution of two sequences is done in time and frequency domain.
Output:Enter x[n] : [1 2 3 4]Rxx = 4.0000 11.0000 20.0000 30.0000 20.0000 11.0000 4.0000
0 0.5 1 1.5 2 2.5 30
1
2
3
4
Time
Am
plitu
deInput sequence x[n]
-3 -2 -1 0 1 2 30
10
20
30
Time
Am
plitu
de
Autocorrelation of x[n]
Output:Enter the first sequence x[n]: [1 2 3 4]Enter the second sequence y[n]: [5 6 7 8]
Rxy =
8.0000 23.0000 44.0000 70.0000 56.0000 39.0000 20.0000
0 0.5 1 1.5 2 2.5 30
2
4
Time
Am
plitu
de
Input sequence x[n]
0 0.5 1 1.5 2 2.5 30
5
10
Time
Am
plitu
de
Input sequence y[n]
-3 -2 -1 0 1 2 30
50
100
Time
Am
plitu
de
Cross correlation of x[n] and y[n]
Result: the autocorrelation and cross correlation of the signals is obtained.
Output:
enter pass band ripple 2enter stop band ripple 10enter pass band frequency 2000enter stop band frequency 8000
n =
1
wn =
1.6755e+004
b =
1.0e+004 *
0 1.6755
a =
1.0e+004 *
0.0001 1.6755
0 100 200 300 400 500 600 700 800 900 10000.98
0.99
1
1.01
normalised frequency
m
magnitude response
0 100 200 300 400 500 600 700 800 900 1000-0.2
-0.15
-0.1
-0.05
0
normalised frequency
a
phase reponse
Result: the response of butterworth low pass filter is obtained.
Output:enter pass band ripple 2enter stop band ripple 10enter pass band frequency 8000enter stop band frequency 2000
n =
1
wn =
3.7699e+004
b =
1 0
a =
1.0e+004 *
0.0001 3.7699
0 2000 4000 6000 8000 10000 12000 14000 160000
0.2
0.4
0.6
0.8
normalised frequency
m
magnitude response
0 2000 4000 6000 8000 10000 12000 14000 160000
0.5
1
1.5
2
normalised frequency
a
phase reponse
Result: the response of butterworth high pass filter is obtained.
Output:
Result:
Hence the IIR low pass filter is designed using Chebyshev approximations.
Output:
Result:
Hence the IIR high pass filter is designed using Chebyshev approximations.
Output:
Result:
Hence an FIR LPF is designed for the given specifications using different window
techniques.
Output:
Result:
Hence an FIR HPF is designed for the given specifications using different
window techniques.
OUTPUT
-0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Time t (sec)
x(t)
Continuous time signal: x(t) = cos(2F1t) + cos(2F
2t)
-0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Time Sample (n)
Am
plitu
de
Discrete Time Signal
Fs < 2Fmax
-0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Time Sample (n)
Am
plitu
de
Discrete Time Signal
Fs > 2Fmax
-0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Time Sample (n)
Am
plitu
de
Discrete Time Signal
Fs = 2Fmax
Result : Under sampling , Nyquist sampling and Over sampling have been observed.
Output:Enter the Decimation factor= 4
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-1
0
1
2original signal
time n
Am
plitu
de
0 5 10 15 20 25 30-2
-1
0
1
2Decimated signal
time n
Am
plitu
de
Result: thus downsampling is performed on the signal with a factor of 4.
Output:Enter the interpolation factor= 4
0 5 10 15 20 25 30-2
-1
0
1
2original signal
time n
Am
plitu
de
0 20 40 60 80 100 120-2
-1
0
1
2Interpolated signal
time n
Am
plitu
de
Result: thus the signal is upsampled by a factor of 4.
Output:
enter the input frequency:100sampling frequency 500
0 0.05 0.1 0.15 0.2 0.25-70
-60
-50
-40
-30
-20
-10
0
Frequency (kHz)
Pow
er/f
requ
ency
(dB
/Hz)
Power Spectral Density Estimate via Periodogram
Result: The output is observed and the graphs are plotted.
OUTPUT:
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5-80
-70
-60
-50
-40
-30
-20
-10
Frequency (kHz)
Pow
er/f
requ
ency
(dB
/Hz)
Power Spectral Density Estimate via Periodogram
Result: The output is observed and the graphs are plotted.
OUTPUT:
enter the frequency of the signal: 100enter the sampling frequency: 500
0 0.05 0.1 0.15 0.2 0.25-30
-28
-26
-24
-22
-20
-18
-16
-14
-12
-10
Frequency (kHz)
Pow
er/f
requ
ency
(dB
/Hz)
Power Spectral Density Estimate via Welch
Result: hence the PSD is observed.
OUTPUT:
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-200
-100
0
100
Normalized Frequency ( rad/sample)
Pha
se (
degr
ees)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-20
0
20
40
Normalized Frequency ( rad/sample)
Mag
nitu
de (
dB)
AR system frequency response
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-25
-20
-15
-10
-5
0
5
10
15
20
25
Normalized Frequency ( rad/sample)
Pow
er/f
requ
ency
(dB
/rad
/sam
ple)
Power Spectral Density Estimate via Yule-Walker
Result: the PSD is calculated.
OUTPUT:
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9-80
-60
-40
-20
0
20
40
60
80
normalised frequency
one
side
d ps
d
Power Spectral Density Estimate via Burg
psd of model output
psd estimate of x
Result : hence the PSD is calculated using Burg method.
Result: The forward predictor of the signal is obtained.
