Discussion of Discrete-time Fourier Transforms. Topics include comparison with analog transforms and discussion of Parseval's theorem.
The Fourier transform of the discrete-time signal
is defined to be
Frequency here has no units. As should be expected, thisdefinition is linear, with the transform of a sum of signals
equaling the sum of their transforms. Real-valued signals haveconjugate-symmetric spectra:
.
A special property of the discrete-time Fourier transform isthat it is periodic with period one:
.
Derive this property from the definition of the DTFT.
Because of this periodicity, we need only plot the spectrum overone period to understand completely the spectrum's structure;
typically, we plot the spectrum over the frequency range
.
When the signal is real-valued, we can further simplify ourplotting chores by showing the spectrum only over
;
the spectrum at negative frequencies can be derived frompositive-frequency spectral values.
When we obtain the discrete-time signal via sampling an analog
signal, the
Nyquist frequency corresponds to the
discrete-time frequency
. To show this, note that a sinusoid having a
frequency equal to the Nyquist frequency
has a sampled waveform that equals
The exponential in the DTFT at frequency
equals
, meaning that discrete-time frequency equals analog
frequency multiplied by the sampling interval
and
represent discrete-time and analog frequency
variables, respectively. The
aliasing figure provides
another way of deriving this result. As the duration of eachpulse in the periodic sampling signal
narrows, the amplitudes of the signal's spectral
repetitions, which are governed by the
Fourier series coefficients of
, become increasingly equal. Examination of the
periodic pulse
signal reveals that as
decreases, the value of
,
the largest Fourier coefficient, decreases to zero:
.
Thus, to maintain a mathematically viable Sampling Theorem, theamplitude
must increase as
, becoming infinitely large as the pulse duration
decreases. Practical systems use a small value of
, say
and use amplifiers to rescale the signal. Thus, the sampledsignal's spectrum becomes periodic with period
.
Thus, the Nyquist frequency
corresponds to the frequency
.
Let's compute the discrete-time Fourier transform of the
exponentially decaying sequence
,
where
is the unit-step sequence. Simply plugging the signal'sexpression into the Fourier transform formula,
This sum is a special case of the
geometric
series .
Thus, as long as
,
we have our Fourier transform.
Using Euler's relation, we can express the magnitude and phase
of this spectrum.
No matter what value of
we
choose, the above formulae clearly demonstrate the periodicnature of the spectra of discrete-time signals.
[link] shows indeed that the spectrum
is a periodic function. We need only consider the spectrumbetween
and
to unambiguously define it. When
,
we have a lowpass spectrum—the spectrum diminishes asfrequency increases from 0 to
—with increasing
leading to a greater low frequency
content; for
,
we have a highpass spectrum(
[link] ).
Derive this formula for the finite geometric series sum.
The "trick" is to consider the difference between theseries' sum and the sum of the series multiplied by
.
which, after manipulation, yields the geometric sum formula.
The ratio of sine functions has the generic form of
,
which is known as the
discrete-time sinc function
.
Thus, our transform can be concisely expressed as
. The discrete-time pulse's spectrum contains many
ripples, the number of which increase with
, the pulse's duration.
The inverse discrete-time Fourier transform is easily derived
from the following relationship:
Therefore, we find that
The Fourier transform pairs in discrete-time are
The properties of the discrete-time Fourier transform mirror
those of the analog Fourier transform. The
DTFT properties table shows similarities and differences. One important common
property is Parseval's Theorem.
To show this important property, we simply substitute theFourier transform expression into the frequency-domain
expression for power.
Using the
orthogonality
relation , the integral equals
,
where
is the
unit sample . Thus, the double sum collapses
into a single sum because nonzero values occur only when
,
giving Parseval's Theorem as a result. We term
the energy in the discrete-time signal
in spite of the fact that discrete-time signals don't consume(or produce for that matter) energy. This terminology is a
carry-over from the analog world.
Suppose we obtained our discrete-time signal from values ofthe product
,
where the duration of the component pulses in
is
. How is
the discrete-time signal energy related to the total energycontained in
?
Assume the signal is bandlimited and that the sampling ratewas chosen appropriate to the Sampling Theorem's conditions.
If the sampling frequency exceeds the Nyquist frequency, thespectrum of the samples equals the analog spectrum, but overthe normalized analog frequency
. Thus, the energy in the sampled signal equals
the original signal's energy multiplied by
.
Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
from theory: distance [miles] = speed [mph] × time [hours]
info #1
speed_Dennis × 1.5 = speed_Wayne × 2
=> speed_Wayne = 0.75 × speed_Dennis (i)
info #2
speed_Dennis = speed_Wayne + 7 [mph] (ii)
use (i) in (ii) => [...]
speed_Dennis = 28 mph
speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5.
Substituting the first equation into the second:
W * 2 = (W + 7) * 1.5
W * 2 = W * 1.5 + 7 * 1.5
0.5 * W = 7 * 1.5
W = 7 * 3 or 21
W is 21
D = W + 7
D = 21 + 7
D = 28
Salma
Devon is 32 32 years older than his son, Milan. The sum of both their ages is 54 54. Using the variables d d and m m to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
please why is it that the 0is in the place of ten thousand
Grace
Send the example to me here and let me see
Stephen
A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg
however, may I ask you some questions about Algarba?
Amoon
hi
Enock
what the last part of the problem mean?
Roger
The Jones family took a 15 mile canoe ride down the Indian River in three hours. After lunch, the return trip back up the river took five hours. Find the rate, in mph, of the canoe in still water and the rate of the current.
Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.
I'm guessing, but it's somewhere around $4335.00 I think
Lewis
12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67.
Check:
Sales = 3542
Commission 12%=425.04
Pay = 500 + 425.04 = 925.04.
925.04 > 925.00
Munster
difference between rational and irrational numbers
Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?