Frequency Domain Representation of Signals

The Discrete Fourier Transform (DFT) of a sampled time domain waveform
1 ,..., xN  1 is a set of N Fourier Coefficients
whose N samples are xn x0, x
N 1
X k  X 0, X 
1 ,..., X N  1 where X k   xn* e
j
2nk
N
n 0


The inverse DFT converts the N Fourier Coefficients to the corresponding N time
2nk
j
1 N 1
domain samples where x[n]   X [k ]e N
N k 0
The Discrete Fourier Transform (DFT) allows us to investigate discrete time
signals in the frequency domain.
The DFT is used to represent a discrete time signal as the superposition of sine
waves.
Fast Fourier Transform (FFT) is an algorithm to compute the DFT efficiently
when the number of points N is a power of 2, i.e., N = 2k where k is an integer.
X k  X R k  jX I k  is complex, and for Real xn, X k  X * N  k .
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X k   X R k   X I k is called the Magnitude or Amplitude Spectrum.
For Real x[n], the Magnitude Spectrum is symmetric about k=N/2.
The Phase Spectrum is:




2
2
 X k 
Phase X k   tan1 I 
 X R k 

The Power Spectrum , S k   X k X k   X k 
*
2
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DFTs and ADCs
If x[n] are samples from an Analog to Digital Converter (ADC) operating at Fs

x[n] is Real.

X[k] represents frequency components at Frequency k 






k * Fs
.
N
F
1
 s . Therefore,
f is the spacing of the points or bins on the x-axis. f 
NTs N
the frequency resolution is inversely proportional to the total sampling time (NTs).
Because the Magnitude Spectrum is symmetric about N/2, we usually only plot
X[k] for k < N/2.
The maximum frequency represented is Fs/2.
Units of Power Spectrum is Watts because we assume that the signal is in volts delivered to a
1 ohm resistor so x[n] = V[n] = I[n] therefore P[n] = x[n]2 Watts.
S k 
The Power Spectral Density, PSDk 
and has units Watts/Hz.
f
Measure Power by integrating the Power Spectral Density. Take square root to obtain Vrms.
That is Pwr = Vrms 2
Note that if two signals have power P1 and P2 and associated Vrms1 and Vrms2 then
P1 + P2 = PTotal and VrmsTotal ≠ Vrms1 + Vrms2. VrmsTotal = sqrt (PTotal).
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Example: 10 periods of 1 Volt, 50KHz sine wave, sampled at 1MHz (200 points)
Note that we have 20 samples per cycle and exactly 10 cycles so no window is
needed (see Windowing section below).
1
 1 us
 xn Asin2f nT  where A=1, f = 50KHz, and T 
Fs
Quiz:
VPeak = _____V VRMS = _____ V = ______ dBVrms
Power1ohm Resistor = ______ W = ______ dBW
Power Spectral Density ( f =5KHz) at 50KHz = _____ W/Hz
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Bandwidth of sampled signal is 500KHz. 500KHz – 1MHz are the image
frequencies.
There are 200 points in the Double Sided DFT.
Amplitude in the Double Sided DFT in the 50KHz and the 950KHz bin is 0.5 V in both
peaks.
Amplitude in the Single Sided DFT in the 50KHz bin is 1.0 V
Bandwidth = 500KHz.
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(Zooming in around peak)

Peak is at 20log (
A
2
)  20log (
1
2
)  -3 dBVrms
(Zooming in around peak)
2


2
 A 
 1 
Peak is at 10log (
 )  10log (
 )  -3 dBW
 2
 2
What is setting the Noise Floor here?
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2

 A 


 2 
Peak is at
f
2
 1 


 2   0.0001 W
1MHz
Hz
200
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Pulse Train

Pulse train has same shape in Time Domain and Frequency Domain.
Square Wave


Square Wave only has odd harmonics at with amplitudes 1/3, 1/5, 1/7, … of the fundamental.
1/3 = -9.5 dB, 1/5 = -14 dB
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White Gaussian Noise (WGN)
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
  Standard Deviation  Vrms
99.7% of values within  3 for White Gaussian Noise (WGN)
PSD is flat for WGN
 2  Variance  Noise Power   PSD df

For WGN,  2   PSDdf  N 0 * BW  N 0

For this example

nW
, Fs  1MHz
Hz
1e6
NoisePower   2  500 *10 9 *
 0.25W
2
Vrms    0.25  0.5V
From Histogram or Time Domain Graph, Maximum Peak-Peak  1.5V  3
BW
BW
Fs
2
N 0  Average of PSD  500
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Windowing
The DFT assumes that the N samples are periodic. In the above examples, there were exactly 10 periods
of the sine wave. However, what if there are 10.5 periods.



Windowing Example, (Installation Instructions)
This discontinuity causes “Spectral Leakage” into all frquencies.
We aren’t getting the –350dB attenuation outside the signal bins anymore (see Frequency
Spectrum for 10.0 periods).
 This causes errors at all frequencies including the signal bins. The peak is now at –4dBVrms
instead of –3.
 Spectral leakage can be virtually eliminated by “windowing” time domain samples prior to
calculating the DFT.
o Windows taper smoothly down to zero at the beginning and the end of the observation
window (eliminating any discontinuities).
o Time domain samples are multiplied by the window on a sample by sample basis.
o Multiplication in the Time Domain is Convolution in the Frequency Domain.
o The effect of the Convolution is to smear our peaks in the Frequency domain.
 We will use the Hodie Window in this class. This window smears the peaks over 17 bins in
the Frequency Domain.
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(Zooming in around peak)

To measure the amplitude after windowing, you have to integrate the Power Spectral Density
over these 17 Signal Bins.
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

Once again we are back to the Power = 0.5W = -3dBW, or A=1.0
Why 10KHz to 90KHz?
References:
 “Mixed-Signal System Design and Modeling”, Eric Swanson, Lecture 13, Fall Semester
2002.
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