301 EECE Signals & Systems Prof. Mark Fowler
Note Set #25
D-T Signals: Relation between DFT, DTFT, & CTFT Reading Assignment: Sections 4.2.4 & 4.3 of Kamen and Heck
1/22
Course Flow Diagram
The arrows here show conceptual flow between ideas. Note the parallel structure between the pink blocks (C-T Freq. Analysis) and the blue blocks (D-T Freq. Analysis). New Signal Models
Ch. 1 Intro
C-T Signal Model Functions on Real Line System Properties LTI Causal Etc D-T Signal Model Functions on Integers
Ch. 3: CT Fourier Signal Models
Fourier Series Periodic Signals Fourier Transform (CTFT) Non-Periodic Signals
Ch. 5: CT Fourier System Models
Frequency Response Based on Fourier Transform
Ch. 6 & 8: Laplace Models for CT Signals & Systems
Transfer Function
New System Model
New System Model
Ch. 2 Diff Eqs
C-T System Model Differential Equations D-T Signal Model Difference Equations Zero-State Response Zero-Input Response Characteristic Eq.
Ch. 2 Convolution
C-T System Model Convolution Integral D-T System Model Convolution Sum
New System Model
New Signal Model Powerful Analysis Tool
Ch. 4: DT Fourier Signal Models
DTFT (for "Hand" Analysis) DFT & FFT (for Computer Analysis)
Ch. 5: DT Fourier System Models
Freq. Response for DT Based on DTFT
Ch. 7: Z Trans. Models for DT Signals & Systems
Transfer Function
New System Model
New System Model 2/22
We can use the DFT to implement numerical FT processing
This enables us to numerically analyze a signal to find out what frequencies it contains!!!
A CT signal "comes in" through a sensor & electronics (e.g., a microphone & amp) N samples are "dumped" into a memory array FFT algorithm computes N DFT values
x(t )
ADC
x[n]
x[0] x[1] x[2]
X[0]
The ADC creates samples (taken at an appropriate Fs)
x[N-1]
memory array
DFT Processing (via FFT)
X[1]
X[2]
X[N-1]
"H/W" or "S/W on processor" memory array
Inside "Computer"
DFT values in memory array (they can be plotted or used to do something "neat")
3/22
If we are doing this DFT processing to see what the original CT signal x(t) "looks" like in the frequency domain... ... we want the DFT values to be "representative" of the CTFT of x(t) Likewise... If we are doing this DFT processing to do some "neat" processing to extract some information from x(t) or to modify it in some way... ... we want the DFT values to be "representative" of the CTFT of x(t)
So... we need to understand what the DFT values tell us about the CTFT of x(t)... We need to understand the relations between... CTFT, DTFT, and DFT
4/22
We'll mathematically explore the link between DTFT & DFT in two cases: 1. For x[n] of finite duration: ...0 0 x[0] x[1] x[2] ... X[N 1] 0 0 N "non-zero" terms
This case hardly ever happens... but it's easy to analyze and provides a perspective for the 2nd case
(of course, we could have some of the interior values = 0) For this case... we'll assume that the signal is zero outside the range that we have captured. So... we have all of the meaningful signal data.
This is the practical case.
2. For x[n] of infinite duration ...or at least of duration longer than what we can get into our "DFT Processor" inside our "computer". So... we don't have all the meaningful signal data. What effect does that have? How much data do we need for a given goal?
5/22
DFT & DTFT: Finite Duration Case
If x[n] = 0 for n < 0 and n N then the DTFT is:
N -1 n =0
X () =
n = -
x[n]e
- jn
= x[n]e - jn
we can leave out terms that are zero
Now... if we take these N samples and compute the DFT (using the FFT, perhaps) we get: N -1
X [k ] = x[n ]e - j 2kn / N
n =0
k = 0,1, 2, ..., N - 1
Comparing these we see that for the finite-duration signal case: X [k ] = X (k
2 N
)
X ()
X [k ]
DTFT & DFT : k
2
6/22
0 1 - -/2
2 3 4 5 6 /2
7
DFT points lie exactly on the finite-duration signal's DTFT!!!
Summary of DFT & DTFT for a finite duration x[n]
TFT X () D
DFT
x[n]
2 X [k ] = X k N
Points of DFT are "samples" of DTFT of x[n]
The number of samples N sets how closely spaced these "samples" are on the DTFT... seems to be a limitation. "Zero-Padding Trick" After we collect our N samples, we tack on some additional zeros at the end to trick the "DFT Processing" into thinking there are really more samples. (Since these are zeros tacked on they don't change the values in the DFT sums) If we now have a total of NZ "samples" (including the tacked on zeros), then the spacing between DFT points is 2/NZ which is smaller than 2/N
7/22
Ex. 4.11 DTFT & DFT of pulse
1, x[n] = 0,
n = 0,1, 2, ...2q otherwise
Then...
1, Recall : pq [n ] = 0,
n = - q, ... ,-1, 0, 1, ..., q otherwise
Note: we'll need the delay property for DTFT
x[n ] = pq [n - q]
sin[( q + 0.5)] sin[ / 2]
From DTFT Table: pq [n ] Pq () = From DTFT Property Table (Delay Property):
X () =
sin[(q + 0.5)] - jq e sin[ / 2]
Since x[n] is a finite-duration signal then the DFT of the N = 2q+1 non-zero samples is just samples of the DTFT:
2 X [k ] = X k N
X [k ] =
sin[( q + .5)2k / N ] - jq 2k / N e sin[k / N ]
8/22
Note that if we don't zero pad, then all but the k = 0 DFT values are zero!!! That doesn't show what the DTFT looks like! So we need to use zero-padding. Here are two numerically computed examples, both for the case of q = 5: For the case of zeropadding 11 zeros onto the end of the signal... the DFT points still don't really show what the DTFT looks like! For the case of zeropadding 77 zeros onto the end of the signal... NOW the DFT points really show what the DTFT looks like! DFTs were computed using matlab's fft command... see code on next slide
9/22
omega=eps+(-1:0.0001:1)*pi; q=5; % used to set pulse length to 11 points X=sin((q+0.5)*omega)./sin(omega/2); subplot(2,1,1) plot(omega/pi,abs(X)); % plot magn of DTFT xlabel('\Omega/\pi (Normalized rad/sample)') ylabel('|X(\Omega)| and |X[k]|') hold on x=zeros(1,22); % Initially fill x with 22 zeros x(1:(2*q+1))=1; % Then fill first 11 pts with ones Make the zero-padded signal Xk=fftshift(fft(x)); % fft computes the DFT and fftshift re-orders points Compute the DFT % to between -pi and pi omega_k=(-11:10)*2*pi/22; % compute DFT frequencies, except make them Compute the DFT point's % between -pi and pi stem(omega_k/pi,abs(Xk)); % plot DFT vs. normalized frequencies frequency values and plot hold off the DFT subplot(2,1,2) plot(omega/pi,abs(X)); xlabel('\Omega/\pi (Normalized rad/sample)') and ylabel('|X(\Omega)| |X[k]|') hold on x=zeros(1,88); x(1:(2*q+1))=1; Xk=fftshift(fft(x)); omega_k=(-44:43)*2*pi/88; stem(omega_k/pi,abs(Xk)); hold off 10/22
Compute the DTFT Equation derived for the pulse. Using eps adds a very small number to avoid getting = 0 and then dividing by 0
Important Points for Finite-Duration Signal Case DFT points lie on the DTFT curve... perfect view of the DTFT
But... only if the DFT points are spaced closely enough
Zero-Padding doesn't change the shape of the DFT... It just gives a denser set of DFT points... all of which lie on the true DTFT
Zero-padding provides a better view of this "perfect" view of the DTFT
11/22
DFT & DTFT: Infinite Duration Case
As we said... in a computer we cannot deal with an infinite number of signal samples. So say there is some signal that "goes on forever" (or at least continues on for longer than we can or are willing to grab samples) x[n] n = ..., -3, -2, -1, 0, 1, 2, 3, ... We've lost some information!
We only grab N samples: x[n], n = 0, ..., N 1
We can define an "imagined" finite-duration signal:
x[n], n = 0,1, 2,..., N - 1 x N [ n] = elsewhere 0,
We can compute the DFT of the N collected samples:
X N [k ] = x N [n ]e - j 2nk / N
n =0
N -1
k = 0,1, ..., N - 1
Q: How does this DFT of the "truncated signal" relate to the "true" DTFT of the full-duration x[n]? ...which is what we really want to see!!
12/22
" True" DTFT : X () =
n = -
x[n]e
- jn
What we want to see
DTFT of truncated signal : X N () =
n = - N -1 n =0
x N [n]e - jn
= x[n]e - jn
A distorted version of what we want to see
DFT of collected signal data : X N [k ] = x[n ]e - j 2kn / N
n =0
N -1
What we can see
DFT gives samples of X N ()
So... DFT of collected data gives "samples" of DTFT of truncated signal "True" DTFT DFT of collected data does not perfectly show DTFT of complete signal. Instead, the DFT of the data shows the DTFT of the truncated signal... So our goal is to understand what kinds of "errors" are in the "truncated" DTFT ...then we'll know what "errors" are in the computed DFT of the data
13/22
To see what the DFT does show we need to understand how XN() relates to X() First, we note that:
xN [n ] = x[n ] pq [n - q]
DTFT
sin[N / 2] - j ( N -1) / 2 Pq () = e sin[ / 2]
with N=2q+1
From "mult. in time domain" property in DTFT Property Table:
X N () = X () Pq ()
Convolution causes "smearing" of X()
So... XN() ...which we can see via the DFT XN[k] ... is a "smeared" version of X() "Fact": The more data you collect, the less smearing ... because Pq() becomes more like ()
14/22
Suppose the infinite-duration signal's DTFT is:
X ()
- 2
-
DTFT of infiniteduration signal
2
Then it gets smeared into something that might look like this:
X N () - 2 -
DTFT of truncated signal
X N [k ]
2
Then the DFT computed from the N data points is:
- 2
-
2
15/22
The DFT points are shown after "upper" points are moved (e.g., by matlab's "fftshift"
Example: Infinite-Duration Complex Sinusoid & DFT Suppose we have the signal
x[n ] = e - jon
n = ..., - 3, - 2, - 1, 0,1, 2, ...
and we want to compute the DFT of N collected samples (n = 0, 1, 2, ..., N-1). This is an important example because in practice we often have signals that consists of a few significant sinusoids among some other signals (e.g. radar and sonar). In practice we just get the N samples and we compute the DFT... but before we do that we need to understand what the DFT of the N samples will show. So we first need to theoretically find the DTFT of the infinite-duration signal. From DTFT Table we have:
X ()
( - 0 ), - < < X () = periodic elsewhere
- 2
-
o
2
16/22
From our previous results we know that the DTFT of the collected data is:
sin[N / 2] - j ( N -1) / 2 X N () = X () e sin[ / 2]
Just Delta's in here Use Sifting Property!!
( - 0 ), - < < X () = periodic elsewhere
Pq()
Just a shifted version of Pq()
N ( - 0 ) sin - j ( N -1)( - ) / 2 2 e 0 , - < < X N () = sin ( - 0 ) 2 periodic elsewhere
This is the DTFT on which our data-computed DFT points will lie... so looking at this DTFT shows us what we can expect from our DFT processing!!!
17/22
True DTFT of Infinite Duration Complex Sinusoid
( - 0 ), - < < X () = periodic elsewhere
0
DTFT of Finite Number of Samples of a Complex Sinusoid
N ( - 0 ) sin - j ( N -1)( - ) / 2 2 e 0 , - < < ( - 0 ) X N () = sin 2 periodic elsewhere
Digital Frequency (rad/sample)
0
The computed DFT would give points on this curve... the spacing of points is controlled through "zero padding"
18/22
So... what effect does our choice of N have??? To answer that we can simply look at Pq() for different values of N = 2q+1 ...
15 10 5 0 -5 -3 60 -2 -1 0 / 1 2 3 N=11
Pq() D(,11)
Pq() D(,41)
40 20 0 -20 -3 -2
N=41
-1
0 /
1
2
3
As N grows... looks more like a delta!! So... less smearing of XN()!!
19/22
Important points for Infinite-Duration Signal Case
1. DTFT of finite collected data is a "smeared" version of the DTFT of the infinite-duration data 2. The computed DFT points lie on the "smeared" DTFT curve... not the "true" DTFT a. This gives an imperfect view of the true DTFT! 3. "Zero-padding" gives denser set of DFT points... a better view of this imperfect view of the desired DTFT!!!
20/22
Connections between the CTFT, DTFT, & DFT
x(t )
ADC X ( f ) CTFT
x[n]
x[0] x[1]
Inside "Computer" XN[0]
f
- Fs / 2 Fs / 2
x[2]
x[N-1]
DFT processing
XN[1] XN [2]
XN [N-1]
X () Full DTFT
Aliasing
-
Look here to see aliased view of CTFT
X N () Truncated DTFT
"Smearing"
X N [k ] Computed DFT
-
-
21/22
Errors in a Computed DFT
CTFT DTFT DTFTN
"Grid" Error control through Aliasing Error control through Fs choice (i.e. through proper sampling) "Smearing" Error control through
See DSP course
N choice "window" choice
N choice "zero padding"
DFT
This is the only thing we can compute from data... and it has all these "errors" in it!! The theory covered here allows an engineer to understand how to control the amount of those errors!!!
Zero padding trick Collect N samples defines XN() Tack M zeros on at the end of the samples Take (N + M)pt. DFT gives points on XN() spaced by 2/(N+M) (rather than 2/N)
22/22